Reteaching 1 Using Properties of Real Numbersbfhskasten2.weebly.com/uploads/8/4/6/8/84688124/... ·...

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© Saxon. All rights reserved. 1 Saxon Algebra 2

You have learned how to solve problems involving real numbers. Now you will learn to identify the properties of real numbers.

Properties of AdditionLet a, b, and c be real

numbers.Examples

Additive Identity

0 is the additive identity.a � 0 � 0 � a � a 4 � 0 � 4

Additive Inverse

The sum of a number and its opposite is 0.

a � (�a) � 0 8 � (�8) � 0

Closure Property

The sum of any two real numbers is a real number.

a � b � R 3 � 5 � 8

Commutative Property

The order does not change the sum.a � b � b � a 6 � 12 � 12 � 6

Associative Property

The grouping does not change the sum.

(a � b) � c � a � (b � c) (2 � 5) � 9 � 2 � (5 � 9)

PracticeComplete the steps to solve each problem.

1. Identify which property of real numbers is 2. Simplify the expression. Identify which 4 � 6 � 6 � 4 � 10 property you used for each step.The order does not change the sum so the 3 � 12 � 7 � 18property being demonstrated is the � 3 � 7 � 12 � 18 Commutative

Commutative Property Property of Addition

of Addition � (3 � 7) � (12 � 18) Associative Property of Addition

� 10 � 30 Add.

� 40 Add.

Solve. Identify which property of real numbers is being demonstrated.

3. 3 � 0 � 3 4. 7 � (�7) � 0

Additive Identity Additive Inverse

5. 4 � (6 � 5) � (4 � 6) � 5 � 15 6. 10 � 4 � 4 � 10 � 14

Associative Property Commutative Property of Addition of Addition

ReteachingUsing Properties of Real Numbers 1


Reteachingcontinued

Properties of MultiplicationLet a, b, and c be real

numbers.Examples

Multiplicative Identity

1 is the multiplicative identity.a � 1 � 1 � a � a �12.1 � �12

Multiplicative Inverse

The product of a number and its inverse is 1.

a � 1 __ a � a, a � 0 2 � � 1 __ 2 � � 1

Closure Property

The product of any two real numbers is a real number.

ab � R 8(�0.5) � �4

Commutative Property

The order does not change the product.

ab � ba 3 � 4 � 4 � 3

Associative Property

The grouping does not change the product.

(ab)c � a(bc) (3 � 7)4 � 3(7 � 4)

Distributive Property

This property involves both addition and multiplication.

a(b � c) � ab � ac �2(4 � 7) � �2(4) � (�2)(7)


7. Identify which property of real numbers 8. Simplify the expression. Identify which is being demonstrated. property you used for each step.

(4 � 5) � 6 � 4 � (5 � 6) � 120 (6 � 18) � 3

The grouping does not change the � (18 � 6) � 3 Commutative Property

of Multiplication

product so the property being � 18 � (6 � 3) Associative Property

of Multiplication

demonstrated is the Associative � 18 � 18 Multiply.

Property of Multiplication. � 324 Multiply.

Simplify the expression. Identify which property you used for each step.

9. (15 � 5) � 3

15 � (5 � 3) Associative Property of Multiplication � 15 � 15 Multiply. � 225 Multiply.

1

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You have learned how to use the properties of real numbers to solve problems. Now you will learn how to evaluate algebraic expressions.

To evaluate an algebraic expression, you substitute numbers for variables then follow the order of operations.

Here is a sentence to help you remember the order of operations.

Please Excuse My Dear Aunt SallyParentheses Exponents Multiply Divide Addition Subtraction

Evaluate x � 2xy � y2 if x � 4 and y � 6.

� 4 � 2(4)(6) � (6)2 Substitute 4 for x and 6 for y.

� 4 � 2(4)(6) � 36 Evaluate exponents: 6 2 � 36.

� 4 � 48 � 36 Multiply from left to right.

� �8 Add and subtract from left to right.

Evaluate x(y � x) � 2xy if x � �2 and y � 3.

� �2(3 � (�2)) � 2(�2)(3) Substitute �2 for x and 3 for y.

� �2(5) � 2(�2)(3) Perform operations inside parentheses.

� �10 � (�12) Multiply in order from left to right.

� �22 Add in order from left to right.

PracticeComplete the steps to evaluate each expression.

1. a 2 � 2 ab 2 � 3a if a � 5 and b � 2 2. c2 � d(c � 3) if c � 7 and d � 6.

� ( 5 )2 � 2( 5 )( 2 )2 � 3( 5 ) � ( 7 )2 � 6 ( 7 � 3)

� 25 � 2( 5 )( 4 ) � 3( 5 ) � ( 7 )2 � 6 (10)

� 25 � 40 � 15 � 49 � 6 (10)

� 40 � 49 � 60

��11Evaluate.

3. 2x � xy � 2 y if x � 3 and y � �2 4. 5xy � 2x2 if x � 3 and y � 1

4 �3

5. 4x2y � xy2 if x � �3 and y � 2 6. 5m � n3 _______

3n if m � 4 and n � 2

60 2

ReteachingEvaluating Expressions and Combining Like Terms 2


Reteachingcontinued

Simplify the expression 3x2 � 5xy � 4x2 � xy by adding like terms.

Add or subtract the coefficients of like terms to simplify an algebraic expression.

Like terms

3x2 � 5xy � 4x2 � xy

Like terms

Like terms: 3x2 and 4x2

5xy and �xy

3x2 � 4x2 � 5xy � xy Group like terms.

� 7x2 � 4xy Add or subtract like terms.

PracticeComplete the steps to simplify each expression.

7. �6x � 2y � 3x � 4y 8. 4c2 � cd � c2 � cd

Group like terms. Group like terms.

�6x � 3x �2y � 4y 4 c 2 � c 2 � cd � cd

Add or subtract like terms: �3x � 2y Add or subtract like terms: 3 c 2 � 2cdSimplify.

9. 4a2 � 5ab � 4a2 � 2ab 10. 3st � 12t2 � 3st � t2

3ab 6st �13 t 2

11. �xy � 10 � 6x � 8xy � 3 12. �3ac � a2 � 7ac � 5a2

7xy � 7 � 6x 4ac � 6 a 2

13. 2rs � 2r � 7 � 4rs � 5r 14. 7xy2 � 4xy � 3xy2 � 4xy

�2rs � 3r � 7 4x y 2

15. 2a2 � 2ab � 6ab � 2a2 16. 3x2 � 2x � 7 � 10x

8ab 3x2 � 12x � 7

17. 5mn � 5 � 3m � 7mn � 2 18. �9 � 3d � 4cd � 10d

12mn � 3m � 3 �4cd � 7d � 9

19. A rectangle is 2x centimeters in length and 5x centimeters in width. What is the perimeter of the rectangle? Use P � 2l � 2w and substitute 2x for l and 5x for w. Simplify.

14x cm

Coefficients of x 2 : 3 and 4Coefficients of xy 5 and �1

Think: 3x2 � 4x2 � 7 x 2

7xy � xy � 4xy

2

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You have learned how to solve problems with real numbers. Now you will learn to solve problems with real numbers using the rules of exponents.

Definition of x �n

If n is any real number and x is any real number that is not zero,

x �n � 1 __ x n

.

Simplify.

a. 5 �2

� 1 ___ 5 2

Move the numerator to the denominator and change the sign of the exponent.

� 1 ___ 25

Simplify.

b. 1 ____ 5 �2

� 5 2 Move the denominator to the numerator and change the sign of the exponent.

� 25 Simplify.

c. � 3 �4

� � 1 ___ 3 4


� � 1 ___ 81

Simplify.

d. (�3 ) �4

� 1 _____ (�3 ) 4


� 1 ___ 81

Simplify.


1. 6 �2 � 1 ___ 6 2

� 1 ___ 36

2. 1 _____ � 6 �2

� � 6 2 � �36

Simplify.

3. 2 �5 1 ___ 32

4. 1 ____ 2 �5

32

5. 1 _____ � 2 �5

�32 6. (�2 ) �5 � 1 ___ 32

ReteachingUsing Rules of Exponents 3


Reteachingcontinued

Rules of Exponents (m, n, and x are real numbers and x � 0.)If exponents have the same base,Product Rule: a m � a n � a m�n Power Rule: ( a m ) n � a m�n

If exponents have different bases, Product Rule: (ab ) m � a m b m

Simplify by using the product rule:

a. w 2 x 5 w 4 x 3

� w 2 w 4 x 5 x 3 Group like bases. To multiply, add exponents.

� w 2�4 x 5�3 Add.

� w 6 x 8 Simplify.

Simplify by using the power rule:

b. ( ab 3 ) 2

� a 2 ( b 3 ) 2 Distribute the exponent.

� a 2 b 3�2 To raise to a power multiply exponents.

� a 2 b 6 Simplify.

Simplify by using the power rule, product rule, and the definition of x �n .

c. x( x �3 ) 2 y(xy ) 2

____________ ( x �2 ) 3 y 2

� xx �6 yx 2 y 2

__________ x �6 y 2

First, use the power rule in both the numerator and the denominator.

� x �3 y 3

______ x �6 y 2

Then, use the product rule in both the numerator and the denominator.

� x �3 x 6 y 3 y �2 Use the definition of x �n .

� x �3�6 y 3�(�2) Use the product rule to multiply like bases.

� x 3 y Simplify.


7. y 4 z 3 ( y �1 z ) 2 8. ( x �3 y ) �2 y 5

___________ x 4 y

� x 6 y �2 y 5

___________ x 4 y

� y 4 z 3 y �2 z 2 � x 6 y 3

_________ x 4 y

� y 2 z 5 � x 6 x �4 y 3 y �1

� x 2 y 2

3

To multiply, add exponents.

To raise to a power, multilply exponents.

Distribute exponents.

Name Date Class


You have learned how to evaluate algebraic expressions for a given value. Now you will learn how to identify functions and use function notation to evaluate functions.

A relation pairs input values x and output values y. The domain is the set of input values or x-values. The range is the set of output values or y-values.

A function is a special type of relation. A function has only one domain element for each range element.

Which of the following depicts a function? Identify the domain and range.

a. 56

34

Domain Range

The relation is not a function. The element 4 of the domain is associated with two elements in the range. The domain is 3 and 4. The range is 5 and 6.

b. x y

3 1

4 1

5 2

The relation is a function. Each element of the domain is associated with only one element of the range. The domain is 3, 4, and 5. The range is 1 and 2.

PracticeComplete the steps to determine if the following depicts a function. Identify the domain and range.

1. x 0 2 4 6

y 0 1 3 3

There is only one unique element of the range for each element of the domain. So, the relation is a function. The domain is: 0, 2, 4 , and 6 .The range is: 0, 1, and 3 .

Which of the following depicts a function? Identify the domain and range.

2. x 0 1 3 0

y 0 2 4 3

3. x 7 5 3 1

y 0 1 0 1

not a function; D: 0, 1, 3; function; D: 1, 3, 5, 7; R: 0, 2, 3, 4 R: 0, 1

ReteachingIdentifying Functions and Using Function Notation 4


Reteachingcontinued

You can use function notation to write a function.

f(x) � 2x � 3

Evaluate f (4) for f (x) � 2x � 3.

f (4) � 2(4) � 3 Replace x with 4.

f (4) � 8 � 3 Multiply.

f (4) � 5 Subtract.

Since the answer is 5 when x � 4, the answer is read as f of 4 equals 5.

If f (x) � x � 5 and g (x) � x � 2, find g (6).

Find g (6) means to substitute 6 for x in g (x).

g (6) � 6 � 2 Substitute 6 for x.

Simplify.

So, g (x) is 4 when x is 6.

PracticeComplete the steps to solve the problem.

4. If g(x) � �2x � 10 and h(x) � 3x � 5, find find g(�3).

Find g(�3) means to substitute �3 for x in g(x).

g(�3) � �2(�3 ) � 10

g(�3 ) � 6 � 10

g(�3 ) � 16 So, g(x) is 16 when x is �3 .

Solve.

5. If s(x) � x 2 � 5 and t(x) � 5x � 2, find s(�4). 11 6. If w(x) � 4 x 2 � 1 and z(x) � 2x � x 2 find z(�1). �3

7. If you buy one CD from a store, the cost is $14. This can be expressed as the ordered pair (1, 14). The store is having a one-day sale on CDs. If you buy one CD at regular price, each additional CD is $10. The table shows the cost for 1, 2, 3, and 4 CDs. Identify the domain and range. Determine if the set of ordered pairs represents a function.

D: 1, 2, 3 4; R: 14, 24, 34 44; function

This is read as f of x equals 2x � 3.

Output f(x) Input (x)

4

x 1 2 3 4

y 14 24 34 44

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You have organized data in tables. Now you will store data in matrices and add and subtract them.

A matrix has rows and columns. The dimensions of a matrix are the

number of rows by the number of columns, or r � c. Matrix A A � [ 2 3 4

5 6

1 ]

shown at the right is a 2 � 3 matrix. Each number in a matrix is calledan element. The location of an element is defined by the name of thematrix, and its row and number. In matrix A, the element in location a 21 is 3.

Matrices with the same dimensions can be added and subtracted. To add or subtract matrices, find the sum or difference of the corresponding elements.

Find the sum.

[ 3 7

9

1

�1

2 ] � [ 1

0 �6

�5

2 0 ]

Add the corresponding elements.

[ 3 7

9

1

�1

2 ] � [ 1

0 �6

�5

2 0 ] � [ 3 � 1

7 � 0 9 � �6

1 � (�5)

(�1) � 2

2 � 0

] � [ 4

7

3

�4

1

2 ]

Find the difference.

[ 8

0

�3

4

�2

5 ] � [ 7

5

6

�1

10 9 ]

Subtract the corresponding elements.

[ 8

0

�3

4

�2

5 ] � [ 7

5

6

�1

10 9 ] � [ 8 � 7

0 � 5 �3 � 6

4 � (�1)

�2 � 10 5 � 9

] � [ 1

�5 �9

5

�12

�4 ]

PracticeComplete the steps to find the sum or difference.

1. [ �4 3 8

0 11

�3 ] � [ 0

15 0

7 �10

6 ] 2. [ �15

�1 3

3 2

6 ] � [ �9

4 0

3 �1

5 ]

� [ �4 � 0 8 � 7 11 � (�10)

3 � 15 0 � 0 �3 � 6 ] � [ �15 � (�9) 3 � 0 2 � (�1)

�1 � 4 3 � 3 6 � 5 ] � [ �4 15 1

18 0 3 ] �

[ �6 3 3�5 0 1

] Find the sum or difference.

3. [ 8 4

2

�6

6

�7

�2

20

11 ] � [ 5

�5 �6

12

7 2

1

�4

9 ] 4. [ 16

4 �3

0

15 22

�8

2 4 ] � [ 12

�3 �5

4

�2

20

�3

12

1 ]

[ 13 6 �1�1 13 16�4 �5 20

]

[ 4 �4 �57 17 �102 2 3 ]

ReteachingUsing Matrices to Organize Data and to Solve Problems 5


Reteachingcontinued

You can solve equations involving matrices. Use corresponding elements to find the unknown matrix or values.

Solve for X.

X � [ 3 7 2

4 2

�3 ] � [ 4

�6 1

2 �2

3 ]

Subtract [ 3 7 2

4 2

�3 ] from each side of the

equation.

X � [ 4 �6

1 2 �2

3 ] � [ 3

7 2

4 2

�3 ]

� [ 1 �13

�1 �2

�4 6 ]

So, X � [ 1 �13

�1 �2

�4 6 ] .

Find the values of a, b, c, and d.

[ 2a �4

5 d � 2

] � [ 14 �c b � 3 � 9

] Write the corresponding elements.

2a � 14

a � 7 Divide each side by 2.

5 � b � 3

8 � b Add 3 to each side.

�4 � �c

4 � c Multiply each side by �1.

d � 2 � �9

d � �11 Subtract 2 from each side.

So, a � 7 b � 8, c � 4, and d � �11.


5. Solve for Y. 6. Find the values of a, b, c, and d.

Y � [ 8 �4

0 17

] � [ �10 3 6

14 ] [ �5a

4 b � 6

�36 ] � [ 15

4 � c �1

9d ]

Y � [ �10 3 6

14 ] � [ 8 0

�4 17 ] �5a � 15 b � 6 � �1

a � �3 b � �7

[ �10 � 8 6 � 03 � �4 14 � 17

] 4 � 4 � c �36 � 9d

[ �2 6�1 31 ] 0 � c �4 � d

7. Solve for A. 8. Find the values of a, b, c, and d.

A � [ 13

�6 2

7

10 �1

] � [ 15

5 2

8

�15 �2

] [ a � 5 7 �2b

�6 ] � [ �14

3 � c �24

6d ]

a � �19 b � 12

A � [ 2 1

11 �250 �1

] c � 4 d � �1

5

Name Date Class


You have found percents of numbers. Now you will solve problems involving percents of change.

A percent of change is a ratio that compares an amount of increase or decrease to an original amount. When the new amount is greater than the original amount, the ratio is a percent increase. When the new amount is less than the original amount, the ratio is a percent decrease.

percent of change � amount of increase or decrease __________________________ original amount

Find the percent increase or decrease.a. increase from 25 to 35

percent of change � amount of increase ________________ original amount

� 35 � 25 _______ 25

� 10 ___ 25

� 2 __ 5

The percent increase is 2 __ 5 , or 40%.

b. decrease from 35 to 25 percent

decrease � amount of decrease _________________ original amount

� 35 � 25 _______ 35

� 10 ___ 35

� 2 __ 7

The percent decrease is 2 __ 7 , or about 29%.

PracticeComplete the steps to find the percent increase or decrease.

1. increase from 8 to 20 2. decrease from 16 to 10

amount of increase ________________ original amount

� 20 � 8 __________

8 amount of decrease _________________

original amount � 16 � 10 _______

16

� 12 ______

8 � 150 % �

6 ______

16 � 37.5 %

Find the percent increase or decrease.

3. increase from 15 to 24 4. increase from 75 to 100

60% �33%

5. decrease from 40 to 32 6. decrease from 124 to 31

20% 75%

ReteachingFinding Percent Increase and Decrease 6


Reteachingcontinued

Stores sometimes use percents to set prices. An increase in price, such as when a store increases the price of an item from the wholesale cost, is called a markup. A decrease in price, such as when an item is on sale, is called a discount.

To find a new price after a markup or discount, use the percent of change and the original price to find the amount of change. For a markup, add the amount of change to the original price. For a discount, subtract the amount of change from the original price.

Find the new price.a. $50 item marked up 250%

amount of markup � percent markup � original price � 2.5(50)

� $125

new price � original price � amount of markup

� 50 � 125

� $175

b. $80 item discounted 20%

amount of discount � percent discount � original price

� 0.2(80)

� $16

new price � original price � amount of discount

� 80 � 16

� $64

PracticeComplete the steps to find the new price.

7. $15 item marked up 300% 8. $250 item discounted 15%

amount of markup � 3(15) amount of discount � 0.15(250)

� $45 � $37.50 new price � 15 � 45 new price � 250 � 37.50 � $60 � $212.50Find the new price.

9. $40 item marked up 275% 10. $150 item marked up 350%

$150 $675 11. $30 item discounted 10% 12. $95 item discounted 40%

$27 $57

13. $5 item marked up 450% 14. $225 item discounted 30%

$27.50 $157.50 15. A store marks up the price of baseball hats 375% from the wholesale

price of $3 each. How much does the store charge for a baseball hat? $14.25 16. Marco buys 3 DVDs that cost $15 each. He has a coupon for 25% off

his total purchase. How much does Marco pay for the 3 DVD’s? $33.75

6

Name Date Class


An equation is a statement that two expressions are equal. Solutions of equations are values that make the statement true. When solving an equation, the goal is to isolate the variable on one side of the equal sign. To solve an equation, you can use properties of equality.

Addition Property of Equality When you add the same number to each side of an equation, the statement remains true.

Subtraction Property of Equality When you subtract the same number from each side of an equation the statement remains true.

Multiplication Property of Equality When you multiply each side of an equation by the same number the statement remains true.

Division Property of Equality When you divide each side of an equation by the same number the statement remains true.

Solve the equation 2x � 5 � 13.2x � 5 � 13

�5 �5 Subtract 5 from each side.2x � 8

2x ___ 2

� 8 __ 2 Divide both sides by 2.

x � 4 Simplify.

The solution is x � 4.

Check: 2(4) � 5 � 13

8 � 5 � 13 13 � 13 ✓

Practice

1. 1 __ 3 x � 4 � �5 Check: 1 __

3 (�3 ) � 4 � �5

� 4 � 4 �1 � 4 � �5

1 __ 3 x � �1 � 5 � �5

(�3 ) 1 __ 3 x � (�3 )(�1 )

x � �3

Solve the equation.

2. 5y � 6 � 19 3. �10n � 7 � �23

y � 5 n � 3

4. � 1 __ 7

a � 9 � 16 5. 1 __ 4 s � 11 � �13

a � �49 s � �8

Add or subtract before multiplying or dividing.

ReteachingSolving Linear Equations 7


Reteachingcontinued

Sometimes you may have to simplify before using the properties of equality to solve an equation.

Solve the equation 3(6 � x) � 12 � 2x � 8. 3(6 � x) � 12 � 2x � 8

3(6) � 3x � 12 � 2x � 8 Use the Distributive Property

6 � 3x � 2x � 8 Simplify.

�6 �6 Subtract 6 from each side.

3x � 2x �2

_ �2x _ �2x Subtract 2x from each side. x � 2

PracticeComplete the steps to solve the equation.

6. 3 __ 4

(2t � 12) � 1 __ 4 (4t � 8) 7. 4(�3n � 9) � 14 � 5n � 6

3 __ 2 t � 9 � t � 2 �12 n � 36 � 8 � 5n

� 9 � 9 �36 �36

3 __ 2 t � t � 11 �12n � �28 � 5n

__ _ �t �t � 5n � 5n

1 __ 2 t � �11 �7 n � �28

2 t � 1 __ 2 t � � 2 (�11 )

�7 n ______ �7

� �28

_____ �7

t � �22 n � 4

Solve the equation.

8. �2(�3p � 7) � p � 6 9. 1 __ 5 (19g � 5) � 2 __

3 (6g � 9)

p � �4 g � �35

10. Tracy and her family are going to her aunt’s house, which is 295 miles away. After traveling 150 miles, they stop for lunch. After lunch, they drive, stopping once for gas before traveling the remaining 45 miles. Solve the equation 150 � d � 45 � 295 to find the distance they traveled between stopping for lunch and stopping for gas.

d � 100 mi

7

Name Date Class


You have used proportions to solve problems. Now you will use directvariation relationships.

When one amount changes proportionally with another, it is called a direct variation. For example, if apples cost $2 per pound, then 2 pounds of apples will cost $4, 3 pounds will cost $6, and so on. The equation c � 2p gives the cost of p pounds of apples. A direct variation can be represented by an equation A � kB, where k is the constant of variation.

The number of pounds varies directly as the number of ounces. If an item weighs 2 pounds, it weighs 32 ounces. If an item weighs 5 pounds, how many ounces does it weigh?

Find the constant of variation and use it to write and solve an equation. Let x be the number of ounces and P the number of pounds. P � kx Write a direct variation.

2 � k(32) Substitute the given values.

2 ___ 32

� 32k ____ 32

Divide each side by 32.

1 ___ 16

� k

There are 80 ounces in 5 pounds.

P � 1 ___ 16

x Substitute in k.

5 � 1 ___ 16

x Substitute 5 for P.

5(16) � 1 ___ 16

x (16) Multiply by 16.

80 � x

PracticeThe number of feet varies directly as the number of yards. If a building measures12 feet, it measures 4 yards. A building measures 21 feet. Complete the steps to find its measure in yards.

1. F � kY P � 3 Y

12 � k( 4 ) 21 � 3 Y

12 ____ 4

� 4 k _____ 4

21 ____ 3

� 3 Y _____ 3

3 � k 7 � Y

7 yards

Find the unknown value.

2. The number of gallons of gas used varies directly with the number of miles driven. If a car travels 110 miles using 5 gallons of gas, how far will it travel using 8 gallons of gas? 176 mi

3. The number of gallons varies directly as the number of quarts. A jug that contains 3 quarts of water contains 3 __

4 gallons. A jug

contains 14 quarts of water. How many gallons does it contain? 3.5 gal

ReteachingFinding Direct Variation 8


Reteachingcontinued

Direct variation problems can also be solved by using proportions. If A

varies directly as B, you can write and solve a proportion A 1

___ A 2

� B 1

___ B 2

to find

an unknown value, where A1 and A2 are the original values and B2 is the other known value.

The number of minutes varies directly as the number of hours. When 3 hours have passed, 180 minutes have passed. How many minutes are in 7 hours?

M 1

___ M 2

� H 1

___ H 2

Write a proportion.

180 ____ M 2

� 3 __ 7 Substitute given values.

180 � 7 � 3 M 2 Cross multiply.

180 � 7 ______ 3 �

3 M 2 ____

3 Divide both sides by 3.

420 � M 2 Simplify.

There are 420 minutes in 7 hours.


4. The perimeter of a square varies directly 5. y varies directly as x. If x � 18, y � 63. as its side length. If a square has a side Find y when x � 10.

length of 8 feet, its perimeter is 32 feet. y 1

__ y 2 � x 1

__ x 2

A square has a side length of 15 feet. Find its perimeter. 63 ____ y 2 � 18 ___

10

P 1

___ P 2

� S 1

___ S 2

10 � 63 � 18 y 2

32 ____ P 2

� 8 ____ 15

10 � 63 _______

18 �

18 y 2 _____

18

15 � 32 � 8 P 2 35 � y 2

15 � 32 ________

8 �

8 P 2 _____

8

60 � P 2

y varies directly as x. Find the unknown value.

6. x 1 � 15; y 1 � 135; x 2 � 25; y 2 � 225 7. x 1 � 16; y 1 � 40; x 2 � 38; y 2 � 95

8. The volume of dirt removed from a certain hole varies directly as its depth. If the hole is 3 feet deep, 27 cubic feet of dirt has been

removed. Find the volume of dirt removed if the hole is 9 feet

deep. 81 f t 3

8

Name Date Class


You have added and subtracted matrices. Now you will multiply matrices. Recall that the dimensions of a matrix are the number of rows by the number of columns. To be able to multiply two matrices A and B, the number of columns of matrix A must be the same as the number of rows of matrix B. The dimensions of the product matrix AB are the number of rows in matrix A by the number of columns in matrix B.

Determine if the product AB is defined. If so, give the dimensions of the product matrix.

A � [ 6 5

9

1

1

7

] B � [ 11 4 5

9 ]

Dimensions of A: Dimensions of B:

Number of rows � 3 Number of rows � 2

Number of columns � 2 Number of columns � 2

Number of columns in A � number of rows in B. The matrices can be multiplied.

Dimensions of the product matrix:

Number of rows in A � Number of columns in B � 3 � 2

PracticeComplete the steps to determine if the product AB is defined. If so, give the dimensions of the product matrix.

1. A � [ �8

0

9

� 2

20

14

] B � [ 1

�5

�8

3 7

6 ]

Dimensions of A: Dimensions of B:

Number of rows � 3 Number of rows � 3

Number of columns � 2 Number of columns � 2

The number of columns in A is not equal to the number of rows in B.

The product matrix is not defined.

Determine if the product AB is defined. If so, give the dimensions of the product matrix.

2. A � [ 4 1

6 1 ] B � [ 3

9 4

2 0

1 ] 3. A � [ 7

�1 3

7

3 19

9

12

�3 ] B � [ 3

9 4

2 ]

yes 2 � 3 no

ReteachingMultiplying Matrices 9


Reteachingcontinued

Each element in a product matrix is equal to the sum of the products of the elements of the corresponding row of the first matrix and corresponding column of the second matrix.

[ a c b d ] � [ e g f g ] � [ ae � bg

ce � dg

af � bh cf � dh

] Find the product.

[ 1

0

�3

�2

5

2

] � [ 4 6

3 0

�1 5 ] � [ 1 � 4 � (�2)6

0 � 4 � 5 � 6 (�3)(4) � 2 � 6

1 � 3 � (�2)0

0 � 3 � 5 � 0 (�3)3 � 2 � 0

1(�1) � (�2)5

0(�1) � 5 � 5 (�3)(�1) � 2 � 5

] � [ �8

30

0

3

0

�9

�11

25

13

] 4. [ �5

10 5

1 0

20 ] � [ 2

�1 �9

8 4

2 ]

�

[ �5 � 2 � 5 � (�1) � 0 � (�9)

10 � 2 � 1 � (�1) � 20 � (�9) �5 � 8 � 5 � 4 � 0 � 2

10 � 8 � 1 � 4 � 20 � 2 ]

� [ �15 �161

�20

124

] Find the product.

5. [ �10 6 5

�8 ] � [ 1

6 �3

9 ] � [ 20

�42

75

�90

] 6. [ 11 7 3

0 �1

1 ] � [ �1

2

12

�8

3

1

] � [ �17

5

�80

�55 ]

7. [ 3

�5

0

3 0

1

�2

6

4

] � [ 8

4 12

2 1

0

�7

�1

6

] 8. [ �16

15

7

4

9 14

2

�6

�8 ] � [ 1

0

0

0 1

0

0 0

1 ]

� [ 12

32 52

9

�10

1

�36

71

23 ] �

[ �16

15

7

4

9 14

2

�6

�8 ]

9. Grace and Carly sold souvenirs at a football game.

Find the product of the matrices to find the total sales for each person.

[ 21 12

31 40

15 13

] � [ 5

10 15

] � [ 640

655 ] Grace: $640 Carly: $655

Mugs ($5) T-shirts ($10)

Sweatshirts ($15)

Grace 21 31 15

Carly 12 40 13

Mugs ($5) T-shirts ($10)

Sweatshirts ($15)

Grace 21 31 15

Carly 12 40 13

9

Name Date Class


You have solved linear equations using the properties of real numbers. Now you will solve and graph inequalities using similar properties.A compound inequality is two inequalities joined by the word and or or.

When the inequalities are joined by and, the solution set is the intersection of the solutions of the two inequalities. You must find solutions that satisfy both inequalities at the same time.

Solve 3 � 2x � 5 � 9 and graph the solution set.

3 � 2x � 5 � 9

3 � 5 � 2x � 5 � 5 � 9 � 5

�2 � 2x � 4

�1 � x � 2

Graph x � �1. 640 2- 2- 4

Graph x � 2. 640 2- 2- 4

Graph �1 � x � 2 640 2-2-4

PracticeComplete the steps to solve the compound inequality and graph the solution set.

1. x __ 3 � 4 � �3 and x __

3 � 4 � 0

Rewrite the compound inequality without and. �3 � x __

3 � 4 � 0

Add 4 to all three parts of the inequality. 1 � x __

3 � 4

Multiply all three parts by 3 . 3 � x � 12

Graph 3 � x. 12108642

Graph x � 12. 12108642

Graph 3 � x � 12. 12108642

Solve the compound inequality and graph the solution set.

2. q � 2 � 2 and q � 2 � 9 121086424 � q � 11

3. �6 � 4c � 10 � 14 40 2-2-4-6�4 � c � 1

Use overlapping regions for compound inequalities with and.

Use overlapping regions for compound inequalities with and.

Subtraction Property of Equality

Simplify.

Division Property

Think:3 � 2x � 5

and 2x � 5 � 9

ReteachingSolving and Graphing Inequalities 10


Reteachingcontinued

When the inequalities are joined by or, the solution set is the union of the solutions of the two inequalities. You must find solutions that satisfy either of the two inequalities.

Solve 2x � 1 � 7 or 3x � 2 � �17 and graph the solution set. 2x � 1 � 7

2x � 1 � 1 � 7 � 1

2x � 6

x � 3

3x � 2 � �17

3x � 2 � 2 � �17 � 2

3x � �15

x � �5

Graph x � 3 40 2-6 -4 -2

Graph x � �5 40 2- 2- 4- 6

Graph x � 3 or x � �5 40 2-2-4-6

PracticeComplete the steps to solve the compound inequality and graph the solution set.

4. 3x � 1 � �10 or 5x � 3 � 13

3x � 1 � �10 5x � 3 � 13

3x � �9 5x � 10

x � �3 x � 2

Graph x � �3 . 4 60 2-2-4

Graph x � 2 . 4 60 2-2-4

Graph x � �3 or x � 2 . 4 60 2-2-4

Solve the compound inequality and graph the solution set.

5. 4s � 3 � 5 or 2s � 10 � 2

s � �4 or s � 2 40 2-2-4-6

6. 3w � 8 � 13 or 6w � 1 � �5

w � �1 or w � 7 40 2 86-2

Solve each inequality separately.

Solve each inequality separately.

Use both regions for compound inequalities with or.

Use both regions for compound inequalities with or.

10

Name Date Class


You have learned about logical reasoning statements. Now you will use truth tables to describe these statements.

A truth table gives the truth values for a logical expression. It shows all possible truth values of the statements that make up the expression.

Construct a truth for the expression p → q.

Step 1: Create a column for each statement and one for the wholeexpression. The table for this expression has 3 columns.

Step 2: Create 2 n rows, where n is the number of basic statements in the expression. The table for this expressions. This table has 2 n � 2 2 � 4 rows.

Step 3: In the first column, fill in the first half of the rows with a T and the second half of the rows with an F. In the second column, fill in the first quarter of the rows with a T and the second quarter of the rows with an F. Repeat this pattern for the whole column.

Step 4: Each possible combination of truth values for the statements is now represented. Fill in a truth value in each row of the expression column. The only time the implication p → q is false is when p is true and q is false.

PracticeComplete the steps to construct the truth table for the expression.

1. (p ∧ q) ∨ r.

Fill in the (p ∧ q) column:

Write T where p and q are true.

Write F where either p or q is false.

Fill in the (p ∧ q) ∨ r column:

Write T where (p ∧ q) or r are true.

Write F where neither p nor q is true.

Construct the truth table for the expressions.

2. p ∨ q

3. (p ∨ q) ∧ r

p q p → qT T

T F

F T

F F

p q r (p ∧ q) (p ∧ q) ∨ r

T T T T TT T F T TT F T F TT F F F FF T T F TF T F F FF F T F TF F F F F

p q p → qT T TT F FF T TF F T

p q r (p ∨ q) (p ∨ q) ∧ r

T T T T TT T F T FT F T T TT T F T FF T T T TF T F T FF F T F FF F F F F

ReteachingLogic and Truth Tables

INV

1


Reteachingcontinued

To compare two expressions, include them both in the same truth table.

Show that ¬(p → q) is logically equivalent to p ∧ (¬q).

Step 1: Create columns for p, q, ¬q, p → q,¬(p → q) and p ∧ (¬q).

Step 2: Create 2 n � 2 2 � 4 rows.

Step 3: Fill in the columns for p and q.

Step 4: Since ¬q is the negation of q, fill in the ¬q column with the opposite of the q column.

Step 5: Fill in the p → q column.

Step 6: Since ¬(p → q) is the negation of p → q fill in the ¬(p → q) column with the opposite of the p → q column.

Step 7: Fill in the p ∧ (¬q) column, using the truth values in the p and the ¬q columns.

The columns for ¬(p → q) and ¬(p → q) are the same. The statements are logically equivalent.

PracticeComplete the steps to show that the set of statements is logically equivalent.

4. p → q and (¬q) → (¬p)

¬p is the negation of p.

¬q is the negation of q.

The columns for p → q and (¬q) → (¬p) are the same. The statements are logically equivalent.

Show that each set of statements is logically equivalent.

5. ¬(p ∧ q) and (¬p) ∨ (¬q)

The columns for ¬(p ∧ q) and (¬p) ∨ (¬q) are the same. The statements are logically equivalent.

6. (¬p) ∧ (¬q) and ¬(p ∨ q)

The columns for (¬p) ∧ (¬q) and ¬(p ∧ q) are the same. The statements are logically equivalent.

p q ¬q p → q ¬(p → q) p∧(¬q)

T T T T F FT F T F T TF T T T F FF F T T F F

p q p ∧ q (¬p) ∧ (¬q) ¬p ¬q (¬q) ∨ (¬p)

T T T F F F FT F F T F T TF T F T T F TF F F T T T T

p q p → q ¬q ¬p (¬q) → (¬p)

T T T F F TT F F T F FF T T F T TF F T T T T

p q p ∨ q ¬(p ∨ q) ¬p ¬q (¬p) ∧ (¬q)

T T T F F F FT F T F F T FF T T F T F FF F F T T T T

INV

1

Name Date Class


You have solved linear equations. Now you will learn about polynomials in preparation for solving polynomial equations.

The degree of a polynomial in one variable is the value of the exponent of the term of the greatest degree.

A polynomial is in standard form when the terms are arranged in order with exponents from greatest to least.

Degree Polynomial in Standard Form0 81 2x � 32 � x 2 � 4x � 53 4 x 3 � x4 6 x 4 � x 3 � 5 x 2 � 3x � 15 9 x 5 � x 3 � 1

Write the polynomial 3 x 2 � x 4 � 2x � 6 x 5 � 7 in standard form. Then identify the leading coefficient, degree, and number of terms of the polynomial.

Order the terms from greatest to least exponent.

6 x 5 � x 4 � 3 x 2 � 2x � 7

The degree of the polynomial is 5, and there are 5 terms.

PracticeComplete the steps to write the polynomial in standard form. Then identify the leading coefficient, degree, and number of terms of the polynomial.

1. 2x � x 3 � x 2 � 5 2. 5 x 2 � 3 x 4 � x

Standard form: x 3 � x 2 � 2x � 5 Standard form: 3 x 4 � 5 x 2 � x,

Leading coefficient: 1 Leading coefficient: 3

Degree: 3 Degree 4 Number of terms: 4 Number of terms: 3

This third degree polynomial has 2 terms.

This fifth degree polynomial has 3 terms.

Constants have degree 0.

6 is the leading coefficient of this polynomial.

ReteachingUnderstanding Polynomials 11


Reteachingcontinued

Like real numbers, polynomials can be added and subtracted.

Add (6x � 2 x 3 � 5 x 2 � 1) � (4 x 2 � 2x � x 3 ).

Step 1: Write each polynomial in standard form. (2 x 3 � 5 x 2 � 6x � 1) � ( x 3 � 4 x 2 � 2x)

Step 2: Align like terms vertically. 2 x 3 � 5 x 2 � 6x � 1

Step 3: Add like terms. __ __ __ __ __ __ __ � x 3 � 4 x 2 � 2x)

3 x 3 � x 2 � 8x � 1Subtract (6x � 2 x 3 � 5 x 2 � 1) � (4 x 2 � 2x � x 3 ).

Step 1: Write each polynomial in standard form.(2 x 3 � 5 x 2 � 6x � 1) � ( x 3 � 4 x 2 � 2x)

Step 2: Add the opposite.(2 x 3 � 5 x 2 � 6x � 1) � (� x 3 � 4 x 2 � 2x)

Step 3: Align like terms vertically. 2 x 3 � 5 x 2 � 6x � 1

Step 4: Add like terms. __ __ � (� x 3 � 4 x 2 � 2x)

x 3 � 9 x 2 � 4x � 1

PracticeComplete the steps to add or subtract the polynomials.

3. (3 x 2 � 2 x 3 � x) � (6x � 2 x 2 � 1) 4. (6 x 2 � 4x � 1) � (2x � x 3 � 1)

2 x 3 � 3 x 2 �x (6 x 2 � 4x � 1) � (�2 x 2 � x 3 � 1)

__ __ � 2 x 2 � 6x � 1 6 x 2 � 4x � 1

2 x 3 � 5 x 2 5x � 1 __ � x 3 � 2x � 1

x 3 � 6 x 2 � 2x � 2

Add or subtract the polynomials.

5. (x � 4 x 3 � 5) � (4 x 2 � x � 2 x 3 ) 6. (4 x 3 � 6) � (3 x 2 � x 3 )

6 x 3 � 4 x 2 � 5

5 x 3 � 3 x 2 � 6

7. The average retail price (in dollars per 1000 cubic feet) of natural gas for the years 2000 through 2005 is modeled by these polynomial functions. The variable x represents the number of years since 2000.

Residential R � 0.224 x 2 � 0.248 x � 8.318 (higher price) Industrial / � 0.244 x 2 � 0.467 x � 4.716 (lower price)

What polynomial function is a model that describes how much higher the average retail price was in the residential sector than in the industrial sector?

f(x) � �0.02 x 2 � 0.219 x � 3.602

11

Name Date Class


You have learned how to identify, write, and solve direct variation equations. Now you will learn how to identify, write and solve inverse variation equations.

When the product of two variables is a constant, the equation is an inverse variation.

x and y are variables. xy � k k is the nonzero constant of variation.

Determine if each data set is an inverse variation. If so, find the constant of variation and the equation. x 1 2 3 4

y 12 6 4 3

Step 1: Multiply the x and y coordinates to see if the products equal the same constant.

x 1 2 3 4y 12 6 4 3xy (1)(12) � 12 (2)(6) � 12 (3)(4) � 12 (4)(3) � 12

Each product is equal to the same constant, so the data set is an inverse variation

Step 2: The constant of variation is 12, the product found above.

Step 3: Write the inverse variation equation.

xy � k

xy � 12

PracticeComplete the steps to determine if the data set is an inverse variation. If so, find the constant of variation and the equation.

1. x 1 2 3 4y 6 3 2 1.5xy 6 6 6 6

constant of variation: 6

equation: xy � 6

Determine if each data set is an inverse variation. If so, find the constant of variation and the equation.

2. x 1 2 3 4

y 1 __ 2 1 __

4 1 __

6 1 __

8

3. x 1 2 3 4y 8 4 2 1

Yes; k � 1 __ 2 ; xy � 1 __

2

No

k � 12

ReteachingSolving Inverse Variation Problems 12


Reteachingcontinued

When one variable depends on changes in two other variables, the equation is a joint variation.

x, y, and z are variables. z __ xy � k k is the nonzero constant of variation.

The area of a triangle (A) is based on the length of its base (b) and its height (h). A 1 3 6 10

b 1 2 3 4h 2 3 4 5

Use the data set to determine if this is a joint variation. If so, find the constant of variation and the equation.

Step 1: Calculate the ratio A ___ bh

for each set of data points.

A 1 3 6 10b 1 2 3 4h 2 3 4 5

A ___ bh

1 __ 2 1 __

2 1 __

2 1 __

2

This is a joint variation.

Step 2: The constant of variation is 1 __ 2 .

Step 2: Write the joint variation equation. A ___ bh

� k

A ___ bh

� 1 __ 2 , or A � 1 __

2 bh

PracticeComplete the steps using the data set to determine if the relation is a joint variation. If so, find the constant of variation and the equation.

4. The volume V of a cylinder is directly proportional to the product of its height (h) and the square of its radius (r ).

Use the data set to determine if the relation is a joint variation.If so, find the constant of variation and the equation.

5. The volume V of a pyramid is directly proportional to the product of its height (h) and the area of its base (B ).

V 2 12 36 80h 2 3 4 5r 1 2 3 4

V ___ h r 2

� � � �

Each ratio is equal to

the same constant, 1 __ 2 .

k � 1 __ 2

This is joint variation. constant of variation: �equation: V ___

h r 2 � �

or V � � r 2 h

V 1 4 8 15h 1 3 4 5B 3 4 6 9

Yes; 1 __ 3 ; V ___

Bh � 1 __

3 or V � 1 __

3 Bh

12

Name Date Class


You have written linear equations in standard form and slope-intercept form. Now you will use these forms to help you graph linear equations.

The points where a graph intersects the axes are the intercepts. You can find the intercepts of a linear equation to help you graph it.

Graph 3x � 6y � 12 using intercepts.

Step 1: The x-intercept is where the graph crosses the x-axis. To find the x-intercept,

set y � 0 and solve for x.

3x � 6y � 12

3x � 6(0) � 12

3x � 12

x � 4

Step 2: The y-intercept is where the graph crosses the y-axis.

To find the y-intercept, set x � 0 and solve for y.

3x � 6y � 12

3(0) � 6y � 12

6y � 12

y � 2

Step 3: Plot the points (4, 0) and (0, 2). Draw a line connecting the points and extend.

PracticeComplete the steps to graph the linear equation using intercepts.

1. 3x � 2y � 6

3x � 2( 0 ) � 6

x-intercept � (2, 0) 3( 0 ) � 2y � 6

y-intercept � (0, 3)

Graph the linear equation using intercepts.

2. 2x � y � 4 3. x � 3y � 3

x

y4

2 4

-2

-2-4

-4

(4, 0)(0, 2)

O

x

y4

2

2 4

-2

-2-4

-4

(2, 0)

(0, 3)

O

x

y4

2

2 4

-2

-2-4

-4

(2, 0)

(0, 4)

O x

y4

2

2

4

-2

-2-4

-4

(3, 0)(0, 1)

O

The x-intercept occurs at the point (4, 0).

The x-intercept occurs at the point (0, 2).

ReteachingGraphing Linear Equations 13


Reteachingcontinued

x

y4

2

2

4

-2

-2-4

(2, - 3)(0, - 4)

O

x

y4

2

2 4

-2

-2-4

-4

(0, 1)

O

x

y4

2

2 4-2-4

-4

-2

(0, 3)

Ox

y4

2

2 4-2-4

-4

(0, -2)

O

A linear equation written in the form y � mx � b is in slope-intercept form. You can find the slope and y-intercept from a linear equation written in this form and use them to graph the equation.

Identify the slope and y-intercept of the line with the equation 2y � x � 8. Graph the line.

Step 1: Write the equation in slope-intercept form.

2y � x � �8

2y � x � x � �8 � x

2y � x � 8

y � 1 __ 2 x � 4

Step 2: Identify the slope and y-intercept.

m � 1 __ 2 , so the slope is 1 __

2 .

b � �4, so the y-intercept is �4.

Step 3: Plot the y-intercept, (0, �4).

Move up 1 and to the right 2 to plot another point at (2, �3).

Step 4: Draw a line through (0, �4) and (2, �3).

PracticeComplete the steps to identify the slope and y-intercept of the line with the equation. Graph the line.

4. 2x � y � �1

y � 2 x � 1

m � 2

b � 1

Identify the slope and y-intercept of the line with each equation. Graph each line.

5. y � 1 __ 2

x � �2 6. x � 3y � 9

m � 1 __ 2 ; b � �2 m �

� 1 __ 3 ; b � 3

Compare y � 1 __ 2 x � 4 form y � mx � b.

13

Name Date Class


You have used matrices to organize data and solve problems. Now you will find the determinants of matrices.

A square matrix has the same number of rows as columns. [ a b c d

] The determinant of a square matrix is shown by using vertical lines. [ a b

c d ]

To find the determinant of a 2 × 2 matrix, find the product of each diagonal, beginning at the upper left corner. Then subtract.

det [ a b c d

] � [ a b c d

] � ad � cb.

Evaluate � 2 3 5 9

� . � 2 3

5 9 � � (2)(9) � (5)(3) � 18 � 15 � 3

Find x. � x � 3 2 5 4

� � 2

(x � 3)(4) � (5)(2) � 2 Write the determinant.

4x � 12 � 10 � 2 Multiply.

4x � 24 Simplify.

x � 6 Divide.

PracticeComplete the steps to find the determinant.

1. � �1 2 �5 4

� � �1 2 �5 4

� � (�1)( 4 ) � (�5)( 2 ) � ( �4 ) � (�10) � 6

Complete the steps to solve the determinant equation.

2. � 2 y � 2

4 3 � � 6 (2)( 3 ) � (4)(y � 2) � 6

6 � 4y � 8 � 6

�4y � �8 y � 2

Find each determinant.

3. � �3 �4 �1 �6

� 4. � 2 �3 6 �4

� 5. � 7 __ 2 � 1 __

4

1 __ 2 1 __

4 �

14 10 1Solve each determinant equation.

6. � 5 3 z � 2 6

� � 0 7. � x � 5 7 �2 4

� � 10 8. � �2 4 5 2b � 3

� � 2

8 4 −7

14ReteachingFinding Determinants


Reteachingcontinued

You can also find the determinant of a 3 × 3 matrix.

Find the determinant of [ 2 1

2

�5

5

�1

4

2

3 ] .

Step 1: Copy the matrix and then repeat the first two columns to the right of the third column.

2 1

2

�5

5

�1

4 2

3

2 1

2

�5

5

�1

Step 2: Multiply each diagonal that has 3 elements.

40 �4 �15

2 1

2

�5

5

�1

4 2

3

2 1

2

�5

5

�1

30 �20 �4

Upper products

Lower products

Step 3: Find the sum of the upper products. 40 � (�4) � (�15) � 21

Step 4: Find the sum of the lower products. 30 � (�20) � (�4) � 6

Step 5: Subtract the upper product sum from the lower product sum. 6 � 21 � �15

PracticeComplete the steps to find the determinant.

9. � 3

4 1

2

1 �3

�2

3 �5

� Find the determinant.

10. � �2

4 2

3

1

�4

1

5

3 � 11. � 6

�4 3

2 3

5

�1

2

1 �

�70 7 12. A triangle with vertices (x1,y1), (x2,y2) and (x3,y3) has an

area equal to the absolute value of 1 __ 2 �

x1 x2 x3

y1 y2 y3

1 1 1

� . Find the x

y4

2

2 4

-2

-2-4

-4

(4, 2)(1, 3)

(-3, -2)

O

area of the triangle shown. 9.5 square units

Upper products: �2 , �27 , �40 Sum: �69 Lower products: �15 , 6 , 24 Sum: 15Determinant: 15 − �69 � 84

Upper products: �2 , �27 , �40 Sum: �69 Lower products: �15 , 6 , 24 Sum: 15Determinant: 15 − �69 � 84

14

Name Date Class


You have learned how to graph linear equations. Now you will learn how to solve systems of linear equations by graphing.

Solve this system by graphing.

y � x � 2 y � 2x � 5

Step 1: Write each equation in slope-intercept form by solving for y.

y � �x � 2 y � 2x � 5

Step 2: Graph each equation using the slope and y-intercept.

x

y4

23

1

21 3 4-4-3-2

-4-3-2-1

O-1

5

5-5

-5

y = 2x + 5y = -x + 2

(-1, 3)On the graph, the point where the lines intersect is the solution. The lines appear to intersect at (�1, 3).

Substitute (−1, 3) into the original equations to check. y � x � 2 y � 2x � 5

3 � (�1) � 2 3 � 2(�1) � 5 2 � 2 5 � 5

PracticeComplete the steps to solve this system by graphing.

1. x � y � 1 2x � y � 5

Write x � y � 1 in standard form: y � �x � 1

x

y4

23

1

21 3 4-4-3-2

-4-3-2-1

O-1

5

5-5

-5

y = -x + 1

y = 2x - 5

(2, -1)

Write 2x � y � 5 in standard form: y � 2x � 5

By inspection of the graph, the solution to the

system is (2, �1).

Solve this system by graphing.

2. 2x � y � 5

x

y45

23

1

21 3 4 5-4-5 -3-2

-4-5

-3-2-1

O-1

x + y = 3x - y = 1 x � y � 1

(2, 1)

ReteachingSolving Systems of Equations by Graphing 15


Reteachingcontinued

Exactly One Solution Independent and Consistent

Infinitely Many SolutionsDependent and Consistent

No SolutionInconsistent

The lines intersect. There is one point in common.

The lines coincide. There are an infi nite number of points in common.

The lines are parallel. There are no points in common.

x

y4

23

1

21 3 4-4-3-2

-2-1

O-1

5

5-5

x + y = 3 x - y = 1

x

y4

23

1

21 3 4-4-3-2

-2-1

O-1

5

5-5

2x = y-14y - 8x = 4

x

y43

1

21 3 4-4-3-2O

5

5-5

y - 1 = -2x

y + 2x = -3

-1-1

-2

2

Determine if the linear system at the right is consistent and independent, consistent and dependent, or inconsistent. If the system is consistent, give the solution.

Solution: Classify by the number of solutions of the system.

The lines intersect at one point. There is one solution (1, 3). The system is independent and consistent.

x

y4

23

1

21 3 4-4-3-2

-4-3-2-1

O-1

5

5-5

-5

y = x + 2

y = 3x

(1, 3)

PracticeComplete the steps to determine if the linear system is consistent and independent, consistent and dependent, or inconsistent. If the system is consistent, give the solution. 3.

x

y4

23

1

21 3 4-4-3-2

-2-1

O-1

5

5-5

y = 3x

y = 3x + 2

The lines intersect at no point(s).

There is no solution.

The system is inconsistent.

Determine if the linear system is consistent and independent, consistent and dependent, or inconsistent. If the system is consistent, give the solution. 4.

x

y4

23

1

21 3 4-4-3-2

-4-3-2-1

O-1

5

5-5

-5

y = -3x - 1 y = -x + 1

5.

x

y4

23

1

21 3 4-4-3-2

-4-3-2-1

O-1-5

5

5

-5

y = -x - 1

2x + 2y = -2

consistent and consistent and dependent; Independent; (–1, 2)

infinite set of ordered pairs

15

Name Date Class


You have learned how to solve systems of linear equations by graphing. Now you will learn how to solve systems of linear equations using Cramer’s rule.

Cramer’s Rule

The solutions of the linear system { ax � by � e

cx � dy � f are

x � � e f b

d � ______

D and y �

� a c e f � ______

D , where D � � a c b

d � , the determinant of the coefficient matrix.

Use Cramer’s Rule to solve { x � y � 2

2x � y � 7 .

Step 1: Write the coefficient matrix of the system of equations.

Then find the determinant of the coefficient matrix.

[ a c b d ] � [ 1

2 1

�1 ]

D � � a c b d

� � � 1 2 1

�1 � � 1(�1) � 2(1) � �3

Step 2: Solve for x and y using Cramer’s rule. Remember to divide by D.

PracticeComplete the steps to use Cramer’s rule to solve the system of linear equations.

1. { 2x � y � �1

4x � y � �5 D � � 2

4 1

1 � � �2

x � � �1 �5

1 1 � _______

D � �2 y �

� 2 4 �1

�5 � _______

D � 3

The solution to the system is (�2, 3).

Use Cramer’s rule to solve the system of linear equations.

2. { x � y � 1

3x � 2y � 4 (2, 1) 3. { y � x � 3

2x � y � 2

(5, 8)

Coefficient matrix

Determinant of the coefficient matrix

The coefficients of x in the coefficient matrix are replaced by the constant terms.

The coefficients of y in the coefficient matrix are replaced by the constant terms.

x � � e f b

d � ______

D �

� 2 7 1

�1 � _______

D � �2 � 7 _______

�3 � �9 ___

�3 � 3

y � � a c e

f � ______

D �

� 1 2 2

7 � ______

D � 7 � 4 _____

�3 � 3 ___

�3 � �1

The solution is (3, �1).

ReteachingUsing Cramer’s Rule 16


Reteachingcontinued

You can use the determinants to classify systems by their solutions.

Exactly One SolutionConsistent and Independent

Infinitely Many SolutionsConsistent and Dependent

No SolutionInconsistent

The determinant of the coefficient matrix is not zero

The determinant of the coefficient matrix is zero and at least one of the numerators is zero.

The determinant of the coefficient matrix is zero, but neither of the numerators are zero.

Use Cramer’s rule to determine if the linear system is consistent and independent, consistent and dependent, or inconsistent. If the system is consistent, give the solution.

{ �x � 3y � 3

�2x � 6y � �12

Solution: Set up and simplify the equations for x and y using the coefficient matrix and the constants.

x � � 3 �12

3 6

� ________

� �1 �2

3 6 � � 18 � 36 _______

�6 � 6 � 54 ___

0 y �

� �1 �2

3 �12

� __________

� �1 �2

3 6 � � 12 � 6 _______

�6 � 6 � 18 ___

0

The determinant of the coefficient matrix is zero, but neither of the numerators are zero. This is because the graphs of the equations are parallel lines and parallel lines never intersect. The system has no solution and is inconsistent.

PracticeComplete the steps to use Cramer’s rule to determine if the linear system is consistent and independent, consistent and dependent, or inconsistent. If the system is consistent, give the solution.

4. { x � 2y � 3

2x � 4y � 6 x �

� 3 6 �2

�4 � _______

� 1 2 �2

�4 � �

�12 � 12 ___________

�4 � 4 �

0 ____

0 y �

� 1 2 3

6 � _______

� 1 2 �2

�4 � �

6 � 6 __________

�4 � 4 � 0 ____

0

The determinant of the coefficient matrix is zero and at least one of the numerators

is zero. So, the system is consistent and dependent . The system

has infinitely many solution(s).

Use Cramer’s rule to determine if the linear system is consistent and independent, consistent and dependent, or inconsistent. If the system is consistent, give the solution.

5. { 3x � y � 0

2x � 2y � �4 6. { 4x � 2y � 8

2x � y � 4

consistent and consistent and dependent;

independent; (1, 3) infinitely many solutions.

16

Name Date Class


You have learned how to solve compound inequalities. Now you will learn how to solve linear equations and inequalities with absolute value.

Definition of Absolute Value

If x � 0, � x � � x. Example: � 5 � � 5. If x � 0, � x � � �x. Example: � �5 � � � � �5 � � 5.

Solve � x � 5 � � 3.

There are two possible cases:

x � 5 � 0, so � x � 5 � � x � 5,

x � 5 � 3 ✓x � 8

OR

x � 5 � 0, so � x � 5 � � �(x � 5).

�(x � 5) � 3

�x � 5 � 3

�x � �2

x � 2

The solutions are 8 and 2.

Check:

� x � 5 � � 3

� 8 � 5 � � 3

� 3 � � 3

3 � 3 ✓

Check:

� x � 5 � � 3

� 2 � 5 � � 3

� �3 � � 3

3 � 3 ✓

PracticeComplete the steps to solve the absolute value equation.

1. � x � 4 � � 9 2. 3 � x � 6 � � 18

x � 4 � 9 � � x � 4 � � 9 x � 6 � 6 � � x � 6 � � 6 x � 5 �x � 4 � 9 x � 12 �x � 6 � 6

x � �13 x � 0 Check: Check:

� 5 � 4 � � � 9 � � 9 3 � 12 � 6 � � 3 � 6 � � 18

� �13 � 4 � � � �9 � � 9 3 � 0 � 6 � � 3 � �6 � � �18

The solutions are 5 and �13. The solutions are 12 and 0 .

Solve.

3. � x � 7 � � 15 4. 2 � x � 3 � � 10

x � 22 or �8 x � 8 or �2

ReteachingSolving Equations and Inequalities with Absolute Value 17


Reteachingcontinued

Solving absolute value inequalities is like solving compound inequalities. Assume a is positive.

Solve: � x � � a

Solution: �a � x � a


Solution: x � �a OR x � aSolve: � x � � a

Solution: �a � x � a


Solution: x � �a OR x � a

Solve � x � 2 � � 3. Graph the solution.

� 3 � x � 2 � 3 Use the solutions from the table to write the inequality.

�3 � 2 � x � 2 � 2 � 3 � 2 Isolate the variable.

� 1 � x � 5 Simplify.

64 1080 2-2

Solve � 2x � 1 � � 5. Graph the solution.

2x � 1 � 5 OR 2x � 1 � �5 Use the Solutions from the table to write the inequality.

2x � 6 OR 2x � �4 Add 1.

x � 3 OR x � �2 Divide by 2.

4 60 2-2-4-6

PracticeComplete the steps to solve the absolute value inequality. Graph the solution.

5. � x � 3 � � 2 6. � 2x � 1 � � 3

�2 � x � 3 � �2 2x � 1 � 3 OR 2x � 1 � �3

�5 � x � �1 2x � 2 OR 2x � �4

x � 1 OR x � �2

4 60 2-2-4-6

4 60 2-2-4-6

Solve. Graph the solution.

7. � x � 5 � � 4 8. � 2x � 3 � � 5

x � �1 or x � �9 �1 � x � 4

0-2-4-6-8-10-12 4 6 80 2-2-4

17

Name Date Class


You have learned how to use ratios to solve problems. Now you will learn how to change one unit of measure to another.

a. Change 45 feet to inches.

Write 45 feet as a fraction: 45 ft ____ 1

There are 12 inches in a foot, so the conversion factor is 12 in _____ 1 ft

or 1 ft _____ 12 in.

.

Write a multiplication sentence so that feet will cancel.

45 ft ____ 1 � 12 in. _____

1 ft � 540 in.

b. Change 360 seconds to hours.

Write 360 seconds as a fraction: 360 s _____ 1

There are 60 seconds in a minute and 60 minutes in an hour, so the conversion factors are

1 min _____ 60 s

and 1 h ______ 60 min

.

Write a multiplication sentence so that seconds and minutes will cancel.

360 s _____ 1 � 1 min _____

60 s � 1 h ______

60 min � 0.1 h


1. Change 1032 inches to feet. 2. Change 6300 seconds toWrite the conversion factor to change Write the conversion factors to change seconds to hours:

inches to feet: 1 ft _____ 12 in.

or 12 in _____ 1 ft

60 s ______ 1 min.

and 60 min ______ 1 h

1032 in. _______ 1

� 1 ft _______ 12 in.

� 86 ft

6300 s ______ 1 � 1 min ________

60 s � 1h _______

60 min � 1.75 h

Solve.

3. Change 672 inches to feet. 4. Change 7200 seconds to

56 ft 2 h

5. Change 80 feet to inches. 6. Change 35 hours to seconds.

960 in. 126,000 s

7. Change 40 yards to feet. 8. Change 2820 seconds to minutes.

120 ft 47 min

ReteachingCalculating with Units of Measure 18


Reteachingcontinued

You can use dimensional analysis to convert rates.Convert 80 miles per hour to feet per second.

Step 1: Write the rate as a fraction.

80 miles per hour � 80 min ______ 1 h

Step 2: Write the conversion factor.There are 5280 feet in a mile, so the conversion factor is 5280 ft ______

1 mi

There are 3600 seconds in 1 hour, so the other conversion factor is 1 h ______ 3600 s

Step 3: Write the multiplication sentence so that hours and miles will cancel.

80 min ______ 1 h

� 5280 ft ______ 1 mi

� 1 h ______ 3600 s

� (80)(5280) ft

___________ (3600) s

� 422,400 ft _________ 3600 s

� 117 feet per second


9. Convert 40 miles per hour to feet per second.

Write the conversion factor to convert miles to feet: 5280 ft ______ 1 mi

or 1 mi ______ 5280 ft

Write the conversion factor to convert hours to seconds: 1 h ______ 3600 s

or 3600 s ______ 1 h

40 mi _____ 1 h

� 5280 ft

_______ 1 mi

� 1 h _______ 3600 s

� (40)(5280 ) ft

___________ (3600 ) s

� 211,200 ft __________

3600 s

� 59 feet per second

Solve.

10. Convert 36 miles per hour to miles 11. Convert 2640 feet per second to milesper second. per second.

0.01 mi/s 0.5 mi/s

12. Convert 50 miles per hour to feet 13. Convert 20 feet per second to milesper second. per hour.

about 73 ft/s about 14 mi/h

14. A soapbox car is a motorless vehicle that can reach speeds of up to 44 feet per second. What is this in miles per hour? 30 mi/h

15. Cullen built a pinewood boxcar and entered it in a local race. His boxcar was clocked at speeds of 60 miles per hour. What is this in feet per second? 88 ft/s

18

Name Date Class


You have added and subtracted polynomials. Now you will multiply polynomials.

To multiply two binomials, you can use the FOIL method. FOIL stands for FIRST, OUTER, INNER, LAST. To find the product of two binomials, find the sum of the products of the first terms, the outer terms, the inner terms, and the last terms.

(a � b)(c � d) � ac � ad � bc � bd

Find the product.

(2x � 4)(5x � 3)

first terms: 2x and 5x outer terms: 2x and 3

inner terms: 4 and 5x last terms: 4 and 3

(2x � 4)(5x � 3) � 2x � 5x � 2x � 3 � 4 � 5x � 4 � 3 Use the FOIL method.

� 10 x 2 � 6x � 20x � 12 Simplify.

� 10 x 2 � 26x � 12 Combine like terms.

PracticeComplete the steps to find the product.

1. (10a � 7)(3a � 1) � 10a � 3a � 10a � (�1) � 7 � 3a � 7 � (�1)

� 30 a 2 � 10a � 21a � 7

� 30 a 2 � 11a � 7Find the product.

2. (8c � 6)(c � 2) 3. (4x � 9)(2x � 3)

8 c 2 � 10c � 12 8 x 2 � 30x � 27

4. (5g � 1)(5g � 7) 5. (3b � 5)(4b � 10)

25 g 2 � 40g � 7 12 b 2 � 10b � 50

6. (y � 10)(6y � 1) 7. (5f � 9)(7f � 8)

6 y 2 � 59y � 10 35 f 2 � 23f � 72

8. (6m � 2)(3m � 3) 9. (�c � 9)(2c � 5)

18 m 2 � 24m � 6 �2 c 2 � 13c � 45

10. (3n � 4)(�2n � 4) 11. (�b � 2)(�5b � 6)

�6 n 2 � 20n � 16 5 b 2 � 16b � 12

ReteachingMultiplying Polynomials 19


Reteachingcontinued

To multiply a trinomial by a binomial, multiply each term in the trinomial by each term in the binomial.

(a � b)(c � d � e) � ac � ad � ae � bc � bd � be

Find the product.

(3a � 5)(5 a 2 � 3a � 2)

(3a � 5)(5 a 2 � 3a � 2) � 3a(5 a 2 ) � 3a(3a) � 3a(2) � 5(5 a 2 ) � 5(3a) � 5(2)

� 15 a 3 � 9 a 2 � 6a � 25 a 2 � 15a � 10

� 15 a 3 � 34 a 2 � 21a � 10

PracticeComplete the steps to find the product.

12. (2c � 4)( c 2 � 6c � 3) � 2c( c 2 ) � 2c(�6c ) � 2c ( 3 ) � 4( c 2 ) � 4(�6c ) � 4 ( 3 )

� 2 c 3 � 12 c 2 � 6c � 4 c 2 � 24c � 12

� 2 c 3 � 8 c 2 � 18c � 12Find the product.

13. (4x � 2)( x 2 � 4x � 1) 14. (5g � 1)(2 g 2 � 3g � 5)

10 g 3 � 17 g 2 � 28g � 5 10 g 3 � 17 g 2 � 28g � 5

16. (10n � 6)(2 n 2 � 4n � 2) 17. (6p � 3)(4 p 2 � 5p � 6)

20 n 3 � 28 n 2 � 4n � 12 5 y 3 � 44 y 2 � 13y � 36

17. (6p � 3)(4 p 2 � 5p � 6) 18. (y � 9)(5 y 2 � y � 4)

24 p 3 � 18 p 2 � 51p � 18 5 y 3 � 44 y 2 � 13y � 36

19. (�m � 3)(2 m 2 � 3m � 1) 20. (2s � 9)(�3 s 2 � 4s � 2)

�6 s 3 � 19 s 2 � 32s � 18 12 x 2 � 30x � 72

21. Mr. Lawrence has a rectangular garden, as shown.

Find the product (6x � 9)(2x � 8) to find its area.

2x + 8

6x - 9

12 x 2 � 30x � 72

19

Name Date Class


You have performed operations with polynomials. Now you will add, subtract, multiply, and divide functions. Let f (x) � x and g (x) � 2x. f (3) � 3 g (3) � 2(3) � 6 f (3) � g (3) � 3 � 6 � 9You can also write the sum algebraically before evaluating. (f � g )(x ) � f(x ) � g (x ) � x � 2x � 3x (f � g )(3) � 3(3) � 9The domain of the sum or difference of two functions is all of the x-values that are in the domain of both functions.

Find the difference of the functions and state the domain.

f (x ) � 3x � 1; D � {Integers} g(x) �5x � 3; D � {Positive integers}

(f � g )(x ) � f (x ) � g (x )

� (3x � 1) � (5x � 3) Substitute in the functions.

� 3x � 1 � 5x � 3

� �2x � 4 Combine like terms.

The domain of f(x) is all integers, but the domain of g(x) is only positive integers, so the domain of (f � g)(x) is {Positive integers}.

PracticeComplete the steps to find the sum of the functions and state the domain.

1. f (x) � 6x � 7; D � {Reals}

g(x) � �2x � 4; D � {Negative reals}

(f � g)(x ) � f (x) � g (x )

� (6x � 7) � (�2x � 4)

� 4x �3

The domain of (f � g)(x) is {Negative reals}.Find the sum or difference of the functions and state the domain.

2. f (x) � 8x � 10; D � {Reals}

g (x) � 4x � 8; D � {Integers}

(f � g)(x) � 12x � 18

Domain � {Integers}

4. f (x) � 12x � 4; D � {Positive integers}

g (x) � x � 3; D � {Reals}

(f � g)(x) � 11x � 7 Domain � {Positive integers}

3. f (x) � 9x � 15; D � {Reals}

g(x) � 10x � 2; D � {Reals}

(f �g)(x) � �x � 13

Domain � {Reals}

5. f (x) � 2x � 6; D � {Integers}

g (x) � �3x � 5; D � {Positive numbers}

(f � g)(x) � �x � 1 Domain � {Positive integers}

ReteachingPerforming Operations with Functions 20


Reteachingcontinued

You can also find products and quotients of functions.

(fg )(x ) � f (x ) � g (x) (f /g )(x ) � f (x) ____ g(x)

The domain of the product of two functions is all of the x-values that are in the domain of both functions. The domain of the quotient of two functions is all of the x-values that are in the domain of both functions, but do not include x-values that make g(x) � 0.

Find the product of the functions and state the domain. Then find (fg)(2).

f (x ) � x � 4; D � {Whole numbers} g (x) � �x � 5; D � {Positive integers}

(fg)(x) � f (x) � g(x)

� (x � 4)(�x � 5) Substitute in the functions.

� �x 2 � 5x � (�4x) � 20 Use the FOIL method.

� �x 2 � x � 20 Combine like terms.

The domain is {Positive integers}.

(fg)(2) � �22 � 2 � 20 Substitute 2 for x.

� 18

PracticeComplete the steps to find the quotient of the functions algebraically and state the domain. Find (f/g)(3).

6. f (x) � �3x � 6; D � {Reals}

g (x) � 2x � 8; D � {Reals}

(f/g)(x) � f (x)

____ g(x)

�3x � 6

_________ 2x � 8

(f/g)(3) � �3( 3 ) � 6

___________ 2( 3 ) � 8

� �9 � 6 _______ 6 � 8

� � 3 ___ 14

Find the value of x where g(x) � 0.

2x � 8 � 0

2x � �8

2x ____ 2

� �8

____ 2

x � �4

The domain is {Reals except �4}.

Find the product or quotient and state the domain. Then evaluate for the given value.

7. f (x) � 5x � 1; D � {Integers} (fg)(x) � 5 x 2 � 9x � 2 g (x) � x � 2; D � {Whole numbers} Domain � {Whole numbers} (fg)(1) � 12

20

Name Date Class


You have learned about and graphed functions. Now you will learn about and graph parametric equations.

Recall that in a function, the output value, y, is found by substituting in an input value, x. In parametric equations, the x- and y-values are both dependent on another variable called the parameter. In the equations below, the parameter is t.

{ x � 2t y � 3t

Graph the parametric equations { x � 2t y � 3t

, where t � 0.

Make a table.

t 0 1 2 3 4

x � 2t 2(0) � 0 2(1) � 2 2(2) � 4 2(3) � 6 2(4) � 8

y � 3t 3(0) � 0 3(1) � 3 3(2) � 6 3(3) � 9 3(4) � 12

Make a list of ordered pairs (x, y).

(0, 0), (2, 3), (4, 6), (6, 9), (8, 12)

Plot and connect the points.

PracticeComplete the steps to graph the parametric equations { x � 0.5t � 1

y � �4t .

where t � 0.1. t 0 1 2 3 4

x � 0.5t � 1 0.5(0) � 1 � 1

0.5(1) � 1 � 1.5

0.5(2) � 1 � 2

0.5(3) � 1 � 2.5

(0.5) � 1 � 3

y � 4t 4(0) � 0 4(1) � 4 4(2) � 8 4 (8) � 12 4 (4) � 16

Graph the parametric equations. Make a list of ordered pairs: (1, 0), (1.5, 4 ), (2, 8 ), (2.5, 12 ), ( 3 , 16 ) Plot and connect the points.

2. { x � 2t y � 2t

, t � 0 3. { x � 2t y � 0.5t � 2

, t � 0

x

y12

8

4

4 8-4O

x

y

12

16

8

4

2 4-4O

x

y6

4

2

2 4 6-2O x

y6

4

2

2 4 6-2O

ReteachingSolving Parametric Equations

INV

2


Reteachingcontinued

Parametric equations can be used to describe the paths of projectiles. In these equations, x is the horizontal distance the object has traveled and y is the height of the object. The parameter t is used to represent time.

An object has an initial horizontal velocity of 100 feet per second and an initial vertical

velocity of 75 feet per second. Use the parametric equations { x � 100t y � �16t2 � 75t

to graph the path of the object.

Make a table. Because t represents time, do not include negative values for t.

t 0 1 2 3 4

x � 100t 100(0) � 0 100(1) � 100 100(2) � 200 100(3) � 300 100(4) � 400

y � �16t 2 � 75t

�16(0)2 � 75(0) � 0

�16(1)2 � 75(1) � 59

�16(2)2 � 75(2) � 86

�16(3)2 � 75(3) � 81

�16(4)2 � 75(4) � 44

Make a list of ordered pairs (x, 1).

(0, 0), (100, 59), (200, 86), (300, 81), (400, 44)


PracticeComplete the steps to graph the path of the object represented by the parametric

equations { x � 125t y � �16t2 � 80t

.

4. t 0 1 2 3 4

x � 125t 125(0) � 0 125(1) � 125 125(2) � 250 125(3) � 375 125(4) � 500y ��16t 2 � 80t

�16(0)2 � 80(0) � 0

�16(1 ) 2 � 80(1) � 64

�16(2 ) 2 � 80(2) � 96

�16(3)2 � 80(3) � 96

�16(4)2 � 80(4) � 64

Make a list of ordered pairs (x, y).

(0, 0), (125, 64 ), (250, 96 ), (375, 96 ), (500, 64 )


x

y

60

80

100

40

20

100 200 300 400O

x

y

60

80

100

40

20

100 200 300 400 500O

INV

2

Name Date Class


You have solved systems of equations by graphing. Now you will solve systems of equations by substitution. To solve a system of equations using the substitution method, solve one of the equations for one of the variables so that it can be substituted in for the variable in another equation.

Solve the system of equations by substitution.

{ y � 4x � �2

2y � 7x � �5

Choose an equation and solve for one of the variables. The first equation requires only 1 step to solve for y. Choose this one to solve first.

y � 4x � �2

y � �2 � 4x

Substitute into the other equation to solve for x.

2(�2 � 4x) � 7x � �5 Substitute.

2(�2) � 2(4x) � 7x � �5 Distributive Property

�4 � x � �5 Combine like terms.

�x � �1 Add 4 to each side.

x � 1 Multiply each side by �1.

Substitute the x-value into one of the equations to solve for y.

y � 4x � �2

y � 4(1) � �2 Substitute for x.

y � 4 � �2 Multiply.

y � �6 Subtract 4 from each side.

The solution is (1, �6).

PracticeComplete the steps to solve the system of equations.

1. { 2y � x � 5

3y � 4x � �5

Solve for a variable:

2y � x � 5

x � 5 � 2y

Substitute in the other equation:

3y � 4 (5 � 2y ) � �5

3y � 20 � 8y � �5

�5y � 20 � �5

�5y � �25

y � 5

Solve for the other variable:

2 ( 5 ) � x � 5

10 � x � 5

x � �5

The solution is (�5, 5 ).

Solve the system of equations.

2. { y � x � 6

2y � 5x � �2

(2, 4)

3. { �3y � x � �3

3y � 4x � �7

� �2, 1 __ 3 �

ReteachingSolving Systems of Equations Using the Substitution Method 21


Reteachingcontinued

When the solution to a system of equations is a point, the graphs of the equations intersect. When you solve a system of equations and the result is a true numerical statement, the lines are coinciding. In other words, the equations describe the same line. When you solve a system of equations and the result is a false numerical statement, the lines are parallel.

Determine if the system of equations form coinciding or parallel lines.

{ y � 2x � 7

�3y � 6x � �21

Choose an equation and solve for one of the variables.

y � 2x � 7

y � 7 � 2x

Substitute into the other equation.

�3(7 � 2x) � 6x � �21

�21 � 6x � 6x � �21

�21 � �21

The statement is true, so the system of equations form coinciding lines.

The solution can be written as {(x, y)�y � 7 � 2x}

PracticeComplete the steps to determine if the system of equations forms coinciding or parallel lines.

4. { 2y � 10x � �8

�4y � 20x � �8

2y � 10x � �8

2y � �8 � 10x y � �4 � 5x

�4(�4 � 5x ) � 20x � �8

16 � 20x � 20x � �8

16 � �8

The statement is false, so the system of equations form parallel lines.

Determine if the system of equations form coinciding or parallel lines.

5. { �y � 3x � 2

5y � 15x � �10

coinciding lines

6. { 2y � 6x � 4

�4y � 12x � 8

parallel lines

7. One hundred people attended a school play. Admission tickets cost $5 for children and $8 for adults. A total of $695 was collected for admission. Find the number of child and adult tickets sold by solving

the system { a � c � 100 8a � 5c � 695

. 65 adult tickets and 35 child tickets

21

Name Date Class


You have graphed functions and identified their domain and range. Now you will learn how to identify different types of functions.

A function with a graph that has no gaps, jumps, or asymptotes is a continuous function.A function with a graph that has gaps, jumps, or asymptotes is a discontinuous function. A discontinuous function that is made up of separate, disconnected points is a discrete function.

Determine if each graph is continuous, discontinuous, and/or discrete. If the function is discontinuous, name the x-value(s) where the discontinuity occurs.

Check the graph for gaps, jumps, and asymptotes.

The graph has many gaps. It is made up of separate, disconnected points, so it is discrete.

The function is discontinuous at x � �1, 1, 1, 3, 4.


It has a jump atx � 1.

The function is discontinuous at x � 1.

PracticeComplete the steps to determine if each graph is continuous, discontinuous, and/or discrete. If the function is discontinuous, name the x-value(s) where the discontinuity occurs.

1. gap(s) nonejump(s) noneasymptote(s) noneContinuous function.

2. gap(s) nonejump(s) noneasymptote(s) x � 0 Discontinuous function.

Discontinuous at x � 0

Determine if each graph is continuous, discontinuous, and/or discrete. If the function is discontinuous, name the x-value(s) where the discontinuity occurs.

3. discontinuous;discontinuous at

x � 0

4. discontinuous anddiscrete;

discontinuous atx � �3, 1, 0, 1, 3

x

y4

2

2 4

-2

-2-4

-4

O x

y4

2

2 4

-2

-2-4

-4

O

x

y4

2

2 4

-2

-2-4

-4

O x

y4

2

2 4

-2

-2-4

-4

O

x

y4

2

2 4

-2

-2-4

-4

O x

y4

2

2 4

-2

-2-4

-4

O

ReteachingAnalyzing Discrete and Continuous Functions 22


Reteachingcontinued

Vertical line test: A relation is a function if any vertical line only intersects the graph of the relation at one point.The vertical line test can be used on continuous, discontinuous, and/or discrete relations.

Determine if each graph is continuous, discontinuous, and/or discrete. Determine if it is a function or a relation.


The graph has many gaps. It is made up of separate, disconnected points, so it is discrete.

The graph passes the vertical line test at every point. It is a function.


The graph has a gap at x � 2.

The graph fails the vertical line test at x � 2. It is a relation.

PracticeComplete the steps to determine if each graph is continuous, discontinuous, and/or discrete. Determine if it is a function or a relation.

5. gap(s) x � 1jump(s) x � 3asymptote(s) noneDiscontinuous graph.

6. gap(s) nonejump(s) noneasymptote(s) noneContinuous graph.

The graph passes the vertical line test.

The graph is a function.

The graph fails the vertical line test.

The graph is a relation.

Determine if each graph is continuous, discontinuous, and/or discrete. Determine if it is a function or a relation.

7. discontinuous;function

8. discontinuous,discrete; relation

x

y4

2

2 4

-2

-2-4

-4

O x

y4

2

2 4

-2

-2-4

-4

O

x

y4

2

2 4

-2

-2-4

-4

O x

y4

2

2 4

-2

-2-4

-4

O

x

y4

2

2 4

-2

-2-4

-4

O x

y4

2

2 4

-2

-2-4

-4

O

22

Name Date Class


You have factored numbers and expressed them as a product of prime factors. Now you will factor polynomials to express them as a product of prime polynomial factors.

Solve the equation ax2 � bx � c � 0 to find the zeros of f(x) � ax2 � bx � c.

Solve x2 � 2x � 15 � 0.

x2 � 2x � 15 � 0

(x � 5)(x � 3) � 0 Factor. Then multiply to check.

(x � 5) � 0 or (x � 3) � 0Set each factor equal to 0.

x � �5 or x � 3

To check the solutions, substitute each solution for the variable in the original equation.

Equation: x2 � 2x � 15 � 0 x2 � 2x � 15 � 0

Solution: x � �5 x � 3

Check: (�5)2 � 2(�5) � 15 � 0 (3)2 � 2(3) � 15 � 0

25 � 10 � 15 � 0 9 � 6 � 15 � 0

0 � 0 ✓ 0 � 0 ✓

The solutions of x2 � 2x � 15 � 0 are −5 and 3.

The zeros of f(x) � x2 � 2x � 15 are −5 and 3.

PracticeComplete the steps to solve each equation.

1. 4x2 � 24x � 0 2. x2 � 4x � 3 �0

4x(x � 6) � 0 (x � 3)(x � 1) � 0

4x � 0 or (x � 6) � 0 (x � 3) � 0 or (x � 1) � 0

x � 0 or x � 6 x � �3 or x � �1

The solutions are 0 and 6 The solutions are �3 and �1Solve each equation.

3. x2 � 5x � 4 � 0 4. 3x2 � 12x � 0

x � 1, x � 4 x � 0, x � �4 5. A rectangular patio is 12 feet long and 8 feet wide. The owner wants to

increase both dimensions by the same amount, resulting in a patio that has an area of 221 square feet. Write and solve an equation to find out how much the owner should increase each dimension.

(x � 12) (x � 8) = 221; 5 ft

Solve each equation for x.

23ReteachingFactoring Polynomials

The solutions of the equation are the zeros of the function.


Reteachingcontinued

Some quadratic equations have special factors.

Difference of Two Squares: a2 � b2 � (a � b)(a � b)

Perfect Square Trinomials: a2 � 2ab � b2 � (a � b)2

a2 � 2ab � b2 � (a � b)2

Always write a quadratic equation in standard form before factoring.

Solve 16x2 � 25.

16x2 � 25

16x2 � 25 � 0

16 and 25 are perfect squares. Use the difference of two squares to factor.

(4x)2 � (5)2 � 0

(4x � 5)(4x � 5) � 0

(4x � 5) � 0 (4x � 5) � 0 Set each factor equal to 0.

x � 5 __ 4 or x � � 5 __

4

Try to factor a perfect square trinomial if the coefficient of x and the constant term are perfect squares.

Solve 4x2 � 12x � 9 � 0.

4x2 � 12x � 9 � 0 4x2 and 9 are perfect squares. (2x)2 � 2(2x)(3) � (3)2 � 0

(2x � 3)2 � 0

(2x � 3) � 0

x � 3 __ 2

The factors are the same.

PracticeComplete the steps to solve each equation.

6. 4 x 2 � 49 7. x 2 1 16 � 8x

4 x 2 � 49 � 0 x 2 �8x � 16 � 0

(2x ) 2 � ( 7 ) 2 � 0 ( x ) 2 � 2( x )( 4 ) � ( 4 ) 2 � 0

(2x � 7 )(2x � 7 ) � 0 (x � 4 ) 2 � 0

(2x � 7 ) � 0 or (2x � 7 ) � 0 (x � 4) � 0

x � � 7 __ 2 or x � 7 __

2 x � 4

Solve each equation.

8. 4 x 2 � 4x � �1 9. 9 x 2 � 64

x � � 1 __

2

x � � 8 __

3 , x � 8 __

3

23

Name Date Class


You have solved systems of equations by graphing and by using substitution. Now you will solve systems of equations using elimination.

Solve the system of equations { 7 � 3x � 2yy � 5x � 3

.

Step 1: Write each equation in standard { 7 � 3x � 2y

y � 5x � 3 { 3x � 2y � 7

5x � y � 3

form.

Step 2: Multiply the second equation by 2. { 3x � 2y � 7

2(5x � y � 3) { 3x � 2y � 7

10x � 2y � 6

Step 3: Add the equations. The y-terms will 3x � 2y � 7 be eliminated. Solve for x. __ �10x � 2y � 6

13x � 13 x � 1

Step 4: Substitute the value for x in either of y � 5x � 3the original equations to find y. y � 5(1) � 3

y � 5 � 3

y � 2

The solution to the system of equations is (1, 2).

Check: Substitute 1 for x and 2 for y in both 7 � 3x � 2y y � 5x � 3original equations: 7 � 3(1) � 2(2) 2 � 5(1) � 3

7 � 3 � 4 2 � 5 � 3

7 � 7 ✓ 2 � 2 ✓

PracticeComplete the steps to solve the system of equations.

1. { 2x � 5 � �3y

11 � 4x � y { 2x � 5 � �3y

11 � 4x � y

{ 2x � 3y � �5

4x � y � 11

2x � 3y � �5

� 12x � 3y � 33

14x � 28

2x � 5 � �3y

2( 2 ) � 5 � �3y x � 2 , y � �3 9 � �3y The solution is (2, �3 ). y � �3 Solve each system of equations.

2. { 3x � 13 � 4y

y � �2x � 12 3. { x � 5y � 6

y � �4x � 18

(7, �2) (�4, �2)

x � 2

ReteachingSolving Systems of equations Using the Elimination Method 24


Reteachingcontinued

Linear systems of two equations are classified by their solutions.

Consistent Inconsistent

Independent Dependent

Exactly One Solution Infinitely Many Solutions No Solution

Two Lines Intersect at a point Two Lines are the same Two Lines are parallel

Solve { x � 3y � 1

�2x � 6y � �2 and classify the system.

Step 1: Multiply the first equation by 2.

{ 2(x � 3y � 1)

�2x � 6y � �2 → { 2x � 6y � 2

�2x � 6y � �2

Step 2: Add the two equations to solve for x and y. 2x � 6y � 2

__ � �2x � 6y � �2

0 � 0

Solve { 2x � 3y � 1

6x � 9y � �3 and classify the system.

Step 1: Multiply the first equation by �3.

{ �3(2x � 3y � 1) 6x � 9y � � 3

→ { �6x � 9y � �3

6x � 9y � �3

Step 2: Add the two equations to solve for x and y. �6x � 9y � �3 __ � 6x � 9y � �3 0 � �6

PracticeComplete the steps to solve and classify each system of equations.

4. { 4x � y � 2

�12x � 3y � 6 → 12x � 3y � 6 5. { �3x � 4y � �1

12x � 16y � 4

→ �12x � 16y � �4 __ � �12x � 3y � 6 __ � �12x � 16y � 4

0 � 12 0 � 0 No solution; inconsistent Infinitely many solutions; consistent and dependentSolve and classify each system of equations.

6. { �2x � 6y � 2 x � 3y � �1

7. { 4x � 2y � 2

2x � y � �1

Infinitely many solutions; No solution; inconsistent consistent and dependent

There are infinitely many solutions. The system is consistent and dependent.

There are no solutions. The system is inconsistent.

0 � �6 is a false statement.

0 � 0 is a true statement.

24

Name Date Class


You have used graphs to analyze linear equations and inequalities. Now you will use measures of central tendency and dispersion to analyze statistical data.

The range and standard deviation are measures of dispersion.

For the data set x1, x2,... xn, _ x is the mean and

� � ��

(x1 �

_ x )2 � (x2 �

_ x )2 � K � (xn �

_ x )2

_______________________________ n

They describe the spread of the data. The standard deviation measures spread from the mean.

is the standard deviation

Find the range, mean, and standard deviation for the data: 6, 8, 9, 8, 7, 5, 6, and 7

Step 1: Find the range. The range is the difference between the largest and smallest data values. The range is 9 � 5 � 4.

Step 2: Find the mean:

_ x � 6 � 8 � 9 � 8 � 7 � 5 � 6 � 7 __________________________

8 � 56 ___

8 � 7

The mean is the sum of the data values divided by the number of data values.

Step 3: Find the standard deviation:

� � ��

(6 � 7)2 � (8 � 7)2 � (9 � 7)2 � (8 � 7)2 � (7 � 7)2 � (5 � 7)2 � (6 � 7)2 � (7 � 7)2

______________________________________________________________________ 8

� ��

1 � 1 � 4 � 1 � 0 � 4 � 1 � 0 __________________________ 8 � �

�

12 ___ 8 � 1.2

PracticeComplete the steps to find the range, mean, and standard deviation for the data.

1. 2, 1, 5, 3, 1, 5, 3, and 4

range � 5 � 1 � 4 _ x �

2 � 1 � 5 � 3 � 1 � 5 � 3 � 4 ______________________________________

8 �

24 ____

8 � 3

� � (2 � 3)

2 � (1 � 3)

2 � (5 � 3)

2 � (3 � 3)

2 � (1 � 3)

2 � (5 � 3)

2 � (3 � 3)

2 � (4 � 3)

2 � 8

� ��

1 � 4 � 4 � 0 � 4 � 4 � 0 � 1 ______________________________________

8 � �

�

18 ______ 8 � 1.5

Find the range, mean, and standard deviation for the data.

2. 9, 4, 6, 5, 4, 8, 5, 7, 6 3. 11, 8, 15, 13, 13, 16, 9, 11, 14, 10

5; 6; 1.6 8; 12; 2.5

4. The annual precipitation amounts for Death Valley, California from 1995 to 2003, rounded to the nearest inch, are 3, 1, 3, 4, 1, 1, 3, 0,

25ReteachingFinding Measures of Central Tendency and Dispersion


Reteachingcontinued

and 3. What is the mean amount of precipitation in Death Valley for those years, and what is the standard deviation?

The mean is 2.1. The standard deviation is 1.3.A box-and-whisker plot can be used to show how the data in a set is distributed.

Draw a box-and-whisker plot to display the data:

7, 2, 1, 9, 6, 10, 6, 5, 3, 8, 3

Step 1: Write the data in ascending order and find the median.

1, 2, 3, 3, 5, 6, 6, 7, 8, 9, 10

Step 2: Find the quartiles. The first quartile (Q1) is the median of the lower half. The third quartile is the median of the upper half (Q3).

1, 2, 3, 3, 5, 6, 6, 7, 8, 9, 10

minimum Q1 median(Q2) Q3 maximum

Step 3: Draw the box-and-whisker plot.

MedianQ1 Q3Max.Min.

3 420 1 5 8 976 10

PracticeComplete the steps to draw a box-and-whisker plot of the data.

5. 6, 2, 4, 2, 1, 3, 4

Find the minimum, maximum, 1, 2, 2, 3, 4, 4, 6

and quartiles.

minimum (Q1) median (Q2) Q3 maximum

Draw the box-and-whisker plot. 3 420 1 5 76

Draw a box-and-whisker plot of the data.

6. 5, 7, 10, 4, 6, 8, 5 7. 3, 2, 5, 1, 2, 3, 0

3 4 8 10 1196 75 3 420 1 5 6

25

Name Date Class


You have solved and graphed linear equations. Now you will write the equation of a line in different forms.You can transform linear equations from slope-intercept form to standard form, and from standard form to slope intercept form.

Slope-intercept form: y � mx � b

m � slope, b � y-intercept

Standard form: Ax � By � C

A, B, C are real numbers. A and B are not both zero.

a. Write 3x � 5y � 20 in slope-intercept form.

Step 1: Isolate the y term on one side of the equation.

5y � �3x � 20

Step 2: Divide both sides of the equation by 5.

5y

___ 5 � �3x � 20 _________

5

y � � 3 __ 5

x � 4

b. Write y � 7 __ 4 x � 6 in standard form.

Step 1: Move all terms that contain variables to � 7 __

4 x � y � �6 one side of the equation.

Step 2: Eliminate all fractions by multiplying Multiply bothboth sides of the equation by a sides by �4factor. Simply.

PracticeComplete the steps to write each equation in the form specified.

1. 5x � 2y � 32 in slope-intercept form 2. y � 2 __ 5 x � 8 in standard form

�2y � �5x � 32 y � 2 __ 5 x � 8

y � �5x � 32 _________ �2

� 2 __ 5 x � y � 8

y � 5 __ 2

x � 16 2x � 5y � �40

Write each equation in the form specified.

3. 6x � 11y � 2 in slope-intercept form 4. y � 1 __ 3 x � 5 __

6 in standard form

y � � 6 ___ 11

x � 2 2x � 6y � 5

�4 � � 7 __ 4 x � y � � �4(�6)

ReteachingWriting the Equation of a Line 26


Reteachingcontinued

The point-slope form of a line is y � y1 � m(x � x1), where m is the slope and (x1, y1) is any point on the line.

a. A line has slope 4 and passes through the point (2, 9). What is the equation of this line written in slope-intercept form?

m � 2, (x1, y1) � (2, 9)

y � y1 � m(x � x1)

y � 9 � 4(x � 2)

y � 4x � 8 � 9 y � 4x � 1

b. A line passes through the points (�1, 8) and (4, �7). What is the equation of this line written in slope-intercept form?Step 1: Find m. Let (x1, y1) � (4, �7) and (x2, y2) � (�1, 8).

m � y2 � y1 ______ x2 � x1

� 8 � (�7)

________ �1 � 4

� 15 ___ �5

� �3

Step 2: Use the point-slope form of a linear equation with m � �3 and

(x1, y1) � (4, �7).

y - y1 � m(x � x1)

y � (�7) � (�3)(x � 4)

y � 7 � �3x � 12

y � �3x � 5

PracticeComplete the steps to find the equation of the line written in slope-intercept form.

5. slope � 1 __ 2 , passes through (�2, �6) 6. passes through (0, �3) and (2, 7)

m � � 1 __ 2 , (x1, y1) � (�2, �6) m �

y2 � y1 ______ x2 � x1 � 5

y � (�6) � � 1 __ 2 (x � (�2)) y � y1 � m(x � x1)

y � (�3) � 5 (x � 0 )

y � � 1 __ 2 x � 7 y � 5x � 3

Find the equation of the line written in slope-intercept form.

7. slope �3, passes through (3, �1) 8. passes through (3, �7) and (9, �3)

y � �3x � 8 y � 2 __ 3 x � 9

You could also use the other point, and let (x1, y1) � (�1, 8.).

26

Name Date Class


You have learned about linear and polynomial functions. Now you will explore quadratic functions, which are one type of polynomial function.

The graph of every quadratic equation is a parabola.

Graph f (x) � x2 � 4x � 3.Step 1: Make a table of

ordered pairs.

Step 2: Plot the ordered pairs from the table.

Step 3: Sketch a smooth curve to connect the points.

PracticeComplete the steps to graph the quadratic function.

1. f(x) � x2 � x � 6

Graph each quadratic function.

2. f(x) � x2 � x � 2 3. f(x) � x2 � 2x � 3

x f (x) � x2 � 4x � 3 (x, f (x))

0 f(0) � 02 � 4(0) � 3 � 3 (0, 3)

1 f(1) � (1)2 � 4(1) � 3 � 1 � 4 � 3 � 0 (1, 0)

2 f(2) � (2)2 � 4(2) � 3 � 4 � 8 � 3 � �1 (2, 1)

3 f(3) � (3)2 � 4(3) � 3 � 9 � 12 � 3 � 0 (3, 0)

4 f(4) � (4)2 � 4(4) � 3 � 16 � 16 � 3 � 3 (4, 3)

x

y4

2

4

-2

-2-4

-4

O x

y

2

4-4 -2

-2

-4

O

x f(x) � x2 � x � 6 (x, f(x))

�3 f(�3) � (�3 )2 � (�3 ) � 6 � 9 � 3 � 6 � 0 (�3, 0)

�2 f(�2) � (�2)2 � (�2) � 6 � 4 � 2 � 6 � �4 (�2, �4)

�1 f(�1) � (�1)2 � (�1) � 6 � 1 � 1 � 6 � �6 (�1, �6)

0 f(0) � (0)2 � (0) � 6 � 0 � 0 � 6 � �6 (0, �6)

1 f(1) � (1)2 � (1) � 6 � 1 � 1 � 6 � �4 (1, 4)

2 f(2) � (2)2 � (2) � 6 � 4 � 2 � 6 � 0 (2, 0)

x

y2

2 4-2-4

-2

-4

O

The curve changes at (2, �1). This point is the vertex of the function.

x

y

2

4

2 4-2-4

-4

Ox

y

2

4

2 4-2-4

-2

O

ReteachingConnecting the Parabola with the Quadratic Function 27


Reteachingcontinued

You can find various characteristics of a quadratic equation from its graph.

Find the intercepts of f(x) � x2 � 4x � 3.

Use the graph of the function from the previous page.

The graph crosses the x-axis at x � 1 and x � 3. These are thex-intercepts.

The graph crosses the y-axis at y � 3. These is the y-intercepts.

Find the vertex of f(x) � x2 � 4x � 3.

The vertex is the lowest point on the graph. Since the graph of a quadratic equation is symmetric, it occurs midway between the

x-intercepts.

Evaluate f(x) at x = 2. f(2) � (2)2 � 4(2) � 3 � 4 � 8 � 3 � �1

The vertex is (2, −1).

Identify the domain and range of f(x) � x2 � 4x � 3.

The domain is all real numbers.

The least possible value of the function occurs at the vertex. The range is all real numbers greater than or equal to −1.

PracticeComplete the steps to find the intercepts, vertex, domain and range of the function.

4. f(x) � x2 � x � 6

The graph crosses the x-axis at x = �3 and x = 2 . The x-intercepts are �3 and 2 .

The graph crosses the y-axis at y = �6. The y-intercept is �6.

The x-coordinate of the vertex is x � �3 � 2

__________ 2 � �0.5.

The y-coordinate of the vertex is f(�0.5) � (�0.5)2 � (�0.5) � 6 � 0.25 � 0.5 � 6

� �6.25. The vertex is (�0.5, �6.25).

D: all real numbers. R: all real numbers greater than or equal to �6.25 .

Find the intercepts, vertex, domain and range of each function.

5. f(x) � x2 � x � 2 x-intercepts: �2, 1; y-intercept: �2;

vertex: (�0.5, �2.25); D: all real numbers;

R: all real numbers greater than or equal to �2.25

6. f(x) � x2 � 2x � 3 x-intercepts: �1, 3; y-intercept: �3; vertex: (1, �4); D: all real numbers; R: all real numbers greater than or equal to �4

x

y

2

4-4 -2

-2

-4

O

The midpoint of the x –intercepts is

x � 3 � 1 _____ 2 � 2

Any real number can be squared.

27

Name Date Class


You have multiplied and factored polynomials. Now you will simplify

rational expressions which are composed of polynomials.

A rational expression is the quotient of two polynomials.

Examples of rational expressions: 3 __ x , x � 1 _____ x � 2

, and x � 3 ______ x 2 � 1

An excluded value for a function is a domain value that makes the range undefined. Therefore, the denominator of a rational function cannot equal zero.

Find the excluded values of 3 __ x .

x � 0Find the excluded values of x � 1 _____

x � 2 .

x � 2 � 0

x � �2Find the excluded values of x � 3 ______

x 2 � 1 .

x 2 � 1 � 0

(x � 1) (x � 1) � 0

x � 1 � 0 or x � 1 � 0

x � 1 or x � �1

PracticeComplete the steps to find the excluded values of each rational expression.

1. x � 5 _____ 2 x 2

2 x 2 � 0 2. 2 � b _______ b 2 � 4b

b 2 � 4b � 0

x 2 � 0 b (b � 4) � 0

x � 0 b � 0 or b � 4

The excluded value is 0. The excluded values are 0 and 4.

Find the excluded values of each rational expression.

3. w 3 _____ w � 5

4. 3 � 2q

______ 9 � q 2

�5 3 and �3

5. 3m � 1 _______ 7

6. 5x ______ 2x � 1

no excluded values 1 __ 2

Set the denominator equal to 0.The excluded value is 0.

Set the denominator equal to 0.The excluded value is �2.

Set the denominator equal to 0.The excluded value are 1 and �1.

ReteachingSimplifying Rational Expressions 28


When simplifying a rational expression:

Factor the numerator and the denominator completely.

Divide out any common factors.

Identify any x-values for which the expression is undefined.

Simplify 24 x 6 _____ 8 x 2

.

24 x 6 ____ 8 x 2

� 8 � 3 _____ 8

x 6 � 2 � 3 x 4

Simplify x 2 � 2x � 8 ___________ x 2 � x � 2

First, factor the numerator and the denominator.

x 2 � 2x � 8 __________ x 2 � x � 2

� (x � 4) (x � 2) ____________ (x � 2) (x � 1)

� ( x � 4) (x � 2) _____________ (x � 2) (x � 1)

� x � 4 _____ x � 1

PracticeComplete the steps to simplify each rational expression. Identify any excluded values.

7. 20 x 9 ____ 4 x 3

� 4 � 5 ______ 4

x 9 � 3 � 5 x 6 8. x 2 � 2x � 3 __________ x 2 � 6x � 5

� (x � 1) (x � 3)

________________ (x � 1) (x � 5)

� x � 3 ______ x � 5

The excluded value is 0 . The excluded values are �1 and �5 .

Simplify each rational expression. Identify any excluded values.

9. 3 s 4 ____ 12 s 8

10. x 2 � 7x � 12 ___________ x 2 � x � 6

1 ___ 4 s 4

; 0 x � 3 _____ x � 2

; �3, 2

11. A soup can is a cylinder. Write a simplified rational expression for the volume to lateral surface area ratio of a soup can. Then find the volume to lateral surface area ratio for a soup can whose diameter is 3 inches. (Hint: for a cylinder, volume � � r 2 h and lateral surface area � 2�rh.)

r __ 2 ; 3 __

4

•

•

•

x � 0, because 8 x 2 is undefined at x � 0.

Use the Quotient of powers Property.

x � �2 and x � 1. Divide out common factors.

Reteachingcontinued 28

Name Date Class


You have solved systems of two linear equations and two variables using the elimination method. Now you will solve systems of three linear equations and three variables using the same method.

Solve the system of equations { x � y � 2z � 8

2x � y � z � �2

x � 2y � z � 2 .

Step 1: Use the first and second equations to eliminate a variable. Add the equations to eliminate y.

Step 2: Use the second and third equations to eliminate y. Multiply the second equation by −2 and add the result to the third equation.

Step 3: Now you have two equations in two variables. Solve using the elimination method for a system of two equations.

Step 4: Substitute the values for x and z in any of the original equations to find y.

x � y � 2z � 8

__ �2x � y � z � �2

3x � z � 6

�2(2x � y � z � �2) �4x � 2y � 2z � 4

__ �x � 2y � z � 2 __ �x �2y � z � 2

�3x � 3z � 6

3x � z � 6 3x � z � 6

__ �3x � 3z � 6 3x � (3) � 6

4z � 12 3x � 3

z � 3 x � 1

x � y � 2z � 8

(1) � y � 2(3) � 8

1 � y � 6 � 8

�y � 1

y � �1The solution to the system of three equations is (1, −1, 3). y � �1

Practice

Complete the steps to solve the system of equations.

1. { 2x � y � z � �3

x � 2y � z � 2

x � 3y � 2z � 3 2x � y � z � �3 �2(x � 2y � z � 2)

__ �x � 2y � z � 2 __ � x � 3y � 2z � 3

3x � y � �1 �x � y � �1

{ 3x � y � �1 �x � y � �1

x � �1, y � 2, z � 1 The solution is � �1, 2, 1 � .Solve the system of equations.

2. { 2x � y � z � 5

x � 2y � z � 6

3x � y � 2z � 5 3. { �x � y � 2z � 0

3x � 2y � z � �5 x � y � 3z � �2

(3, �2, �1) (1, �3, 2)

ReteachingSystems of Three Equations and Three Variables 29


Reteachingcontinued

Linear systems in three variables are classified by their solutions.

Exactly One Solution Consistent

Infinitely Many Solutions Consistent

No Solution Inconsistent

Three plane intersect at one point

Three planes intersect at a line.

All three planes never intersect.

Classify { x � z � 1

x � y � z � 2

x � y � z � 1 .

Solve: { x � z � 1 2x � 2z � 3

Since 0 does not equal 1, the system has no solution and is inconsistent.

x � y � z � 2

__ x �y � z � 1

0 � 1✗

�2x � 2z � �2

__ � 2x � 2z � 3

0 � 1✗

Classify { x � 2y � 4z � 3

4x � 2y � 6z � 2

2x � y � 3z � 1 .

Now you have a system with two identical equations. Subtracting the equations gives 0=0.The system has infinitely many solutions and is consistent.

x � 2y � 4z � 3

__ �4x � 2y � 6z � 2

5x � 2z � 5

4x � 2y � 6z � 2

__ � x � 2y � 4z � 3

5x � 2z � 5

{ 5x � 2z � 5 5x � 2z � 5

PracticeComplete the steps to classify the system of equations.

4. { x + z � 0

x � y � 2z � 3

y � z � 2 5. { y � z � 0

x � 3z � �1 �x � 3y � 1

x � z � 0 x � y � 2z � 3 x � 3z � �1

__ � � y � z � �2 __ � � 2y � 2z � �4 __ � � x � 3y � 1 { y � z � 0

3y � 3z � 0

x � y � �1 x � y � �2 3y � 3z � 0

No solution; inconsistent Infinitely many solutions; consistent

Classify each system of equations.

6. { x � y � 2

x � 3y � z � 5

x � y � z � �1 7. { x � 2y � z � 2

�y � z � 1 x � 3y � 2z � �1

Infinitely many No solution; inconsistent solutions; consistent

Add the second and third equations to eliminate y.

Multiply the first equation by �2, then add.

Add the first and second equations.

Multiply the third equation by 2, Add to the first equation.

29

Name Date Class


You have learned that the graph of a quadratic function is a parabola.

Now you will learn how to shift parabolas by applying transformations to their related quadratic functions.

Suppose g(x) � a(x � h) 2 � k. The vertex of the parabola has a vertical shift of k units compared to the vertex of the parabola of the parent function f(x) � x 2 . If k � 0, the shift is upward; if k � 0, the shift is downward. The vertex of the parabola has a horizontal shift of h units compared to the vertex of the parabola of the parent function f(x) � x 2 . If h � 0, the shift is to the right; if h � 0, the shift is to the left.

If a � 1, the shape of the graph is the same as the shape of the graph of the parent function f(x) � x 2 .

Graph g(x) � (x � 2 ) 2 � 1.

Step 1 Graph the parent function f(x) � x 2 .

Step 2 Horizontal shift.

h � �2, shift the graph to the left two units.

Step 3 Vertical shift

k � �1, shift the graph down one unit.

PracticeComplete the steps to graph the function.

1. g(x) � (x � 3 ) 2 � 1.

Shift the graph 3 units to the left .

Shift the graph up 1 unit.

Graph the function.

2. g(x) � (x � 2 ) 2 � 4

x

y

6

4

2

2-4 -2-6-2

O

y = (x + 2)2- 1y = x2

x

y

6

4

2

2-4 -2-6-2

O

y = x2

x

y

6

8

4

2

2 4 6-2 O

10

ReteachingApplying Transformations to the Parabola and Determining the Minimum or

Maximum

30


Reteachingcontinued

The sign of a in g(x) � a(x � h ) 2 � k tells whether the parabola opens upward or downward. When a is positive, the parabola opens upward. When a is negative, the parabola opens downward.

The absolute value of a tells whether the parabola is stretched or compressed. If �a� � 1, then the parabola is stretched vertically and is narrower than the parabola of the parent function. If �a� � 1, then the parabola is compressed vertically and is wider than the parabola of the parent function.

Match the graph with its function.

a. g(x) � 0.3 x 2

b. g(x) � �0.3 x 2

c. g(x) � 3 x 2

d. g(x) � �3 x 2

The graph opens downward so a is negative. The graph is stretched vertically so, �a� � 0. The function that matches the graph is d.

PracticeComplete the steps to match the graph with its function.

3. a. g(x) � 0.3 x 2

b. g(x) � �0.3 x 2

c. g(x) � 3 x 2

d. g(x) � �3 x 2

The graph opens downward, so a is negative.

The graph is compressed so 0 � �a� � 1 .

The function that matches the graph is b .

Match the graph with its function.

4. a. g(x) � �10 x 2

b. g(x) � 10 x 2

c. g(x) � 0.1 x 2

d. g(x) � �0.1 x 2

b

x

y4

2

2 4

-2

-2-4

-4

O

y = x2

xy

2

2 4

-4

-2

-2-4

-6

O

x

y

6

4

2

2 4-2-4-2

O

30

Name Date Class


You have graphed linear equations in two variables. Now you will graph linear equations in three variables.

In three-dimensional space, the graph of a linear equation is a plane. You can graph the plane by finding its x-, y-, and z-intercepts.

Refer to the equation 2x � 4y � 3z � 12.

a. Find the intercepts.

To find the intercepts, set the other variables equal to 0.

Find the x-intercept.

Set y � z � 0.2x � 4(0) � 3(0) � 122x � 12x � 6The x-intercept is at (6, 0, 0)

Find the y-intercept.

Set x � z � 0.2(0) � 4y � 3(0) � 12�4y � 12y � �3

The y-intercept is at (0, �3, 0).

Find the z-intercept.

Set x � y � 0.2(0) � 4(0) � 3z � 123z � 12z � 4

The z-intercept is at (0, 0, 4).

b. Graph the equation.

Plot each point. Use a dashed line to connect the points. The triangle represents the plane.

y

xz8

4

4 8

8

4

-8

-4

-4-8

-8

(0, 0, 4)

(6, 0, 0)

O(0, -3, 0)

PracticeComplete the steps to solve the problem.1. Refer to the equation 3x � 4y � 6z � 12.

a. Find the intercepts.

x-intercept is at (4, 0 , 0 )

y-intercept is at ( 0 , 3, 0 )

z-intercept is at ( 0 , 0 , 2)

b. Graph the equation.

y

xz8

4

4 8

-4

-4-8

-8

O

8

4

-8

Solve.

2. Refer to the equation 2x � 2y � 5z � 10.

a. Find the intercepts. (5, 0, 0); (0, �5, 0); (0, 0, 2)b. Graph the equation.

y

xz8

4

4 8

-4

-4-8

-8

O

8

4

-4

-8

INV

3Reteaching

Graphing Three Linear Equations in Three Variables


Reteachingcontinued

A complex number can be written as a � bi, where a is the real part and bi is the imaginary part. Graphing complex numbers is like graphing real numbers. The real axis corresponds to the x-axis and the imaginary axis corresponds to the y-axis.

Refer to the complex number 3 � i.

a. Name the real part and the imaginary part.

The real part of the complex number is 3.

The imaginary part of the complex number is �i.

b. Write the coordinates for the point.

The real part of the complex number corresponds to the x-axis. The imaginary part corresponds to the y-axis. So write (3, �i ).

x

y4

2

2 4

-2

-2-4

-4

O

Imaginary axis

Rea

l axi

s

c. Plot the point.

d. Find � 3 � i � .

To find the absolute value of a complex number use

� a � bi � � ��

a2 � b2 .

� 3 � i � � ��

32 � (�1)2

� ��

9 � 1

� ��

10


3. Refer to the complex number 5 � 2i.

a. Name the real part b. Write the c. Plot the point. d. Find � 5 � 2i � .

and the imaginary part. coordinates for the

x

y4

2

2 4

-2

-2-4

-4

O

Imaginary axis

Rea

l axi

s

� 5 � 2i � � ��

5 2 � (�2) 2

point. � ��

25 � 4

Real part: 5. Imaginary ( 5 , �2i ) � ��

29 part: �2i

Solve.

4. Refer to the complex number �4 � 3i.

a. Name the real part and the imaginary part.

Real part: �4,imaginary part: 3i

b. Write the coordinates for the point.

(�4, 3i )

c. Plot the point.

x

y4

2

2 4

-2

-2-4

-4

O

Imaginary axis

Rea

l axi

s

d. Find � �4 � 3i � .

5

Think: 3 � i � 3 � 1i; so a � 3 and b � �1.

INV

3

Name Date Class


You have factored polynomials and simplified rational expressions. Now you will multiply and divide rational expressions.

To multiply rational expressions, first factor. Then multiply the numerators and multiply the denominators. Finally, divide out common factors.

Multiply 5x � 10 _______ x2 � 16

� x � 4 _______ x2 � 2x

.

5x � 10 _______ x2 � 16

� x � 4 _______ x2 � 2x

� 5(x � 2)

____________ (x � 4)(x � 4)

� x � 4 _______ x(x � 2)

� 5(x � 2) (x � 4)

____________________ x(x � 4) (x � 4) (x � 2)

� 5 _______ x(x � 4)

PracticeComplete the steps to multiply the rational expression. Identify all values of x that make the expression undefined.

1. 2x � 2 ______ x � 4

� x2 � 4x __________ x2 � 3x � 2

� 2(x � 1)

________ x � 4

� x(x � 4)

____________ (x � 1)(x � 2)

� 2x (x � 1) (x � 4)

___________________ (x � 4) (x � 1) (x � 2)

� 2x _____ x � 2

x � �4, x � 1, x � 2

Multiply. Identify all values of x that make the expression undefined.

2. 8x � 16 _______ x2 � 1

� x � 1 ______ 4x � 8

3. x � 3 ______ 6x � 6

� x � 1 ______ x2 � 9

2 _____ x � 1

; �2, �1, 1 1 ________ 6(x � 3)

; �3, 1, 3

4. x2 � 2x � 3 __________

2x � 4 � x2 � 2x __________

x2 � 6x � 5 5. x

2 � 9 ______ 2x � 6

� 2x2 __________

x2 � 6x � 9

x(x � 3)

________ 2(x � 5)

; �5, �2, �1 x2 _____

x � 3 ; �3, 3

The factors in the denominators cannot be zero, or the expression will be undefined.So, x � �4, x � 4, x � 0, and x � �2.

Factor.

Multiply and divide out common factors.

Simplify.

Factor.


Simplify.

ReteachingMultiplying and Dividing Rational Expressions 31


Reteachingcontinued

To divide by a rational expression, first multiply by its reciprocal. Then simplify the multiplication.

Divide x � 7 _____ x � 2

� x2 � 49 _______

2x � 4 .

x � 7 _____ x � 2

� x2 � 49 _______ 2x � 4

� x � 7 _____ x � 2

� 2x � 4 _______ x2 � 49

� x � 7 _____ x � 2

� 2(x � 2)

____________ (x � 7)(x � 7)

� 2(x � 7) (x � 2)

___________________ (x � 2) (x � 7) (x � 7)

� 2 ______ (x � 7)

PracticeComplete the steps to divide the rational expression. Identify all values of x that make the expression undefined.

6. 4x � 8 ______ x2 � 4

� 2x _____ x � 2

� 4x � 8 ______ x2 � 4

� x � 2 ______ 2x

� 4(x � 2)

_____________ (x � 2) (x � 2)

� x � 2 ______ 2x

� 42 (x � 2) (x � 2)

_______________ 2x (x � 2) (x � 2)

1

.

� 2 _____ x

Divide. Identify all values of x that make the expression undefined.

7. x2 � x � 12 __________

x2 � 9 � x

2 � 3x � 4 __________ 3x � 9

8. x2 � 2x � 3 __________

x2 � 9 � x

2 � 3x � 4 __________ x2 � 2x � 3

3 _____ x � 1

; �4, �3, 1, 3 x � 1 _____ x � 4

; �4, �3, �1, 1, 3

9. A square prism and a sphere have the same volume if the diameter of the sphere is equal to

the length s of one side of the prism’s base.

Solve the equation 4r2h � 4�r 3 ____ 3

for h to find the

height of the prism in terms of the radius r of the sphere.

h � �r ___ 3

When dividing a __ b � c __

d � a __

b � d __

c ,b, c, and d

must all be nonzero. So, x � 2, x � �7, and x � 7.

Multiply by the reciprocal.


Simplify.

Factor.

Multiply by the reciprocal.


Factor.

x � 2 , x � �2, x � 0

r

s

s

h

31

Name Date Class


You have multiplied matrices and found the determinant of a matrix. Now you will use inverses of matrices to solve systems of linear equations.

The identity matrix of a 2 � 2 matrix is [ 1 0 0 1

] . If a square matrix A has an

inverse A�1, then the product of A and A�1 is the identity matrix.

Inverse of a 2 � 2 matrix:

The inverse of A � [ a b c d

] is A�1 � 1 _____ det A

[ d �b �c a ] .

Find the inverse of A � [ 2 1 4 1

] .First find the determinant: det [ 2 1

4 1 ] � � 2 1

4 1 � � 2 � 4 � �2

Then switch ops and multiply by 1 ___ det

� 1 ___ �2

� � 1 __ 2 .

A�1 � � 1 __ 2 [ 1 �1

�4 2 ] � [ � 1 __

2 (1) � 1 __

2 (�1)

� 1 __ 2 (�4) � 1 __

2 (2)

] � [ � 1 __ 2 1 __

2

2 �1 ]

PracticeComplete the steps to find the inverse of the matrix.

1. A � [ 2 7 �1 �2

] det [ 2 7 �1 �2

] � [ 2 7 �1 �2

] � 3 1 _____ det A

� 1 __ 3

A�1 � 1 __ 3 [ �2 �7

1 2 ] � [ � 2 __

3 � 7 __

3

1 __ 3 2 __

3 ]

Find the inverse of the matrix.

2. [ 6 1 8 2

] 3. [ �4 6 1 �2

] 4. [ �5 �2 3 1

]

[ 1 __ 2 � 1 __

4

�2 3 __ 2 ] [ �1 �3

� 1 __ 2 �2

] [ 1 2 �3 �5

]

Think: “Switch ops.”Switch a and d and take the opposites of b and c.

If the determinant is 0, the matrix has no inverse.

The determinant exists, so the matrix has an inverse

ReteachingSolving Linear Systems with Matrix Inverses 32


Reteachingcontinued

The inverse of a matrix can be used to solve a system of equations.

Solve { 2x � y � �1

4x � y � �5 .

Step 1: Write the matrix equation.

A X � B

[ 2 1 4 1

] [ x y ] � [ �1 �5

] Step 2: Find the determinant of the coefficient matrix A.

det [ 2 1 4 1

] � [ 2 1 4 1

] � 2 � 4 � �2

Step 3: Find A�1.

A�1 � � 1 __ 2

[ 1 �1 �4 2

] � [ � 1 __ 2

1 __ 2

2 �1 ]

Step 4: Solve AX � B by multiplying both sides by A�1.

A�1AX � A�1B

A�1A � I, so X � A�1B

[ x y

] � [ � 1 __ 2 1 __

2

2 �1 ] [ �1

�5 ] � [ � 1 __

2 (�1) � 1 __

2 (�5)

2(�1) � (�1)(�5) ] � [ 1 __

2 � 5 __

2

�2 �1 ] � [ �2

3 ]

The solution is (�2, 3).

PracticeWrite the matrix equation for the system. Then complete the steps to solve the system using the inverse of the matrix.

5. { x � y � 2

x � 2y � 6 AX � B: [ 1 1

1 2 ] [ x y ] � [ 2

6 ] det A: 1

A�1 � [ 2 �1 �1 1

] X � A�1B � [ 2 �1 �1 1

] [ �1 5 ] � [ �2

4 ] [ x y ] � [ �2

4 ] The solution

is (�2, 4).

Write the matrix equation for the system. Then solve using the inverse of the matrix.

6. { 2x � 3y � �1

x � 2y � 1 x � �5, y � 3 7. { 3x � y � 2

x � y � 4

x � 3 __

2 , y � � 5 __

2

Constant matrix

Variable matrix

Coefficient matrix

32

Name Date Class


You have added and multiplied real numbers. Now you will use these operations to apply counting principles.

Addition Counting Principle: To count the number of possible outcomes of mutually exclusive events, add the number of outcomes of each event.

Find the number of possible outcomes of the event.

Draw a red face card or a club by drawing a card at random from a standard deck. Number of outcomes � 6 � 13

� 19

Fundamental Counting Principle: To count the number of possible outcomes when choosing more than one item, multiply the number of ways you have of choosing each item.

Find the number of possible outcomes of the event.

Choose a 4-character password that must be two letters followed by two numbers.

Number of outcomes � 26 � 26 � 10 � 10

� 67,600

PracticeComplete the steps to find the number of possible outcomes of the given events.

1. Randomly choose a multiple of 4 or a prime number from the whole numbers 1 through 20.

Multiples of 4: 4, 8, 12, 16, 20 Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19 5 � 8 � 13

2. Choose from 7 types of bread and 9 types of fillings when ordering a sandwich.

Number of bread choices: 7 Number of filling choices: 9

5 � 8 � 13

Find the number of possible outcomes of the given events.

3. Draw a face card or a number greater than or equal to 7 by drawing a card at random from a standard deck. 28

4. Randomly choose a multiple of 7 or a prime number from the whole numbers 10 through 30. 9

5. Choose a 5 digit password that begins and ends with a letter, and has three numbers in the middle. 676,000

There are 13 cards in each suit.

There are 2 red jacks, 2 red queens, and 2 red kings.�

�

There are 10 possible numbers.

There are 26 possible letters.

ReteachingApplying Counting Principles 33


Reteachingcontinued

Two events can be independent or dependent.

Two events are independent if the occurrence of one does not affect the probability of the occurrence of the other.

A card is drawn from a standard deck and then placed back in the deck.A second card is then drawn.The second card drawn could be any of the 52 cards in the deck, just like the first card.These events are independent.

Two events are dependent if the occurrence of one affects the probability of the occurrence of the other.

A card is drawn from a standard deck. It is not replaced.A second card is then drawn.The second card drawn can only be one of the 51 remaining cards in the deck.These events are dependent.

PracticeA bag contains a blue marble, a red marble, a green marble, and a white marble. A marble is chosen at random and then another marble is chosen at random. Complete the steps to determine whether the following events are independent or dependent.

6. Choose a white marble and then choose a red marble, if the first marble is not replaced before the second marble is chosen.

The probability of choosing the red marble does depend on which marble is chosen first.

The events are dependent .

7. Choose a white marble and then choose a red marble, if the first marble is replaced before the second marble is chosen.

The probability of choosing the red marble does not depend on which marble is chosen

first. The events are independent.

There are 10 balls in a bag. Five are green, three are yellow, and 2 are orange. A ball is chosen at random then a second ball is chosen at random. Determine whether the following events are independent or dependent.

8. Choose a green ball, replace it, and choose another green ball.

independent 9. Choose a green ball, don’t replace it, and choose a yellow ball.

dependent 10. Choose a yellow ball, don’t replace it, and choose an orange ball.

dependent 11. Choose an orange ball, replace it, and choose a green ball.

independent

33

Name Date Class


You have graphed linear equations using tables, intercepts, and slope-intercept form. Now you will identify graphs of linear functions and graph linear equations using transformations.

A linear function has a constant rate of change, and its graph is a line.

Determine whether the graph repesents a linear function.

The graph passes the vertical line test, so it is a function. The function does not have a constant rate of change. It is not linear.

The graph passes the vertical line test, so it is a function. The function has a constant rate of change. It is linear.

PracticeComplete the steps to determine whether the graph repesents a linear function.

1.

x

y4

4

2-2

2-1-2-4

-4

-2

O

2. y

2

2

-1-1 3

5

4-2-4

-4

-2

O x

Yes, the function is linear . No, the function is not linear .

Determine whether the graph repesents a linear function.

2.

x

y4

2

32

5

4-4 -2

-4

O

3.

x

y4

2

23

1

3

14-4 -2

-4

O

no yes

Rate of change: 3 __ 2 Rate of change: 3 __

2


2 Rate of change: 3 __


2

Rate of change: 3 __ 2

Rate of change: 3 __ 2

Rate of change: 2 __ 4 � 1 __


4 � 1 __

2

Rate of change:

5 __ 1 � 5

Rate of change:

5 __ 1 � 5


2

Rate of change:

3 __ 1 � 3

Rate of change:

3 __ 1 � 3

x

y4

2

5

2

2

4-4 -2

-2

-4

O

3

2

x

y4

2

22

3

32

4-2-4

-4

O

ReteachingGraphing Linear Equations II 34


Reteachingcontinued

Any linear function can be graphed by transforming the basic parent function of all linear functions y � x.

Graph g (x) � �2x � 3 by transforming the parent function f (x) � x.

Stretch or Compression? Refl ection? Shift?

g(x) � 2 x g(x) � �2x g(x) � �2x � 3

stretch vertically by a factor of 2 reflect over the x-axis shift vertically 3 units up

xf(x) = x

y4

2

2 4-4 -2

-4

-2

O

g(x) = 2x

x

y4

2

2 4-4 -2

-4

-2

O

g(x) = -2x

f(x) = 2xx

y4

2

2 4-4 -2

-4

-2

O

g(x) = -2x+3f(x) = -2x

PracticeComplete the steps to graph by transforming the parent function f (x) � x.

4. g(x) � � 1 __ 2

x � 2

compress vertically by

a factor of 1 __ 2

reflect over thex-axis

shift vertically 2 units down

x

y4

2

2 4-4 -2

-4

-2

O

f(x) = x

g(x)2

x1=

x

y4

2

2 4-4 -2

-4

-2

O

f(x) =2

x1

g(x) = -

2x1

x

y4

2

2 4-4 -2

-4

O

g(x) = -

2x-21

f(x) = -

2x1

Graph by transforming the parent function f (x) � x.

5. y � �x � 4 6. y � 3x � 1

x

y

2

4

2 4-4 -2

-4

-2

O

x

y

2

4

2 4-4 -2

-4

O

34

Name Date Class


You have factored quadratic equations. Now you will use the Zero Product Property to find the zeros, or roots, of quadratic equations.

Zero Product Property: If a product is zero, then at least one of its factors must be zero.

Find the roots of x 2 � 2x � 3 � 0 Graph the related function f (x) � x 2 � 2x � 3 and fi nd its zeros. x 2 � 2x � 3 � 0

(x � 3) (x � 1) � 0

� 3) � 0 or (x � 1) � 0

x � �3 or x � 1

The roots of x 2 � 2x � 3 � 0 are �3 and 1.

The zeros of f (x) � x 2 � 2x � 3 are 3 and 1.

The x-intercepts of the graph are �3 and 1.

PracticeComplete the steps to find the roots of the equation. Graph the related function and find its zeros. 1. x 2 � 4x � 4 � 0

x 2 � 4x � 4 � 0

(x � 2) (x � 2) � 0

(x � 2) � 0

x � 2roots: 2zeros: 2x-intercepts: 2

Find the roots of the equation. Graph the related function and find its zeros.

2. x 2 � 2x � 8 � 0 3. x 2 � 2x � 1 � 0

�2, 4 �1

Solve each equation for x.

Set each factor equal to 0.

x

y4

2

2 4-4 -2

-2

O(-3,0) (1,0)

x

y

4

2

4 6-2

-2

O

(2,0)

This equation hasa double root.

x

y8

4

8-8 -4

-8

O(-2,0) (4,0)

x

y

4

6

2

2 4-4

-2(-1,0)

O

Factor, then multiply to check.

ReteachingSolving Quadratic Equations I 35


If you know the zeros of a quadratic function, you can work backwards and write an equation for the function.

Write a quadratic function that has zeros � 1 __ 2 and 3.

x � � 1 __ 2 or x � 3 � 0

x � 1 __ 2 � 0 or x � 3 � 0

The quadratic function f (x) � 2 x 2 � 5x � 3 has zeros � 1 __ 2 and 3.

PracticeComplete the steps to write a quadratic function that has the given zeros.

4. −2, 1 __ 3

x � �2 or x � 1 __ 3

x � 2 � 0 or x � 1 __ 3 � 0

The quadratic function f (x) � 3 x 2 � 5x � 2 has zeros �2 and 1 __ 3 .

Write a quadratic function that has the given zeros.

5. −4, 5 6. −1, 2 __ 3

f (x) = x 2 � x � 20 f (x) = 3 x 2 � x � 2

7. The height of a free-falling object is given by the function h(t) � �16t 2 � v 0t � h0 where h0 is the initial height in feet, v0 is the initial velocity in feet per second, and h(t) is the height in feet at time t seconds. How long does it take for an object to hit the ground after it is thrown straight up at a speed of 24 feet per second from a height of 72 feet? (Hint: Substitute the given values into the equation. Then set the equation equal to zero to find the zeros, where the function crosses the x-axis.)

3 s

The factors of the quadratic equation.The factors of the quadratic equation.

� �

(x � 2) (x � 1 __ 3 ) � 0

3 (x � 2) (x � 1 __ 3 ) � 3 (0)

(x � 2) (3x � 1) � 0

3 x 2 � x � 6x � 2 � 0

3 x 2 � 5x � 2 � 0

Multiply the factors.

Multiply both sides by 3 to clear the fraction.Distribute the 3.Multiply using FOIL.Simplify.


� x � 1 __ 2 � (x � 3 � 0 Multiply the factors.

2 � x � 1 __ 2 � (x � 3) � 2(0) Multiply both sides by 2

to clear the fraction.

(2x � 1)(x � 3) � 0 Distribute the 2.

2x 2 � 6x � x � 3 � 0 Multiply using FOIL.

2x 2 � 5x � 3 � 0 Simplify.

Name Date Class


You have written and graphed linear equations. Now you will write an equation for a line that is parallel or perpendicular to another line and passes through a given point. You will also determine if two lines are parallel or perpendicular to each other.

Parallel lines have the same slope, but different y-intercepts. The slopes of perpendicular lines are negative reciprocals of each other.

Write the equation of the line with the given characteristics.

a. parallel to y � 2x � 3 and passesthrough (3, 1)

Step 1: Find the slope. m � 2

Step 2: Substitute the point and the slope into the point-slope formula.

y � y1 � m(x � x1)

y � 1 � 2(x � 3)

y � 1 � 2x � 6

y � 2x � 5

b. perpendicular to y � 1 __ 2 x � 4 and

passes through (2, �1)

Step 1: Find the slope. m � � 1 __

1 __ 2 � �2

Step 2: Substitute the point and the slope into the point-slope formula.

y � y1 � m(x � x1)

y � (�1) � �2(x � 2)

y � 1 � �2x � 4

y � �2x � 3

PracticeComplete the steps to write the equation of each line with the given characteristics.

1. parallel to y � 1 __ 3 x � 4 and passes through (�3, �6)

point-slope form: equation:

slope: m � 1 __ 3

y � (�6) � 1 __ 3 [ x � (�3) ] y � � 1 __

3 x � 4

2. perpendicular to and passes through (4, �4)

point-slope form: equation:

slope: m � � 1 __ 4 y � (�4) � � 1 __

4 (x � 4) y � � 1 __

4 x � 3

Write the equation of each line with the given characteristics.

3. parallel to y � 3x � 1 and passes through (1, �3) y � 3x � 6

4. perpendicular to y � �2x � 10 and passes through (4, 3) y � 1 __ 2 x � 1

5. parallel to y � �4x � 3 and passes through (�1, �1) y � �4x � 5

ReteachingUsing Parallel and Perpendicular Lines 36


Reteachingcontinued

x

y

A

B

C

6

4

2

42

(0,1)

(2,5)

(6,3)

6

slope of line 1

Compare the slopes of two lines to determine if they are parallel or perpendicular.

Determine if the lines are parallel or perpendicular.a. line 1: passing through (3, 2) and (1, �4)

line 2: passing through (�3, 1) and (�1, 7)

slope of line 1:

m1 � y2 � y1 _______ x2 � x1

� �4 � 2 _______ 1 � 3

� �6 ___ �2

� 3

slope of line 2:

m2 � y2 � y1 _______ x2 � x1

� 7 � 1 __________ �1 � (�3)

� 6 __ 2 � 3

The slopes are the same.

The lines are parallel.

b. line 1: 2x � y � 1

line 2: x � 2y � 6

slope-intercept from of line 1:

2x � y � 1

�y � �2x � 1

y � 2x � 1

slope-intercept from of line 2:

x � 2y � 6

2y � �x � 6

y � � 1 __ 2 x � 3

m1m2 � (2) � � 1 __ 2 � � �1

The product of the slope is �1.

The lines are perpendicular.

PracticeComplete the steps to determine if the lines are parallel or perpendicular.

6. line 1: passing through (3, 1) and (5, 9) 7. line 1: �2x � 3y � � 1line 2: passing through (1, 3) and (5, 2) line 2: 4x � 6y � 7

m1: 4

m2: � 1 __ 4 m1: 2 __

3 m2: 2 __

3

The lines are perpendicular. The lines are parallel.

Determine if the lines are parallel or perpendicular.

8. line 1: 4x � 5y � 3 9. line 1: passing through (�3, �2) and (0, 2)line 2: 10x � 8y � 5 line 2: passing through (4, 5) and (�5, �7)

perpendicular parallel

10. Use what you have learned about the properties of perpendicular lines to prove that �ABC is a right triangle.

See student work.

slope of line 2

36

Name Date Class


You have multiplied, divided and simplified rational expressions. Now you will add and subtract rational expressions.

If two rational expressions have the same denominator, you can combine the numerators.

Add 6x � 4 ______ x � 5

� 2x � 8 ______ x � 5

.

Step 1: Add.

6x � 4 ______ x � 5

� 2x � 8 ______ x � 5

� 6x � 4 � 2x � 8 ______________ x � 5

The denominators are the same. Add the numerators.

� 6x � 2x � 4 � 8 ______________ x � 5

Group like terms.

� 8x � 4 ______ x � 5

Combine like ter ms.

Step 2: Find any values of x for which the expression is undefined.

x � �5 because �5 makes the denominator 0.

PracticeComplete the steps to subtract the rational expressions. Find any values of x for which the expression is undefined.

1. 4x � 3 ______ 2x � 1

� 8x � 2 ______ 2x � 1

Step 1: Subtract.

4x � 3 ______ 2x � 1

� 8x � 2 ______ 2x � 1

� (4x � 3) � (8x � 2) _________________________

2x � 1 The denominators are the

same. Subtract the numerators.

� 4x � 3 � 8x � 2

___________________ 2x � 1

Use the Distributive Property.

� �4x � 5

_______________ 2x � 1

Combine like ter ms.

Step 2: Identify x-values for which the expression is undefined.

x � 1 __ 2 because 1 __

2 makes the denominator equal 0.

Add or subtract. Find any values of x for which the expression is undefined.

2. 7x � 5 ______ x � 3

� 4x � 1 ______ x � 3

3. 2x � 1 ______ x � 1

� 5x � 4 ______ x � 1

11x � 6 _______ x � 3

; �3 �3x � 5 ________ x � 1

; 1

ReteachingAdding and Subtracting Rational Expressions 37


Reteachingcontinued

Use the least common denominator (LCD) to add rational expressions with different denominators. The process is the same as adding fractions.

Add x � 4 __________ x2 � 2x � 3

� 2x _____ x � 1

.

Step 1: Factor denominators completely. x � 4 __________ x2 � 2x � 3

� 2x _____ x � 1

� x � 4 ____________ (x � 3)(x � 1)

� 2x _____ x � 1

Step 2: Find the LCD. The LCD is the least common multiple of the denominators:(x � 3)(x � 1)

Step 3: Write each term of the expression using the LCD.

x � 4 ____________ (x � 3)(x � 1)

� 2x _____ x � 1

� x � 4 ____________ (x � 3)(x � 1)

� 2x _____ x � 1

� � x � 3 _____ x � 3

� � x � 4 ____________

(x � 3)(x � 1) � 2x2 � 6x ____________

(x � 1)(x � 3)

Step 4: Add. x � 4 ____________ (x � 3)(x � 1)

� 2x2 � 6x ____________ (x � 3)(x � 1)

� x � 4 � 2x2 � 6x _______________ (x � 3)(x � 1)

� 2x2 � 7x � 4 ___________ (x � 3)(x � 1)

Step 5: Find any values of x for which the expression is undefined.x � �3 or because both values make the denominator equal 0.

PracticeComplete the steps to add the rational expressions. Find any values of x for which the expression is undefined.

4. x � 1 ______ x2 � 4

� 3x _____ x � 2

x � 1 ______ x2 � 4

� 3x _____ x � 2

� x � 1 ____________ (x � 2)(x � 2)

� 3x _____ x � 2

� � (x � 2) ________

(x � 2) �

LCD � (x � 2)(x � 2) � x � 1 ____________ (x � 2)(x � 2)

� 3x2 � 6x

____________ (x � 2)(x � 2)

� x � 1 � 3x2 � 6x

_____________________ (x � 2)(x � 2)

� 3x2 � 5x � 1

__________________ (x � 2)(x � 2)

x � �2 or 2

Add. Find any values of x for which the expression is undefined.

5. 4x � 1 __________ x2 � 3x � 2

� 3 _____ x � 1

6. 2x ______ x2 � 9

� 7 _________ x2 � x � 6

�7x � 5 _____________ (x � 2)(x � 1)

; �2, �1 �2x2 � 11x � 21 __________________ (x � 2)(x � 3)(x � 3)

;�3,�2, 3

7. On a ferry trip, it took 20 minutes longer for the westbound lake crossing than it did for the eastbound crossing. Find the difference in the speeds of the two crossings.

(Hint: Simplify the expression d __ t � d _____ t � 1 __

3 .) d ________

t(3t � 1)

37

Name Date Class


You have added, subtracted and multiplied polynomials. Now you will divide polynomials using long division.

In arithmetic long division, you follow these steps: divide, multiply, subtract, and bring down. Follow these same steps to use long division to divide polynomials by monomials.

Divide (8x3 � 2x2 � 5x � 1) � 2x.

Step 1: Divide the fi rst term of the dividend, 8x3, by the divisor, 2x.

4x2

2x � �

8x3 � 2x2 � 5x � 1 Divide: 8x3 � 2x � 4x2.

_ �(8x3) Multiply the divisor: 4x2(2x) � 8x3.

0 � 2x2 � 5x � 1 Subtract and bring down.

Step 2: Continue dividing the first term of the current dividend by the divisor.

4x2 � x � 2

2x � �

8x3 � 2x2 � 5x � 1

_ �(8x3)

0 � 2x2 � 5x � 1 Divide: �2x2 � 2x � �x. _ �(�2x2) Multiply the divisor: �x(2x) � �2x2.

0 � 5x � 1 Subtract and bring down. Divide: 5x � 2x � 2.5.

_ �(4x) Multiply the divisor: 2(2x) � 4x.

x � 1 Subtract and bring down.

Step 3: Write the quotient including the remainder.

(8x3 � 2x2 � 5x � 1) � 2x � 4x2 � x � 2 � x � 1 _____ 2x

PracticeComplete the steps to divide.

1. (�18x2 � 23x � 5) � 3x

�6x � 7 The quotient with remainder:

3x � �

18x2 � 23x � 5 (�18x2 � 23x � 5) � 3x

�(�18x2) � �6x � 7 � 2x � 5 ______

3x

0 � 23x � 5

�(21x)

2x � 5Divide.

2. (12x3 � 32x2 � 28x � 7) � 4x 3. (15x3 � 20x2 � 7x � 3) � 5x

3x2 � 8x � 7 � 7 ___ 4x

3x2 � 4x � 1 � 2x � 3 ______ 5x

Use only the integer portion of the quotient.

ReteachingDividing Polynomials Using Long Division 38


Reteachingcontinued

You can divide any polynomial by a polynomial of a lesser degree using long division.

Divide (2x3 � 3x � 4) � (x � 1).

Step 1: Divide the first term of the dividend, 2x3, by the first term of the divisor, x.

2x2

x � 1 � �

2x3 � 0x2 � 3x � 4 Divide: 2x3 � x � 2x2.

_ �(2x3 � 2x2) Multiply the complete divisor: 2x2(x � 1) � 2x3 � 2x2

2x2 � 3x � 4 Subtract and bring down.

Step 2: Continue dividing the first term of the current dividend by the first term of the divisor.

2x2 � 2x � 5

x � 1 � �

2x3 � 0x2 � 3x � 4 _ �(2x3 � 2x2) 2x2 � 3x � 4 Divide: 2x2 � x � 2x. _ �(2x2 � 2x) Multiply the complete divisor: 2x(x � 1) � 2x2 � 2x. 5x � 4 Subtract and bring down. Divide: 5x � x � 5. _ �(5x � 5) Multiply the complete divisor: 5(x � 1) � 5x � 5. 1 Subtract.

Step 3: Write the quotient including the remainder.

(2x3 � 3x � 4) � (x � 1) � 2x2 � 2x � 5 � 1 _____ x � 1

PracticeComplete the steps to divide.

4. (3x3 � 4x2 � 5) � (x � 2)

Include a place for the x, or linear, term.

Write the long division: x � 2 � �

3x3 � 4x2 � 0x � 5

The quotient with remainder: (3x3 � 4x2 � 5) � (x � 2) � 3x2 � 2x � 4 � 3 _____

x � 2

Divide.

5. (x3 � 2x2 � 4x � 5) � (x � 1) 6. (4x3 � 9x2 � 3x � 15) � (x � 3)

x2 � x � 5 4x2 � 3x � 6 � 3 _____ x � 3

7. The area of a rectangular garden is (2x3 � 45x � 27) square feet. The width of the garden is (x � 5) feet. Find an expression for the length of the garden in feet.

2x2 � 10x � 5 � 2 _____ x � 5

ft

Include a place for the squared term

38

Name Date Class


You have solved systems of linear equations by graphing. Now you will solve systems of linear inequalities by graphing.

Graphing a linear inequality is similar to graphing a linear function. The intercepts can be used to graph a linear inequality.

Graph 2x � y � 4 using the intercepts.

Step 1: Write the corresponding equation. Then identify the x-intercept and the y-intercept.

2x � y � 4 When y � 0, x � 2. Plot (2, 0). When x � 0, y � 4. Plot (0, 4).

Step 2: Draw a solid boundary for � or �.

Draw a dashed boundary for � or �.

Draw the graph of 2x � y � 4 using a dashed line.

Step 3: Choose a point to check which half-plane to shade.Use (0, 0).

2x � y � 4

2(0) � (0) �?

4

(0) �?

4 ✘

Step 4: The inequality is false, so shade the half-plane above the line.

PracticeComplete the steps to graph the inequality using the intercepts.

1. 2x � 3y � 6

x-intercept: _ (3, 0)

y-intercept: (0,2) boundary line: _ solid

test (0, 0): _ true

shade: below the line

Graph the inequality using the intercepts.

2. �3x � y � �3 3. x � 2y � �4

x

y

O

(0,4)

(2,0)

-2

-4

4

-4

2

-2

2 4

x

y

O

4

(0, 2)

(3, 0)

-4 -2 2

-2

-4

x

y4

-2-4

2

(1, 0)

(0, -3)-2

O2 4

x

y

O

-2

-4

-2-4 4

4

2

(0, 2)(-4, 0)

ReteachingGraphing Linear Inequalities in Two Variables 39


Reteachingcontinued

The slope-intercept form can also be used to graph a linear inequality.

Graph 3y � 2x � 3 using the slope-intercept form.

Step 1: Write the corresponding equation in slope-intercept form. Then identify the slope and the y-intercept.

3y � 2x � 3

3y � 2x � 3

y � 2 __ 3

x � 3

m � 2 __ 3

and b � 1

Step 2: Draw the graph of y � 2 __ 3 x � 1 using a solid line.

Step 3: Choose a point to check which half-plane to shade.Use (0, 0).

y � 2 __ 3

x � 1

0 � 2 __ 3 (0) � 1

0 � 1 ✓

Step 4: The inequality is true, so shade the half-plane below the line.

PracticeComplete the steps to graph the inequality using the slope-intercept form.

4. �x � y � 2

slope: 1

y-intercept: (0, 2) boundary line: dashed test (0, 0): true

shade: below the line

Graph the inequality using the slope-intercept form.

5. 4x � y � 3 6. 2x � 4y � 8

x

y

O-4

-2

-4

4

4

2

2-2

2

x

y

O2 4

4

-2

-2

-4

-4

x

y

O4-2-4

-2

-4

2

2

x

y

O-4

-4

-2 4

4

2

-2

39

Name Date Class


You have learned to simplify expressions with exponents. Now you will learn to simplify radical expressions.

Simplify each expression. Assume any variables are positive.

If the number under the radical sign contains a perfect power, the expression is not in simplest form. Factor out perfect powers from the radicand to simplify the radical expression.

a. 3 ��

�27 Use the definition of nth roots.If an� b, n ��

b � a.

3 ��

�27 � then 3 � �27 What number raised to the third power is �27?

3 ��

�27 � �3 Because (�3)3 � �27.

b. ��

x10 Use the rule: If n is even, ��

xn � x n __ 2

.

��

x10 � x 10

___ 2

� x5 Because ��

x10 � ��

x5 � x5 � ��

(x5)2 � x5.

c. ��

18 � ��

3 Use the product rule for radicals: n �� a � n ��

b � n ��

ab .

��

18 � ��

3 � ��

54 Look for a perfect square in the radicand.

��

9 � 6 � �� 9 � �

� 6 � 3 �

� 6 Because 9 � 6 � 32 � 6 � 54.

PracticeComplete the steps to simplify each expression. Assume any variables are positive.

1. 3 ��

�8 2. ��

x6 3. ��

12 � ��

5

3 ��

�8 � a ��

x6 � x 6 __ 2 �

� 12 � �

� 5 � �

� 12 � 5

�8 � a3

� x 3 � ��

60

�8 � �23

� ��

4 � 15

� 2 ��

15


4. 3 ��

�216 �6 5. ��

x8 x 4

6. ��

15 � ��

3 3 ��

5 7. 6 ��

64 2

8. ��

20x5 � ��

3x3 2x 4 ��

15 9. ��

20x15 2x 7 ��

5x

ReteachingSimplifying Radical Expressions 40


Reteachingcontinued


If a radical expression contains like terms, the expression is not in simplest form. Combine like terms to simplify the expression.

a. 2 3 ��

8 � 7 3 ��

8 � 5 ��

8 � 2 ��

8 Identify the like terms.

� � 2 3 ��

8 � 7 3 ��

8 � � � 5 ��

8 � 2 ��

8 � Like terms have the same radicand and root.

� 9 3 ��

8 � 3 ��

8 Combine the like terms.

� 9(2) � 3(2) ��

2 3 ��

8 � 2 and ��

8 � ��

4 ��

2 � 2 ��

2

� 18 � 6 ��

2 Simplify.

If a radical expression contains a radical in the denominator, the expression is not in simplest form. Multiply both the numerator and the denominator by the radical that allows the denominator to have a radical with a perfect root.

b. 1 ________ �

� 6 � �

� 2

1 ________ �

� 6 � �

� 2 � �

� 6 � �

� 2 ________

��

6 � ��

2 � �

� 6 � �

� 2 ______________________

��

36 � ��

12 � ��

12 � ��

4

� ��

6 � ��

2 ________ �

� 6 � �

� 2

� ��

6 � ��

2 ________ 4

PracticeComplete the steps to simplify each expression. Assume any variables are positive.

10. 3 ____ �

� 10 3 ____

��

10 � �

�

10 ______

��

10

11. 5 �� x � 2 ��

x � 5 3 �� x � 3 �� x 12. 1 ________ �

� 5 � �

� 3

� 3 ��

10 ______

��

10

� 7 �� x � 4 3 �� x 1 ________

��

5 � ��

3 � �

� 5 � �

� 3 ___________

��

5 � ��

3

� ��

5 � ��

3 ________ 2


13. 4 ___ �

� 5 4 �

� 5 ____

5 14. � 5 �� y � �

� y � 3 5 �� y � 2 �

� y 15. 1 _________

��

10 � ��

7 �

� 10 � �

� 7 __________

3

2 5 �� y � 3 ��

y

16. The length of the base of a square pyramid can be represented by ��

3V ____ �

� h

where V is the volume and h is the height. Simplify this expression. ��

3Vh ______ h

Because the denominator is a difference, multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of �

� 6 � �

� 2

is ��

6 � ��

2 . Use the FOIL method to find the product of the denominator. Then simplify.

Because the denominator is a difference, multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of �

� 6 � �

� 2

is ��

6 � ��

2 . Use the FOIL method to find the product of the denominator. Then simplify.

40

Name Date Class


Cryptography is the practice and study of hiding information. Encryption is the process of translating a message into secret code. Decryption is the process of translating secret code into a message. A cipher is the key to encryption and decryption. One of the simplest ciphers is a substitution cipher, where a number replaces each letter of the alphabet. A spreadsheet program automates the process of using a substitution cipher.

1. Input the letters A-Z in cells A1 to A26, and input the numeric cipher ABCDE

102147

11

from the table below in cells B1 to B26.

Check students’ work.

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

10 21 4 7 11 25 1 9 22 13 20 15 6 19 12 16 2 23 14 5 26 8 17 3 24 18

The first 5 rows are shown in the figure at the right.

2. Input the letters of the words BLETCHLEY PARK in cells ABCDE

BLETC

102147

11

D1 through D13. Do not leave a space between the words.

Check students’ work. 3. The VLOOKUP formula searches for a value in the leftmost column of the table (A) and then returns a value in the same row

ABCDE

BLETC

102147

11

21151154

from a column you specify (B). Input this formula into cell E1: = VLOOKUP(D1,$A$1:$B$26,2). Then click on the lower right

hand corner of cell E1 and drag the cursor to highlight cells

E2 to E13. Check students’ work. 4. Write the encrypted message as a string of numbers.

21 15 11 5 4 9 15 11 24 16 10 23 20 5. Use the spreadsheet to encrypt the following messages. List the

message as a string of numbers. If necessary, copy and paste the spreadsheet formula

to more cells.

a. ENIGMA 11 19 22 1 6 10 b. ALAN TURING 10 15 10 19 5 26 23 22 19 1 6. Create a new spreadsheet and input the numeric cipher in ascending

12345

GQXCT

order to cells A1 to A26, and input the alphabet in B1 to B26. The letters should match to their corresponding number and will not be in alphabetical order. Input the encrypted message below to cells D1 to D20. Input this formula into cell E1: =VLOOKUP(D1,$A$1:$B$26,2). Then click on the lower right hand corner of cell E1 and drag the cursor to highlight cells E2 to E20.

17 9 12 17 10 14 7 11 19 19 22 14 21 10 21 21 10 1 11

a. What is the message? WHO WAS DENNIS BABBAGE By first encrypting with the substitution cipher and then encrypting

with matrix multiplication an additional level of security is added.

ReteachingUnderstanding Cryptography

INV

4


Reteachingcontinued

7. Use the substitution cipher from problems 1 – 6 to encrypt the following message: A CRYPTOGRAPHER

10 4 23 24 16 5 12 1 23 10 16 9 11 23 8. Write the string of numbers in two 3 x 3 matrices. The first number

goes in the first row first column, the second number goes in the second row first column, and the extra entries are filled in with zeros.

[ 10 24 12

4 16 1

23 5 23 ] _____________

X

[ 10 11 0

16 23 0

9 0 0

] ___________

Y

9. Complete the multiplication of the encoding matrix and cipher matrices.

[ 1

�2

0

�1

0

0

0

1

2

] X � [ 6

3

46

8

�43

10

11

�1

46

] [ 1

�2

0

�1

0

0

0

1

2

] Y � [ 6

3

46

8

�43

10

11

�1

46

] 10. Write the newly encrypted message as a string of numbers.

6 3 46 8 �43 10 11 �1 46 �6 �11 18 �12 �22 0 0 0 0 11. Follow the steps to decrypt the following message. �5 14 1 �6 �19 11 �4 21 10 �10 �12 �27 9 16 �21 5 �38 �20 20 52 38 22 48 22 38 12 0

a. Store the decryption matrix in your calculator as matrix [A]. [A] is the inverse of the encryption matrix used in problem 9.

[ 0

�1

0

�0.5

�0.5

0

0.25

0.25

0.5

] b. Store the message as three separate matrices [B], [C], and [D], in your calculator. c. Multiply each message matrix by [A].

[ A ] [ B ] � [ 10

15

10

19

5

26

23

22

19

] [ A ] [ C ] � [ 1

7

11

4

23

24

16

5

11

] [ A ] [ D ] � [ 7

11

19

22

1

6

10

0

0

] d. Write the partially decrypted message as a string of numbers.

10 15 10 19 5 26 23 22 19 1 7 11 4 23 24 16

5 11 7 11 19 22 1 6 10 e. Use the decryption spreadsheet to decipher the message. What

is the message?

ALAN TURING DECRYPTED ENIGMA

INV

4

Name Date Class


You have learned how to solve problems using the Pythagorean Theorem. Now you will use the Pythagorean Theorem to develop the Distance Formula.

The distance between any two points on a coordinate plane can be found using the Pythagorean Theorem.

a 2 � b 2 � c 2

Find the value of z. Give the answer in simplest radical formand to the nearest tenth.

Step 1: Count the number of units to find the length of the vertical and horizontal legs. The vertical leg is 6 units long.The horizontal leg is 4 units long.

Step 2: Use the Pythagorean Theorem.

a 2 � b 2 � c 2 4 2 � 6 2 � c 2 Substitute 4 for a and 6 for b.16 � 36 � c 2 Square 4 and 6.

52 � c 2 Add. �

� 52 � �

� c 2 Take the square root of both sides.

2 ��

13 � c Simplify. ��

52 � ��

4 � ��

13 � ��

4 � ��

13 � 2 ��

13

The Value of z is exactly 2 ��

13 units, or about 7.2 units long.

PracticeComplete the steps to find the value of z. Give the answer in simplest radical form and to the nearest tenth.

1.

x

y

O

4

2

2 4 6

-2

-2-4-6

-4

z

a 2 � b 2 � c 2 → 5 2 � 3

2 � c 2

25 � 9 �

34 � c 2

��

34 � ��

c 2

��

34 � c

5.8 � c

Find the value of z. Give the answer in simplest radical form and to the nearest tenth. 2.

x

y

O

2

2 4 6

-2

-2-4-6

-4

z

4

��

89 � 9.4

ReteachingApplying the Pythagorean Theorem and the Distance Formula 41

x

y

zO

4

2

2 4 6

-2

-2-4-6

-4


Reteachingcontinued

b2

a2 c2

b

a da = y2- y1

(x1, y1)

(x1, y2) (x2, y2)

b = x2- x1

c =√a2+ b2 d =√(x2- x1)2 + (y2- y1)2

The distance between any two points with coordinates � x 1 , y 1 � and

� x 2 , y 2 � is d � ��

� x 2 � x 1 � 2 � � y 2 � y 1 � 2 .

Find the distance between the pair of points. Give the answer insimplest radical form and to the nearest tenth.

A(3, 5), B(11, 7)

Step 1: Label (3, 5) as, and (11, 7) as � x 2 , y 2 � .

d � ��

� x 2 � x 1 � 2 � � y 2 � y 1 �

2 → d � �� 11 � 3 � 2 � � 7 � 5 � 2

Step 2: Evaluate.

d � �� 11 � 3 � 2 � � 7 � 5 � 2 � ��

� 8 � 2 � � 2 � 2 � ��

64 � 4

� ��

68 � ��

4 ��

17 � 2 ��

17 � 8.2

PracticeComplete the steps to find the distance between each pair of points. Give the answer in simplest radical form and to the nearest tenth. 3. A(2, 6), B(7, 7)

d � ��

� 7 � 2 � 2 � � 7 � 6 � 2 � ��

� 5 � 2 � � 1 � 2

� ��

25 � 1

� ��

26 � 51

4. A(�1, 8), B(3, 5)

d � ��

� 3 � (�1) � 2 � � 5 � 8 � 2

� ��

� 4 � 2 � � �3 � 2

� ��

16 � 9

� ��

25 � 5

Find the distance between each pair of points. Give the answer insimplest radical form and to the nearest tenth.

5. A(0, 0), B(6, 8)

10 6. A(�5, 2), B(4, 5) 7. A(1, 5), B(3, 1)

3 ��

10 � 9.5 2 ��

5 � 4.5 8. A cartographer constructs a map of Texas over a grid system such that

Paris is at (125, 180) and Italy is at (200, 50). Find the distance between the cities to the nearest mile if each unit represents one mile. 150 mi

41

Name Date Class


You have learned how to use the Fundamental Counting Principle to determine all possible outcomes in a given situation. Now you will learn to find permutations and combinations of sets of objects.

A permutation is a selection of items from a group in which the order is important. In a permutation, AB is not the same as BA.

The number of permutations of n items taken r at a time is shown by the following formula.

P (n, r ) � n! ______ (n � r)!

Find the permutation of 10 items taken 4 at a time.

To find the number of permutations of 10 items taken 4 at a time, use n � 10 and r � 4 in the permutation formula. Then evaluate.

P (n, r ) � 10! _______ (10 � 4)

� 10! ___ 6!

� 10 � 9 � 8 � 7 � 6 � 5 � 4 � 3 � 2 � 1 __________________________ 6 � 5 � 4 � 3 � 2 � 1

Cancel common factors.

� 10 � 9 � 8 � 7 Simplify.

� 5040 Multiply.

There are 5040 permutations.


1. Find the permutation of 8 objects taken 5 at a time.

P (8, 5) � 8! _______ (8 � 5 )!

� 8! __ 3!

� 8 � 7 � 6 � 5 � 4 � 3 � 2 � 1 ______________________ 3 � 2 � 1

� 8 � 7 � 6 � 5 � 4 � 6720Solve.

2. Find the permutation of 6 objects taken 4 at a time.

360 3. Find the permutation of 12 objects taken 5 at a time.

95,040 4. A magazine editor has 3 different spaces to arrange articles in a

magazine. She must choose from 7 different articles. How many different arrangements are possible? Use the formula for finding the permutation.

210

ReteachingFinding Permutations and Combinations 42


Reteachingcontinued

A combination is a selection of items from a group in which the order isNOT important. In a combination, AB is the same as BA.

The number of combinations of n items taken r at a time is shown by

C (n, r ) � n! _______ (n�r)!r !

Find the combination of 10 objects taken 4 at a time.

To find the number of combinations of 10 items taken 4 at a time, use n � 10 and r � 4 in the combination formula, then evaluate.

C (10, 4 ) � 10! _________ (10 � 4)!4!

� 10! ____ 6!4!

� 10 � 9 � 8 � 7 � 6 � 5 � 4 � 3 � 2 � 1 __________________________ (6 � 5 � 4 � 3 � 2 � 1)(4 � 3 � 2 � 1)

Divide out common factors.

� 10 � 3 9 � 8 � 7 ___________ 4 � 3 � 2 � 1

Divide out common factors.

� 10 � 3 � 7 Simplify.

� 210 Multiply.

There are 210 combinations of 10 objects taken 4 at a time.


5. Find the combination of 8 objects taken 3 at a time.

C(8, 3)� 8! ________ (8�3)!3!

� 8! ____ 5!3!

� 4 8 � 7 � 6 � 5 � 4 � 3 � 2 � 1 ______________________ (5 � 4 � 3 � 2 � 1)(3 � 2 � 1)

� 4 � 7 � 6 ________ 3�1

� 56

Solve.

6. Find the combination of 9 objects taken 5 at a time. 126

7. Find the combination of 12 objects taken 4 at a time. 495

8. The track team has 6 runners in a race. The coach selects 2 runners to run in the first heat. In how many ways can the runners be selected? Use the formula for finding the combinations. 15

9. A newspaper reporter has the names of 7 people available to answer her questions.She must select 3 of the people to interview. How many different combinations of respondents can she select? 35

42

Name Date Class


You have learned how to solve and graph the solution to a single linear inequality. Now you will learn to solve a system of linear inequalities.

To use graphs to find the solution to a system of inequalities:

1. Draw the graph of the boundary line for the first inequality.Remember to use a solid line for � or � and a dashed line for � or �.

2. Shade the region above or below the boundary line that is asolution of the inequality.

3. Draw the graph of the boundary line for the second inequality.

4. Shade the region above or below the boundary line that is a solution of the inequality using a different pattern.

5. The region where the shadings overlap is the solution region.

Graph { y � x � 2

x � 1 .

Step 1: Graph y � x � 2.

Use a solid line for the boundary.Shade the region below the line.

Step 2: Graph x � 1 on the same plane.Use a dashed line for the boundary.Shade the region to the right of the line.

Step 3: Describe the solution set.

The solution set of the system consists of all points that are the solutions to both inequalities. It is represented by the dark shaded region and the solid boundary line, not including the point (1, 3).

PracticeComplete the steps to graph the system of linear inequalities and describe the solution set.

1. { y � �x � 1

y � 2

Graph the system of linear inequalities. Describe the solution set.

2. { y � x � 2

x � �1

43ReteachingSolving Systems of Linear Ineqalities

x

y

-4

-2

2

4

2 4 6-2-4-6O

x

y

-2

4

4 6-2-4-6O

-4

2

2

x

y

-4

-2

4

2 4 6-2-4-6O

2

a. Shade above the line for y � �x � 1.

b. Shade below the line for y � 2.

The solution set of the system consists of all points in the dark shaded region including the solid boundary line,

except the point (�1, 2).

The solution set of the systemconsists of all points in thedark shaded region includingthe solid boundary line,except the point (�1, 1).


Reteachingcontinued

The solution set of a system may create a geometric figure.

Graph { y � x � 2

y � �2

x � 3 .

Step 1: Graph y � x � 2 and y � �2.

Use solid boundary lines. Shade the region below y � x � 2 and above y � �2.

Step 2: On the same plane, graph x � 3.

Use a solid boundary line. Shade the region to the left of x � 3.

The figure created by the overlapping pattern is a triangle. The solution set of the system consists of all points on the triangle formed by the three boundary lines and all the points inside the triangle.

PracticeComplete the steps to graph the system of linear inequalities and describe the solution set.

3. { y � 2x � 1

y � �x � 1

x � 3

a. Shade below the line for y � 2x � 1.

b. Shade above the line for y � �x � 1.

c. Shade to the left of the line for x � 3.

The solution set of the system consists of all points on the triangle formed by the three solid boundary lines and all the points inside the triangle.

x

y

-4

2

4

6

2 4 6-2-4-6O

-2

Graph the system of linear inequalities. Describe the solution set.

4. { y � �x � 3

y � 0

x � 2

The solution set of the system consists of all points on the triangle formed by the three solidboundary lines and all the pointsinside the triangle.

5. { y � x � 4

y � �x � 5

x � 0

The solution set of the system consists of all points on the triangle formed by the three solidboundary lines and all the pointsinside the triangle.

x

y

-4

2

4

2 4 6-2-4-6O

-2

43

x

y

-4

-2

2

4

42 6-2-4-6Ox

y

-4

2

4

4 6-2-4-6O

-2

2

Name Date Class


You have learned to simplify rational expressions by finding square or higher order roots. Now you will learn to simplify a radical expression by rationalizing a denominator.

An expression containing square root radicals is in simplest form when no radicand has a perfect square root factor and there are no irrational radical expressions in any denominator.

Simplify 3 ____ 5 �

� 2 .

The expression has an irrational denominator. Multiply both the numerator and the denominator by �

� 2 to eliminate the radical from

the denominator.

3 ____ 5 �

� 2 � �

� 2 ____

��

2 � 3 �

� 2 _______

5 ��

2 ��

2 Multiply both the numerator and the denominator by �

� 2 .

� 3 ��

2 ____ 5 � 2

Simplify the denominator: 5 � ��

2 � ��

2 � 5 ��

4 � 5 � 2.

� 3 ��

2 ____ 10

The expression is in simplest form.

Simplify ��

5 ___ 12

.

The radicand is a fraction. Multiply both the numerator and the denominator of the radicand by 3 to get a perfect square in the denominator.

��

5 ___ 12

� 3 __ 3 � �

�

15 ___ 36

� ��

15 _____ �

� 36 � �

� 15 ____

6


1. 5 ____ 4 �

� 3 5 _____

4 ��

3 � �

� 3 ___

��

3 � 5 �

� 3 ___________

4 ��

3 ��

3 2. �

�

3 __ 5 �

�

3 __ 5 �

5 _____

5 � �

�

15 _____ 25

� 5 ��

3 _____ 4 � 3

� ��

15 _______ �

� 25

� 5 ��

3 ____ 12

� ��

15 ____ 5

Simplify.

3. 3 ____ 2 �

� 2

3 ��

2 _______

4 4. �

�

1 __ 3

��

3 _______

3

44ReteachingRationalizing Denominators


Reteachingcontinued

Multiply by a conjugate to simplify radical expressions that have binomialdenominators.

Radical Conjugates

If a, b, c, and d are rational numbers, then a ��

b � c ��

d and a ��

b � c ��

d are conjugates of each other, and their product is a rational number.

If the denominator has the form a ��

b � c ��

d multiply numerator and denominator by a �

� b � c �

� d .

If the denominator has the form a ��

b � c ��

d multiply numerator and denominator by a �

� b � c �

� d .

Simplify 1 ______ 5 � �

� 2 .

The expression has an irrational binomial denominator. Multiply by aconjugate to rationalize the denominator. The conjugate of 5 � �

� 2

is 5 � ��

2 .

1 _______ 5 � �

� 2 � 5 � �

� 2 _______

5 � ��

2 � 5 � �

� 2 ____________

(5) 2 � � �� 2 � 2

Use the Distributive Property to simplify the numerator.Use (a � b)(a � b) � a 2 � b 2 in the denominator.

� 5 � ��

2 _______ 25 � 2

Simplify the denominator: 5 2 � � �� 2 � 2 � 25 � 2.

� 5 � ��

2 _______ 23

The expression is in simplest form.


5. 1 _______ 5 � �

� 2 6. 2 _______

5 � ��

6

1 _______ 5 � �

� 2 � 5 � �

� 2 __________

5 � ��

2 � 5 � �

� 2 ______________

( 5 ) 2 � � �

� 2 �

2 2 _______

5 � ��

6 � 5 � �

� 6 ________

5 � ��

6 � 2 � 5 � �

� 6 � ______________

( 5 ) 2 � � �

� 6 �

2

� 5 � ��

2 __________ 25 � 2

� 10 � 2 ��

6 __________ 25 � 6

� 5 � ��

2 _______ 23

� 10 � 2 ��

6 _________ 19

Simplify.

7. 1 _______ 6 � �

� 5 6 � �

� 5 _______

31 8. 2 _______

5 � ��

8 10 � 2 �

� 8 _________

17

44

Name Date Class


You have created scatter plots and graphed linear equations. Now you will find the line of best fit for data that appear to have a linear relationship.

Use a scatter plot to identify a correlation. If the variables appear to be correlated, then find a line of best fit.

Create a scatter plot of the data and describe the correlation. Sketch a line of best fit, write the equation of the line, and describe the slope.

Step 1: Make a scatter plot of the data. As x increases, y decreases. The data is negatively correlated.

Step 2: Use a straightedge to draw a line.There will be some points above and below the line.

Step 3: Choose two points on the line to find the equation.Use (1, 16) and (7, 2).

Step 4: Use the points to find the slope.

m � change in y

__________ change in x

� 16 � 2 ______ 1 � 7

� 14 ___ � 6

� � 7 __ 3

Step 5: Use (7, 2) and the point-slope form to find the equation of a line that models the data.

Step 6: Describe the slope. The y-value decreases by 7 for every increase of 3 in the x-value.

PracticeComplete the steps to describe the correlation of the data in the scatter plot. Write the equation of the line of best fit and describe the slope 1. correlation: positive

slope of line: 3 __ 2

equation of line: y � 3 __ 2 x � 2

describe the slope: The y-valueincreases

by 3 for everyincrease

of 2 in the x-value.

Describe the correlation of the data in the scatter plot. Write the equation of the line of best fit and describe the slope.

2. negative correlation; y � �x � 10; The y-value decreases by 1 for every increase of 1 in the x-value.

16

12

8

4

02 4 6 8 x

y

x 1 2 3 4 5 6 7 8

y 16 14 11 10 5 2 3 2

45ReteachingFinding the Line of Best Fit

y � y 1 � m � x � x 1 �

y � 2 � � 7 __ 3

� x � 7 �

y � � 7 __ 3 x � 18 1 __

3

16

12

8

4

02 4 6 8 x

y

16

12

8

4

02 4 6 8 x

y


Reteachingcontinued

A line of best fit can be used to predict data. Use the correlation coefficient r to measure how well the data fit.

Make a scatter plot of the data. Then use a graphing calculator to find the correlation coefficient. Predict y when x � 3.5.

x 1 2 3 4 5 6 7 8

y 16 14 11 10 5 2 3 2

Step 1: Plot the data.

Step 2: Use STAT EDIT to enter the data.

Step 3: Use LinReg from the STAT CALC menu to find the line of best fit and the correlation coefficient. Round your answers as shown in Steps 4–6.

Step 4: The correlation coefficient is �0.9648. The data is very close to linear with a negative slope.

Step 5: Write the linear regression equation. y � �2.2x � 17.79

Step 6: Find y when x � 3.5. y � �2.2x � 17.79 y � �2.2 � 3.5 � � 17.79 y � 10.09

PracticeComplete the steps using a calculator and the scatter plot of the data to find the correlation coefficient and to predict y when x � 2.6.

3. Read the points from the graph.

x 1 2 3 4 5 6 7 8y 4 5 5 8 10 11 13 12

correlation coefficient, r : 0.9651

equation of the line of best fit: y � 1.4x � 2.29 predict y when x � 2.6: y � 5.9Use a calculator and the scatter plot of the data to find the correlation coefficient, the equation of the line of best fit, and to predict y when x � 1.4.

4. 16

12

8

4

02 4 6 8 x

y r � �0.8149 y � �0.8x � 8.86 y � 7.74

45

LinRegy � ax � ba � �2.202b � 17.786 r 2 � .9308r � �.9648

16

12

8

4

02 4 6 8 x

y

16

12

8

4

02 4 6 8 x

y

Name Date Class


You have used the Pythagorean Theorem to find the lengths of the sides of a right triangle. Now you will use trigonometric functions to describe the relationship between the lengths of the sides of a right triangle.

A trigonometric ratio compares the lengths of two sides of a right triangle. The values of the ratios depend upon one of the acute angles of the triangle.

sin�A � Opposite

___________ Hypotenuse

� a __ c

cos�A � Adjacent

__________ Hypotenuse

� b __ c

tan�A � Opposite

________ Adjacent

� a __ b

Use the definitions of each ratio and the corresponding values from a given right triangle to find the values of the sine, cosine, and tangent of �A.

sin�A � Opposite


� 14 ___ 50

� 7 ___ 25

cos�A � Adjacent


� 48 ___ 50

� 24 ___ 25

tan�A � Opposite

________ Adjacent

� 14 ___ 48

� 7 ___ 24

PracticeComplete the steps to find the values of the sine, cosine, andtangent of �A.

1. B

C A

9

15

12

Find the values of the sine, cosine, and tangent of �A.

2. A

C B

10

26

24

46ReteachingUnderstanding Trigonometric Functions and their reciprocals

Use SOHCAHTOA to remember the relationships between the sides of a right triangle that correspond to the trigonometric ratios sine, cosine, and tangent.

B

C A

a c

b

Hypotenuse is opposite the right angle

Leg adjacent to ∠A

Leg opposite ∠A

B

C A

14

50

48

Sine is Opposite over Hypotenuse,

Cosine is Adjacent over Hypotenuse,

Tangent is Opposite over Adjacent.

sin�A � Opposite __________ Hypotenuse

� 9 ___ 15

� 3 __ 5

cos�A � Adjacent __________ Hypotenuse

� 12 ___ 15

� 4 __ 5

tan�A � Opposite ________ Adjacent

� 9 ___ 12

� 3 __ 4

sin�A � 12 ___ 13

cos�A � 5 ___ 13

tan�A � 12 ___ 5



The reciprocals of the sine, cosine, and tangent ratios are also trigonometric ratios.

The cosecant function (csc) is the reciprocal of the sine function.

csc�A � 1 _____ sin�A

� Hypotenuse

__________ Opposite

� c __ a

The secant function (sec) is the reciprocal of the cosine function.

sec�A � 1 ______ cos�A

� Hypotenuse

__________ Adjacent

� c __ b

The cotangent function (cot) is the reciprocal of the tangent function.

Cot�A � 1 ______ tan�A

� Adjacent

________ Opposite

� b __ a

Use the reciprocal relationship of the ratios to find the values of the cosecant, secant, and cotangent of �A.

sin�A � 3 __ 5 csc�A � 1 ______

sin�A � 5 __

3

cos�A � 4 __ 5 sec�A � 1 ______

cos�A � 5 __

4

tan�A � 3 __ 4

cot�A � 1 ______ tan�A

� 4 __ 3

PracticeComplete the steps to find the values of the six trigonometric functions for �A.

3. C B

A

8

15

17

Find the values of the six trigonometric functions for �A.

4.

B C

A

15

3639

5.

B C

A

7

2425

B

C A

ac

b

B

C A

35

4

sin�A � 15 ___ 17

cos�A � 8 ___ 17

tan�A � 15 ___ 8

csc�A � 1 ________ sin�A

� 17 ___ 15

sec�A � 1 ________ cos�A

� 17 ___ 8

cot�A � 1 ________ tan�A

� 8 ___ 15

sin �A � 5 ___ 13

csc�A � 13 ___ 5

cos�A � 12 ___ 13

sec�A � 13 ___ 12

tan�A � 5 ___ 12

cot�A � 12 ___ 5

sin �A � 7 ___ 25

csc�A � 25 ___ 7

cos�A � 24 ___ 25

sec�A � 25 ___ 24

tan�A � 7 ___ 24

cot�A � 24 ___ 7

Name Date Class


You have graphed linear functions. Now you will graph exponential functions.

When two exponential functions have bases that are reciprocals, the graphs of the functions are reflection images of each other over the y−axis.

Graph y 1 � � 1 __ 4 � x and y 2 � 4 x . Identify the domain, the asymptote,

and the range of each function.Step 1: Make a table of values.

Step 2: Plot the ordered pairs and draw smooth curves through them.

The domain of both functions is the set of all real numbers. The asymptote is the line y � 0. The range is the set of all positive real numbers.

PracticeComplete the steps to graph the functions. Identify the domain, the asymptote, and the range of each function.

1. y 1 � � 1 __ 3 � x and y 2 � 3 x

The domain for both functions is the set of all real numbers.

The asymptote is the line y � 0 . The range is the set of all positive real numbers.

Graph the functions on the same plane. Identify the domain, the asymptote, and the range of each function.

2. y 1 � � 2 __ 5 � x and y 2 � � 5 __

2 � x

Domain: the set of all real numbers Asymptote: the line y � 0 Range: the set of all positive real numbers

47ReteachingGraphing Exponential Functions

x

y

6

1

4

2

2-2 -1O

y1 = (4)

x1 y2 = 4x

x

y

6

1

4

2

2-2 -1O

y1 = (3)

x1 y2 = 3x

x

y

6

1

4

2

2-2 -1O

y1 = (5)

x2 y2 = (2)

x5

x �2 �1 0 1 2

y 1 � � 1 __ 4

� x 16 4 1 1 __ 4

1 ___ 16

y 2 � 4 x 1 ___ 16

1 __ 4

1 4 16

x �2 �1 0 1 2

y 1 � � 1 __ 3

� x 9 3 1 1 __ 3 1 __

9

y 2 � 3 x 1 __ 9 1 __

3 1 3 9


Reteachingcontinued

The transformations of the exponential function are vertical stretches and compressions, reflections over the x−axis, and vertical and horizontal shifts.

Graph the functions y 1 � 2 x , y 2 � 1 __ 4

. 2 x and y 3 � 1 __ 4

. 2 x � 1. Describe each graph.

Step 1: Make a table of values.

x �2 �1 0 1 2 3 4

y 2 � 2 x 1 __ 4

1 __ 2

1 2 4 8 16

y 2 � 1 __ 4 � 2 x 1 ___

16 1 __

8 1 __

4 1 __

2 1 2 4

y 3 � 1 __ 4 � 2 x � 1 � 15 ___

16 � 7 __

8 � 3 __

4 � 1 __

2 0 1 3

Step 2: Plot the ordered pairs and draw smooth curves through them.

Step 3: Describe the graphs.

y 1 � 2 x is the parent exponential function with b � 2.

y 2 � 1 __ 4

. 2 x is a vertical compression of the graph of y1 by a factor of 1 __ 4 .

y 3 � 1 __ 4

. 2 x � 1 is a vertical shift 1 unit down of the graph of y2.

PracticeComplete the steps to graph the functions.

3. y 1 � � 0.2 � x , y 2 � 3 � 0.2 � x , and y 3 � 3 � 0.2 � x � 2

x �2 �1 0 1 2 3 4

y 1 � � 0.2 � x

y 2 � 3 � 0.2 � x

y 3 � 3 � 0.2 � x � 2

25 5 1 0.2 0.008 0.00160.4

75 15 3 0.6 0.024 0.00480.12

17 5 2.6 2.024 2.00482.1277

Graph each function on the same plane. Describe each graph.

4. y 1 � � 1 __ 3

� x reflection of y � 3 x over the y-axis

5. y 2 � 3 � 1 __ 3

� x vertical stretch of y � � 1 __ 3 � x

6. y 3 � 3 � 1 __ 3

� x � 3 vertical shift 3 units down of y � � 1 __ 3 � x

47

x

y

2

4

2

4-4 -2

-2

-4

y1 = 2x

y2 = 4

. 2x1

y3 = 4

. 2x-11

O

x

y

2

4

2

4-4 -2

-2

-4

y1 = (0.2)x

y2 = 3(0.2)x

y3 = 3(0.2)x+ 2

O

x

y

2 4-4 -2

-2

-4

y1 = (3)

x1y2 = 3(3)

x1

y3 = 3(3)

x- 31

O

Name Date Class


You have added and subtracted rational expressions by rewriting them with the least common denominator. Now you will simplify complex fractions using a similar method.

A complex fraction is a fraction that contains one or more fractions in its numerator or denominator.

To simplify a complex fraction, first simplify the numerator and denominator so that they are each a single term. Then divide.

Simplify 2 � 1 __

3 _____

1 __ 3 � 1 __

4 .

2 � 1 __

3 _____

1 __ 3 � 1 __

4 �

6 __ 3 � 1 __

3 _______

4 ___ 12

� 3 ___ 12

Add the terms in the numerator and add the terms in the denominator.

� 7 __ 3 ___

7 ___ 12

Simplify.

� 7 __ 3

� 12 ___ 7

Write the division as multiplication by the reciprocal.

� 12 ___ 3 Cancel out common factors.

� 4 Simplify.

Practice

Complete the steps to simplify the complex fraction.

1. 2 __ 5

� 1 __ 3 _____

5 __ 6

� 1 __ 2

LCD of the numerator: 15 LCD of the denominator: 6

2 __ 5

� 1 __ 3 _____

5 __ 6

� 1 __ 2

�

6 ___

15 �

5 ___

15 ________

5 __ 6

� 3 ___ 6 �

11 ___ 15

____

2 ___ 6 �

11

___ 15

___

1 ___ 3

�

11 ___

15 �

3 ___

1 �

11 ___

5

Simplify.

2. 3 __ 8

� 1 __ 3 _____

1 __ 4

� 5 __ 6 3.

1 __ 4 � 7 ___

10 ______

2 � 3 __ 4 4.

2 __ 3 � 1 __

2 ______

3 ___ 10

� 3 __ 5

17 ___ 26

19 ___ 25

5 ___ 27

48ReteachingUnderstanding Complex Fractions

The LCD of 2 __ 1 and 1 __

3 is 3

The LCD of 1 __ 3 and 1 __

4 is 12



You can simplify the numerator and the denominator at the same time by multiplying them by their least common denominator.

Simplify 1 � a __

b ______

3 ___ 2a

� 3 �

1 � a __

b ______

3 ___ 2a

� 3

� � 1 � a __

b � � 2ab

____________

� 3 ___ 2a

� 3 � � 2ab Multiply the numerator and �

1 � 2ab � a __ b � 2ab _______________

3 ___ 2a

� 2ab � 3 � 2ab Distribute 2ab.

the denominator by 2ab. � 2ab � 2 a 2 _________ 3b � 6ab

Simplify.

PracticeComplete the steps to simplify the complex fraction.

5. m __ n � 1 _____

n � 2 _________

2 _____ n � 2

�

� m __ n � 1 _____ n � 2

� �n (n � 2) ____________________

� 2 _____ n � 2

� �n(n � 2)

� � m __ n � �n (n � 2) � � 1 _____

n � 2 � �n (n � 2)

_______________________________

� 2 _____ n � 2

� �n (n � 2) Distribute n(n � 2).

� mn � 2m � n _____________ 2n

Simplify.

Simplify.

6. j � 6

_____ 3j

______

k � 6 k __ j

1 ___ 3k

7.

2 __ x � 2 __ y _____

x � y

_____ x

2 __ y

8. A hollow metal pipe has a length of 1 __ 2 meter. Express the

formula for the volume of the metal in simplest form in terms of R if the small radius r is one third the large radius R.

(Hint: substitute R __ 3

for r.) 4 � R 2 ______

9

R

r ½ m

The LCD of a __ b and 3 ___

2a is 2ab.

V � � � R 2 � r 2 � _________ 2

Multiply the numerator and

the denominator by n(n � 2).

Name Date Class


You have learned that a combination is a selection of items when order does not matter. Now you will relate combinations to the Binomial Theorem.

The coefficients of the expansion of (x � y) n are the numbers in Pascal’s Triangle. The binomial coefficients are also combinations.

Pascal’s Triangle Combinations Binomial Expansion

1 0 C 0 (x � y) 0 � 1

1 1 1 C 0 1 C 1 (x � y) 1 � x � y

1 2 1 2 C 0 2 C 1 2 C 2 (x � y) 2 � x 2 � 2xy � y 2

1 3 3 1 3 C 0 3 C 1 3 C 2 3 C 3 (x � y) 3 � x 3 � 3x 2 y � 3xy 2 � y 2

The Binomial Theorem summarizes the relationship.

(x � y) n � n C 0 x n y 0 � n C 1 x n�1 y 1 � n C 2 x n�2 y 2 � . . . � n C n�1 x 1 y n�1 � n C n x 0 y n

You can use the Binomial Theorem to expand any power.

Expand (x � y) 5 .

Step 1: Use the Binomial Theorem with n � 5.

(x � y) 5 � 5 C 0 x 5 y 0 � 5 C 1 x 4 y 1 � 5 C 2 x 3 y 2 � 5 C 3 x 2 y 3 � 5 C 4 x 1 y 4 � 5 C 5 x 0 y 5

Step 2: Use the fifth row Pascal’s Triangle to find the coefficients. Add pairs of terms from the fourth row to find the numbers in the fifth row.

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

Step 3: Write the coefficients.

(x � y) 5 � 1 x 5 y 0 � 5 x 4 y 1 � 10 x 3 y 2 � 10 x 2 y 3 � 5x 1 y 4 � 1 x 0 y 5

Step 4: Simplify.

(x � y) 5 � x 5 � 5 x 4 y � 10 x 3 y 2 � 10 x 2 y 3 � 5 xy 4 � y 5

PracticeComplete the steps to use the Binomial Theorem to expand the power.

1. (2x � y) 3 � 3 C 0 (2x) 3 y 0 � 3 C 1 (2x) 2 y 1 � 3 C 2 (2x) 1 y 2 � 3 C 3 (2x) 0 y 3

� 3 C 0 � 8 x 3 y 0 � 3 C 1 � 4 x 2 y 1 � 3 C 2 � 2 xy 2 � 3 C 3 � y 3

� 1 � 8 x 3 � 3 � 4 x 2 y � 3 � 2 xy 2 � 1 � y 3

� 8 x 3 � 12x 2 y � 6 xy 2 � y 3

49

The sum of the exponents for each term is n.

The sum of the exponents for each term is 5. The exponents of x decrease as the exponents of y increase.

ReteachingUsing the Binomial Theorem


Reteachingcontinued

The Binomial Theorem can also be used to find probabilities for a binomial experiment.

If a binomial experiment has n trials with probability of success p and probability of failure q for any trial, then the binomial probability of exactly r successes is given by the following formula.

P(r ) � n C r p r q n�r

In a game with a six-sided number cube, a person wins if the number 1 or 2 is rolled. The person loses if the number 3, 4, 5, or 6 is rolled. In 4 rolls, find the probability of exactly 2 wins.

Step 1: Find the probability of success (win) and failure (loss).

p � 2 __ 6

� 1 __ 3 , so q � 1 � p � 1 � 1 __

3 � 2 __

3

Step 2: Use the binomial probability formula with r � 2 successes, or wins, n � 4 rolls, or trials,

p � 1 __ 3

, and q � 2 __ 3 .

P(2) � 4 C 2 � 1 __ 3 � 2 � 2 __

3 � 4�2

� 4 C 2 � 1 __ 3 � 2 � 2 __

3 � 2

� 6 � 1 __ 9 � � 4 __

9 �

� 8 ___ 27

So, the probability is 8 ___ 27

that 2 out of 4 rolls of a number cube will result in a 1 or a 2.

PracticeDiane and Ed are playing a game with a six-sided number cube. Diane wins if the number 6 is rolled. She loses if any other numberis rolled. Complete the steps to find the probability.

2. Diane has exactly 2 wins in 3 rolls.

p � 1 __ 6 q � 5 __

6 n � 3 r � 2 P(r) � n C r p r q n�r �

5 ___ 72

Rob and Jenny are playing a game with a six-sided number cube. Rob wins if the number 1, 2, 3, or 4 is rolled. He loses if a 5 or 6 is rolled.

3. Find the probability Rob has exactly 1 win in 4 rolls. 8 ___ 81

4. Find the probability Rob has exactly 3 wins in 4 rolls. 32 ___ 81

49

Remember, a binomial experiment must meet these conditions:

It has a fixed number of trials, n.

The trials are independent.

The outcomes of the trials must fall into two categories (success and failure).

The probabilities remain constant for each trial.

•

•

•

•

q � 1 � p

Name Date Class


You have learned about relations and how to determine whether a relation is a function. Now you will learn about the inverses of relations and functions.

Not all relations are functions and not all relations have inverse functions. To determine whether the inverse of a relation is a function, use the horizontal-line test.

If any horizontal line passes through more than one point on the graph of a relation, the inverse of the relation is not a function.

PracticeComplete the steps to determine whether the inverse of each relation is a function.

1. You can draw a horizontal line that passes through more than one point on the graph.

2. You cannot draw a horizontal line that passes through more than one point on the graph.

The inverse is not a function. The inverse is a function.

Determine whether the inverse of each relation is a function.

3. 4.

The inverse is a function. The inverse is not a function.

50ReteachingFinding Inverses of Relations and Functions

Can you draw a horizontal line that passes through more than one point on the graph? Yes. So the inverse of the relation is not a function.

x

y45

23

1

21 3 54-4-3-5 -2

-5-4-3-2-1

O-1

x

y45

23

1

21 3 54-4-3-5 -2

-5-4-3-2-1

O-1

Can you draw a horizontal line that passes through more than one point on the graph? No. So the inverse of the relation is a function.

x

y45

23

1

21 3 54-4-3-5 -2

-5-4-3-2-1

O-1

x

y45

23

1

21 3 54-4-3-5 -2

-5-4-3-2-1

O-1

x

y45

23

1

21 3 54-4-3-5 -2

-5-4-3-2-1

O-1

x

y45

23

1

21 3 54-4-3-5 -2

-5-4-3-2-1

O-1


Reteachingcontinued

The inverse of a function,f � x � , is denoted f � 1 � x � .

To find the inverse of a function f � x � , write y � f � x � . Then switch x and y and solve the equation for y.

Find the inverse of f (x) � x 2 � 1.

Step 1: Write y � f � x � .

y � x 2 � 1

Step 2: Switch x and y.

x � y 2 � 1

Step 3: Solve for y.

x � y 2 � 1 Isolate the variable, y.

x � 1 � y 2 Take the square root of both sides.

�� x � 1 � ��

y 2 The domain is { x � x � � 1 } . ��

x � 1 � y

Because y could have 2 different x-values, �� x � 1 and � ��

x � 1 , this inverse is not a function. Check by graphing the original function on a graphing calculator and use the horizontal-line test.

The domain of the inverse is the range of f � x � : { x � x � � 1 } . The range of the inverse is the domain of f � x � : all real numbers.

PracticeComplete the steps to find the inverse of the function and state its domain and range.

5. f � x � � �� x � 3

y � �� x � 3 Domain of f �1 � x � and range of f � x � :

x � ��

y � 3 all real numbers x such that x � 0

x 2 � y � 3 Range of f �1 � x � and domain of f � x � :

f �1 � x � � (x 2 � 3) all real numbers x such that x � �3

Find the inverse of each function and state its domain and range.

6. f � x � � 2x � 1 7. f � x � � 1 __ 8 x 3 � 1

f �1 � x � � 1 __ 2 x � 1 __

2 f �1 � x � � 2 3

�� x � 1

Domain: all real numbers; Domain: all real numbers;

range: all real numbers range: all real numbers

50

Remember, �� x � 1 can be either positive or negative.

Name Date Class


You have learned about probability experiments. Now you will learn about a specific kind of probability experiment called a binomial experiment.

A binomial experiment is a probability experiment that meets all of the following conditions:

1. There must be a fixed number of trials.

2. Each trial can have only two outcomes or the outcomes can be reduced to only two.

3. The outcomes of each trial must be independent of each other.

4. The probability of success must be the same for each trial.

Determine if the experiment is a binomial experiment.

a. Taking a quiz with 10 true-false questions and guessing on all of them.

This is an example of a binomial experiment, because all four conditions are met:

1. There are a fixed number of trials: 10 questions.

2. There are only two outcomes for each trial (true or false).

3. The answer to one question does not affect the answer to any of the other questions.

4. The probability of success is 1 __ 2 for each trial.

b. Rolling a six-sided number cube until the number 6 appears.

Since the cube is rolled until a 6 appears, there are not a fixed number of trials. Therefore, this is not an example of a binomial experiment.

PracticeComplete the steps to determine if the experiment is a binomial experiment.

1. Rolling a six-sided number cube three times to see if the number 6 appears.

How many trials are there? 3 How many outcomes are there for each trial? 2; either a 6 appears or it does not Is each outcome independent of the others? Yes, the outcome of one roll does not affect the others. Is the probability of success the same for each

trial? Yes, the probability of rolling a 6 is 1 __ 6 for each trial. Is this an example of a

binomial experiment? yes

Determine if the experiment is a binomial experiment. If it is not, explain why.

2. Asking 100 people how old they are. 3. Asking 25 high school students if they have

no; There are more than two outcomes. ever traveled outside the United States.

yes

ReteachingFinding the Binomial Distribution

INV

5


Reteachingcontinued

Display the results of the binomial experiment in a bar graph.

Twenty students took a quiz made up of 10 true/false questions. All 20 students guessed the answer to each question. The quiz results are shown in the table.

Number of correct answers

10 9 8 7 6 5 4 3 2 1 0

Frequency 0 0 1 2 4 6 3 2 1 0 1

To display the results in a bar graph, place the number of possible correct answers from 0 to 10 along the x-axis and the frequency from 0 to 6 along the y-axis. Draw the bars representing each quantity. Label each axis and give the graph a title.

PracticeComplete the steps to create a bar graph that shows the results of the binomial experiment.

4. Five students each flipped a coin five times and counted the number of heads for each trial. The results of the experiment are shown in the table.

Number of heads

0 1 2 3 4 5

Frequency 0 1 1 2 1 0

To display the results in a bar graph, show the number of heads on the x -axis and the frequency on the y -axis. The range of the x-axis should be from 0 to 5 . The range of the y-axis should be from 0 to 5. Label the x-axis Number of heads. Label the y-ais Frequency. A possible title for the graph is Results of Coin Tosses.

Results of Guessing the Correct Answer

Number of Correct Answers

6

5

4

3

2

1

0

Freq

uenc

y

0 1 2 3 4 5 6 7 8 9 10

7

6

5

4

3

2

1

0

Freq

uenc

y

0 1 2 3 4 5

Number of Heads

Results of Coin Tosses

INV

5

Name Date Class


You have divided polynomials using long division. Now you will divide polynomials using synthetic division.

You can use long division to divide a polynomial by another polynomial. When the divisor is in the form � x � a � use synthetic division to divide.

Divide: � 2 x 2 � x � 10 � � � x � 3 � .

Step 1: Find a. The divisor is � x � 3 � . So, a � 3.

Step 2: Write a on the left.

Then write the coefficients of the dividend.

2 �1 �103

Step 3: Draw a horizontal line. Copy the first coefficient below the line.

2 �1 �103

2

Step 4: Multiply the first coefficient by a, or 3. Write the product in the second column.

Add the numbers in the column.

2 �1 �10 6

3

2 5

Step 5: Multiply that sum by a, or 3. Write the product in the third column.

Add the numbers in the column. The last number is the remainder.

2 �1 �10 6 15

3

2 5 5

Step 6: Write the quotient.

� 2 x 2 � x � 10 � � � x � 3 � � 2x � 5 � 5 _____ x � 3

PracticeComplete the steps to divide using synthetic division.

1. � 4 x 2 � 7x � 10 � � � x � 2 � 4 7 10 �8 2

�2

4 �1 12

4x � 1 � 12 _____ x � 2

a � �2

Use synthetic division to divide.

2. � 2 x 2 � 6x � 12 � � � x � 5 � 3. � 3 x 2 � 7x � 9 � � � x � 3 �

2x � 4 � 8 _____ x � 5

3x � 2 � �3 _____ x � 3

2, −1, and −10 are coefficients of 2 x 2 − x −10.

2 a � 2(3) � 6

5 a � 5(3) � 15

The numbers in the bottom row are the coefficients of the quotient.

51ReteachingUsing Synthetic Division


Synthetic substitution is the process of using synthetic division to evaluate a polynomial.

The Remainder Theorem says that if a polynomial f � x � is divided by

x�k, the remainder is r � f (k).

Find: f � 4 � for f � x � � x 3 � 2 x 2 � 9x � 6.

Step 1: Divide f � x � by x�4.

� x 3 � 2 x 2 � 9x � 6 � � � x � 4 �

Step 2: Find a. The divisor is � x � 4 � . So, a � 4.

Step 3: Use synthetic division.

1 �2 �9 6 4 8 �4

3

1 2 �1 2

Step 6: The remainder is the value of f(4).

f(4) � 2

PracticeComplete the steps to find f (3) using synthetic substitution.

4. f (x) � 4 x 3 � 11 x 2 � 2x � 8 4 �11 2 �8 12 3 15

3

4 1 5 7

f(3) � r � 7 a � 3

Find f (5) using synthetic substitution.

5. f � x � � x 3 � 3 x 2 � 9x � 1 6. f � x � � 3 x 3 � 18 x 2 � 11x � 9

6 �11

7. f � x � � 2 x 3 � 7 x 2 � 19x � 6 8. f � x � � � x 3 � 4 x 2 � 6x � 2

�14 7

9. Based on local census results for the years 1970 through 2000, the population of a city can be modeled by the function P � x � � 3 x 3 � 15 x 2 � 20x � 126, where x is the number of decades since 1970 and P (x) is the population in thousands. The table shows the relationship between the year and the number of decades since 1970. Use P (x) to predict the population of the city in 2010.[Hint: Use synthetic substitution to evaluate P (x) at x � 4.

Year 1970 1980 1990 2000 2010

Number of decades since 1970 0 1 2 3 4

In 2010, the population of the city will be about 158,000.


The coefficients of the polynomial form the top row of the synthetic division.

The last numberis the remainder.

Name Date Class


You have used trigonometric ratios to find the length of a side of a right triangle. Now you will use the particular ratios of two special right triangles.

When two triangles are similar, the ratios of the lengths of their corresponding sides are all the same. This ratio is the scale factor. To find the length of an unknown side, set up a proportion that includes the unknown side and solve.

Find the length of x given the triangles shown.

Step 1: Identify the corresponding sides.

• The sides opposite the right angles.

• The sides opposite the 30° angles.

Step 2: Write a proportion.

x __ 9 � 6 ___

12

Step 3: Solve for x.

12x � 54 Cross multiply.

x � 54 ___ 12

Divide both sides by 12.

x � 9 __ 2 Simplify.

PracticeComplete the steps to find the length of x given the triangles shown.

1. Corresponding

sides: sides opposite the right angles; sides oppositethe 60°angles

Proportion: x __ 6 � 6 �

� 3 ____

9

x � 4 ��

3

Find the length of x given the triangles shown.

2.

x � 6

ReteachingUsing Two Special Right Triangles 52

x

30°30°

60°60°12 9

6

x

30°

30°

60°

60°

9

6

6 3

x

30°

30°

60°

60°

143

7

Recall: multiply the diagonal terms: 12 � x and 9 � 6


Reteachingcontinued

The trigonometric ratios for triangles with angle measures of 30°, 45°, and 60° are known.

Use these ratios to find the length of an

unknown side.

Use a trigonometric ratio to find the length of s.

Step 1: Identify the relevant ratio. Side s is opposite the 30° angle.

The hypotenuse is given. Sine is the ratio opposite

__________ hypotenuse

.

Step 2: Write a proportion.

sin 30� � opposite __________ hypotenuse

� s ___ 16

Step 3: Solve for s.

sin 30� � s ___ 16

1 __ 2 � s ___

16 sin 30� � 1 __

2

2s � 16 Cross multiply.

s � 8 Divide by 2.

PracticeComplete the steps to find the length of s using a trigonometric ratio.

3. Trigonometric ratio: sine or cosine

Proportion: s __ 6 � �

� 2 ___

2

s � 3 ��

2

Solve.

4. The diagram shows a roof antenna supported by a guy wire. The guy wire forms a 60° angle as shown. Find the height of the antenna. (Hint: The support wire is attached below the top of the antenna. Remember to add the distance between the top of the antenna and the point where the wire is attached.)

26 ft

52

Angle Sine Cosine Tangent

30� 1 __ 2 �

� 3 ___

2 �

� 3 ___

3

45� ��

2 ___ 2 �

� 2 ___

2 1

60� ��

3 ___ 2 1 __

2 �

� 3

S

30°

60°16

S

45°

6

Look up sin 30� in the table.

60°

16 3 ft

2 ft

Name Date Class


You have learned how to use function rules and how to evaluate functions. Now you will learn to perform and evaluate a composition of functions.

The composition of functions allows two processes to be combined into one.

Let f � x � � 2x � 7 and g � x � � x 2 . Evaluate the composition f � g � 3 � � using two different methods.

Method 1: Working from the inside out, evaluate each function separately. g (x) is the innermost function. Evaluate g (x) first.

g � x � → g � 3 � � 3 2 � 9

Use the output of g(3) as the input for f (x).

f � x � → f � 9 � � 2 � 9 � � 7 � 18 � 7 � 25

So, f � g � 3 � � � 25.

Method 2: Replace the function g(x) for the input for the function f(x).

f � g � x � � � f � x 2 � Substitute g (x) with x 2 .

f � g � x � � � 2 � x 2 � � 7 Replace the x in f (x) with x 2 .

f � g � x � � � 2 � x 2 � � 7 Simplify.

Now, evaluate the function for x � 3.

f � g � 3 � � � 2 � 3 2 � � 7 � 2 � 9 � � 7 � 18 � 7 � 25

So, f � g � 3 � � � 25.


1. Let f � x � � x � 3x and g � x � � 2x. Evaluate the composition f � g � 5 � � using two different methods.

Method 1:

g � 5 � � 2 � 5 � � 10 f � 10 � � 10 � 3 � 10 � � 10 � 30 � 40f � g � 5 � � � 40

Method 2:

f � g � x � � � 2x � 3 � 2x � � 2x � 6xf � g � 5 � � � 2( 5 ) � 6 � 5 � �10 � 30 � 40f � g � 5 � � � 40

53ReteachingPerforming Compositions of Functions

gxDomain of g

Domain of f g(x)Range of g

f Range of f (g(x))

Function g Function f


Reteachingcontinued

Solve.

2. Let f � x � � 3x � 4 and g � x � � x 2 . Evaluate the composition f � g � 2 � � using two different methods.

Method 1: g � 2 � � 2 2 � 4; f � g � 2 � � � 3 � 4 � � 4 � 16.

Method 2: f � g � 2 � � � 3 � 2 2 � � 4 � 16 .

Composition can be used to combine three or more functions. Remember to work from the inside out.

Let f � x � � �2x � 3, g � x � � 5x, and h � x � � x 2 . Find the compositefunction f � g � h � x � � � .Step 1: Apply the function rule for h(x).

f � g � h � x � � � � f � g � x 2 � � The output of h(x) is the input for g(x).

Step 2: Apply the function rule for g(x).

f � g � h � x � � � � f � g � x 2 � � � f � 5 � x 2 � � The output of g(x) is the input for f(x).

� f � 5 x 2 � Simplify.

Step 3: Apply the function rule for f(x).

f � g � h � x � � � � f � g � x 2 � � � f � 5 x 2 � � �2 � 5 x 2 � � 3

� �10 x 2 � 3 Simplify.

So, f � g � h � x � � � � �10 x 2 � 3.


3. Let f � x � � x 2 � 5, g � x � � 4x � 1, and h � x � � �x. Find the compositefunction f � g � h � x � � � .

f � g � h � x � � � � f � g � �x � � � f � 4 � �x � �1 � � � �4x � 1 � 2 � 5

f � g � h � x � � � � 16 x 2 � 8x � 6Solve.

4. Let f � x � � x 2 � 5, g � x � � 4x � 1, and 5. Let f � x � � x 2 � 5, g � x � � 4x � 1, and h � x � � �x. Find the composite h � x � � �x. Find the composite function h � g � f � x � � � . function g � h � f � x � � � .

�4 x 2 � 19 �4 x 2 � 21

6. Let f � x � � 3x� 1, g � x � � 2 x 2 , and 7. Let f � x � � 3x� 1, g � x � � 2 x 2 ,h � x � � x � 2. Find the composite h � x � � 8. Find the compositefunction f � g � h � x � � � . function f � g � h � x � � � .

6 x 2 � 24x � 23 383

53

Name Date Class


You have learned how to graph linear inequalities. Now you will learn how to solve problems using linear programming.

The objective function is the best combination of values to maximize or minimize a function subject to the constraints graphed in the feasible region. The maximum or minimum occurs at one or more of the vertices of the feasible region. Evaluate the objective function for each vertex to find the maximum or minimum.

Given the objective function P � 5x � 7y and the graph and vertices shown, find the minimum and maximum values.

Evaluate the objective function at each vertex.

Find the minimum and maximum values.

x y P � 5x � 7y

0 0 0

0 1 7

4 3 41

6 0 30

The objective function is minimized at (0, 0) and maximized at (4, 3). The minimum and maximum values are 0 and 41.


1. Given the objective function P � 2x � 5y and the graph and vertices shown, find the minimum and maximum values.

P( 0 , 0 ) � 0 P( 0 , 1 ) � 5

P( 2 , 4 ) � 24 P(6, 0) � 12

The minimum value is 0.

The maximum value is 24.

Solve.

2. Given the objective function P � 3x � 6y and the graph and vertices shown, find the minimum and maximum values.

Minimum: 0; Maximum: 36.

x

y

O

56789

10

4

123

4 5 61 2 3 7 8

Feasibleregion

Vertices(0,0)(0,1)(4,3)(6,0)

9 10

x

y

O

56789

10

4

123

4 5 61 2 3 7 8

Feasibleregion

9 10

Vertices(0,0)(0,1)(2,4)(6,0)

x

y

O

56789

10

4

123

4 5 61 2 3 7 8

Feasibleregion

9 10

Vertices(0,0)(0,1)(2,5)

(4.5,0)

54ReteachingUsing Linear Programming


Reteachingcontinued

Linear programming has many applications in business and economics. It is often used either to maximize the income or minimize the cost of a production scheme.

A company makes two models of MP3 players. The M10 takes 3 hours tomanufacture and the M15 takes 1 hour. The company has multiple shifts and has a20−hour day for manufacturing the players. The M10 generates a profit of $15, andthe M15 a profit of $9. The M10 uses 15 special chips, while the M15 uses 10. Forthe next 21−day manufacturing cycle, there are 3000 chips available. How many ofeach type of MP3 player should the company produce to maximize its profits?

Step 1: Choose variables to represent the products.

Let x � the number of M10 models.

Let y � the number of M15 models.

Step 2: Write the objective function. The objective function is profit � 15x � 9y.

Step 3: Identify the constraints.

x � 0, y � 0 The number of each type of MP3 player is not negative.

3x � y � 420 The sum of the number of hours needed to manufacture each type of player must be less than or equal to the total number of manufacturing hours: 20 � 21 � 420.

15x � 10y � 3000 The sum of the number of chips needed for each player must be less than or equal to the number of chips available.

Step 4: Graph the constraints. Find the feasible region and

the vertices. Then find the profit at each vertex.

P(0, 0) � 0

P(80, 180) � 2820

P(140, 0) � 2100

P(0, 300) � 2700

The solution that offers the greatest profit is to produce

80 M10s and 180 M15s.


3. Using the information in the Example above, suppose that for the next manufacturing cycle, consisting of 42 days, there are 8000 chips available. How many of each type of MP3 player should the company produce to maximize its profits?

P � 15 x � 9 y Vertices: The company should produce

x � 0 , y � 0 � 0, 0 � � 0, 800 � 26 M10s and 760 M15s.

3x � y � 840 � 280 , 0 � � 26 2 __ 3 , 760 �

15x � 10y � 8000

54

x

y

O

300

400

200

100

200 300100 400

Vertices(0, 0)

(80,180)(140, 0)(0, 300)

Name Date Class


You have learned how to use the Fundamental Counting Principle to determine the number of possible outcomes in a sample space. Now you will learn to find the probability of random events.

Probability measures the likelihood of an event, or outcome. All of the possible outcomes make up the sample space.

For equally likely outcomes, the theoretical probability P of an event occurring is defined as follows,

P (event) � number of favorable outcomes _________________________ total number of outcomes

.

A calculator is programmed to generate random integers from 1 to 100 inclusive when the enter button is pressed. What is P (multiple of 10) when the enter button is pressed one time?

Step 1: Find the sample space.

There are 100 possible outcomes.

Step 2: Find the number of favorable outcomes.

There are 10 favorable outcomes (10, 20, 30, 40, 50, 60, 70, 80, 90, 100).

Step 3: Use the definition for theoretical probability.

P (multiple of 10) � number of multiples of 10 _____________________ total number of outcomes

� 10 ____ 100

� 1 ___ 10

� 0.1

The probability of getting a multiple of 10 is 0.1.


1. A calculator is programmed to generate random integers from 1 to 50inclusive when the enter button is pressed. What is P(prime number) when the enter button is pressed one time?

P (prime) � number of primes � 50 _______________________ total number of outcomes

� 15 ___ 50

� 3 ___ 10

� 0.3

Solve.

2. A calculator is programmed to generate random integers from 1 to 50inclusive when the enter button is pressed. What is P(composite number) when the enter button is pressed one time?

0.7

3. Estimate the probability that a point chosen randomly inside the rectangle is in the shaded region. Use the formula:

P (shaded region) � area of rectangle � area of triangle _____________________________ area of rectangle

0.8

55ReteachingFinding Probability

10 in.

15 in.

8 in.


Reteachingcontinued

Two events, A and B, are dependent if the occurrence of one affects theprobability of the occurrence of the other.

Dependent case: Event A: A marble is chosen from a bag and not replaced.

Event B: A second marble is then chosen.

Events A and B are dependent.

Multiplication Rule for the Probability of Dependent Events:P(A and B) � P(A) � P(B⏐A) where P(B⏐A) is the probability of B given that A has already occurred.

Two events, A and B, are independent if the occurrence of one does NOT affect the probability of the occurrence of the other.

Independent case: Event A: A marble is chosen from a bag and replaced.

Event B: A second marble is then chosen.

Events A and B are independent.

Multiplication Rule for the Probability of Independent Events:P (A and B) � P(A( � P (B)

A bag contains 4 green and 5 blue marbles. What is the probability ofpicking a green marble, not replacing it, and then picking ablue marble?

Not replacing the first marble affects the probability of picking the next marble because the sample space has one less outcome. Use the rule for probability of dependent events.

Step 1: Find the probability of picking a green marble on the first pick.

P (green) � 4 __ 9

Step 2: Find the probability of picking a blue marble on the second pick.

P (blue) � 5 __ 8

Step 3: Use the rule for the probability of dependent events.

P (green then blue) � P(green marble) . P(green marble⏐blue marble)

P (green then blue) � 4 __ 9

� 5 __ 8 � 20 ___

72 � 5 ___

18


4. A bag contains 5 green and 8 blue marbles. What is the probability ofpicking a green marble, not replacing it, and then picking a blue marble?

P(green then blue) � 5 ___ 13

� 8 ___ 12

� 40 ____ 156

� 10 ___ 39

Solve.

5. A bag contains 5 green and 8 blue marbles. What is the probability ofpicking a green marble, replacing it in the bag, and then picking a blue marble?

40 ____ 169

55

Name Date Class


You have learned how to identify rotational symmetry and measure angles. Now you will learn how to draw angles of rotation.

To draw an angle with a given measure, start with the initial side on the x-axis and rotate the terminal side to show the angle measure.

• Rotate counterclockwise for a positive angle measure.

• Rotate clockwise for a negative angle measure.

Draw the indicated angle in standard position.a. 210°

Start at the initial side and rotate the terminal side counterclockwise past 180° to 210°. Draw and label the angle.

x

y

0°180°

270°

210°

90°

Terminal side

Initial side

b. �145°

Start at the initial side and rotate the terminal side clockwise past �90° to �145°. Draw and label the angle.

x

y

0°-180°

-145°

-270°

-90°

Terminal side

Initial side

PracticeComplete the steps to draw the indicated angle in standard position.

1. 175° 2. �225°

Rotate the terminal side Rotate the terminal side

counterclockwise clockwise past

past 90° to 175°. �180° to −225°.

Draw the indicated angle in standard position.

3. 125°

x

y

0°180°

270°

90° 4. �65°

56

x

y

0°-180°

-270°

-90°

ReteachingFinding Angles of Rotation

x

y

0°180°

270°

90°

x

y

0°-180°

-270°

-90°


Reteachingcontinued

Coterminal angles are angles that share the terminal side.

• To find a positive coterminal angle to a given angle, add a multiple of 360°.

• To find a negative coterminal angle to a given angle, subtract a multiple of 360°.

The reference angle for an angle � in standard position is the positive acute angle formed by the terminal side of � and the x-axis. To find the reference angle for an angle � :

• Draw the angle in standard position.

• Find the reference angle formed by the terminal side of � and the x-axis.

• Find the measure of the reference angle.

a. Find the measure of one positive and one negative coterminal angle to 65�.

65� � 360� � 425� 65� � 360� � �295�

So, 425° and �295° are coterminal with 65�.

b. Find the measure of the reference angle for 130�.

180� � 130� � �50�

The reference angle for 130� is 50�.


5. Find the measure of one positive and one 6. Find the measure of the reference angle for negative coterminal angle to 135°. 75°.

For a positive angle, add 360° to 135° The measure of the acute angle formed to get 495° . by the terminal side of 75° and the x-axis is 75° .

For a negative angle, subtract 360° from 135° to get �225° .

Solve.

7. Find the measure of one positive and one negative 8. Find the measure of the coterminal angle to 115°. reference angle for 140°.

Sample: �245° and 475° 40°

56

x

y

θ =130°

Reference angle 50°

Name Date Class


You have learned how to evaluate exponential functions. Now you will learn to use exponential growth and exponential decay functions to solve problems.

Exponential Function

f(x) � kb x , where k and b are constantsGrowth Decay

1. k � 0 and b � 1

2. Increasing

3. Graph rises from left to right

1. k � 0 and 0 � b � 1

2. Decreasing

3. Graph falls from left to right

A savings account earns interest at an annual rate of 6% compounded monthly. If the account begins with a principal amount of $1,500, what will its value be after 2 years?

Interest is compounding monthly, so the rate at each compounding is 1 ___ 12 of 6% , or 0.5%.

Therefore each month the value is multiplied by 1.005. Two years consist of 24 months. The exponential function that represents the situation is f � 24 � � 1500 � 1.005 � 24 where,

k � 1500

b � 1.005; b � 1, so the function shows growth.

x � 24

Use a calculator to find the value of the account.

The value is about $1,690.74.

A mass of carbon-14 atoms decays in such a way that half disappear every 5,730 years. What portion of a 20-kg mass of atoms will remain after 11,460 years?

The rate of decay is 1 � 0.5 � 0.5. So, each half-life cycle the mass of the sample is multiplied by 0.5. The amount of time elapsed, 11,460 years, consists of 2 half-life cycles. The exponential function that represents the situation is f � 2 � � 20 � 0.5 � 2 where,

k � 20

b � 0.5; 0 � b � 1, so the function shows decay.

x � 2

The amount of carbon-14 left after 11,460 years is 5 kg.


1. A savings account earns interest at an 2. A mass of carbon-14 atoms decays inannual rate of 8% compounded annually. such a way that half disappear everyIf the account begins with a principal amount 5,730 years. What portion of a 50-kg massof $2,000, what will its value be after 5 years? of atoms will remain after 10,000 years?

k � 2000, b � 1.08 , x � 5 k � 50 , b � 0.5 , x � 1.75

The value is about $2,938.66. The amount of carbon-14 left after 10,000

years is about 14.87 kg .

ReteachingFinding Exponential Growth and Decay 57


Reteachingcontinued

Natural Logarithm

The natural logarithm of a positive number x is the power to which e is raised to get x. That is, if e y � x, then y is called the natural logarithm of x. The abbreviation is y � In (x).

Applications in exponential growth and decay often have exponentialfunctions that use the irrational number e as the base.

Because the exponential and logarithmic functions are inverses, apply the definition of the natural logarithm to both sides of an equation to“bring down” the exponent.

A doctor gives A milligrams of medicine to a patient. The functionC(t) � Ae �0.36t gives the amount of medication in the bloodstream after t hours. If the initial dose is 250 mg, after how many hours will it be 25 mg?

Step 1: Write an equation.

The initial dose, given at t � 0, is 250 mg. So, the initial amount A � 250.

C(t) � e �0.36t Replace A in the equation with 250.

Step 2: Find the time t at which the amount equals 25 mg.

25 � 250 e �0.36t Replace C(t ) with 25.

0.1 � e �0.36t Divide both sides by 250.

In (0.1) � �0.36t Apply the definition of natural logarithm.

�2.30 � �0.36t Approximate In (0.1) with a calculator.

6.40 � t Divide both sides by �0.36.

The amount of medication will be 25 mg after about 6.4 hours.


3. A doctor gives A milligrams of medicine to a patient. The function C(t) � Ae �0.36t gives the amount of medication in the bloodstream after t hours. If the initial dose is 400 mg, after how many hours will it be 50 mg?

C � t � � Ae �0.36t C � t � � 400 e �0.36t

50 � 400 e �0.36t

0.125 � e �0.36t

In � 0.125 � � �0.36 t

The amount of medication will be 50 mg after about 5.78 hours.

57

Name Date Class


You have solved quadratic equations by factoring. Now you will solve quadratic equations by completing the square.

You can use the square root property to solve some quadratic equations.

Square Root Property

If x 2 � a, where a � 0, then x � �� a or x � � ��

a .

In general,

If x 2 � a, then x � � �� a for any a � 0.

Solve: x 2 � 12x � 36 � 50.

Step 1: Factor the perfect square trinomial.

x 2 � 12x � 36 � 50

x 2 � 2 � 6 � x � 6 2 � 50

� x � 6 � 2 � 50

Step 2: Apply the square root property.

� x � 6 � 2 � 50

x � 6 � � ��

50 Take the square root of both sides.

x � 6 � � 5 ��

2 Simplify: ��

50 � ��

2.25 � 5 ��

2 .

Step 3: Solve for x.

x � 6 � � 5 ��

2

x � 6 � 6 � � 5 ��

2 � 6 Subtract 6 from both sides.

x � �6 � 5 ��

2

x � �6 � 5 ��

2 or x � �6 � 5 ��

2


1. x 2 � 8x � 16 � 18 x 2 � 8x � 16 � 18

� x � 4 � 2 � 18

x � 4 � �18

x � 4 � � 3 ��

2

x � 4 � 3 ��

2 or x � 4 � 3 ��

2

Solve.

2. x 2 � 2x �1 � 10

x � 1� ��

10

3. x 2 � 10x � 25 � 3

x � 5� ��

3

58ReteachingCompleting the Square


Reteachingcontinued

You can solve equations by completing the square.

If you have an equation of the form x 2 � bx � c � 0, always begin by subtracting c from both sides of the equation to isolate the binomial. Then add the square of half the coefficient of the x term to both sides. Once the square of the binomial x 2 � bx is completed, the equation can be solved by using square roots.

Solve by completing the square: x 2 � 4x � 10 � 0.

Step 1: x 2 � 4x � 10 � 10 � 0 � 10 Add 10 to both sides.

x 2 � 4x � 10 Isolate the binomial x 2 � 4x.

Step 2: � 4 __ 2 � 2 � 2 2 � 4 Find � b __

2 � 2 .

x 2 � 4x � 4 � 10 � 4 Add � b __ 2 � 2 to both sides.

x 2 � 4x � 4 � 14 Simplify.

Step 3: � x � 2 � 2 � 14 Factor the perfect square binomial.

Step 4: x � 2 � � ��

14 Apply the square root property.

Step 5: x � 2 � 2 � � ��

14 � 2 Subtract 2 from both sides.

x � � 2 ��

14 Solve for x.

PracticeComplete the steps to solve by completing the square.

4. x 2 � 8x � 12 � 0

x 2 � 8x � �12 x 2 � 8x � 16 � �12 � 16 � x � 4 � 2 � 4 x � 4 � �2 x � 6 or 2

5. x 2 � 6x � 4 � 0

x 2 � 6x � �4 x 2 � 6x � 9 � �4 � 9 � x � 3 � 2 � 5 x � 3 � � �

� 5

x � 3 � ��

5

Solve by completing the square.

6. x 2 � 4x � 1 � 0

2 � ��

3 7. x 2 � 8x � 10 � 0

4 � ��

6

8. x 2 � 6x � 40

�3 � ��

13

9. x 2 � 12x � 4 � 0

�6 � 4 ��

2

10. A toy rocket is launched into the air from ground level with an upward velocity of 96 feet per second. The height of the rocket at

any given time t can be described by the equation h � �16 t 2 � 96t. To find when the rocket is at a height of 112 feet, solve the equation

t 2 � 6t � �7 by completing the square. 3 � ��

2 sec

58

Name Date Class


You have learned to simplify radical expressions. Now you will learn to simplify radical expressions with fractional exponents.

Rational Exponents Property

For m and n integers and n positive:

a m __ n �

n ��

a m

Radical expressions of the form n �� a are called n th roots. In this form n is the index and a is the radicand.

Write the rational exponent as a radical expression.

a. 5 1 __ 3

Apply the rational exponent property:

a 1 __ n � n �� a

In this case, a � 5 and n � 3.

5 1 __ 3 �

3 ��

5

b. 7 3 __ 5

Apply the rational exponent property:

a m __ n �

n ��

a m

In this case, a � 7, n � 5, and m � 3.

7 3 __ 5

� 5 �

�� 7 3 or � 5 ��

7 � 3

PracticeComplete the steps to write the rational exponent as a radical expression.

1. 6 1 __ 4 a � 6, n � 4

6 1 __ 4

� 4

��

6

2. 5 3 __ 4

a � 5, n � 4, m � 3

5 3 __ 4

� 4

��

5 3 Write the rational exponent as a radical expression.

3. 5 1 __ 5

5

��

5

4. 8 2 __ 3

3

��

8 2

5. 15 1 __ 3

3

��

15

6. 10 3 __ 5

5

��

10 3 7. The area of a side of a cube is 50 3 square inches. The side length

of the cube is given by the expression 50 3 __ 2

inches. Write the

rational exponent as a radical expression.

��

50 3 in.

59ReteachingUsing Fractional Exponents


Reteachingcontinued

Simplify. Assume that all variables are positive.

a. 3 ��

125 x 6

� 3 ��

5 3 � x 2 � 3 Factor the radicand into perfect cubes.

� 3 �

�� 5 3 �

3 ��

� x 2 � 3 Apply the product property: n ��

ab � n �� a � n ��

b .

� 5 � x 2 � 5 x 2 Evaluate the radicals and simplify.

b. 81 5 ___

12 � 81

1 __ 2

� 81 5 ___

12 � 1 ___

12 � 81

6 ___ 12

� 81 1 __ 2

Apply the product of powers property: a m � a n � a m � n .

� ��

81 � 9 Apply the rational exponents property: a 1 __ n � n �� a .

c. � 256 1 ___

12 � 3

� 256 3 ___

12 � 256

1 __ 4

Apply the power of a power property.

� 4 ��

256 � 4 Apply the rational exponents property and simplify.

d. Solve x 3 __ 4 � 8

� x 3 __ 4 �

4 __ 3 � 8

4 __ 3

Raise both sides to the reciprocal power of x : 4 __ 3 .

x � 3 �

�� 8 4 �

3 ��

4096 � 16 Apply the rational exponents property and simplify.

PracticeComplete the steps to simplify each expression or solve the equation. Assume that any variables are positive.

8. 4 ��

625 x 12

� 4 ��

5 4 � � x 3 � 4

� 4 ��

5 4 �

4 ��

� x 3 � 4

� 5 x 3

9. 16 3 ___

20 � 16

2 ___ 20

� 16 3

___ 20

� 2

___ 20

� 16 1 _ 4

� 4 �� 16

� 2

10. � 27 1 __ 3

� 4

� 27 4

__ 3

�

3

��

27 14

� 3 ��

531,441 � 81

11. � x 5 __ 2

� � 32

� x 5 __ 2

� 2 _ 5 � 32

2 _ 5

x �

5

��

32 2

x � 4

Simplify or solve. Assume that all variables are positive.

12. 4 ��

625 x 8

5 x 2

13. 25 2 ___

14 � 25

5 ___ 14

5

14. � 1024 3 __ 5

� 1 __ 3

4

15. � x 3 __ 2

� � 64

16

59

Name Date Class


You have learned how to calculate the probability of certain events. Now you will differentiate between inclusive and mutually exclusive events and between independent and dependent events. Also, you will learn how to calculate their probabilities.

Determine whether an event is inclusive or mutually exclusive and calculate its probability.

a. A bag contains 10 tiles, numbered from 1 to 10. What is the probability of choosing an even number or a 6?

Solution: 6 is an even number, so choosing a 6 or choosing an even number are inclusive events. Inclusive events are events that can occur at the same time.

To calculate the probability of an inclusive event, use the following formula.

P � A or B � � P � A � � P � B � � P � A and B �

Let A represent the event of choosing a 6. Let B represent the event of choosing an even number.

P � A � � 1 ___ 10

PB � 5 ___ 10

P � A and B � � 1 ___ 10

P � A or B � � 1 ___ 10

� 5 ___ 10

� 1 ___ 10

= 5 ___ 10

or 1 __ 2

b. A bag contains 10 tiles, numbered from 1 to 10. What is the probability of choosing an even number or a 9?

Solution: 9 is not an even number, so choosing a 9 or choosing an even number are mutually exclusive events. Mutually exclusive events are events that cannot occur at the same time.

To calculate the probability of a mutually exclusive event, use the following formula.

P � A or B � � P � A � � P � B �

Let A represent the event of choosing a 9. Let B represent the event of choosing an even number.

P � A � � 1 ___ 10

P � B � � 5 ___ 10

P � A or B � � 1 ___ 10

� 5 ___ 10

� 6 ___ 10

or 3 __ 5

PracticeComplete the steps to determine whether an event is inclusive or mutually exclusive and calculate its probability.

1. Sheree has a number cube. What is the probability that she will roll a 1 or an odd number?

Rolling a 1 or an odd number are inclusive events.

P � A or B � � P � A � � P � B � � P � A and B � P � A � � 1 __ 6 P � B � � 5 ___

10 P � A and B � � 1 __

6

Let A represent the event of rolling a 1 and B represent the event of rolling P � A or B � � 1 __

6 � 5 ___

10 � 1 __

6 � 5 ___

10 or 1 __

2

an odd number.

Determine the probability of the event.

2. A bag contains 6 tiles, numbered from 1 to 6. What is the probability of choosing a 5 or an odd number?

3 __ 6 or 1 __

2

3. Derek rolls a number cube. What is the probability that he will roll a 2 or a number greater than 3?

4 __ 6 or 2 __

3

60ReteachingDistinguishing Between Mutually Exclusive and Independent Events


Reteachingcontinued

Calculate the probability of independent events.

a. Janice has two number cubes. One is blue and one is yellow. What is the probability that she will roll a 3 on the yellow cube and an even number on the blue cube?

Solution: Rolling a 3 on one cube does not affect the number rolled on the other cube. These are independent events. Two events are independent if the outcome of one does not affect the probability of the outcome of the other.

To calculate the probability of independent events, use the following formula.

P � A 1 , A 2 , . . . , A n � � P � A 1 � � P � A 2 � � . . . � P � A n � Let A 1 represent the event of rolling a 3.

Let A 2 represent the event of rolling a 3.

P � A 1 � � 1 __ 6

P � A 2 � � 3 __ 6 = 1 __

2

P � A 1 � � P � A 2 � � 1 __ 6 � 1 __

2 = 1 ___

12

b. Donna rolls three number cubes. What is the probability that she will roll an odd number on each of them?

Solution: The number rolled on one cube does not affect the number rolled on the other cubes. These are independent events. Use the independent events formula.

P � A 1 , A 2 , . . . , A n � �

P � A 1 � � P � A 2 � � . . . � P � A n � Let A 1 represent the event of rolling an odd number on the first cube.

Let A 2 represent the event of rolling an odd number on the second cube.

Let A 3 represent the event of rolling an odd number on the third cube.

P � A 1 � � 3 __ 6 or 1 __

2 P � A 2 � � 1 __

2 P � A 3 � � 1 __

2

P � A 1 � � P � A 2 � � P � A 3 � � 1 __ 2 � 1 __

2 � 1 __

2 � 1 __

8

PracticeComplete the steps to determine the probability.

4. The weather forecast on a certain TV station correctly predicts the high temperature for the day 62% of the time. What is the probability that the forecasted high temperature will be correct three days in a row?

Use the independent events formula. Round your answer to the nearest percent.

Let A 1 represent the probability of being correct on Day 1.



P � A 1 � � 0.62 P � A 2 � � 0.62 P � A 3 � � 0.62

P � A 1 � � P � A 2 � � P � A 3 � � 0.62 � 0.62 � 0.62 � 0.24 or 24 %

Determine the probability of the event.

5. Trang flips three coins. What is the probability that all three coins will come

up tails? 1 __ 8

6. Art rolls two number cubes. What is the probability that he will roll an even number on one and an odd number on

the other? 1 __ 4

60

Name Date Class


You have learned how to solve some quadratic equations. Now you will learn more about quadratic expressions.

The algebra tiles show that the area of the large square is both � a � b � 2 and a 2 � 2ab � b 2 . Therefore, � a � b � 2 � a 2 � 2ab � b 2 . The trinomial a 2 � 2ab � b 2 is known as a perfect-square trinomial.

Determine whether each trinomial is a perfect square.

a. x 2 � 8x � 16

Not all trinomials are perfect squares. Ask yourself these three questions to determine whether a trinomial is a perfect-square trinomial.

1) Is the first term, x 2 , a perfect square? Yes, x 2 � x � x.

2) Is the last term, 16, a perfect square? Yes, 16 � 4 � 4.

3) Is the middle term, 8x, twice the product of the square roots of the first and last term? Yes, 8x � 2(4x)

If you can answer “yes” to all three questions, the trinomial is a perfect square.

b. 4 x 2 � 36x � 9

1) Is the first term, x 2 , a perfect square? Yes, x 2 � x � x.

2) Is the last term, 9, a perfect square? Yes, but it has a negative sign in front of it. To be a perfect-square trinomial, the last term must be a positive perfect square.

Therefore 4 x 2 � 36x � 9, is not a perfect-square trinomial.

PracticeComplete the steps to determine whether the trinomial is a perfect square.

1. x 2 � 24x � 144 2. 4 x 2 � 50x � 25

Is the first term a perfect square? Yes Is the first term a perfect square? Yes Is the last term a perfect square? Yes Is the last term a perfect square? Yes

Is the middle term twice the product of the Is the middle term twice the product of thesquare roots of the first and last terms? square roots of the first and last terms?

Yes No

Therefore, x 2 � 24x � 144 is Therefore, 4 x 2 � 50x � 25 is not a a perfect-square trinomial. perfect-square trinomial.

Determine whether the trinomial is a perfect square.

3. x 2 � 16x � 64 No 4. 2 x 2 � 6x � 9 No

b

a

ab

a2

a

b2

ab

b

ReteachingDeriving the Quadratic Formula

INV

6


Reteachingcontinued

The trinomials a 2 � 2ab � b 2 and a 2 � 2ab � b 2 are perfect-square trinomials.

a 2 � 2ab � b 2 � � a � b � 2

a 2 � 2ab � b 2 � � a � b � 2

Factor each perfect-square trinomial.

a. 9 x 2 � 24x � 16

Take the square root of the first and last terms.

��

9 x 2 � 3x ��

16 � 4

Since 9 x 2 � 24x � 16 is a perfect-square trinomial of the form a 2 � 2ab � b 2 , you can write 9x 2 � 24x � 16 � � 3x � 4 � 2 .b. x 2 � 10x � 25


��

x 2 � x ��

25 � 5Since x 2 � 10x � 25 is a perfect-square trinomial of the form a 2 � 2ab � b 2 , you can write x 2 � 10x � 25 � � x � 5 � 2 .

PracticeComplete the steps to factor each perfect-square trinomial.

5. 16 x 2 � 16x � 4


��

16 x 2 � 4x ��

4 � 2 Since 16 x 2 � 16x � 4 is a perfect-square trinomial of the form

a 2 � 2ab � b 2 , you can write 16 x 2 � 16x � 4 � (4x � 2) 2

6. 36 x 2 � 84x � 49


��

36 x 2 � 6x ��

49 � 7

Since 36 x 2 � 84x � 49 is a perfect-square trinomial of the form

a 2 � 2ab � b 2 , you can write 36 x 2 � 84x � 49 � (6x � 7) 2

Factor each perfect-square trinomial.

7. x 2 � 18x � 81 8. x 2 � 20x � 100

(x � 9) 2 (x � 10) 2

9. 4 x 2 � 12x � 9 10. 14 x 2 � 28x � 49

(2x � 3) 2 (2x � 7) 2

INV

6

Name Date Class


You have learned how to use synthetic division to divide polynomials. Now you will learn how to use synthetic division and other methods to factor polynomials of degree 3 or higher.

Factor Theorem

The polynomial � x � a � is a factor of the polynomial P � x � if and only if P � a � � 0.

Determine whether the linear binomial is a factor of the polynomial.

a. P � x � � x 3 � x 2 � 2x � 8 � x � a � � x � 2 � x � � �2 �

By the factor theorem, x � 2 is a factor of x 3 � x 2 � 2x � 8 if P � �2 � � 0.

Use synthetic division to see if P � �2 � � 0.

�2 1 �1 �2 8 __ �2 6 �8

1 �3 4 0

The remainder is 0, therefore P � �2 � � 0. So, x � 2 is a factor of x 3 � x 2 � 2x � 8. Seeing the remainder is 0, the quotient of the division problem, x 2 � 3x � 4, is also a factor.

b. P � x � � x 3 � 4 x 2 � 1 � x � a � � x � 2

By the factor theorem, x � 2 is a factor of x 3 � 4 x 2 � 1 if P � 2 � � 0.

Use synthetic division to see if P � 2 � � 0.2 1 �4 0 1 Use 0 to represent coefficient of the x-term

__ 2 �4 �8 that does not exist.

1 �2 �4 �7

The remainder is –7, therefore P � 2 � � 0. So, x � 2 is not a factor of x 3 � 4 x 2 � 1.

PracticeComplete the steps to determine whether the linear binomial is a factor of the polynomial.

1. P � x � � x 3 � 2 x 2 � x � 3 � x � a � � x � 2

By the factor theorem, x � 2 is a factor of x 3 � 2 x 2 � x � 3 if P � 2 � � 0.

Use synthetic division to see if P � 2 � � 0.

2 1 �2 1 �3 2 0 2

1 0 1 �1The remainder is �1 , therefore P � 2 � � 0. So, x � 2 is not a factor of x 3 � 2 x 2 � x � 3.

Determine whether the linear binomial is a factor of the polynomial.

2. P � x � � 2 x 3 � 3 x 2 � 4 � x � a � � x � 2 yes

3. P � x � � 2 x 3 � 3 x 2 � 4x + 1 � x � a � � x � 1 no

61ReteachingUnderstanding Advanced Factoring


Reteachingcontinued

The sum of two cubes can be represented by a 3 � b 3 � � a � b � � a 2 � ab � b 2 � . The difference of two cubes can be represented by a 3 � b 3 � � a � b � � a 2 � ab � b 2 � .

Factor each expression.

a. x 3 � 64 Notice that the expression is the sum of two cubes.

� x � 4 � � x 2 � 4x � 16 � Use the sum of two cubes, substituting x for a and 4 for b.

The factors of x 3 � 64 are � x � 4 � � x 2 � 4x � 16 � .

b. 8 x 6 � 27 y 9 Notice that the expression is the difference of two

cubes, � 2 x 2 � 3 � � 3 y 3 � 3 .

� 2 x 2 � 3 y 3 � � 4 x 4 � 6 x 2 y 3 � 9 y 6 � Use the difference of two cubes, substituting 2 x 2 for a and 3 y 3 for b.

The factors of 8 x 6 � 27 y 9 are � 2 x 2 � 3 y 3 � � 4 x 4 � 6 x 2 y 3 � 9 y 6 � .c. x 3 � 6 x 2 � 8x � 48

x 2 � x � 6 � � 8x � 48 Factor x 2 from the first and second terms.

x 2 � x � 6 � � 8 � x � 6 � Factor 8 from the third and fourth terms.

� x 2 � 8 � � x � 6 � Factor out the GCF of � x � 6 � and combine x 2 and 8

The factors of x 3 � 6 x 2 � 8x � 48 are � x 2 � 8 � � x � 6 � .

PracticeComplete the steps to factor the expression.

4. 27 x 3 � 64 y 3 This expression is the difference of two cubes. � 3x � 4y � � 9 x 2 � 12xy � 16 y 2 � Substitute 3x for a and 4y for b.

The factors of 27 x 3 � 64 y 3 are � 3x � 4y � � 9 x 2 � 12xy � 16 y 2 � . 5. x 3 � 4 x 2 � 5x � 20

x 2 � x � 4 � � 5x � 20 Factor x 2 from the first and second terms.

x 2 � x � 4 � � 5 � x � 4 � Factor 5 from the third and fourth terms.

� x 2 � 5 � � x � 4 � Factor out the GCF of � x � 4 � and

combine x 2 and 5

The factors of x 3 � 4 x 2 � 5x � 20 are � x 2 � 5 � � x � 4 � .Factor each expression.

6. x 3 � 216 7. x 3 � 7 x 2 � 8x � 56

� x � 6 � � x 2 � 6x � 36 � � x 2 � 8 � � x � 7 �

61

Name Date Class


You have found square roots of real numbers. Now you will find square roots of complex numbers.

The equation x 2 � �1 has no solutions that are real numbers. It does however have a solution that is an imaginary number. Square roots of negative numbers are imaginary numbers. The imaginary unit i is the number such that i 2 � �1, so ��

�1 � i. Imaginary numbers have the form bi, where b is a real number.

Simplify the expression 3 �� 64 .

3 �� 64 � 3 �

� 64 � (�1) Write �64 as the product of 64 and �1.

� 3 ��

64 � �� 1 The square root of a product is the product of the square roots.

� 3 ��

8 2 � ��

i 2 Rewrite the numbers under the radical as squares.

� 3(8)i Simplify the square roots.

� 24i Multiply.

PracticeComplete the steps to simplify the expression.

1. �8 �� 100 � �8 ��

100 � ( �1 )

� �8 ��

100 � ��

�1

� �8 ��

( 10 ) 2 � ��

( i ) 2

� �8( 10 ) i � �80i

2. 10 �� 144 � 10 ��

(144) � (�1 )

� 10 ��

144 � ��

�1

� 10 ��

( 12 ) 2 � ��

( i ) 2

� 10( 12 ) i

� 120i

Simplify the expression.

3. 9 �� 81 4. 20 ��

�16

81i 80i

5. �50 ��

�1 6. 26 ��

�36

�50i �36i

7. 7 ��

�121 8. 218 ��

�400

77i �360i

9. �12 ��

�4 10. 15 ��

�169

�24i 195i

62ReteachingUsing Complex Numbers



Some quadratic equations have imaginary solutions. For example, the equation x 2 � �1 has solutions i and �i. Sometimes solutions will have a real part as well as an imaginary part. A number of the form a � bi, where a and b are real numbers, is called a complex number.

Solve the equation (x � 5) 2 � �49

(x � 5) 2 � �49

��

(x � 5) 2 � � �� 49 Take the square root of both sides.

x � 5 � �7i Simplify the square root.

x � 5 � 7i Add 5 to both sides.

The solutions are 5 � 7i and 5 � 7i.


11. (x � 12) 2 � �144

��

(x � 12) 2 � _ � ��

�144

x � 12 � �12i x � �12 � 12i

The solutions are �12 � 12i and �12 � 12i .

Solve the equation.

12. x 2 � �225 13. x 2 � �4

�15i �2i

14. x 2 � �9 15. x 2 � �16

�3i �4i

16. (x � 6) 2 � �121 17. (x � 15) 2 � �36

�6 � 11i �15 � 6i

18. (x � 9) 2 � �64 19. (x � 22) 2 � �169

9 � 8i 22 � 13i

20. Solve the equation (x � 2i ) 2 � �1 using the same method as above. Combine the i terms.

�i and �3i

Name Date Class


You have evaluated trigonometric functions using right triangles and angle measurements in degrees. Now you will evaluate trigonometric functions using the unit circle and angle measurements in radians.

Radians are real number measures of rotation. Radians are based on arc length. Recall that an arc is an unbroken piece of a circle.

To convert from degrees to radians, multiply the number of degrees

by � π radians ________ 180�

� .To convert from radians to degrees, multiply the

number of radians by � 180� _________ π radians � .

Convert 30� to radians.

30� � 30� � π radians ________ 180� 6

� � π __ 6

radians

So, 30� equal π __ 6 radians.

Convert 3π ___ 2 radians to degrees.

3π ___ 2

radians � 3π radians _________ 2

� 180� 90° ________ π radians � � 3 � 90° � � 270�

So, 3π ___ 2

radians equal 270�.

PracticeComplete the steps to convert between radians and degrees.

1. Convert 120° to radians. 2. Convert π __ 4 radians to degrees.

120° � 2

120° � π radians ________ 180°3

� = 2π

___ 3

radians

So, 120� equal 2π

___ 3

radians. π __ 4 radians � π radians ________

14 � 180° 45�

________ π radians �

� 45� _____ 1

� 45�

So, π __ 4 radians equal 45� .

Convert between radians and degrees.

3. Convert 60� to radians. π __ 3 radians 4. Convert π __

3 radians to degrees. 60°

5. Convert 135� to radians. 3π ___ 4 radians 6. Convert 5π ___

4 radians to degrees. 225°

63ReteachingUnderstanding the Unit Circle and Radian Measures


The unit circle has a radius of 1 unit.

A reference angle is a positive acute angle formed by the terminal side of an angle and the x-axis.

Use a reference angle to find the exact value of the sine, cosine, and tangent of an angle in any quadrant.

Find the sine, cosine, and tangent of 120�. Find exact values.

Step 1: Draw the angle.

Find the measure of the reference angle.

180� � 120� � 60�

Step 2: Draw the triangle with the reference angle.

Step 3: Find the sine, cosine, and tangent of 60°

in the first quadrant.

sin 60° � ��

3 ___ 2

, cos 60� � 1 __ 2

, tan 60� � ��

3

Step 4: Adjust the signs for 120�.

sin 120� � ��

3 ___ 2 , cos 120� � � 1 __

2 , tan 120� � − �

� 3

Complete the steps to find the sine, cosine, and tangent. Find exact values.

7. Find the three trigonometric values of 150�.

reference angle � 180° � 150� � 30�

sin 30° � 1 __ 2

, cos 30° � ��

3 ____ 2

, tan 150° � � 1 ____

��

3

The triangle with this reference angle will be in the second quadrant.

So, sin 150� � 1 __ 2 , cos 150° �

��

3 ____ 2

, tan 150° � � 1 ____

��

3 .

8. Find the sine, cosine, and tangent of 210�

sin 210� � � 1 __ 2 , cos 210� � �

��

3 ___

2 , tan 210� �

��

3 ___

3

63Reteachingcontinued

x

y

60°120°

1

23

21

x

y

30°150°

121

23

30°

121

x

y

210°23

Name Date Class


64You have worked with equations in exponential form. Now you will convert equations from exponential form to logarithmic form.

A logarithm is another way to define an exponent in an equation.

If b x � a, then log b a � x.

Use the definition of the logarithm to write exponential equations in logarithmic form and to write logarithmic equations in exponential form.

Write the exponential equation, 3 4 � 81inin logarithmic form.

The base b � 3, the exponent x � 4, and the value a � 81.

exponential form → logarithmic form

3 4 � 81 → log 3 81 � 4

Write the logarithmic equation, log 5 125 � 3 in exponential form.


logarithmic form → exponential form

log 5 125 � 3 → 5 3 � 125

PracticeComplete the steps to write each exponential equation in logarithmic form and to write each logarithmic equation in exponential form.

1. 7 2 � 49

b � 7, x � 2 , a � 49 log 7 49 � 2

2. log 2 32 � 5

b � 2, x � 5 , a � 32 2 5 � 32

Write each exponential equation in logarithmic form and write each logarithmic equation in exponential form.

3. 6 3 � 216

log 6 216 � 3 4. log 2 64 � 6

2 6 � 64

5. 4 3 � 64

log 4 64 � 3

6. log 3 27 � 3

3 3 � 27

ReteachingUsing Logarithms

This is read as “If b raised to the x power equals a, then the log base b of a equals x.



Logarithms in base 10 are sometimes written without a base.

Example: log100 � 2

In this equation, the base is assumed to be 10. Instead of writing log 10 100 � 2, the equation is written as log100 � 2 where b � 10, x � 2, and a � 100.

This is true seeing that 10 2 � 100.

Write the logarithm equation, log1000 � 3 as an exponential equation.


logarithmic form → exponential form

log1000 � 3 → 10 3 � 1000

A logarithm in base 10 may not always be a whole number. It is helpful to use the [log] button on your calculator to evaluate theselogarithms.

Find x.

log34 � x

Select the LOG button on calculator, then enter 34.

log34 � 1.53

So, log34 � 1.53 or 10 1.53 � 34.

Practice Complete the steps to write the logarithmic equation as an exponential equation. 7. log10000 � 4

b � 10, x � 4 , a � 10,000

10 4 � 10,000

8. log10 � 1

b � 10 , x � 1 , a � 10

10 1 � 10

Write each logarithmic equation as an exponential equation. Use your calculator to find answers that are not whole numbers.

9. log1,000,000 � 6

10 6 � 1,000,000

10. log51 � x

10 1.71 � 51

11. Acidity is measured in pH and is given by the function pH � �log [ H � ] , where [ H � ] represents the hydrogen ion concentration in moles per liter. The hydrogen concentration for a certain solvent is 0.0000629. What is the pH of the solvent to the nearest tenth?

pH � 4.2

Name Date Class


65

You have solved quadratic equations by factoring and completing the square. Now you will use the quadratic formula to solve quadratic equations.

The quadratic formula can be used to solve any quadratic equation.

Solve 3 x 2 � 26x � 9 � 0.

Step 1: Write the Quadratic Formula: x � �b � ��

b 2 �4ac _____________ 2a

.

Step 2: Substitute values for a, b, and c into the Quadratic Formula. a � 3, b � 26, and c � �9

x � �b � ��

b 2 �4ac _____________ 2a

� � � 26 � � ��

� 26 � 2 �4 � 3 � � �9 � ________________________ 2 � 3 �

Step 3: Simplify.

x � � � 26 � � ��

� 26 � 2 �4 � 3 � � �9 � _____________________ 2 � 3 �

� �26 � ��

676 �108 _________________ 6 � �26 � �

� 784 ___________

6

Step 4: Write in simplest form.

x � �26 � ��

784 ___________ 6 � �26 � 28 ________

6

x � �26 � 28 _________ 6 or x � �26 � 28 _________

6

x � 2 __ 6

or x � �54 ____ 6

x � 1 __ 3

or x ��9


1. 4 x 2 �13x�12 � 0 a � 4 b � �13 c � �12

x � �b � ��

b 2 �4ac _____________ 2a

� � � �13 � � ��

� �13 � 2 �4 � 4 � � �12 � _________________________________ 2 � 4 �

� 13 � �� 361 _____________

8

x � 13 � �� 361 __________

8 or x � 13 � ��

361 _____________ 8

x � 4 or x � � 3 __ 4


2. 3 x 2 � 17x � 10 � 0 3. 4 x 2 � 20x � 25 � 0

x � 2 __ 3 or x � 5 x � 2.5

The ± sign means there are two possible solutions. Write them separately with “or”

ReteachingUsing the Quadratic Formula



The solutions of a quadratic equation are not always real numbers.

Solve x 2 �2x�2 � 2. Write the solutions as complex numbers in standard form.

Step 1: Write the Quadratic Formula: x � �b � ��

b 2 �4ac _______________ 2a

.

Step 2: Substitute values for a, b, and c into the Quadratic Formula. a � 1, b � 2, and c � 2

x � �b � ��

b 2 �4ac _____________ 2a

� � � 2 � � ��

2 2 �4 � 1 � � 2 � __________________ 2 � 1 �

Step 3: Simplify.

x � � � 2 � � ��

2 2 �4 � 1 � � 2 � __________________ 2 � 1 �

� �2 � ��

4�8 ___________ 2 � �1 � �

� �4 _____

2

Step 4: Write in simplest form.

x � �1 � ��

�4 _____ 2 � �1 � �

� � 4 � � �1 � _________

2 � �1 � 2 �

� �1 ______

2 � �1�i

x � �1 � i or x � �1�i

PracticeComplete the steps to solve the equation. Write the solutions as complex numbers in standard form.

4. 2 x 2 � 4x � 3 � 0 a � 2 b � �4 c � 3

x � �b � ��

b 2 �4ac _____________ 2a

� � � �4 � � ��

� �4 � 2 �4 � 2 � � 3 � ___________________________ 2 � 2 �

� 1 � ��

�8 ______ 4

x � 1 � ��

�8 ______ 4

� 1 � ��

(4)(�2) __________ 4

� 1 � 2 ��

�2 _______ 4

� 1 � ��

2 ___ 2 i

x � 1 � ��

2 ___ 2 i or x � 1 � �

� 2 ___

2 i

Solve each equation. Write the solutions as complex numbers in standard form.

5. x 2 � 2x � 4 � 0 6. x 2 � x � 2 � 0

�1 � ��

3 i, �1 � ��

3 i 1 __ 2 � �

� 7 ___

2 i, 1 __

2 � �

� 7 ___

2 i

Remember to divide both terms of the numerator by 2 to simplify.

Name Date Class


66You have used factoring and the Quadratic Formula to solve quadratic equations. Now you will solve polynomial equations using other methods.

Factoring and the Zero Product Property can be used to solve apolynomial equation.

Solve 2 x 5 � 6 x 4 � 8 x 3 .

Step 1: To set the equation equal to 0, rearrange the equation so that all the terms are on one side.

2 x 5 � 6 x 4 � 8 x 3

2 x 5 � 6 x 4 � 8 x 3 � 0

Step 2: Look for the greatest number and the greatest power of x that can be factored from each term.

2 x 5 � 6 x 4 � 8 x 3 � 0

2 x 3 � x 2 � 2 x 3 � 3x � 2 x 3 � 4 � 0

2 x 3 ( x 2 � 3x � 4) � 0

Step 3: Factor the quadratic.

2 x 3 ( x 2 � 3x � 4) � 0

2 x 3 (x � 4)(x � 1) � 0

Step 4: Set each factor equal to 0.

2 x 3 � 0 x � 4 � 0 x � 1 � 0

x � 0 x � �4 x � 1

The solutions are �4, 0, and 1.


1. 3 x 6 � 9 x 5 � 30 x 4 � 0

GCF 3 x 4 3 x 4 ( x 2 � 3x � 10) � 0

3 x 4 (x � 5)(x � 2) � 0

3 x 4 � 0 x � 5 � 0 x � 2 � 0

x � 0 x � 5 x � �2Solve each equation.

2. 2 x 3 � 6 x 2 � 36x � 0 3. 2 x 6 � 32 x 4 � 0

�3, 0, 6 �4, 0, 4

4. x 4 � 5 x 3 � 6 x 2 � 0 5. 4 x 5 � 4 x 4 � 48 x 3 � 0

0, 2, 3 �3, 0, 4

ReteachingSolving Polynomial Equations

The GCF is 2 x 3 .

Use the Zero Product Property



You can use the Rational Root Theorem to find rational roots.

Identify all the real roots of x 3 � 3 x 2 � 6x � 8 � 0.

Step 1: Identify possible roots. The constant term is �8. The leading coefficient is 1.

p: factors of �8. are �1, �2, �4, �8 q: factors of 1 are �1

Possible roots, : p __ q : �1, �2, �4, �8

Step 2: Test some possible roots to find an actual root using synthetic division. Use a synthetic substitution table. The first column lists possible roots. The last column represents the remainders. An actual root will have a remainder of 0.

Step 3: Use the coefficients from the table to write the other factor.

(x � 2)( x 2 � 5x � 4) � 0

(x � 2)(x � 4)(x � 1) � 0

x � 2 or x � �4 or x � �1

The roots are 2, �4, and �1.

PracticeComplete the steps to identify all the real roots of the equation.

6. x 3 � 3 x 2 � 13x � 15 � 0

Possible roots: �1, �3, �5, �15 (x � 3) x 2 � 6x � 5 � 0

(x � 3)(x � 1)(x � 5) � 0

Roots: 3, �1, �5Identify all the real roots of the equation.

7. x 3 � 2 x 2 � 5x � 6 � 0 8. x 3 � x 2 � 14x � 24 � 0

�2, 1, 3 �3, �2, 4

Rational Root Theorem

If a polynomial has integer coefficients, then every rational

root can be written in the form p __ q , where p is a factor of the

constant term and q is a factor of the leading coefficient.

p __ q Coefficients of the Equation

1 3 �6 �8

1 1 4 �2 �102 1 5 4 0

Factor the quadratic factor to find the other factors.

p __ q Coefficients of the Equation

1 3 �6 �8

1 1 4 �9 �243 1 6 5 0

Name Date Class


67ReteachingFinding Inverse Trigonometric Functions

u is the angle whose sine is a.

u is the angle whose cosine is a.

u is the angle whose tangent is a.

Trigonometric Function Inverse Relation

sin u � a Sin �1 a � u

cos u � a Cos �1 a � u

tan u � a T an �1 a � u

Evaluate the inverse of a trigonometric function.

Step 1: Each inverse trigonometric relation has multiple values.

Think: “What reference angle has the given trigonometric value?”

Find the two angles between 0 and 2� radians that have the given value for the trigonometric function.

Step 2: Add (2�)n to each angle to represent all the coterminal angles, where n is an integer.

Describe all the values of Cos �1 ��

3 ___ 2 .

Step 1: The value of the cosine is positive. Cosine is positive in Quadrants I and IV.

In Quadrant I, the reference angle is the angle itself.

cos � __ 6 � �

� 3 ___

2 , so Cos �1 �

� 3 ___

2 � � __

6

In Quadrant IV, subtract the reference angle from 2�.

2� � � __ 6 � 12� ____

6 � � __

6 � 11� ____

6

Step 2: (2�)n to each angle.

� __ 6 � � 2� � n or 11� ____

6 � � 2� � n

PracticeComplete the steps to describe all the values of the angle.

1. Tan �1 � � ��

3 �

Tangent is negative in Quadrants: II and IV

Two angles that have a tangent of : � ��

3 : 2� ___ 3 and 5� ___

3

Add (2�)n to each angle: 2� ___ 3 � (2�)n and 5� ___

3 � (2�)n

Describe all the values of each angle.

2. Sin �1 � � 1 __ 2 � 3. Cos �1 �

� 2 ___

2

7� ___ 6 � (2�n) and � __

4 � � 2� � n

� __ 4 � (2�)n and 7� ___

4 � � 2� � n

Think: What reference angle

has a cosine of ��

3 ____ 2 ?

x

y

30° 12

6

3

π or

x

y

1

23

?


Reteachingcontinued

Inverse trigonometric relations can be defined as functions when their domains are restricted. These inverse trigonometric functions can be distinguished from the inverse trigonometric relations by the capital letters used to represent the functions.

Use the range of the inverse functions to help you evaluate the functions.

• For Sin �1 a � u, u lies in Quadrants I and IV and � � __ 2 � u � � __

2 .

• For Cos �1 a � u, u lies in Quadrants I and II and 0 � u � �. • For Tan �1 a � u, u lies in Quadrants I and IV and � � __

2 � u � � __

2 .

For example: Sin �1 � �1 � � � � __ 2 or �90�.

Evaluate tan �1 � � ��

3 ___ 3 � . Give your answer(s) in both radians and degrees.

Step 1: Decide which quadrant � lies in.

The value of tangent is negative. Tangent is negative in Quadrant IV.Step 2: Evaluate the function.

tan � � � __ 6 � � � �

� 3 ___

3

So Tan �1 � � ��

3 ___ 3 � �� __

6 � �30�

PracticeComplete the steps to evaluate the inverse trigonometric function. Give your answer(s) in both radians and degrees.

4. Cos �1 � � ��

2 ___ 2

� Is the value of cosine negative or positive? negative

In which quadrant does � lie? Quadrant II

What angle in the range has a cosine of ��

2 ___ 2 ? 3� ___

4 , or 130°

Evaluate each inverse trigonometric function. Give your answer(s) in both radians and degrees.

5. Sin �1 � � 1 __ 2 � � � __

6 or �30° 6. Tan �1 �

� 3 � __

3 or 60°

7. A hiker plans to walk from a campsite to a cave. The cave is 0.6 miles east and 0.2 miles north of the campsite. To the nearest degree, in what direction should the hiker head? (Hint: Use the tangent ratio to write an inverse trigonometric function. Evaluate it on your calculator.)

18° north of east

67

Recall that tan � is not defined at � � __

2 or � __

2

Think: What angle has a tangent of � �

� 3 ___

3 and lies in

Quadrant IV?

x

y

12

3

π-

6

0.2 mi

0.6 mi

Cave

Campsite

θ

N

Name Date Class


You have calculated the probability of two independent events. Now you will find the probability of two dependent events, which requires finding a conditional probability.

The probability of B given that A has occurred is called the conditional probability of B, given A. It is denoted P(B | A).

Dependent Events

If A and B are dependent events, then P(A and B) � P(A) � P(B | A).

A deck of cards has 12 face cards and 40 number cards. A card is drawn from a deck and NOT replaced. A second card is then drawn. What is the probability of drawing two face cards from the deck?

Step 1: Find the probability of drawing a face card on the first draw.

P(face card) � P(A) � 12 ___ 52

� 3 ___ 13

Step 2: Find the probability of drawing a face card on the second draw.

P(face card | face card) � P(B | A) � 11 ___ 51

Step 3: Use the rule for the probability of dependent events.

P(2 face cards) � P(A and B) � P(A) � P(B | A) � 3 ___ 13

� 11 ___ 51

� 33 ____ 663

� 11 ____ 221

PracticeComplete the steps to find the probability.A deck of cards has 12 face cards and 40 number cards. A card is drawn from a deck and NOT replaced. A second card is then drawn.

1. A face card is drawn, and then a number card is drawn.

P(face card) � P(A) � 12 ___ 52

� 3 ___ 13

P(number card | face card) � P(B | A) � 40 ___ 51

P(face card and number card) � P(A and B) � P(A) � P(B | A) � 3 ___ 13

� 40 ___ 51

� 40 ____ 221

Find each probability.

2. A number card is drawn, and then a face card is drawn. 40 ____ 221

3. A number card is drawn, and then another number card is drawn. 10 ___ 17

ReteachingFinding Conditional Probability 68

Event A.

Event B.

If a face card was drawn; 11 face cards remain and 51 total cards remain.


Reteachingcontinued

You can use conditional probabilities to analyze survey results organized in a table.

The table shows the results of a survey that asked students from three grade levels to pick the sport they prefer to play from those listed.

Find P(Soccer | 11th grade)Step 1: Find the total number of 11th-grade students who

participated in the survey.

15 � 13 � 8 � 9 � 45

Step 2: Find the number of 11th-grade students who prefer soccer.

The number at the intersection of the 11th-grade column and the soccer row is 9.

Step 3: Find P(Soccer | 11th grade).

P(Soccer | 11th grade) � 11th grade soccer

_________________ 11th grade all sports

� 9 ___ 45

� 1 __ 5

Find P (9th grade | Basketball)Step 1: Find the total number of students who prefer basketball.

12 � 11 � 13 � 36

Step 2: Find the number of students who prefer basketball and are in the 9th grade.

The number in the basketball row and the 9th-grade column is 12.

Step 3: Find P(9th grade | Basketball).

P(9th grade | Basketball) � 9th grade basketball

_________________ all grades basketball

� 12 ___ 36

� 1 __ 3

PracticeComplete the steps to find the probability.The table shows the results of a survey which asked students from three grade levels to pick their favorite animal from those listed.

4. P(dinosaur | 1st grade)

total number of 1st-grade students 35

number of 1st-grade students who prefer dinosaurs 10

P(dinosaur | 1st grade) � 10

___ 10

� 2 __ 7

Find each probability.

5. P(2nd grade | horse 1 __ 3 6. P(cat | 3rd grade) 1 __

4

1st 2nd 3rd

cat 6 8 9

dinosaur 10 9 7

dog 7 11 12

horse 12 10 8

9th 10th 11th

Baseball 13 10 15

Basketball 12 11 13

Football 9 7 8

Soccer 14 16 9

68

Name Date Class


69You have added and subtracted complex numbers. Now you will graph complex numbers and simplify complex expressions.

Graphing complex numbers is like graphing real numbers. The real axis corresponds to the x-axis and the imaginary axis corresponds to the y-axis. The absolute value of a complex number is its distance from the origin. The distance can be found using the Pythagorean Theorem.

� a � bi � � ��

a 2 � b 2

Graph (2, −3i ) and find its absolute value.

Step 1: Find the real and imaginary parts of the complex number.

a � bi � 2 � 3i � 2 � � �3 � i

a � 2 and b � �3

Step 2: Plot (2, �3) on the coordinate plane. From the origin move 2 units right and 3 units down.

Step 3: Find the absolute value of the complex number.

� a � bi � � ��

a 2 � b 2

� ��

� 2 � 2 � � �3 � 2

� ��

4 � 9

� ��

13

PracticeComplete the steps to graph each complex number and find its absolute value.

1. 3 �i 2. 4i a � 3 and b � �1 a � 0 b � 4 Plot � 3 , �1 � Plot � 0 , 4 � � a � bi � = �

� a 2 � b 2 � a � bi � � �

� a 2 � b 2

� ��

3 2 � �1 2 � ��

� 0 � 2 � 4 2

� ��

10 � 4

Graph each complex number and find its absolute value.

3. 5 � 2i 4. �4 � 3i

��

29 5

ReteachingSimplifying Complex Expressions

x

y4i

2i

24-2-4

-4i

-3i

O real

imaginary

|2 - 3i|

x

y4i

2i

42-2-4

-4i

3 - i

O real

imaginary

-2i

x

y4i 4i

2i

42-2-4

-4i

O real

imaginary

-2i

x

y4i

2i

42-2-4

-4i

O real

imaginary

-2i5 - 2i

x

y4i

2i

42-2-4

-4i

O real

imaginary

-2i

-4 + 3i

2

322 + 32



You can also multiply and divide complex numbers.

Use the Distributive Property to multiply complex numbers.Multiply 3i � 2 � i � .

3i � 2 � i � � 6i � 3 i 2 � 6i � 3 � �1 � � 3 � 6i

Distribute. Use i 2 � �1Write in a � bi form

Multiply � 4 � 2i � � 5 � i � . � 4 � 2i � � 5 � i � � 20 � 4i � 10i � 2i 2 � 20 � 6i � 2i 2 � 20 � 6i � 2i � �1 � � 22 � 6i

Distribute using FOIL.Combine imaginary parts.Use i 2 � �1Combine real parts.

Dividing complex numbers is similar to rationalizing an irrational denominator.

Divide 6 _______ 4 � i �

� 2 .

6 _______ 4 � i �

� 2 � 6 _______

4 � i ��

2 � 4 � i �

� 2 ________

4 � i ��

2 �

� 24 � 6i ��

2 ___________ 16 � � i ��

2 � 2

� 24 � 6i ��

2 __________ 16 � 2 � �1 �

� 24 � 6i ��

2 _________ 18

� 4 __ 3 � i �

� 2 ____

3

Multiply the numerator and the denominator by the complex conjugate.

Distribute and simplify.

Use i 2 � �1

Simplify the denominator.

Simplify.

PracticeComplete the steps to multiply or divide.

5. � 2�i � � 2 � 3i � � 4 � 6i � 2i � 3 i 2 6. 1 � 4i ______ 2i

� 1 � 4i ______ 2i

� �2i ____ �2i

� � 4 � 4i �3 � �1 � � �2i � 8i 2 ____________

�4i 2

� 4 � 4i � 3

� 7�4i � �2i � 8 ___________ 4

�

2� i __

2

Multiply or divide.

7. � 3 � i � � 2 � 2i � 8. 3 _______ 2 � i �

� 3 9. 3 � i _____

5i

4 � 8i 6 __ 7 � 3i �

� 3 _____

7

� 1 __ 5 � 3i __

5

Name Date Class


70You have solved linear and polynomial equations. Now you will solve radical equations. A radical equation is an equation that contains at least one radical with a variable in the radicand.

Solve radical equations by raising both sides of the equation to the power of the index of the radical. For example,

�� x � 3

( �� x ) 2 � 3 2

x � 9

Solve 3 ��

x � 2 � 18.Step 1: Isolate the radical. Divide both sides of the equation by 3 and simplify.

3 ��

x � 2 ________ 3 � 18 ___

3

��

x � 2 � 6

Step 2: Square both sides of the equation and simplify.

� �� x � 2 � 2 � 6 2 x � 2 � 36

Step 3: Solve.

x � 38

Step 4: Check.

3 ��

x � 2 � 18

3 ��

38 � 2 � 3 ��

36 � 3 � 6 � � 18

PracticeComplete the steps to solve the equation and check the solution.

1. 4 3 ��

2x � 11 � 12 � 3 �� 2x � 11 � 3

� 3 3

3 ��

2x � 11 � 3

4 3 ��

2x � 11 __________ 4

� 12 _____ 4

2x � 11 � 27 3 ��

2 � 8 � � 11 � 3

3 ��

2x � 11 � 3 2x � 16 3 ��

27 � 3

x � 8 3 � 3


2. 2 ��

2x � 4 � 10 21 ___ 2 3. 2 �

3 ��

5x � �1 �27 ____ 5

4. 2 3 ��

5x � 6 � 8 14 5. 5 � ��

x � 3 � 9 19

ReteachingSolving Radical Equations

The index of �� x is 2. Raise

both sides to the power of 2.

Always check for extraneous solutions when solving radical equations

Remember � n �� a � n � a.

Check:



Radical equations can have more than one radical. If the indices are the same and the radicals are on opposite sides of the equation, raise each side of the equation to the power of the index, as shown previously. If the indices are different, estimate the solution on a graphing calculator.

Solve ��

�3x � 3 �� 4x � 5 � 0.

Step 1: Move one of the radicals to the right side of the equation.

��

�3x � 3 �� 4x � 5 � 0

��

�3x � 5 � 3 �� 4x

Step 2: Use one side of the equation for Y1 and the other side for Y2.

Y1 � ��

�3x � 5

Y2 � 3 ��

4x

Step 3: Enter the two equations into the graphing calculator.

Step 4: Find the intersection of the two graphs using the intersect function on the CALC menu.

The curves intersect at x � �2.63.


6. 3 ��

x � 3 � �� 5x � 1 � 0 Y1 � 3 ��

x � 3 � 1

3 ��

x � 3 � 1 � ��

5x Y2 � ��

5x Intersection x � 1.39


7. ��

2x � 5 � 3 �� x � 6 � 0 8. 3 ��

x � 1 � ��

3x � 2 � 0

x � 6.18 x � 4.819. The period of a pendulum, T, is the time it takes for the pendulum to make one back-and-forth swing, and is given

by T � 2� �� L ___

9.8 , where L is the length of the pendulum in meters.

Find the length of a pendulum whose period is 4 seconds. (Hint: Solve 4 � 2� �� L ___

9.8 .)

about 3.97 m

Use the cubed root function on the MATH menu, or enter the equation as (4x) ^ (1/3).

Name Date Class


You have found measures of central tendency given a data set. Now you will learn how to collect data from a sample.

A sample is a small group of individuals taken from a larger population.A sample is biased if the group of individuals does not accurately represent the larger population. A survey is biased if there is a chance the questions asked could persuade people to answer in a specific way.

Determine whether the survey questions are biased. Explain your answer. If the answer is biased, rewrite it so it is not.

a. Do you agree with the unfair policy that students must take part in at least two extra-curricular activities each year to graduate?

Solution: Biased; The phrase “unfair policy” could influence a student to disagree with the policy. It could be rewritten as: Do you agree with the policy that students must take part in at least two extra-curricular activities each year to graduate?

b. Do you think recycling is an important part of saving the environment?

Solution: Not biased; There are no phrases that would influence a person’s answer.

Determine whether the following samples are biased. Explain your answer.

c. A teacher places all 25 of her students’ names in a hat. She selects 10 names and has those students complete a survey.

Solution: Not biased; The names are being chosen at random.

d. The president of a company with 300 employees and 10 different departments surveys everyone in the accounting department to determine how satisfied employees are withtheir jobs.

Solution: Biased; The accounting department may not accurately represent all the employees. The president should ask people from each department to have an unbiased sample.

PracticeComplete the steps to determine if the samples are biased or not biased.

1. A teacher asks every female student in 2. A store manager asks every tenththe class how many hours she spends on customer who enters the store whathomework each night. he or she thinks of the newly designed entrance.

This sample is biased. This sample is not biased not biased.

Determine whether the following survey questions are biased or not biased.

3. Would you rather visit the cold-weather 4. Which sport do you enjoy watchingstate of Alaska or the warm-weather state more-football or baseball?of Florida?

biased not biased

ReteachingCollecting Data

INV

7


Reteachingcontinued

Measures of central tendency are useful when describing a data set. Themean is the average of a set, the median is the middle value in a set,and the mode is the value that occurs most frequently.

Each student in a class of 25 is surveyed about how many hours per day they spend playing video games.

Hours 0 1 2 3 4 5 6Frequency 5 4 9 4 0 3 0

a. Find the mean.

0(5) � 1(4) � 2(9) � 3(4) � 4(0) � 5(3) � 6(0) ________________________________________ 25

� 1.96

b. Find the median.

List the hours in numeric order: 0,0,0,0,0,1,1,1,1,2,2,2,2,2,2,2,2,2,3,3,3,3,5,5,5

The value in the middle (the 13th place) is 2, so the median is 2.

c. Find the mode.

The mode is 2 because it represents the response that was given most frequently.

PracticeComplete the steps to determine the mean, median, and mode of the data set.

5. A survey is done in a neighborhood to determine how many TVs are in each household.

Number of TVs 0 1 2 3 4 5 6 7Frequency 0 4 8 10 5 4 3 1

a. Mean � 0(0) � 1(4) � 2(8) � 3(10 ) � 4(5) � 5(4) � 6(3) � 7(1) _______________________________________________ 35

� 3.29

b. Median: List the number of TVs in order. The middle value (or the value in the 18th place) is 3, so the median is 3.

c. Mode: Ten people responded with 3 TVs. This was the most frequent response so, the mode is 3.

Find the mean, median and mode of the data set.

6. A survey was done to find the number of pets that people own.

Number of Pets 0 1 2 3 4 5

Frequency 6 5 5 3 1 1

a. Mean � 1.57 b. Median � 1 c. Mode � 0

INV

7

Name Date Class


71You have used the trigonometric functions to find an unknown side in a right triangle. Now you will use the Law of Sines to find unknown sides and angles in triangles.

If you know the lengths of two sides of a triangle and the measure of the angle between the two sides, you can find the area of the triangle.

Find the area of the triangle. Round to the nearest tenth.

Step 1: Write the formula.

Area � 1 __ 2

bc sin A

Step 2: Substitute the known values into the formula.

A � 42�, b � 9, c � 14

Area � 1 __ 2

(9)(14) sin 42�

Step 3: Use a calculator to evaluate the area.

Area � 42.1552282

Step 4: Round the answer to the nearest tenth.

The area is about 42.2 ft 2 .

PracticeComplete the steps to find the area of the triangle. Round to the nearest tenth. 1.

Area � 1 __ 2 ac sin B

B � 64�

a � 6 m c � 15 m

Area � 40.4 m 2 Find the area of each triangle. Round to the nearest tenth. 2. 3.

217.3 yd 2 68.0 in. 2

ReteachingUsing the Law of Sines

Area of a Triangle

1 __ 2 bc sin A 1 __

2 ac sin B 1 __

2 ab sin C

The two sides must form the angle.

Check that your calculator is set to degree mode to compute the sine.

C

A B

b a

c

A42°

14 ft

9 ft

C

A

B 64°

15 m

6 m

B

C

A76°

16 yd

28 yd

AC

B

54°

14 in.

12 in.



Use the Law of Sines to solve a triangle for missing side lengths and anglemeasures if you know two angle measures and any side length or two side lengths and an angle measure that is not between them.

Solve the triangle.

Step 1: Decide what you know. m�B � 102� m�C � 53� c � 10

Step 2: Find the missing angle measure.

53� � 102� � m�A � 180� m�A � 25�

Step 3: Use the Law of Sines to find the unknown side lengths.

sin A _____ a � sin C _____ c sin B _____ b � sin C _____ c

sin 25� ______ a � sin 53� ______ 10

sin 102� _______ b � sin 53� ______

10

a sin 53� � 10sin 25� b sin 53� � 10sin 102�

a � 10sin 25� ________ sin 53�

b � 10sin 102� _________ sin 53�

a � 5.3 b � 12.2

PracticeComplete the steps to solve the triangle.

4. m�A � 34� m�B � 180� � m�A � m�C � 79�

m�C � 67� sin A _____ a � sin B _____ b sin A _____ a � sin C _____ c

a � 12 b � 21.1 c � 19.8

Solve each triangle.

5.

m�A � 59�, a � 6.8, b � 7.6

6. A surveyor stands on a road and finds the angle formed by the road and the line of sight with the corner of a barn is 58°. The surveyor then walks 50 yards west and finds the angle formed by the road and the barn to be 37°. Estimate the distance from the road to the barn. (Hint: Find the height of the triangle.) about 26 yd

Law of Sines

For � ABC: sin A _____ a � sin B _____ b � sin C _____ c

Use sin C _____ c since you know the measures of C and c.

The sum of the measures of the angles of a triangle is 180�.

B

A C

c a

b

A

B

C b

a

53°

102° 10

A

B

C b

c

34°67°

12

C

A

Ba

b

49°72°

6

CA

B

37° 58°

50 yd road

barn

h a

Name Date Class


72You have learned how to use the properties of exponents to solve problems. Now you will learn how to use properties of logarithms.

Product Property of Logarithms

The logarithm of a product can be written as the sum of the logarithms of the numbers.

log b mn � log b m � log b n

where m, n, and b are all positive numbers and b � 1

Quotient Property of Logarithms

The logarithm of a product can be written as the sum of the logarithms of the numbers.

log b m __ n � log b m � log b n

where m, n, and b are all positive numbers and b � 1

Approximate the value of log 8 60. Use log 8 4 � 0.6667 and log 8 15 � 1.3023.

log 8 60 � log 8 (4 � 15) Write 60 as the product of 4 and 15.

� log 8 (4) � log 8 (15) Use the Product Property of Logarithms.

� 0.6667 � 1.3023 Substitute 0.6667 for log 8 (4) and 1.3023 for log 8 (15).

� 1.9690 Add.

Approximate the value of log 3 5. Use log 3 30 � 3.0959 and log 3 6 � 1.6309.

log 3 5 � log 3 � 30 ___ 6 � Write 6 as the quotient of 36 and 6.

� log 3 (30) � log 3 (6) Use the Quotient Property of Logarithms.

� 3.0959 � 1.6309 Subsitute 3.0959 for log 3 (30) and 1.6309 for log 3 (6).

� 1.4650 Subtract.

PracticeComplete the steps to approximate the logarithmic expression. Use log 6 6 � 1 and log 6 9 � 1.2263.

1. log 6 54

� log 6 ( 6 � 9 )

� log 6 6 � log 6 9 � 1 � 1.2263

� 2.2263Approximate the logarithmic expression. Use log 3 24 � 2.8928 and log 3 3 � 1.

2. log 3 8 1.8928

ReteachingUsing the Properties of Logarithms



Power Property of Logarithms

The logarithm of a power can be written as the product of the exponent and the logarithm of the base.

log b a p � p log b a

for any real number p where a, and b are positive numbers and b ≠ 1

Logarithms and exponents undo each other when their bases are the same.

Inverse Properties of Exponents and Logarithms

The logarithm of b x to the base b is equal to x.

log b b x � x

The logarithm undoes the exponent when the bases are the same.

b raised to the logarithm of x to the base b is equal to x.

b lo g b x � x

The exponent undoes the logarithm when the bases are the same.

Simplify log 4 64 5 .

log 4 64 5 � 5 log 4 64 Use the Power Property of Logarithms.

� 5 log 4 4 3 Think 4 to what power is equal to 64, or 4? � 64.

� 5 � 3 The bases are the same, so log 4 4 3 � 3.

� 15 Multiply.

Simplify log 7 7 4x .

The bases are the same. The logarithm undoes the exponent. So,

log 7 7 4x � 4x

Simplify 3 log 3 64 .

The bases are the same. The exponent undoes the logarithm. So,

3 log 3 64 � 64

PracticeComplete the steps to simplify each logarithmic expression.

3. log 5 125 2

Write 125 as a power of 5: 5 3 Apply the Power Property:

log 5 ( 5 3 ) 2

→ 2 log 5 5 3 Apply the Inverse Property: 2 � 3 � 6

4. 4 log 4 75

Apply the Inverse Property: 4 log 4 75 � 75

Simplify each logarithmic expression.

5. log 2 16 4

16 6. 2 log 2 3x

3x 7. log 9 81 3

6 8. 5 log 5 5

5

Name Date Class


73You have found probabilities given a sample space. Now you will identify populations and different types of sampling.

A population is a group of individuals from which specific information is gathered. A sample is a small part of a population. Information gathered from a sample can be applied to the entire population.

Describe the individuals in the population and in the sample.

a. A baseball coach surveys 10 of his players to determine how they feel about having practice on Saturdays.

Solution: The population is all the players on the team. The sample is the 10 players whom the coach surveyed.

b. Nick selects four of his friends and asks them what they think of a new video game. Solution: The population is all of Nick’s friends. The sample is the four friends Nick asks.

PracticeComplete the steps to describe the individuals in the population and in the sample. 1. All the registered voters in

New Hampshire receive a questionnaire in the mail about the upcoming election and are asked to return it within 10 days.

2. For one week, every 10th person who enters a supermarket is asked to complete a survey.

The population is all the registered voters in New Hampshire. The

sample is the individuals who return the questionnaire within 10 days.

The population is all the individuals who enter the supermarket during the week. The sample is all the individuals who complete the survey.

Describe the individuals in the population and in the sample.

3. The president of a company surveys 5 employees from each department to determine what their feelings are on the new company logo.

population: all employees of the company; sample: employees surveyed

4. Michaela wants to know her neighbors’ opinions on a new town regulation. She knocks on every door in her neighborhood and asks her neighbors a series of questions about the regulation.

population: all people in Michaela’s neighborhood; sample: all people who were home and answered Michaela’s questions

ReteachingUsing Sampling



The process of choosing a sample is called sampling. It is important that a sample accurately represents the population being studied. Simple random sampling (SRS) involves choosing a sample so that every individual in the population has an equal chance of being selected.

One method for choosing a SRS is to use a table of random numbers. First assign a two-digit number to every individual in the population.

01 Nicholas 02 Frank 03 Steve 04 John 05 Ryan 06 Jerry 07 Alex

08 Austin 09 Beth 10 Nicole 11 Sarah 12 Bill 13 Dave 14 Kate

15 Nora 16 Finn 17 Cassie 18 Lily 19 Mike 20 Jess 21 Alison

Which 3 names will be picked for the sample if the numbers are chosen by starting at the beginning of the random number table shown?

39067 47892 39656 83217 73385 57663 59229 08886 47279 41433 37901

Solution: Look at each consecutive 2-digit number. Ignore column breaks, numbers that are not labeled 01-21, and numbers that repeat. The first number from the table is 06. Continue until two more numbers from 01-21 have been identified.

39067 47892 39656 83217 73385 57663 59229 08886 47279 41433 37901

The three numbers are 06, 17, and 14. These numbers correspond with Jerry, Cassie, and Kate.

PracticeComplete the steps to select a simple random sample using a random number table.

5. A teacher assigns each student in her class a two-digit number.

20 Richard 21 Lillian 22 Joe 23 Ron 24 Jake

25 Kelly 26 Patti 27 Emily 28 Max 29 Isabella

30 Mackenzie 31 Jackie 32 Franklin 33 Maya 34 Claire

Use the random number table to choose six students for the sample.

46693 45473 07714 02605 88856 64557 60278 27615

71954 35916 84532 17660 12337 02263 71353 33326

The six numbers chosen are 34, 26 , 27, 21 , 33, and 22 .

These correspond to the students Claire, Patti , Emily, Lillian, Maya, and Joe .

Determine whether the following are simple random samples.

6. A teacher surveys only the female students in his class.

not a simple random sample

7. The class president puts every student’s name in a hat and chooses 20 students randomly.

simple random sample

Name Date Class


You have learned how to use the quadratic formula to find the roots of a quadratic equation. Now you will learn how to use the discriminant to describe the roots of a quadratic equation.

The discriminant of ax 2 � bx � c � 0 (a � 0) is b 2 � 4ac.

Use the discriminant to determine the number of roots of a quadratic equation. A quadratic equation can have 2 real roots, 1 real root, or 2 complex roots.

Use the discriminant to describe the roots of each equation.a. 2x 2 � 5x � 3

Write the equation in standard form:

2 x 2 � 5x � 3 � 0

a � 2, b � �5, c � �3

Evaluate the discriminant:

b 2 � 4ac

� (�5) 2 � 4(2)(�3)

� 25 � 24

� 49

When b 2 � 4ac � 0, the equation has 2 real roots.

b. x 2 � 10x � �25


x 2 � 10x � 25 � 0

a � 1, b � 10, c � 25


b 2 � 4ac

� (10) 2 � 4(1)(25)

� 100 � 100

� 0

When b 2 � 4ac � 0, the equation has 1 real root.

c. 3 x 2 � 4x � �2


3 x 2 � 4x � 2 � 0

a � 3, b � �4, c � 2


b 2 � 4ac

� (�4) 2 � 4(3)(2)

� 16 � 24

� �8

When b 2 � 4ac � 0, the equation has 2 complex roots.

PracticeComplete the steps to use the discriminant to describe the roots of each equation.

1. x 2 � 12x � �36 2. x 2 � 4x � �7 3. x 2 � 7x � �3

x 2 � 12x � 36 � 0 x 2 � 4x � 7 � 0 x 2 � 7x � 3 � 0

a � 1 , b � �12, c � 36 a � 1 , b � �4, c � 7 a � 1 , b � �7, c � 3

b 2 � 4ac � 0 b 2 � 4ac � �12 b 2 � 4ac � 37

The equation has The equation has The equation has

1 real root(s). 2 complex root(s). 2 real root(s).

Use the discriminant to describe the roots of each equation.

4. x 2 � 12x � �4 5. �2x 2 � x � 5

2 real roots 2 complex roots

6. x 2 � 8x � �16 7. 3x 2 � 5x � 2

1 real roots 2 real roots

ReteachingFinding the Discriminant 74



Sometimes it is not necessary to find exact solutions to a quadratic equation. The discriminant can be evaluated to determine the outcome of a real world situation.

A football is kicked straight up at a velocity of 50 feet per second from a hill 10 feet above ground. The function that gives the height of the ball after x seconds is given by y � � 16x 2 � 50x � 10. Use the discriminant to explain why the ball will not reach a height of 70 feet above ground.

Step 1: In the equation, y represents the height of the ball. Write the equation that represents the situation.

70 � � 16x 2 � 50x � 10 Substitute 70 for y in the equation.

Step 2: Write the equation in standard form.

�16x 2 � 50x � 60 � 0

Step 3: Evaluate the discriminant.

b 2 � 4ac � (50) 2 � 4(�16)(�60)

� 2500 �3840

� �1340

Step 4: Apply the discriminant to the situation.

The discriminant is negative. There are no real outcomes for the situation. The ball never reaches a height of 70 feet.

PracticeComplete the steps to solve the problem. 8. A soccer ball is kicked straight up at a velocity of 48 feet per second from

a hill 8 feet above ground. The function that gives the height of the ball after x seconds is given by y � � 16x 2 � 48x � 8. Use the discriminant to explain why the ball will reach a height of 44 feet above ground.

Write the equation that represents the situation: 44 � � 16x 2 � 48x � 8

Write the equation in standard form: �16x 2 � 48x � 36 � 0

b 2 � 4ac � 48 2 � 4(�16)( �36 )

� 2304 � 2304

� 0 Because the discriminant is 0 , there is one real solution(s).

The ball reaches the height of 44 feet once .

Solve. 9. A tennis ball is hit straight up at a velocity of 52 feet per second from

a height of 5 feet. The function that gives the height of the ball after x seconds is given by y � � 16x 2 � 52x � 5. Use the discriminant to explain why the ball will reach a height of 45 feet above ground.

The discriminant is positive, so the ball will reach the height twice.

Name Date Class


You have learned how to solve radical equations. Now you will learn how to graph radical functions.

Graph the square root function y � ��

x � 1 and its inverse. Determine the domain and the range of both functions.

Step 1: Find the domain of the function.

Set the radicand equal to or greater than zero, then solve for x to find the domain.

x � 1 � 0

x � �1

The domain of the function is x � �1, so the range is y � 0.

Step 2: Graph the function.

Use your calculator to graph the function.

Step 3: Find the inverse function.

Interchange x for y and solve for y.

The inverse function is y � x 2 � 1.

x � ��

y � 1

x 2 � y � 1

x 2 � 1 � y

Step 4: Find the domain and range of the inverse function. Then graph the inverse function on the same graph.

Since the range of the original function is y � 0, the domain of the inverse function is x � 0. The range of the inverse function is y � �1.

PracticeComplete the steps to graph the square root function and its inverse. Determine the domain and range of both functions.

1. y � ��

3x Domain: x � 0; Range: y � 0

Inverse function: y � 1 __ 3 x 2

Domain: x � 0 ; Range: y � 0

75ReteachingGraphing Radical Functions

x

y

y = x + 1

O

6

8

10

2

2-2 4 6 8 10

4

x

y

y = x2- 1

O

6

8

10

2

2 4 6 8 10

4

-2

y = x + 1

x

y

O

6

8

10

2

2 4 6 8 10

4

-2

y = 3x

y = 13x2


Reteachingcontinued

Graph the cube root function y � 3 ��

x � 1 and its inverse.


Use your calculator to graph the function.

Step 2: Find the inverse function.

Interchange x for y and solve for y.

The inverse function is y � x 3 � 1.

Step 3: Graph the inverse function on the same graph.

x � 3 ��

y � 1

x 3 � y � 1

x 3 � 1 � y

PracticeComplete the steps to graph the cube root function and its inverse.

2. y � 3 ��

2x � 1

Interchange x and y to find the inverse:

x � 3 ��

2 y � 1

x 3 � 2 y �1

1 __ 2 x 3 � 1 __

2 � y

The inverse of the function is y � 1 __ 2 x 3 � 1 __

2 .

Graph the cube root function and its inverse.

3. y � 3 ��

3x

Inverse function: y � 1 __ 3 x 3

75

x

y8

4

4 8-8 -4

-8

-4

O

y = 3 x - 1

x

y8

4

4 8-8 -4

-8

-4

O

y =3 x - 1

y = x3+1

x

y8

4

4 8-8-4

-8

-4

O

y = 3 2x + 1

y = 12x3

-

12- -

x

y8

4

4 8-8 -4

-8

-4

O

y = 3 3x

y = 13x3-

x

y8

4

4 8-8 -4

-8

-4

O

y = 3 3x

y = 13x3-

x

y8

4

4 8-8 -4

-8

-4

O

y = 3 3x

y = 13x3-

Name Date Class


76You have factored polynomials. Now you will factor polynomials to find the roots or zeros.

The polynomial zeros of a polynomial equation are values of a variable that make an equation true. To find the zeros of a polynomial equation, factor the polynomial and use the Zero Product Property to solve for the variable.

Zero Product Property

If a and b are expressions and ab � 0, then a � 0 or b � 0.

Example: If (x � 5)(2x � 7) = 0, then x � 5 � 0 or 2x � 7 � 0.

Find the zeros of the polynomial function y � (x � 2)(2x � 5).


x � 2 � 0 or 2x � 5 � 0

Step 2: Solve each equation.

x � 2 or 2x � �5

x � � 5 __ 2

The zeros of y � (x � 2)(2x � 5) are

2 and � 5 __ 2 .

Find the zeros of the polynomial function y � 4x(3x � 1)(x � 7).


4x � 0 or 3x � 1 � 0 or x � 7 � 0


x � 0 or 3x � �1 or x � �7

x � � 1 __ 3

The zeros of y � 4x (3x � 1)(x � 7) are

0, � 1 __ 3 , and �7.

PracticeComplete the steps to find the zeros of the polynomial function.

1. y � (x � 8)(4x � 9)

x � 8 � 0 or 4x � 9 � 0 x � �8 or 4x � 9

x � 9 __ 4

2. y � 7x (x � 4)(5x � 6)

x � 0 or x � 4 or 5x � �6 x � � 6 __

5

Find the zeros of the polynomial function.

3. y � (x � 2)(x � 6) �2, 6 4. y � (2x � 7)(3x � 8) � 7 __ 2 , 8 __

3 ,

5. y � �4x(x � 5)(3x � 7) 0, �5, 7 __ 3 6. y � 9x (3x � 1)(2x � 3) 0, � 1 __

8 , 3 __

2

ReteachingFinding Polynomial Roots I



Find the zeros of the polynomialfunction y � x 3 � 4 x 2 � 5x.

Step 1: Factor: The GCF is x.

y � x( x 2 � 4x � 5) � x(x � 1)(x � 5)


x � 0 or x � 1 � 0 or x � 5 � 0


x � 0 or x � �1 or x � 5

The zeros of y � x 3 � 4 x 2 � 5x

are 0, �1, and 5.

Find the zeros of the polynomial function

y � 10 x 3 � 34 x 2 � 24x.

Step 1: Factor: The GCF is 2x.

y � 2x (5 x 2 � 17x � 12)


2x � 0 or 5 x 2 � 17x � 12 = 0

Step 3: Solve using the quadratic formula.

x � �17 �

��

(17) 2 � 4(5)(�12) ________________________

2(5) � �17 �

��

289 � 240 _________________ 10

� �17 � 23 _________ 10


x � �17 � 23 _________ 10

� 3 __ 5 or x � �17 � 23 _________

10 � �4

or 2x � 0 → x � 0

The zeros of y � 10 x 3 � 34 x 2 � 24 x 2 � 24x are

0, �4, and 3 __ 5 .

PracticeComplete the steps to find the zeros of the polynomial function.

7. y � 2 x 3 � 2 x 2 � 24x

2x ( x 2 � x � 12 ) � 2x(x � 4 )(x � 3 ) � 0

2x � 0 or x � 4 � 0 or x � 3 � 0

x � 0 or x � �4 or x � 3

8. y � 12 x 3 � 15 x 2 � 18x

3x (4 x 2 � 5 x � 6 ) � 0

x � �5 �

��

5 2 � 4( 4 )(�6) _____________________

2( 4 )

� �5 � ��

25 � 96 _______________ 8

� �5 � 11 ________ 8

x � �5 � 11 ________

8 or x �

�5 �11 ________ 8 or 3x = 0

x � 3 __ 4 or x � �2 or x � 0

Find the zeros of the polynomial function.

9. y � 4 x 3 � 16 x 2 � 48x 0, �2, 6 10. y � 3 x 2 � 13x � 56 �7, 8 __ 3

11. y � �4 x 3 � 4 x 2 � 8x 0, �1, 2 12. y � 6 x 3 � 12 x 2 � 18x 0, �3, 1

Name Date Class


77Now you will use the Law of Cosines to solve triangles.

The Law of Cosines can be used to find the side length of a triangle if you know two side lengths and the angle measure between them.

For �ABC : a 2 � b 2 � c 2 � 2bc cos A

b 2 � a 2 � c 2 � 2ac cos B

c 2 � a 2 � b 2 � 2ab cos C

Find the length b in the triangle shown.

Round your answer to the nearest tenth.

B

ACb

6 10

82°

Use the Law of Cosines with a � 6, c � 10, and m�B � 82�

b 2 � a 2 � c 2 � 2ac cos B Choose the formula.

b 2 � 6 2 � 10 2 � 2(6)(10)cos 82� Substitute the known values.

b 2 � 119.3 Simplify using your calculator.

b � 10.9 Take the positive square root.

PracticeComplete the steps to solve for the third side of the given triangle. Round your answer to the nearest tenth.

1. Find a. m�A � 46� , b � 9 , c � 7 Find a using a 2 � b 2 � c 2 � 2bc cos A

a 2 � (9) 2 � � 7 � 2

� 2 � 9 � � 7 � cos 46�

a 2 � 130 � 126 cos 46�

a 2 � 42.5 a � 6.5

Solve for the third side of the given triangle. Round your answer to the nearest tenth. 2. Find a.

a � 9.4

ReteachingUsing the Law of Cosines

B

A C

c a

b

B

A C

a7

9

46°

C

A

Ba

410

70°



The Law of Cosines can also be used to find the angle measures of a triangle when all three side lengths are given.

Find the measure of �A in the triangle shown. Round your answer to the nearest tenth.

Use the Law of Cosines with a = 10, b = 15, and c = 12.

a 2 � b 2 � c 2 � 2bc cos A Choose the formula.

10 2 � 15 2 � 12 2 � 2(15)(12)cos A Substitute the known values.

100 � 369 � 360 cos A Simplify.

�269 � �360 cos A Subtract 369 from both sides.

0.7472 � cos A Divide each side by �360.

A � 41.7� Use cos�1.

PracticeComplete the steps to solve for the given angle in the triangle. Round your answer to the nearest tenth.

3. Find the measure of �B.

b 2 � a 2 � c 2 � 2ac cos B

5 2 � 9 2 � 7 2 � 2(9)( 7 )cos B

25 � 130 � 126 cos B

0.83333 � cos B

B � 33.6�

Solve for the given angle in the triangle. Round your answer to thenearest tenth.

4. Find the measure of

�C � 68�

5. An airplane leaves Island A and flies 91 miles to Island B. It stops briefly and then flies 63 miles to Island C. After another brief stop, the plane returns to Island A. What is the distance between Island C and Island A?

about 150 mi

C

A

B

12

15

10

A

C B

5 7

9

A

C B

12 14

13

B

A C

91 mi 63 mi153.6°

Name Date Class


You have found solved quadratic equations by using the quadratic formula. Now you will solve quadratic equations by factoring.

Some quadratic equations are solved by factoring the difference of two squares and then applying the Zero Product Property.

Difference of Two Squares Zero Product Property

Algebraic For m: a 2 � b 2 � (a � b)(a � b)

Example: x 2 � 121 � (x � 11)(x � 11)

If a and b are expressions and ab � 0, thena � 0 or b � 0.

Example: � x � 9 � � x � 9 � � 0, then

x � 9 � 0 or x � 9 � 0.

Solve the quadratic equation 16 x 2 � 1 � 0.

Step 1: Write as a difference of two squares.

(4x) 2 � 1 2 � 0

Step 2: Factor.

(4x � 1)(4x � 1) � 0

Step 3: Use Zero Product Property.

4x � 1 � 0 or 4x � 1 � 0


4x � �1 or 4x � 1

x � � 1 __ 4 or x � 1 __

4

The zeros of 16x 2 � 1 � 0 are � 1 __ 4 and 1 __

4 .

Solve the quadratic equation 5 x 2 � 20.

Step 1: Solve for zero.

5x 2 � 20 � 0

Step 2: Factor: The GCF is 5.

5 ( x 2 � 4) � 0

Step 3: Factor the difference of squares.

5 (x � 2)(x � 2) � 0


x � 2 � 0 or x � 2 � 0


x � �2 or x � 2

The zeros of 5x 2 � 20 are �2 and 2.

PracticeComplete the steps to solve the quadratic equation.

1. 25x 2 � 64 � 0 2. 3 6x 2 � 16

� 5x � 2 � 8 2

� 0 36x 2 � 16 � 0

� 5x � 8 � � 5x � 8 � � 0 4 � 9x 2 � 4 � � 0

5x � 8 � 0 or 5x � 8 � 0 4 � 3x � 2 � � 3x � 2 � � 0

5x � �8 or 5x � 8 3x � �2 or 3x � 2

x � � 8 __ 5

or x � 8 __ 5 x � � 2 __

3 or x � 2 __

3

Solve the quadratic equation using the difference of squares.

3. 81x 2 � 49 � 0 4. 3x 2 � 48

� , �4, 4

78ReteachingSolving Quadratic Equations II

7 __ 9 7 __

9


Reteachingcontinued

You can solve cubic equations using the same method of factoring.

Solve 4x 3 � 36x � 0.


4x ( x 2 � 9) � 0


4x (x � 3)(x � 3) � 0


4x � 0 or x � 3 � 0 or x � 3 � 0


x � 0 or x � �3 or x � 3

The zeros of 4x 3 � 36x � 0 are 0, �3 and 3.

Solve 5x 3 � 80x � 0.


5x ( x 2 � 16) � 0


5x (x � 4)(x � 4) � 0


5x � 0 or x � 4 � 0 or x � 4 � 0


x � 0 or x � �4 or x � 4

The zeros of 5x 3 � 80x � 0 are 0, �4 and 4.


5. 2x 3 � 162x � 0 6. 3x 3 � 75x � 0

2x � x 2 � 81 � � 0 3x � x 2 � 25 � � 0

2x � x � 9 � � x � 9 � � 0 3x � x � 5 � � x � 5 � � 0

2x � 0 or x � 9 � 0 or x � 9 � 0 3x � 0 or x � 5 � 0 or x � 5 � 0

x � 0 or x � �9 or x � 9 x � 0 or x � �5 or x � 5

Solve the equation by factoring.

7. 6x 3 � 6x � 0 8. 4x 3 � 400x � 0

0, �1, 1 0, �10, 10

9. 8x 3 � 32x � 0 10. 28x 3 � 63x � 0

0, �2, 2 0,� ,

11. The expression � 4.9t 2 � 19.6 represents the height of a free falling object t seconds after it is dropped from a height of 19.6 meters. How long does it take for the object to hit the ground? Find the zeros of the equation � 4.9t 2 � 19.6 � 0.

2s

78

3 __ 2 3 __

2

Name Date Class


You have graphed and evaluated linear and quadratic functions. Now you will graph and evaluate piecewise functions.

A piecewise function is a function that has different rules for different parts of its domain. To evaluate a piecewise function for a given value of x find the interval that x belongs to. Then find the corresponding definition of the function for that interval.

Evaluate the piecewise function, f (x) for x � � 5 and x � 5.

f (x) � { 4x � 1 if x � 2 x 2 � 3 if x � 2

At x � �5, lies in the interval x � 2, At x � 5, lies in the interval x � 2,

since �5 � 2. since 5 � 2.

Use f (x) � 4x � 1 to evaluate f (�5). Use f (x) � x 2 � 3 to evaluate f(5)

f(�5) � 4(�5) � 1 � �21 f (5) � (5) 2 � 3 � 28

PracticeComplete the steps to evaluate the piecewise function for x � � 2 and x � 4.

1. h(x) = { 2x 2 � 1 if x � �1 3x � 2 if x � �1

At x � � 2, x is in the interval At x � 4, x is in the interval

x � � 1, since �2 � �1. x � �1 , since 4 � � 1.

Use h(x) � 2x 2 � 1 to find Use h(x) 5 3x � 2 to find

h(�2). h(4).

h(�2) � (�2 ) 2 � 1 h(4) � 3( 4 ) � 2

� 4 � 1 � 12 � 2

� 3 � 14

Evaluate each piecewise function for x � � 8 and x � 5

2. f(x) � { 2x � 4 if x � �8

�1 if �8 � x � 5

x 2 if x � 5 3. g(x) � { 2 � x if x � 5

� x 2 if 5 � x � 8 6 if x � 8

f (�8) � �12 and f (5) � 25 g(�8) � 10 and g(5) � �3

79ReteachingUnderstanding Piecewise Functions

When x � 2, f(x) � 4x � 1.

When x � 2, f(x) � x 2 � 3.


Reteachingcontinued

When graphing a piecewise function make a table of values and then draw the graph.

Graph the function.

f(x) � { 2x � 4 if x � �1 �x � 3 if x � �1

Make a table of values.

x f (x) � 2x � 4 f(x) � �x � 3�4 2(�4) � 4 � �4

�3 2(�3) � 4 � �2

�2 2(�2) � 4 � 0�1 2(�1) � 4 � 2 �(�1) � 3 � 4

0 �(0) � 3 � 3

1 �(1) � 3 � 2

2 �(2) � 3 � 1

PracticeComplete the steps to fill in the table and graph the function.

4. g(x) � { �2x if x � 1 x � 4 if x � 1

Graph the function.

5. f(x) � { 12 � x if x � 5 x � 2 if x � 5

79

x g(x) � �2x g(x) � x � 4�1 �2(�1) � 2

0 �2(0) � 0

1 �2(1) � �2 1 � 4 � �3 2 2 � 4 � �2 3 3 � 4 � �1

x f(x) � 12 � x f(x) � x � 2

3 9 4 8

5 7 7 6 8 7 8

Evaluate both 2x � 4 and �x � 3 � �1 to graph.

x

y45

23

1

21 3 54-4-3-5 -2

-5-4-3-2-1

O-1

Check the endpoints.At x � �1, f (x) �� x � 3. Put a closed circle at (�1, 4). Put a open circle at (�1, 2).

x

y45

23

1

21 3 54-4-3-5 -2

-5-4-3-2-1

O-1

x

y

O

6

8

10

21

3

5

7

-1

9

21-1-2

-2

3 5 7 94 6 8 10

4

Name Date Class


The normal curve below shows how data is distributed about the mean.

Answer the questions about the data.

A set of test scores is normally distributed with a mean of 70 and a standard deviation of 6.

a. What percent of the scores are between 64 and 76?

Solution: 64 is 6 less than the mean, 70, and 76 is 6 greater than 70. 6 � 6 � 1. Therefore each score is 1 standard deviation away from the mean. According to the normal curve, 68% of the data lie within 1 standard deviation of the mean. So, 68% of the scores are between 64 and 76.

b. What percent of the scores are between 76 and 82?

Solution: 76 is 6 greater than 70 and 82 is 12 greater than 70. That means that 76 is 1 standard deviation and 82 is 12 � 6, or 2 standard deviations from the mean. However, both scores fall to the right of the mean. 34% of scores are between 70 and 76 and 47.5% of the scores are between 70 and 82. So, 47.5% � 34% � 13.5% of the scores are between 76 and 82.

PracticeComplete the steps to answer the questions about the data.

1. A set of test scores is normally distributed with a mean of 80 and a standarddeviation of 4.

a. What percent of the scores are between 72 and 88?

72 is 8 less than 80, the mean. 88 is 8 greater than 80. 8 � 4 � 2 , so both scores are 2 standard deviations from the mean. According to the normal curve, 95 % of all data lie within 2 standard deviations of the mean. 95 % of the scores

are between 72 and 88.

b. What percent of the scores are less than 76?

According to the normal curve, 50 % of the scores are less than the mean and 34 % of the scores are between the mean and 76. 50 % � 34 % = 16 % of the scores are less than 76.


2. A set of test scores is normally distributed b. What percent of the scores are between 65 with a mean of 75 and a standard and 85? 95%deviation of 5.

a. What percent of the scores are between 70 c. What percent of the scores are greater than and 80? 68% 70? 84%

80ReteachingFinding the Normal Distribution

320

34%

13.5%2.35%

0.15%2.35%

0.15%13.5%

34%

1-1-2-3




The weights of bottles of water produced by one company have a mean of 18.5 ounces and a standard deviation of 0.15 ounces.

a. A randomly selected bottle weighs 18.575 ounces. How many standard deviations is this above the mean? Find the z−score, the standardized weight.

Solution: To compare data from normal distributions, use the z−score. The z−score converts the data into a common scale so they can be compared.

Use the equation z−score � x � mean _______________ standard deviaiton

, where x is the value in a normally

distributed set of data.

z−score � 18.575 � 18.5 ____________ 0.15

� 0.5

The standardized weight is 0.5. The bottle is 0.5 standard deviations above the mean.

b. A randomly selected bottle weighs 18.2 ounces. How many standard deviations is this below the mean?

z−score � 18.2 � 18.5 __________ 0.15

� �0.3 _____ 0.15

� �2

The standardized weight is �2. The bottle is 2 standard deviations below the mean.

PracticeComplete the steps to answer the questions about the data.

3. Paul received a score of 72 on a test where the scores are normally distributed with a mean of 68 and a standard deviation of 6. Erik received a score of 45 on a test where the scores are normally distributed with a mean of 37.5 and a standard deviation of 1.5. Who scored better compared to the rest of his class?

To compare the scores, convert them both to standardized scores, or z−scores.

Paul z-score � 72 � 68 _________

6 � 2 __

3

Erik z-score � 45 � 37.5 __________

1.5 � 5

Erik has the greater standardized score and did better compared to the rest of his class than did Paul.


4. The weights of cordless phones produced in a factory have a mean of 5.38 ounces and a standard deviation of 0.02 ounce. A randomly selected phone weighs 5.4 ounces. How many standard deviations is this above the mean? 1 standard deviation above

5. Chloe received a score of 88 on a test where the scores are distributed normally with a mean of 76 and a standard deviation of 4. Faye received a score of 34 on a test where the scores are distributed normally with a mean of 28 and a standard deviation of 3. Who scored better compared to the rest of her class? Chloe


Name Date Class

You have graphed several types of functions. Now you will estimate the area enclosed by part of the graph of a function and the x-axis.

One way of estimating the area of an irregular shape is to divide it into rectangles and add the areas of the rectangles.

Estimate the area under the curve shown from 0 � x � 4 using 4 intervals.

Step 1: Divide the region 0 � x � 4 into four evenly spaced intervals of 1 unit each. Draw four rectangles as shown.

Step 2: Find the height of each rectangle. This is the value of f(x) at the midpoint of each interval.

x f(x) � � x 2 � x � 120.5 12.25 units

1.5 11.25 units

2.5 8.25 units

3.5 3.25 units

Step 3: Add the heights and then multiply by the rectangle width.

Area � (12.25 � 11.25 � 8.25 � 3.25) � 1 � 35 square units

PracticeComplete the steps to estimate the area under the curve from 0 � x � 4 using 4 intervals. 1.

area 26 square unitsEstimate the area under the curve from 0 � x � 4 using 4 intervals.

2. 3.

25 square units 43 square units

x f(x) � 2x 2 � 5x � 6 area � f(x) � 1 unit0.5 4 units 4 square units

1.5 3 units 3 square units2.5 6 units 6 square units3.5 13 units 13 square units

x

y

y = -x2+ x + 12

O

12

16

4

1 2 3 4

8

x

y

y = -x2+ x + 12

O

12

16

4

1 2 3 4

8

x

y

y = 2x2- 5x + 6

O

12

16

4

1 2 3 4

8

x

y

y = -x2+ 3x + 10

O

12

16

4

1 2 3 4

8

x

y

y = x2- x + 3

O

12

16

4

1 2 3 4

8

INV

8Reteaching

Finding the Area Under a Curve


Reteachingcontinued

Now estimate the same areas using the rectangle technique with more rectangles.

Estimate the area under the curve from 0 � x � 4 using 8 intervals. Compare your estimate with the exact area 34 2 __

3 .

Step 1: Divide the region and draw eight rectangles as shown.Step 2: Find the height of each rectangle.

x f (x) � � x 2 � x � 12 x f (x) � � x 2 � x � 120.25 12.1875 units 2.25 9.1875 units

0.75 12.1875 units 2.75 7.1875 units

1.25 11.6875 units 3.25 4.1875 units

1.75 10.6875 units 3.75 1.1875 units

Step 3: Sum the heights and then multiply by the rectangle width.

Area � (12.1875 � 12.1875 � 11.6875 � 10.6875 � 9.1875 �7.1875 � 4.6875 � 1.6875) � 0.5 � 34.75 square units

Step 4: This estimate is closer to the exact area than the estimate from the previous page, which was 35 square units.

PracticeComplete the steps to estimate the area under the curve from 0 � x � 4 using 8 intervals. Compare your estimate with the exact area of 26 2 __

3 .

4.

area 26.5 square units This estimate is closer to the exact area than the previous estimate.

Estimate the area under the curve from 0 � x � 4 using 8 intervals. Compare your estimate with the exact area.

5.

exact area � 25 1 __ 3

25.25 square units

This estimate is closer to the exact area than the previous estimate.

x f(x) � 2x 2 � 5x � 6 x f(x) � 2x 2 � 5x � 6

0.25 4.875 units 2.25 4.875 square units0.75 3.375 units 2.75 7.375 square units1.25 2.875 units 3.25 10.875 square units1.75 3.375 units 3.75 15.375 square units

x

y

y = -x2+ x + 12

O

12

16

4

1 2 3 4

8

x

y

y = 2x2- 5x + 6

O

12

16

4

1 2 3 4

8

x

y

y = x2- x + 3

O

12

16

4

1 2 3 4

8

x

y

y = -x2+ x + 12

O

4

1 2 3 4

8

12

16

INV

8

Name Date Class


You have learned to use common logarithms. Now you will learn how to use logarithms that have e as a base. This kind of logarithm is called a natural logarithm, and is written as ln.

Properties of Natural Logarithms

Product Property ln ab � ln a � ln b When a and b are positive.

Quotient Property ln a __ b � ln a � ln b When a and b are positive.

Power Property ln a p � p � ln a When a is positive and p is a real number.

Simplify the logarithmic expression ln e 4m�2 .

In the expression e 4m�2 , 4m � 2 is the exponent. Recall that a logarithm is an exponent.

Therefore, ln e 4m�2 � 4m � 2.

Simplify the exponential expression e lnv .

e lnv

ln � e lnv � � lnv Recall that a logarithm is an exponent.

e lnv � v If ln � e lnv � � lnv, then e lnv � v.

PracticeComplete the steps to simplify the expressions.

1. e ln d

ln � e ln d � � lnd A logarithm is an exponent.

e ln d � d If ln � e lnd � � lnd , then e lnd � d .

2. ln e p�1 A logarithm is a(n) exponent.

ln e p�1 � p�1 In the expression e p�1 , the exponent is p�1 .

Therefore, ln e p�1 � p�1 .

Simplify the expressions.

3. e ln j j 4. ln e 2h�7 2h�7

5. e ln � b�4 � b�4 6. ln e z�10 z�10

7. ln e 2x�1 2x�1 8. e ln 2t 2t

81ReteachingUsing Natural Logarithms


Reteachingcontinued

Use the properties of natural logarithms to rewrite each expression. Then simplify.

a. ln 10 e

ln 10 e � ln 10 � ln e Use the Product Property to rewrite the expression.

� ln 10 � 1 e � e 1 , ln e � 1.

b. ln 2e ___ x

ln 2e ___ x � ln 2e � ln x Use the Quotient Property to rewrite the expression.

ln 2e � ln x � ln 2 � ln e � ln x Use the Product Property to rewrite ln 2e.

� ln 2 � 1 � ln x e � e 1 , ln e � 1.

� ln 2 � ln x � 1 Use the Commutative Property to reorder the expression.

c. ln e 3x

ln e 3x � 3x � ln e Use the Power Property to rewrite the expression.

� 3x � 1 e � e 1 , ln e � 1.

� 3x Simplify.


9. Radioactive substances decay over time, at a rate known as a half-life. Half-life is the time it takes for half the atoms of a radioactive substance to decay. The radioactive substance sodium-26 has a half-life of 1.07 seconds. How long will it take 1 gram of sodium-26 to decay to 0.1 gram? Use the natural decay function.

N � t � � N 0 e �kt N 0 is the initial amount of the substance, N � t � is the amount leftafter time t, and k is the decayconstant for the sustance.

0.1 � 1 e �0.65t Substitute 0.1 for N � t � , 1 for N 0 , and 0.65 k.

ln 0.1 � ln e �0.65t Take the natural logarithm of both sides.

ln 0.1 � � 0.65t In e raised to a power is equal to that power,so ln e �0.65t � �0.65t.

t � ln 0.1 ______ �0.65

� 3.5 Solve for t. Use a calculator to find In 0.1.

It will take about 3.5 seconds for 1 gram of sodium-26 to decay to 0.1 gram.

Solve each problem.

10. Rewrite and simplify.

ln 6e ln 6 � 1 11. Rewrite and simplify. ln 7e ___ x ln 7 � ln x � 1 12. Chlorine-38 has a half-life of 37.2 minutes. How long will it take 1 gram of it to decay to 0.1

gram? Use the equation 0.1 = 1 e � 0.019t and solve for t. �121 min

81

Name Date Class


You have learned how to use the law of sines. Now you will learn how to graph the sine function.

For a sine function y � asin(bx ) � c the period

is 2� ___ b

and the amplitude is a.

Graph the following function. Determine its domain, range, period, and amplitude.

y � 3sin(x) � 2

The domain, or x-values, of the sine function is the set of all real numbers.

Seeing a � 3 and b � 1, the amplitude is 3 and

the period is 2� ___ 1 � 2�.

Compare this graph to the graph of the parent function, y � sin(x). Multiplying the parent function by 3, multiplies the maximum and minimum values by 3. Adding 2 translates the graph vertically 2 units up.

The range is from the greatest y-value to the least, or 5 to �1.

Practice 1. Graph the function y � 2sin(x). Determine its domain,

range, period, and amplitude.

The parent function, y � sin(x), is multiplied by 2, so its maximum and minimum values are multiplied by 2 .

The new maximum is 2 and the new minimum is �2 , so the range is from 2 to �2 . The domain of a sine function is all real numbers .

In this function, a � 2 , so the amplitude is 2 , and b � 1,

therefore the period is 2� ___ 1

or 2� .

Graph and analyze the sine function.

2. y � 4sin(x) � 1

domain � all real numbers period � 2�

range � 5 to �3

amplitude � 4

ReteachingGraphing the Sine Function 82

x

yy = 3sin(x) + 2

y = sin(x)

O

456

1

1-1-2

2 43 5 6 87 9

23

-1-2-3

x

y

O1

1-1-2-3

2 43 5 6 87 9

23

-1-2-3

x

y

O

456

1

1-1-2-3

2 43 5 6 87

23

-1-2-3-4


Reteachingcontinued

Graph the following function. Determine its domain, range, period, and amplitude.

y � sin(3x) � 4

For a sine function, the domain is the set of all real numbers.

Compare to the graph of y � sin(x). Multiplying x in the parent function by 3 compresses the period. For this function, the

period is 2� ___ b � 2� ___

3 or 2 __

3 �.

The c term of the function translates the maximum and minimum values up or down the y-axis, depending on the sign and value of c. Here, c � �4, so the function translates down 4 units.

The range is now �3 to �5 and the amplitude is a, which is 1.

PracticeComplete the steps to graph and analyze the sine function.

3. Graph the function y � sin � 1 __ 2 x � . Determine its domain,

range, period, and amplitude.

The domain is the set of all real numbers .

In this function,

b � 1 __ 2 , so the period is 2� ___

b � 2� ___

1 __ 2 � 4�

The period is expanded compared to the parent function. It takes twice as long for the function to complete one period.

The range is 1 to �1 and the amplitude is a, which is 1 .Graph and analyze the sine function. 4. y � sin(4x) � 1

domain � all real numbers

period � 1 __ 2 �

range � 0 to 2

amplitude � 1

82

x

y

y = sin(3x) - 4

y = sin(x)

O

1

1

-1

-2

-3

-4

-5

2 3-1-2-3

x

y

O1

1-1-2-3

2 43 5 6 87 9 10 11 1213

23

-1-2-3

x

y

O

1

1

-1

-2

-3

2 3-1-2

Name Date Class


You have learned how to find the roots of quadratic equations. Now you will learn how to write a quadratic equation when given its roots.

Use the converse of the Zero−Product Property.

Zero−Product Property If ab � 0, then a � 0 or b � 0.

Converse of the Zero−Product Property If a � 0 or b � 0, then ab � 0.

Write three quadratic equations so that the root of each is 3.

x � 3 or x � 3 Write the roots as solutions of two equations.

x � 3 � 0 or x � 3 � 0 Add or Subtract the root to make one side of the equation � 0.

(x � 3) (x � 3) � 0 Use the converse of the Zero-Product Property to combine the

two equations into one.

x 2 � 6x � 9 � 0 Simplify by multiplying the two binominals.

The equation x 2 � 6x � 9 � 0 has one real root of 3.

To find two other equations with the same root,

multiply the two binomials by a constant.

�(x � 3)(x � 3) � 0 2 � x � 3 � � x � 3 � � 0

� x 2 � 6x � 9 � 0 2x 2 � 12x � 18 � 0

The root of the equations x 2 � 6x � 9 � 0,

� x 2 � 6x � 9 � 0, and 2x 2 � 12x � 18 � 0

is 3. Check this by graphing the equationsand seeing whether their Zeros occur at 3.

Practice

1. Write a quadratic equation whose roots are 2 and �6.

x � 2 or x � �6 Write the roots as solutions.

x �2 � 0 or x �6 � 0 Add or Subtract the roots to make one side of the equation � 0

� x �2 � � x �6 � � 0 Use the converse of the Zero-Product

Property to combine the two equations into one.

x 2 �4x � 12 � 0 Simplify by multiplying the two binomials.

The roots of the equation x 2 �4x � 12 � 0 are 2 and �6. You can check this by graphing the equation and seeing whether its zeros occur at 2 and �6.

2. Write a quadratic equation whose roots are �4 and 1. x 2 � 3x � 4 � 0

83ReteachingWriting Quadratic Equations from Roots

x

y

O

1

1

-1

-2

-3

2 43 5

2

3

-1


Reteachingcontinued

Write a quadratic equation whose roots are 2 � 4i and 2 � 4i.

x � 2 � 4i x � 2 � 4i Write the roots as solutions to two equations.

x � 2 � 4i Combine the equations into one using the � sign.

x � 2 � � 4i Subtract 2 from both sides.

x � 2 � � �� 16 Rewrite 4i as a square root.

� x � 2 � 2 � �16 Square both sides.

x 2 � 4x � 4 � �16 Multiply out the binomial.

x 2 � 4x � 20 � 0 Add 16 to both sides

You can check your equation by graphing the related function y � x 2 � 4x � 20 to confirm that it has no zeros.

Practice

3. The bases of an arch are 100 feet apart. The arch has a height of 60 feet. Write a quadratic function to approximate the arch.

Step 1 Think of the arch as a parabola. The general form of an equation of a parabola is y � a � x � h � 2 � k. Let h represent half the distance between the bases, where the arch will be at its highest point. Let k represent the height of the arch.

h � 50 k � 60 Step 2 Set x and y equal to 0 and solve y � a � x � h � 2 � k for a.

0 � a � 0 � 50 � 2 � 60

�60 � a � 0 � 50 � 2 Subtract 60 from both sides

�60 � a � 2500 � Square the expression inside parentheses

a � �0.024 Divide both sides by 2500 Step 3 Substitute the value for a into the related function y � a � x � 50 � 2 � 60.

y � �0.024 � x � 50 � 2 � 60

y � �0.024 x 2 � 2.4x Square the binomial and multiply both terms by �0.024.

Step 4 Graph the function to check. 4. Write a quadratic equation whose roots are �5i and �5i x 2 � 25 � 0

83

x

y

O1

-1 2 43 5 6-1-2-3

1415161718192021

Name Date Class


You have learned about rational expressions. Now you will learn how to solve rational equations containing linear polynomials and rational equations containing quadratic polynomials.

Solve the equation.

5x _____ x � 2

� 12

� x � 2 � 5x _____ x � 2

� 12(x � 2) Multiply both sides by (x � 2) to remove all fractions.

5x � 12x � 24

�7x � �24 Subtract 12x from both sides.

x � 3.43 Solve for x and round to the nearest hundredth.

Make sure the value you found for x does not make the denominator of the equation 0.In this case, it does not. The solution to the equation is x � 3.43.

You can check your solution by graphing Y1 � 5x _____ x � 2

and Y2 � 12 on a graphing calculator and finding their intersection.


1. �10 ____ x � 7 � x

x �10 ____ x � 7 � x � x � To clear the fraction, multiply both sides by x.

�10 � 7x � x 2 Use the Distributive Property on the left side and simplify the right.

x 2 � 7x � 10 � 0 Rewrite the equation, setting it equal to 0 .

� x � 2 � � x � 5 � � 0 Factor the polynomial.

x � 2 � 0 or x � 5 � 0 Solve the two equations for x.

x � 2 or x � 5 Check to make sure the x-values do not make the denominator 0.

Check your solution by graphing Y1 � �10 ____ x � 7 and Y2 � x and finding their intersection.


2. 6x _____ x � 1

� 10

x � 2.5

4. 2x _____ x � 8

� 6

x � 12

3. �9 ___ x � 6 � x

x � �3

5. �4 � 3 __ x � x

x � �1 or x � �3

84ReteachingSolving Rational Equations



Solve the equation.

12 _______ x 2 � 36

� 4 _____ x � 6

The denominator on the left side is a difference of two squares.

x 2 � 36 � (x � 6)(x � 6) Factor the squares. Use the two factors as the LCD.

(x � 6)(x � 6) � 12 ____________ (x � 6)(x � 6)

� � 4 _____ x � 6

� x � 6 � � x � 6 �

12 � 4 � x � 6 � Cancel out the common factors.

12 � 4x � 24 Use the Distributive Property.

�12 � 4x Simplify.

x � �3

Check to make sure the x-value does not make the denominator 0. In this case, it does not. The solution is x � �3.

You can check your solution by graphing Y1 � 12 _______ x 2 � 36

and Y2 � 4 _____ x � 6

and finding the points where the two graphs intersect.

PracticeComplete the steps to solve the problem. 6. An airplane flying with the wind travels 200 miles. When it flies against the wind, it travels

150 miles in the same amount of time. If the wind speed is 20 miles per hour, what is the speed of the plane when there is no wind?

From the distance formula, distance � rate � time, we know time � distance

__________ rate

.

Let the flying time with the wind � 200 ______ r � 20

and let the flying time against the wind � 150 ______ r � 20

.

The flying time is the same whether the plane flies into the wind or against the wind.

200 ______ r � 20

� 150 ______ r � 20

� r � 20 � � r � 20 � � 200 ______ r � 20

� � � 150 ______ r � 20

� � r � 20 � � r � 20 � Multiply by the LCD.

200 � r � 20 � � 150 � r � 20 � Cancel out the common factors.

200r � 4000 � 150r � 3000 Use the Distributive property.

50r � 7000 Combine like terms.

r � 140 Solve for r.The speed of the plane when there is no wind is 140 mph.

Solve each problem.

7. 3 _______ x 2 � 16

� 1 _____ x � 4

x � 7

8. An express train travels 100 miles in the same time a local train travels 50 miles. The express train travels 25 miles per hour faster than the local train. How fast does

each train travel? Use 100 ______ r � 25

to represent the express

train. Use 50 ___ r to represent the local train.

express train 50 mph; local train 25 mph

Name Date Class


You have found polynomial roots by factoring. Now you will find roots by using patterns of sums and differences of cubes and by the rational root theorem.

The rules below can be used to factor sums and differences of cubes.

Sum of cubes: x 3 � a 3 � (x � a)( x 2 � ax � a 2 )

Difference of cubes: x 3 � a 3 � (x � a)( x 2 � ax � a 2 )

After you have rewritten the expression and determined any real roots of the first term, check if the quadratic term has any real roots by using the discriminant or by graphing.

Remember, if an polynomial has the form a x 2 � bx � c , the discriminant is b 2 � 4ac. If the discriminant is positive, the polynomial has 2 real roots. If it is 0, the polynomial has 1 real root. If it is negative, there are no real roots.

Find the roots of y � x 3 � 8.

x 3 � 8 � (x � 2)( x 2 � 2x � 4) Sum of cubes pattern.

�2 is a root of x � 2

You can determine if there are any other roots by checking the discriminant of x 2 � 2x � 4. b 2 � 4ac � (�2) 2 � 4(1)(4) � �4

The discriminant is negative. There are no other real solutions.

� x 3 � 8 has 1 root, �2.

PracticeComplete the steps to find the roots of y � x 6 � 27.

1. y � x 6 � 27

� ( x 2 ) 3 � 3

3

� ( x 2 � 3)( x 4 � 3 x 2 � 9)

��

3 and � ��

3 are the roots of x 2 � 3 .

Graph x 4 � 3 x 2 � 9 to see if there are any other roots.

The roots of the polynomial are ��

3 and � ��

3 .

Find the roots of the polynomial.

2. y � x 3 � 25 3. y � x 3 � 64

�5 4

4. y � x 6 � 216 5. y � x 6 � 8

no real roots ��

2 and � ��

2

85ReteachingFinding Polynomial Roots II

x

y

O

12

16

18

4

4 8

8

-12 -8 -4

-4

-8


Reteachingcontinued

The Rational Root Theorem is a way to check if a polynomial has any rational roots. The steps are:

1: Write the polynomial in the form a n X n � a n � 1 X n � 1 � � � � � a 0 .

2: Find all of the factors of a n and a 0 .

3: List all possible rational numbers p __ q , where p is a factor of a 0

and q is a factor of a n . These are all of the possible roots.

4: Check if x � p __ q is a factor of the polynomial using synthetic

division.

Find the roots of y � x 3 � 2 x 2 � x � 2 using the rational roots theorem.

Values for p: �1, �2; values for q: �1; values for p __ q : �1, �2.

Use synthetic division to test the roots.

1 1 �2 �1 2

1 �1 �2

1 �1 �2 0

x 3 � 2 x 2 � x � 2 � (x � 1)( x 2 � x � 2)

� (x � 1)(x � 2)(x � 1) Factor x 2 � x � 2.

The roots are 1, 2, and �1.

PracticeComplete the steps to find the roots of y � x 4 � x 3 � 27x � 27.

6. values for p: �1, �3, �9 , �27 ; values for q : �1; values for p __ q : �1, �3 , �9 , �27

(x � 1) is not a root (x � 1) is not a root

x 4 � x 3 � 27x � 27 � (x � 1 )( x 3 � 27)

� (x � 1)(x � 3 )( x 2 � 3x � 9 ) Factor the sum of cubes

The roots are �3 and �1 .

Find the roots of the polynomial. 7. x 3 � 3x � 2

�2 and 1 8. x 4 � 3 x 3 � x � 3

�1 and 3

9. A box has a volume of 15 cubic feet. It has a length that is 5 times its width and a height that is 2 feet greater than its width. Find the roots of the polynomial 5 x 3 � 10 x 2 � 15 to find the dimensions. x is the width. (Hint : Use the positive roots.)

length � 5 ft, width � 1 ft, height � 3 ft

85

1 1 1 0 27 27

1 2 2 29 1 2 2 29 56

�1 1 1 0 27 27

�1 0 0 �27 1 0 0 27 0

Name Date Class


ReteachingGraphing the Cosine Function 86

You have graphed the sine function. Now you will graph the cosine function. Cosine is a periodic function, meaning that its graph has a repeating pattern. The period is the length of the pattern. The amplitude of the graph is equal to half the distance between the maximum and minimum values.

The parent function y � cos(x) has a period of 2� and an amplitude 1. If you have a function of the form y � a � cos(bx), the period equals 2� ___

� b �

and the amplitude equals � a � .

Graph the function y � 4cos(2x) and identify its period and amplitude.

The function is of the form y � a � cos(bx), where a � 4 and b � 2.

From the graph, you can see that the pattern starts at x � 0 and begins to repeat at x � �, so the period is �. You can also find the period using the expression above.

2� ___ � b �

� 2� ___ � 2 �

� �

The amplitude is � a � � 4. Note that this is equal to half the distance between the maximum and minimum values, or

� 4 � (�4) �

_________ 2

� 4.

PracticeComplete the steps to identify the period and amplitude and graph the function.

1. y � 5cos � x __ 2 �

The period of the function � 2� ____ � b �

� 2� ____ � 1 __ 2 �

� 4π

The amplitude of the function � � a � � � 5 � � 5 .

Identify the period and amplitude of the function.

2. y � 15cos(10x)

Period � π __ 5 Amplitude � 15

3. y � �6cos(4x)

Period � π __ 2 Amplitude � 6

x

y

O

1

2

3

4

-1

-2

-3

-4

π3π4

π

4π

2

x

y

O

1

π 2π 3π 4π

2

3

4

5

-1

-2

-3

-4

-5



Sometimes the graph of the cosine function is translated horizontally and/or vertically. If a function has the form y � a � cos(b(x � c)) � d, where b is greater than 0, then the graph is shifted c units horizontally and d units vertically. The distance the graph is shifted horizontally is called the phase shift.

Graph the function y � 3 cos � 2 � x � � __ 4 � � � 2. Identify its period, amplitude, phase

shift and vertical shift.

The function is of the form y � a � cos(b(x � c)) � d

where a � 3, b � 2, c � � __ 4 , and d � �2

The period equals 2� ___ � b �

� 2� ___ � 2 �

� �.

The amplitude is � a � � 3. Notice that the maximum

of the graph is 1 and the minimum is –5. Using the

definition, the amplitude is � 1� (�5) �

_________ 2 � 6 __

2 � 3.

The phase shift equals c � � __ 4

. The vertical shift equals d ��2.

PracticeComplete the steps to graph the function and identify its period, amplitude, phase shift, and vertical shift.

4. y � 12cos � 0.5 � x� � __ 4 � � � 3

period � 2� ___ � b �

� 2� ______ � 0.5 �

� 4π , amplitude � � a � � � 12 � � 12

phase shift � c � π __ 4

, vertical shift � d � 3

Graph the function and identify its period, amplitude, phase shift, and vertical shift.

5. y � 10cos � x � � __ 8 � � 1

Period � 2π Amplitude � 10

Phase shift � π __

8 Vertical shift � 1

6. y � 0.5cos(0.5(x � � ))� 8

Period � 4π Amplitude � 0.5

Phase shift � π Vertical shift � �8

x

y

O-π

1

2

3

4

-3

-4

-5

-1

-2

π3π4

π

4π

4π

2

x

y

O

3

π-π 2π 3π 4π

6

9

12

15

-3

-6

-9

Name Date Class


ReteachingEvaluating Logarithmic Expressions 87

Two rules that you have used to evaluate expressions are:

log b (mn) � log b (m) � log b (n) log b ( m n ) � n log b (m)

Sometimes you may have to use more than one rule to evaluate an expression.

Evaluate log 3 (27x) 5 when x � 9.

log 3 (27x) 5 � 5 log 3 (27x) Use the second rule to rewrite the expression.

� 5( log 3 27 � log 3 x) Use the first rule to rewrite the expression.

� 5(3 � l og 3 x) log 3 27 simplifies to 3.

� 15 � 5 log 3 x Use the distributive property.

� 15 � 5 log 3 9 Substitute 9 for x.

� 15 � 5(2) log 3 9 simplifies to 2.

� 15 � 10 Multiply.

� 25 Add.

Practice 1. log 2 (32x) 10 when x � 128

log 2 (32x) 10 � 10 log 2 (32x)

� 10( log 2 32 � log 2 x)

� 10 (5 � log 2 x)

� 50 � 10 log 2 x

� 50 � 10 log 2 128

� 50 � 10(7)

� 50 � 70 � 120

2. ln (8e) 3 � 3 ln(8e)

� 3 (ln 8 1 ln e)

� 3 ln 8 � 3 ln e

� 3 (2.08) � 3

� 9.24

Evaluate the expression.

3. log 4 ( 64x) 9 when x � 64 54 4. log 10 (100x) 4 when x � 1000 20

5. log 5 (125x) 2 when x � 625 14 6. log 3 (81x) 6 when x � 729 60

7. ln (20e) 5 � 20 8. ln (4e) 12 � 28.6



Sometimes it can be helpful to use the change of base function when solving equations with variables in the exponents. If you change to a base of e or 10, you can evaluate the expression. Recall that the

change of base formula is log b x � log a x

_____ log a b

, where a is the base you are changing from.

Evaluate log 8 (1 5x) 7 when x � 12.

log 8 ( 15x) 7 � 7 log 8 (15x) Use the second rule to rewrite the expression.

� 7 � ln 15x _____ ln 8

� Use the change of base formula.

� 7 � ln 15 � ln x _________ ln 8

� Use the first rule to rewrite the expression.

� 7 � ln 15 � ln 12 __________ ln 8

� Substitute 12 for x.

� 7 � 2.71 � 2.48 __________ 2.08

� Evaluate logs.

� 17.5 Simplify within parentheses.

PracticeComplete the steps to evaluate the expression.

9. log 7 (9x) 15 when x � 20

log 7 (9x) 15 � 15 log 7 (9x)

� 15 � log 9x ______

log 7 �

� 15 � log 9 � log x ___________

log 7 �

�15 � log 9 � log 20 ____________

log 7 �

� 15 � 0.95 � 1.3 ___________

0.85 �

� 39.7

10. log 13 (12x) 5 when x � 10

log 13 (12x) 5 � 5 log 13 (12x)

� 5 � ln 12x ______

ln 13 �

� 5 � ln 12 � ln x __________ ln 13

� � 5 � ln 12 � ln 10

____________ ln 13

� � 5 � 2.48 � 2.3

___________ 2.56

� � 9.34

Evaluate the expression.

11. log 20 (45x) 6 when x � 20 � 13.6 12. log 22 (30x) 24 when x � 15 � 47.4

Name Date Class


You have solved linear, polynomial, radical, and rational equations. Now you will use these methods to solve abstract equations.

An abstract equation, or a literal equation, has two or more variables. When working with abstract equations such as formulas, you may need to solve for one of the variables.

The area of a trapezoid is given by the formula A � 1 __ 2 � b 2 � b 2 � h.

Solve the formula for b1.

A � 1 __ 2 � b 1 � b 2 � h

A � 2 __ h � � 1 __

2 � b 1 � b 2 � h � � 2 __

h � Multiply both sides by � 2 __

h � .

b2

b1

h

2A ___ h � b 1 � b 2 Simplify.

2A ___ h

� b 2 � b 1 � b 2 � b 2 Subtract b 2 from both sides.

2A ___ h � b 2 � b 1 Simplify.

PracticeComplete the steps to solve the formula for the surface area of a right circular cone for l.

1. S � � r 2 � �rl

S � � r 2 � � r 2 � �rl � � r 2

h

r

l S � � r 2 � �rl

S � � r 2 ________

�r � �rl ___ �r

S � � r 2 _______ �r � l

Solve the equation for the given variable.

2. The volume of a cylinder is given by the formula V � � r 2 h. Solve the formula for h.

h � V ____ � r 2

3. The area of a triangle is given by the formula A � 1 __ 2 bh. Solve the

formula for b.

b � 2A ___ h

4. Solve the equation y � mx � b for m.

m � y � b _____ x

88ReteachingSolving Abstract Equations


Reteachingcontinued

The length, c, of the hypotenuse of a right triangle is given by the

equation c � ��

a 2 � b 2 , where a and b are the lengths of its legs. Solve the equation for a.

c � ��

a 2 � b 2

c 2 � � ��

a 2 � b 2 � 2

c 2 � a 2 � b 2 a

b

c

c 2 � b 2 � a 2 � b 2 � b 2

c 2 � b 2 � a 2

��

c 2 � b 2 � ��

a 2

��

c 2 � b 2 � a

PracticeComplete the steps to solve the formula for the volume of a right square pyramid for s.

5. V � 1 __ 3 s 2 h

V � � 3 ______

h � � 1 __

3 s 2 h � � 3

______ h

�

3 V _______ h

� s 2

��

3V ______

h � �

� s 2

��

3V ______

h � s

Solve the equation for the given variable.

6. The volume of a right circular cone is given by the formula V � 1 __ 3 � r 2 h.

Solve the formula for r.

r � ��

3V ___ �h

7. The equation E � m c 2 relates energy to mass times the velocity of light squared. Solve the equation for c.

c � ��

E __ m

8. A playground ball has a surface area of 100 square inches. Solve the formula for the surface area of a sphere,S � 4� r 2 , for r to find the radius of the ball.

5 in.

88

h

ss

Name Date Class


You can graph a quadratic inequality by graphing its related function as the boundary. If the parabola includes solutions (� or �), use a solid line. If it does not include solutions (� or �), use a dashed line. Then shade the area that contains the solutions.

Graph y � x 2 � 4x � 3.

Graph y � x 2 � 4x � 3.

vertex: x−coordinate � �b ___ 2a

� �(�4)

______ 2 1

� 2

y−coordinate � 2 2 � 4(2) � 3 � �1

y−intercept: (0, c) � (0, 3)

x−intercepts: Solve x 2 � 4x � 3 � 0; (x � 3)(x � 1) � 0.

The x−intercepts are (3, 0) and (1, 0).

The parabola passes through the vertex and intercepts.

The inequality uses �, so draw the parabola with a solid curve. Shade the area above the curve.

PracticeComplete the steps to graph the inequality.

1. y � x 2 � 6x � 8

vertex: x−coordinate � �b ___ 2a

� �6 _____ 2 � 1

� �3

y−coordinate � � �3 � 2 � 6 � �3 � � 8 � �1

y−intercept: (0, c) = (0,8)

x−intercepts: Solve x 2 � 6x � 8 � 0; (x � 2)(x � 4) � 0.

The x−intercepts are ( �2, 0) and ( �4, 0).

Draw a parabola using a dashed curve and shade the area below the curve.

Graph the inequality.

2. y � � x 2 � 2x � 3 3. y � x 2 � x � 2

89

x

y

2

4

4 6-4-6 -2

-2

-4

-6

O

6

x

y

2

4

6

8

10

42-4-6-8 -2O

6

ReteachingSolving Quadratic Inequalities

x

y

1

2

3

21-2-3 -1

-2

-1

O3

x

y

1

2

3

4

21 3-2-3 -1

-1

O


Reteachingcontinued

To solve a quadratic inequality, you can solve the related equation. Find the critical values and graph them on a number line. Then test a value on each region separated by the values to see if it satisfies the inequality.

Solve 3 x 2 � 17x � 6.

Write and solve the related equation.

3x 2 � 17x � 6

3x 2 � 17x � 6 � 0

(x � 6)(3x � 1) � 0

x � 6 � 0 3x � 1 � 0

x � 6 3x � �1

x � � 1 __ 3

The critical values are 6 and � 1 __ 3 . Graph the values on a number line.

Choose a value from each region to test.

The solution to the inequality is � 1 __ 3 � x � 6.

PracticeComplete the steps to solve the inequality.

4. 2x 2 � x � 3

2x 2 � x � 3

2x 2 � x � 3 � 0

(2x � 3)(x � 1) � 0

2x � 3 � 0 x � 1 � 0

x � 3 __ 2 x � �1 The solution is x � �1 or x � � 3 __

2 .

Solve the inequality.

5. x 2 � 3x � 45 6. x 2 � 5x � �6

x � �5 or x � 8 2 � x � 3

89

32 40

3(-1)2 - 17(-1) = 2020 > 6

not a solution

1 65 7-1

3(0)2 - 17(0) = 00 ≤ 6

solution

3(7)2 - 17(7) = 2828 > 6

not a solution

0-1.5

2(-2)2 - (-2) = 1010 > 3

solution

-1 2-2

2(0)2 - 0 = 00 < 3

not a solution

2(2)2 - (2) = 66 > 3

solution

-0.5 0.5 1 1.5

Name Date Class


You have graphed the sine function and the cosine function. Now you will graph the tangent function.

Transformations of the tangent function change the period and/or asymptotes of the graph.

For y � tan(bx):

• the period is � __ b

.

• the asymptotes are located at x � � ___ 2b

� n� ___ b , where n is an

integer.

Graph y � tan(2x).

Step 1: Find b to identify the period.

b � 2, and � __ b � � __

2 , � __

2 , so the period is � __

2 . This is one half of the period of y � tan(x).

Step 2: Use the period to identify the x-intercepts. The first x-intercept of the function occurs at 0. Because the period is � __

2 ,

the intercepts

occur at n� ___ 2 � � � __

2 , 0, and � __

2 are x-intercepts.

Step 3: Identify the asymptotes, where the function is undefined.

b � 2, so the asymptotes are

x � � ____ 2(2)

� n� ___ 2 � � __

4 � n� ___

2 .


PracticeComplete the steps to graph y � tan � 1 __

4 x � on the interval [�5�, 5�].

1. Find the period. � __ b � 4�

Find the x-intercepts. �4�, 0, 4�

Find the asymptotes. x � �2�, x � 2�

Graph each function.

2. y � tan(3x) 3. y � tan � 1 __ 3 x �

90ReteachingGraphing the Tangent Function

x

y4

2

π

-2

-π

-4

Oπ

2-

2π

x

y4

2

4π

-2

-4π

-4

O-2π 2π

x

y4

2

Oπ

32π

-

3π

-

32π3

x

y4

2

4π

-2

-4π

-4

O-2π 2π

cos(bx) � 0 when bx � � __ 2 � n� or x � � ___

2b � n� ___

b .

There will be one full cycle between � � __ 4 and � __

4 .


Reteachingcontinued

A phase shift is a horizontal shift of the graph of the tangent function.

For y � tan [ b(x � c) ] :• the period is � __

b .

• the asymptotes are located at x � � ___ 2b

� c � n� ___ b , where n is an integer.

Graph: y � tan � x � � __ 4 � .

Step 1: Find b to identify the period.

b � 1, and � __ b � �, so the period is �.

Step 2: Use the period to identify the x-intercepts. The first

x-intercept of the function occurs at c, which is � __ 4 .

Because the period is �, the intercepts occur at intervals

of � from � __ 4

� � 3� ___ 4 , � __

4 , and 5� ___

4 . are x-intercepts.

Step 3: Identify the asymptotes. b � 1, so the asymptotes are

x � � ____ 2(1)

� � __ 4 � n� ___

1 � 2� ___

4 � � __

4 � 4n� ____

4 �

�(3 � 4n) _________

4 .

Step 4: Graph the function

PracticeComplete the steps to graph y � tan � y � � __

2 � on the interval [ � 5� ___ 2 , 5� ___

2 ] .

4. Find the period. � __ b � 4�

Find the phase shift. c � � � __ 2

Find the x-intercepts. � 3� ___ 2 , � � __

2 , � __

2 , 3� ___

2

Find the asymptotes. x � �2�, x � ��, x � 0,

x � �, x � 2�

Graph each function.

5. y � tan � x � � __ 3 � 6. y � tan � x � � __

3 �

90

cos[b(x � c)] � 0 when b(x � c) � � __ 2 � n� or x = � ___

2b � c � n� ___

b .

There will be one full cycle between c � � __

2 and c � � __

2 .

tan(x � c) � 0, so x � c � 0, x � c

x

y4

2

Oπ

2π

-

2π-π

x

y4

2

-2

-4

O-π π 2π-2π

x

y4

2

-2

-4

O-π π

x

y4

2

-2

-4

-π πO

Name Date Class


INV

9

You have graphed absolute value functions. Now you will use the greatest integer function and least integer function to solve problems.

greatest integer function

The greatest integer function, f (x) � ⎣x⎦, is a function where the output, f(x), is the greatest integer that is either less than or equal to the input, x.

Think: ⎣ ⎦ means down to the floor.

Example, if x � 3.65 then f(x) � ⎣3.65⎦ � 3.

Graph the function, f(x) � ⎣x⎦ for the domain [�1, 5].

x f(x) � ⎣x⎦ �1 � x � 0 �1

0 � x � 1 01 � x � 2 12 � x � 3 23 � x � 4 34 � x � 5 4

Practice

1. A cell phone company charges customers for each phone call that they make. A call costs $0.60 for a connection fee plus $0.20 for each minute of the call. The company rounds the time of the call down to the nearest whole minute. Complete the following steps to find the cost of each phone call.

a. a 5-minute call b. a 5.85 minute call c. a 0.56 minute call

0.60 � 5(0.20) � $1.60 0.60 � 5(0.20) � $1.60 0.60 � 0(0.20) � $0.60

2. Complete the table below. Then graph the function as a step function.

x f(x)

0 � x � 1 $0.60

1 � x � 2 $0.80

2 � x � 3 $1.003 � x � 4 $1.204 � x � 5 $1.405 � x � 6 $1.60

ReteachingUnderstanding Step Functions

x

y4

2

2 4

-2

-2-4

-4

O

x

y

0.20.40.60.8

11.21.41.6

2 4 6O

Reteachingcontinued

INV

9


The least integer function f(x) � ⎡x⎤, is a function where the output, f(x), is the least integer that is either greater than or equal to the input, x.

Think: ⎡ ⎤ means up to the ceiling.

Example, if x � 1.15 then f(x) � ⎡1.15⎤ � 2.

Graph the function f(x) � ⎡x⎤ for the domain [�1, 5].

x f(x) � ⎡x⎤ �1 � x � 0 0

0 � x � 1 1

1 � x � 2 2

2 � x � 3 3

3 � x � 4 4

4 � x � 5 5

Practice

3. Parking lot customers are charged by the hour. The length of time that a car is parked is rounded up to the nearest hour. The function for the cost for parking is f(x) � 8⎡x⎤ where x is the time in hours and f(x) is the cost of parking in dollars. Complete following steps to find the cost of parking for each given amount of time.

a. 0.5 hours b. 3 hours c. 3.25 hours

8 ⎡0.5⎤ � 8 � 1 � 8 8 ⎡3⎤ � 8 � 3 � $24 8 ⎡3.25⎤ � 8 � 4 � $32

4. Complete the table below. Then graph the function as a step function.

x f(x)

x

y

10

20

40

2 4 6O

30

0 � x � 1 $8.00

1 � x � 2 $16.00

2 � x � 3 $24.003 � x � 4 $32.004 � x � 5 $40.005 � x � 6 $48.00

x

y4

2

2 4

-2

-2-4

-4

O

Name Date Class


You have learned how to use the distance and midpoint formulas. Now you will learn how to use these formulas to write the equation of a circle.

Equation of a Circle Distance Formula

The equation of a circle with center (h, k) and radius r is

(x � h ) 2 � (y � k ) 2 � r 2 .

The length of a segment whose endpoints are ( x 1 , y 1 ) and ( x 2 , y 2 ) is

d � ��

� x 2 � x 1 � 2 � � y 2 � y 1 �

2 .

Write the equation of a circle with center (�1, 3) that contains the point (2, 7).

Step 1: Find the length of the radius of the circle. Use the distance formula to find the distance between the center and the point on the circle.

r � ��

� x 2 � x 1 � 2 � � y 2 � y 1 �

2 Substitute r for d in the distance formula.

r � ��

� 2 � (�1) � 2 � � 7 � 3 � 2 Substitute (2, 7) and (�1, 3) for ( x 2 , y 2 ) and ( x 1 , y 1 ), respectively.

r � ��

3 2 � 4 2 r � �

� 9 � 16

r � ��

25 r � 5

Step 2: Write the equation for the circle with center (�1, 3) and radius 5.

(x � h ) 2 � (y � k ) 2 � r 2 Equation of a circle with center (h, k) and radius r.(x � (�1) ) 2 � (y � 3 ) 2 � 5 2 Substitute (�1, 3) for (h, k) and 5 for r.

(x � 1 ) 2 � (y � 3 ) 2 � 25 Simplify.

PracticeComplete the steps to write the equation of a circle with the given center and point on the circle.

1. Center (3, �7); contains the point (8, 5) r � ��

(8 � 3 ) 2 � (5 � (�7) ) 2

r � ��

5 2 � 12

2

r � ��

169 � 13 (x � h ) 2 � (y � k ) 2 � r 2

(x � 3 ) 2 � (y � (�7) ) 2 � 13 2

(x � 3 ) 2 � (y + 7 ) 2 � 169

Write the equation of a circle with the given center and point on the circle. 2. Center (4, 5); contains the point (10, 13) 3. Center (�3, 5); contains the point (4, 5)

(x � 4 ) 2 � (y � 5 ) 2 � 100 (x � 3 ) 2 � (y � 5 ) 2 � 49

91ReteachingMaking Graphs and Using Equations of Circles


Reteachingcontinued

For a segment whose endpoints are at ( x 1 , y 1 ) and ( x 2 , y 2 ),

Midpoint Formula Distance Formula

M � � x 1 � x 2 _______

2 ,

y 1 � y 2 _______

2 � . d � �

�� x 2 � x 1 �

2 � ( y 2 � y 1 ) 2 .

Write an equation of the circle whose diameter has endpoints (�5, 2) and (3, 6).

Step 1: Find the center of the circle. Use the midpoint formula.

M � � �5 � 3 _______ 2 , 2 � 6 _____

2 � Substitute (�5, 2) for ( x 1 , y 1 ) and (3, 6) for ( x 2 , y 2 ), respectively.

M � � �2 ___ 2

, 8 __ 2 � � (�1, 4) Simplify.

Step 2: Find the radius of the circle. Use the distance formula.

r � ��

(�1 � 3 ) 2 � (4 � 6 ) 2 Substitute the center (�1, 4) for ( x 2 , y 2 ) and one of the

r � ��

(�4 ) 2 � (�2 ) 2 other points on the cricle for ( x 1 , y 1 ).

r � ��

16 � 4

r � ��

20 � 2 ��

5

Step 3: Write the equation of the circle.

(x � h ) 2 � (y � k ) 2 � r 2 Use the equation for a cricle.

(x � (�1) ) 2 � (y � 4 ) 2 � � 2 ��

5 � 2 Substitute (�1, 4) for (h, k) and 2 ��

5 for r.

(x � 1 ) 2 � (y � 4 ) 2 � 20 Simplify.

PracticeComplete the steps to write an equation of the circle with the given diameter.

4. Diameter with endpoints (2, 7) and (�6, 15)

M � � x 1 � x 2 _______

2 ,

y 1 � y 2 _______

2 � r � �

�� x 2 � x 1 �

2 � � y 2 � y 1 � 2 (x � h ) 2 � (y � k ) 2 � r 2

M � � 2 � � �6 � _________ 2 , 7 � 15

_______ 2 � r � �

��

� �2 � � �6 � � 2 � � 11 � 15 � 2

r � ��

4 2 � � �4 � 2

M � � �2, 11 � r � ��

32 � 4 ��

2

Write an equation of the circle whose diameter has the given endpoints.

5. (3, 4) and (�1, �6) 6. (�6, 8) and (2, �4)

(x � 1 ) 2 � (y � 1 ) 2 � 29 (x � 2 ) 2 � (y � 2 ) 2 � 52

91

� x � � �2 � � 2 � � y � 11 � 2 � � 4 ��

2 � 2

� x � 2 � 2 � � y � 11 � 2 � 32

Name Date Class


You have learned how to find patterns and identify functions. Now you will learn how to find arithmetic sequences.

If the difference between each pair of consecutive terms in a sequence is the same, then the sequence is an arithmetic sequence. To determine whether a sequence is an arithmetic sequence, check for a common difference.

Determine if this sequence is arithmetic: � 7, �3, 1, 5, 9, …Calculate the difference between each pair of consecutive terms:

�3 � � �7 � � 4

1 � � �3 � � 4

5 � 1 � 4

9 � 5 � 4

The common difference is 4. Therefore, the sequence is arithmetic.

If you know the first term of an arithmetic sequence a 1 and the common difference d, then you can find the nth term a n using the following formula:

a n � a 1 � � n � 1 � d.

Find the 15th term of the arithmetic sequence 10, 4, �2, �8, �14, …Step 1: Find the common difference, d. The sign of d is important, so be careful to find the difference as you move from left to the right in the sequence.

d � 4 � 10 � �6

Step 2: Identify the first term, a 1 .

a 1 � 10

Step 3: Use the formula with n = 15 to find the 15th term, a 15 .

a n � a 1 � � n � 1 � d Write the formula.

a 15 � a 1 � � 15 � 1 � d Substitute n �15.

a 15 � 10 � � 14 � � �6 � Substitute a 1 � 10 and d � �6.

a 15 � �74 Simplify.

The 15th term of the sequence is �74.

PracticeComplete the steps to find the 10th term of each arithmetic sequence.

1. 5, 13, 21, 29, 37, … 2. 7, 4, 1, �2, �5, … d � 8 a 1 � 5 n � 10 d � �3 a 1 � 7 n � 10 The 10th term is 77 . The 10th term is �20 .Find the 12th term of each arithmetic sequence.

3. 3, 15, 27, 39, 51, … 4. 10, 2, �6, � 14, � 22, … 135 �78

ReteachingFinding Arithmetic Sequences 92


Reteachingcontinued

If you know any two terms in an arithmetic sequence, you can find any other term in the sequence.

• Find the common difference by using the two terms and the formula for the n th term.

• Then use the formula for the n th term to find the first term and the n th term.

Find a 12 for an arithmetic sequence, given that a 3 � 33 and a 9 � 117.

Step 1: Use the known terms and the formula for the n th term to find the common difference.

a n � a 1 � � n�1 � d Write the formula.

a 9 � a 3 � � 9 � 3 � d Let a n � a 9 and a 1 � a 3 .

a 9 � a 3 � 6d Simplify.

117 � 33 � 6d Substitute a 9 � 117 and a 3 � 33.

14 � d Solve for d.

Step 2: Use one of the known terms and the common difference, d � 14, to find a 1 .Use a 3 � 33 and the formula for the nth term.

a n � a 1 � (n � 1)d Write the formula.

a 3 � a 1 � (3 � 1)(14) Let a n � a 3 , so n � 3 and d � 14.

a 3 � a 1 � (2)(14) Simplify.

33 � a 1 � 28 Substitute a 3 � 33.

5 � a 1 Solve for a 1 .

Step 3: Use a 1 � 5, d � 14, n � 12 in the formula for the n th term to find a 12 .

a n � a 1 � (n � 1)d Write the formula.

a 12 � 5 � (12 � 1)(14) Substitute a 1 � 5, d � 14, and n � 12.

a 12 � 5 � (11)(14) Simplify.

a 12 � 159 Solve for a 12 .

PracticeComplete the steps to find a 10 for the arithmetic sequence, given that a 4 � 34 and a 6 � 52. 5. Find d. Let a n � a 6 and a 1 � a 4 .

a 6 � a 4 � (6 � 4)d

d � 9

6. Find a 1 .

Let a n � a 4 a 4 � a 1 � (4 � 1)( 9 )

a 1 � 7

7. Find a 10 .

n � 10 a 10

� 7 � ( 10 � 1)(9 )

a 10 � 88

8. Find for a 8 the arithmetic sequence, given that a 4 � 27 and a 10 � 75.

59

92

Since a n � a 9 , n � 9 in the formula. Replace 1 with 3 since a 1 � a 3 .

Name Date Class


You have solved expressions with exponents. Now you will solve exponential equations and inequalities.

When an initial amount, a, increases or decreases by a constant rate, r, over a number of time periods, t, this formula shows the final amount, A(t).

A(t ) � a (1 � r ) t

A population is increasing by 12% per year and is now 15,000. If it continues to increase at the same rate, how long will it take to reach 25,500?

Step 1: Let y represent the population at the end of t years. Identify values for a, r, and y.

a � 15,000 r � 12% � 0.12 y � 25,500

Step 2: Substitute values for a, r, and y into the formula.

y � a (1 � r ) t

25,500 � 15,000 (1 � 0.12) t

25,500 � 15,000 (1.12) t Simplify (1 � 0.12).

1.7 � (1.12) t Divide both sides by 15,000.

log 1.7 � log (1.12) t Take the common logarithm of both sides.

log 1.7 � t log(1.12) Apply the Power Property of Logarithms.

log1.7

_______ log1.12

� t Divide both sides by log 1.12

t � 4.68 years Use a calculator to approximate.

PracticeComplete the steps to solve the problem. 1. The population of a city is increasing a � 40,000 r � 0.08 y � 60,000 by 8% per year and is now 40,000. If it y � a(1 � r ) t

continues to increase at the same rate, 60,000 � 40,000(1 � 0.08 ) t

how long will it take to reach 60,000? 1.5 � (1.08 ) t

log1.5 � t log(1.08)

t � 5.27 yearsSolve.

2. The enrollment at a high school is increasing 5% per year and is now 4000. If it continues to increase at the same rate, how long will it take to reach 5000? approximately 4.57 years

93ReteachingSolving Exponential Equations and Inequalities

a is the initial amount.

The rate, r, is usually a percent.

Time, t, is measured in years

A(t ), the final amount, isa function of time t.



The base of the natural logarithm, e, appears in the formula for interest compounded continuously.

A � Pe rt

A � total amount

P � principal, or initial amount

r � annual interest rate

t � time in yearsIf $2000 is invested at 8% compounded continuously, how long would it take for the value of the investment to reach $3500?

Step 1: Identify the values that correspond to the variables in the formula.

P � 2000 r � 8% � 0.08 A � 3500

Step 2: Substitute the known values into the formula.

A � Pe rt

3500 � 2000e 0.08t Substitute the known values.

1.75 � e 0.08t Divide both sides by 2000.

ln1.75 � ln e 0.08t Take the natural logarithm of both sides.

ln 1.75 � 0.08t Apply the Power Property ln e x � x.

ln 1.75 ______ 0.08

� t Divide both sides by 0.08

t � 7 years Use a calculator to approximate.

PracticeComplete the steps to solve the problem. 3. If $6000 is invested at 8% compounded continuously, how long

would it take for the value of the investment to reach $9600?

P � 6000 r � 0.08 A � 9600 A � Pe rt

9600 � 6000 e 0.08t

1.6 � e 0.08t

ln1.6 � ln e 0.08t

ln1.6 � 0.08t

t � 5.88 years

Solve.

4. If $5000 is invested at 6% compounded continuously, how long would it take for the value of the investment to reach $6000?approximately 3.04 years

5. What is the total amount for an investment of $4000 invested at 5% interest and compounded continuously for 5 years? $5136.10

Name Date Class


You have solved linear inequalities. Now you will solve rational inequalities. A rational inequality is an inequality that contains at least one rational expression.

When a rational inequality is solved using the LCD, two cases result: the LCD is either positive or negative.

Solve 10 _____ x � 5

�� 2 using the LCD.

Solution: Case 1: The LCD is positive.

Step 1: The LCD is x � 5.

Since it is positive, x � 5 � 0. So x � 5.

Step 2: Multiply both sides of the inequality by x � 5.

(x � 5) 10 ______ (x � 5)

� 2(x � 5)

10 � 2x � 10 Simplify.

20 � 2x Solve for x. x � 10

The solution must satisfy x � 5 from Step 1 and x � 10 from Step 2, so, 5 � x � 10.

Case 2: The LCD is negative.

Step 1: The LCD is x � 5.

Since it is negative, x � 5 � 0. So x � 5.

Step 2: Multiply both sides of the Inequality by x � 5 and reverse the sense of the inequality.

(x � 5) 10 _____ x � 5

� 2(x � 5)

10 � 2x � 10

20 � 2x x � 10The solution must satisfy x � 5 from Step 1 and x � 10 from Step 2. However, no number can be both smaller than 5 and larger than 10, so there is no solution for Case 2.

Combining the solutions from both cases, the solution to 10 _____ x � 5

� 2 is 5 � x � 10.

PracticeComplete the steps to solve the rational equation using the LCD.

1. 3 _____ x � 1

� 3 The LCD is x � 1.

Case 1: x � 1 � 0, so x � 1 Case 2: x � 1 � 0, so x � 1

(x � 1) 3 _____ x � 1

� 3(x � 1) (x � 1) 3 _____ x � 1

� 3(x � 1)

3 � 3x � 3 3 � 3x � 3 6 � 3x 6 � 3x x � 2 x � 2 x must satisfy x � 1 and x � 2, so x � 2. x must satisfy x � 1 and x � 2, so x � 2.

Combine the solutions from Case 1 and Case 2 get the solution: x � 2 or x � 2.

Solve the following rational inequalities using the LCD.

2. 6 _____ x � 4

� 2 �4 � x � �1 3. 6 _____ x � 1

� �3 �3 � x � �1

94ReteachingSolving Rational Inequalities



The sign of a rational expression can be determined by finding the sign of the numerator and the sign of the denominator. This can be used as another method for solving rational inequalities.

Use a sign table to solve the rational inequality x � 2 _____ x � 9

�� 0.

Solution:

The right side of the inequality is already zero. Find the values for which the numerator and denominator are equal to zero. The numerator is zero when x � 2. The denominator is zero when x � �9. This gives three intervals to test: less than �9, between �9 and 2 inclusive, and greater than or equal to 2. We exclude x � �9 because the expression is undefined there.

x � �9 �9 � x � 2 x � 2

Numerator: x � 2 � � �

Denominator: x � 9 � � �

Sign of x � 2 _____ x � 9

� � �

Since the rational expression is positive for x � �9 and x � 2, and , the solution to the rational inequality is x � �9 or x � 2.

PracticeComplete the steps to solve the rational inequality using a sign table.

4. x � 1 _____ x � 8

� 0 The numerator is zero when x � �1 and the denominator is zero when x � �8.

x � �8 �8 � x � �1 x � �1

Numerator: x � 1 � � �

Denominator: x � 9 � � �

Sign of x � 2 _____ x � 9

� � �

Since the rational expression is negative when �8 � x � �1, the solution to the rational inequality is �8 � x � �1.

Use a sign table to solve the following rational inequalities.

5. 5x _____ x � 2

� 0 6. 2x � 4 ______ x � 5

� 0

x � 0 0 � x � 2 x � 2

Numerator � � �

Denominator � � �

Expression � � �

x � �2 �2 � x � 5 x � 5

Numerator � � �

Denominator � � �

Expression � � �

The solution is 0 � x � 2. The solution is x � �2 or x � 5.

Name Date Class


You have used synthetic division to divide polynomials. Now you will evaluate polynomials and find roots of polynomials using the Remainder and Factor Theorems.

Remainder Theorem

For a polynomial P(x),

if P(x) � (x � a)Q(x) � r(x) where Q(x) is the quotient and r(x) is the remainder,

P(x) can easily be evaluated for any value of a because when x � a,

P(a) � r(a).

Use the Remainder Theorem to evaluate the polynomial P(x) � 3x 4 � 6x 3 � x 2 � 8x � 2 for x � 4.

Solution: Use synthetic division to divide P(x) by (x � 4) and find r(x). Use the Remainder Theorem to evaluate r(4).

4 3 �6 �1 8 �2 12 24 92 400 By the Remainder Theorem, P(4) � r(4) � 398.

3 6 23 100 398

Practice Complete the steps to evaluate the polynomial using the Remainder Theorem.

1. Evaluate the polynomial P(x) for x � 3. 2. Evaluate the polynomial P(x) for x � 2

P(x) � 6x 4 � 4x 3 � 3x 2 � x �2 P(x) � 2x 4 � 4x 3 � 6x 2 � 5x � 10

3 6 4 �3 1 �2

18 66 189 570 6 22 63 190 568

2 2 4 �6 �5 10

4 16 20 30 2 8 10 15 40

P(3) � r(3 ) � 568 P(2) � r(2 ) � 40

Use the Remainder Theorem to evaluate the polynomial for the given value of x.

3. Evaluate the polynomial P(x) for x � �2. 4. Evaluate the polynomial P(x) for x � 3.

P(x) � � 4x 4 � 5x 3 � 2x 2 � 6x � 12 P(x) � � 3x 4 � 4x 3 � 3x 2 � 5x � 3

P(�2) � �72 P(3) � �312

95ReteachingUsing the Remainder and Factor Theorems


Reteachingcontinued

Factor Theorem

For a polynomial P(x),

if x � a is a root of P(x) then P(a) � 0

Use the Factor Theorem to determine if the given values are roots of the polynomials.

a. Determine if x � �4 is a root of P(x) � 5x 4 � 16x 3 � 15x 2 � 8x � 16.

Solution:

Use synthetic division to test if r (�4) � 0. Divide P(x) by x � 4.

�4 5 16 �15 8 16 �20 16 �4 �16 5 �4 1 4 0

Since P(�4) � r(�4) � 0, x � �4 is a root ofP(x).

b. Determine if x � 1 is a root of P(x) � 2 x 4 � 3 x 3 � 3 x 2 � 5x � 7.

Solution:

Use synthetic division to test if r(1) � 0 . Divide P(x) by x � 1.

1 2 �3 3 �5 7 2 �1 2 �3 2 �1 2 �3 4

Since P(1) � r(1) � 0, x � 1 is not a root of P(x).

PracticeComplete the steps to determine if x � 3 is a root of P(x).

5. P(x) � �3x 4 � 5x 3 �2x 2 � 5x �4 6. P(x) � �x 4 � 3x 3 � 4x 2 � 4x � 24

3 �3 5 �2 5 �4 �9 �12 �42 �111 �3 �4 �14 �37 �115

3 �1 3 4 �4 �24

�3 0 12 24

�1 0 4 8 0

P(3) � r(3) � 0 , so 3 is not a root of P(x). P(3) � r(3) � 0, so 3 is a root of P(x).

Determine if x � 2 is a root of P(x).

7. P(x) � 3x 4 � 8x 3 � 5x 2 �4 8. P(x) � � 4x 4 � 4x 3 � 8x 2 � 2x � 3

P(2) � r(2) � 0, so 2 is a root of P(x) P(2) � r(2) � 0, so 2 is not a root of P(x)

9. The total number of dollars donated each year to small charitable organizations has followed the trend d(t) � 2t 3 � 10t 2 � 2000t � 10,000, where d is dollars and t is the number of years since 1990. How much money was donated in the year 2000? (Hint: Find d(10).)

$33,000

95

Name Date Class


You have used the unit circle to find radian measure. Now you will use radian measure to write and plot polar coordinates.

A polar coordinate is a point P written (r, �), where r is the directed distance from O to P and � is the angle measure from the polar axis to _

OP .

Converting Coordinates

Polar to Cartesian

x � r cos �

y � r sin �

Cartesian to Polar

tan � � y __ x

r 2 � x 2 � y 2

Convert (3, �) to Cartesian coordinates.

Solution:

x � r cos � � 3 cos � � 3(�1) � �3

y � r sin � � 3 sin � � 3(0) � 0

The Cartesian coordinates are (�3, 0).

Convert (2, 2) to polar coordinates.

Solution:

tan � y __ x � 2 __

2 � 1

(2, 2) is in quadrant I, so � � � __ 4

r 2 � x 2 � y 2 � (2) 2 � (2) 2 � 8

r � ��

8 � 2 ��

2

The polar coordinates are � 2 ��

2 , � __ 4

� .

PracticeComplete the steps to convert between Cartesian and polar coordinates.

1. Convert � �2, � __ 2 � to Cartesian coordinates.

x � r cos � � �2 cos � __ 2 � �2(0) � 0

y � r sin � � �2 sin � __ 2 � �2(1) � �2

The Cartesian coordinates are (0, �2).

2. Convert (�5, 0) to polar coordinates.

tan � � y __ x � 0 ___

�5 � 0

(�5, 0) is on the negative x-axis, so � � �

r 2 � x 2 � y 2 � (�5) 2 � 0 2 � 25

r � 5

The polar coordinates are (5, � ).

Convert between Cartesian and polar coordinates.

3. Convert � 1, 5� ___ 4 � to Cartesian coordinates.

� � ��

2 ___ 2 , � �

� 2 ___

2 �

4. Convert (2 ��

3 , 2) to polar coordinates.

� 4, � __ 6 �

96ReteachingUsing Polar Coordinates


Reteachingcontinued

Graph the given polar equations. Check by converting to a Cartesian equation.

a. Graph r � 5.

Solution: The graph consist of all the points 5 units from π

2

5 0π

3π2

the pole. The pole is at the origin, so the graph is a circle centered at origin with a radius of 5.

Check: r � 5

r 2 � 25 Square both sides.

x 2 � y 2 � 25 Use r 2 � x 2 � y 2 .

This is the Cartesian equation of the same circle.

b. Graph � � � __ 4

.

Solution: The graph consist of all the points on the line that π

2

π

4 0π

3π2

makes an angle of � __ 4 radians with the positive x-axis.

Check: � � � __ 4

tan � � tan � __ 4 Take the tangent of both sides.

y __ x � 1 tan � �

y __ x and tan � __

4 � 1

y � x Multiply both sides by x.

y � x is the Cartesian equation of the line.

PracticeComplete the steps to graph the polar equation.

5. Graph r � 1. π

2

0π

3π2

1

The graph consists of all points 1 unit(s) from the origin. Therefore, it is a circle of radius 1 that is

centered at the origin.

Convert the following polar equations to Cartesian equations.

6. � � 3� ___ 4 7. r � �

� 2

y � �x x 2 � y 2 � 2

96

Name Date Class


You have found arithmetic sequences. Now you will find geometric sequences. In a geometric sequence the ratio between consecutive terms is the same. This is called the common ratio, or r where r � 1.

Determine whether the following sequences are geometric. If so, then determine the common ratio and find the next 3 terms.

a. 3, 9, 27, 81, 243, …

Solution:

Find the ratios of consecutive pairs of terms:

9 __ 3 � 3 27 ___

9 � 3 81 ___

27 � 3 243 ____

81 � 3

The common ratio is 3. The sequence is geometric.

The next three terms in the sequence are 3(243) � 729, 3(729) � 2187, and 3(2187) � 6561.

b. 10, 60, 110, 160, 210, …

Solution:

Find the ratios of consecutive pairs of terms:

60 ___ 10

� 6 110 ____ 60

� 1.83 160 ____ 110

� 1.45

210 ____ 160

� 1.31

There is no common ratio, so the sequence is not geometric.

PracticeComplete the steps to determine whether the given sequence is geometric. If so, state the common ratio.

1. 5, �10, 20, �40, 80, �160, …

Ratios: �10 ____ 5 � �2 20 ____

�10 � �2 �40 ____

20 � �2

80 _____

�40 � �2 �160 _____

80 � �2

The sequence is geometric. The common ratio is �2 . Determine whether the following sequences are geometric. If so, state the common ratio.

2. 1, 2, 4, 7, 11, 16, …

not geometric

3. 2, �6, 18, �54, …

geometric; common ratio is �3

4. 100, 50, 25, 12.5, 6.25, …

geometric; common ratio is 1 __ 2

5. 21, 13, 5, �3, �11, …

not geometric

97ReteachingFinding Geometric Sequences


Reteachingcontinued

Geometric Sequences

If you know the first term of a geometric sequence a 1 and the common ratio r, then you can

find the nth term a n using the following rule:

a n = a 1 r n � 1

Find the 10 th term of the geometric sequence: 3, 12, 48, 192, 768, …

Step 1: Find the common ratio, r.

r � 12 ___ 3

� 4


a 1 � 3

Step 3: Use the formula with r � 3 to find the 10 th term, a 10 .

a n � a 1 r n � 1 Write the rule.

a 10 � a 1 r 10 � 1 Substitute n � 10.

a 10 � 3 (4) 9 Substitute a 1 � 3 and r � 4.

a 10 � 3(262,144) � 786, 432 Simplify.

The 10 th term of the sequence is 786,432.

PracticeComplete the steps to find the 8 th term of the geometric sequence.

6. 8, 24, 72, 21, 648, …

r � 24 ___ 8 � 3 and a 1 � 8

a n � a 1 r n � 1

a 8 � a 1 r 8 � 1

a 8 � 8 (3) 7

a 8 � 8(2187)

a 8 � 17,496

The 8 th term of the sequence is 17,496.

Find the 10 th term in the following geometric sequences.

7. �7, 14, �28, 45, �112, …

a 10 � 3584

8. 1600, 800, 400, 200, …

a 10 � 3.125

9. In his first year on the job, Matthew earned a salary of $20,000. He earned a 5% raise each year. How much money did he earn in his tenth year? Round your answer to the nearest hundredth. (Hint: His yearly salaries can be written as a geometric sequence with a 1 � 20,000 and r � 1.05.)

$31,026.56

97

Name Date Class


You have created graphs and solved equations of circles. Now you will work with ellipses. An ellipse is the set of points in the plane for which the sum of the distances from two fixed points—called foci—is the same number.

The standard form for the equation of an ellipse with center (0,0) is

x 2 ___ a 2

� y 2 ___ b 2

� 1.

The vertices are (a, 0), (�a, 0), (0, b), and (0, �b).

x

(a, 0)

(0, b)(-a, 0)

(0, -b)

y

O

Write the equation of the ellipse with center (0, 0) and vertices (2, 0) and (0, 5).

Step 1: Identify the values of a and b from the vertices.

The vertices are (2, 0) and (0, 5), so a � 2 and b � 5.

Since 5 � 2, the major axis of this ellipse is vertical.

Step 2: Substitute a � 2 and b � 5 into the equation in standard form, x 2 ___ a 2

� y 2

___ b 2

� 1.

x 2 ___ 2 2

� y 2

___ 5 2

� 1 → x 2 __ 4 �

y 2 ___

25 � 1

PracticeComplete the steps to write an equation in standard form for each ellipse with center at (0, 0).

1. Vertices (4, 0) and (0, 6) 2. Vertices at (7, 0) and (0, 3)

a � 4 and b � 6 a � 7 and b � 3

x 2 ___ 4 2

� y 2

___ 6 2

� 1 → x 2 _____ 16

� y 2

___ 36

� 1 x 2 ___ 7 2

� y 2 ___ 3

2 � 1 → x 2 ______

49 �

y 2 ____

9

� 1

Since 6 � 4, the major axis is vertical . Since 7 � 3, the major axis is horizontal.

Write an equation in standard form for each ellipse with center (0, 0).

3. Vertices (10, 0) and (0, 8) 4. Vertices (3, 0) and (0, 1)

x 2 ____

100 � y 2 ___

64 � 1

x 2 __ 9 � y 2 __

1 � 1

98ReteachingMaking Graphs and Solving Equations of Ellipses

x

(a, 0)

(0, b)

(-a, 0)

(0, -b)

y

O

If a � b, then the major axis is horizontal.

If b � a, then the major axis is vertical.


Reteachingcontinued

The standard form for the equation of an ellipse with center (h, k) is

(x � h) 2 _______ a 2

� (y � k) 2

_______ b 2

� 1.

The vertices are (h, k � b) and (h � a, k).

Graph the ellipse (x � 4) 2

_______ 25

� (y � 3) 2

_______ 49

� 1.

Step 1: Write the equation in standard form.

(x � 4) 2

_______ 25

� (y � 3) 2

_______ 49

� 1 → (x � 4) 2

_______ 5 2

� (y � 3) 2

_______ 7 2

� 1

Step 2: Identify h, k, a, and b from the equation.

(x � h) 2

_______ a 2

� (y � k) 2

_______ b 2

� 1 → (x � 4) 2

_______ 5 2

� (y � 3) 2

_______ 7 2

� 1

So, h � 4, k � 3, a � 5, and b � 7.

Step 3: Find the vertices and the center of the ellipse.

Center (h, k) is (4, 3).

Vertices:

(h, k � b):(4, 3 � 7) → (4, 10) and (4, �4)

(h � a, k):(4 � 5, 3) → (9, 3) and (�1, 3)Step 4: Graph.

PracticeComplete the steps to graph the ellipse.

5. (x � 2) 2

_______ 16

� (y � 1) 2

_______ 9 � 1

h � 2 , k � 1, a � 4, and b = 3

Center: (h, k) � (2, 1)

Vertices:

(h, k � b) � (2, 4) or (2, �2)

(h � a, k) � (6, 1) or (�2 ,1)

Graph the ellipse.

6. (x � 4) 2

_______ 36

� (y � 1) 2

_______ 25

� 1

98

x(9, 3)

(4, 3)

(4, 10)

(-1, 3)

(4, -4)

y

O

8

4

-4

-4-8

-8

Plot the center and plot the points for the vertices.Then draw the ellipse through the vertices.

x

y

O

8

4

4 8

-4

-4-8

-8

(6, 1)

(2, 4)

(2, 1)(-2, 1)

(2, -2)

x

y

O

8

4

84 12

-8

-4

-4

(10, 1)

(4, 6)

(4, 1)(-2, 1)

(4, -4)

Name Date Class


You have plotted points on the Cartesian plane. Now you will use vectors to describe points.

A vector is a quantity that has magnitude and orientation.

A vector can be expressed as a complex number and as a matrix. Vectors can be graphed on both the Cartesian and complex planes.

Graph a vector _ › z whose endpoints are the origin and (5, 6). Express the Cartesian

version as a coordinate matrix and the complex version as a complex number. Find the angle of orientation of the vector.

Solution

Cartesian Coordinate System Complex Coordinate System

x

y

θ= 50.19°

z= [56]

O

6

8

21

21 4 6 73 5 8

(5, 6)

4

7

3

5→

x

y

θ= 50.19°

z= 5 + 6i

O

6

8

21

21 4 6 73 5 8

(5, 6)

4

7

3

5→

_ › z � [ 5

6 ]

_ › z � 5 � 6i

The angle of orientation is � � tan �1 y __ x , so � � tan �1 6 __

5 � 50.19�.

PracticeComplete the steps to express the vector with the given endpoints as a matrix and a complex number. Find the angle of orientation.

1. Endpoints at origin and (5, 2). 2. Endpoints at origin and (3, 8).

matrix complex number matrix complex number

[ x __ y ] � [ 5 __ 2

] x + yi � 5 � 2i [ x y ] � [ 3 __ 8 ] x + yi � 3 � 8i

� � tan �1 y __ x � tan �1 2 __

5 � 21.8� � � tan �1

y __ x � tan �1

8 __ 3 � 69.44�

Express the vector with the given endpoints in the indicated form. Then find the angle of orientation.

3. (1, 3) as a matrix 4. (2, 7) as a complex number

_ › z � [ 1

3 ] ; � � 71.57�

_ › z � 2 � 7i; � � 74.05�

99ReteachingUsing Vectors


Reteachingcontinued

Vector Addition Vector Subtraction

[ x 1 y 1 ] � [ x 2

y 2 ] � [ x 1 � x 2 y 1 � y 2 ]

(a � bi ) � (c � di ) � (a � c) � (b � d )i

[ x 1 y 1 ] � [ x 2

y 2 ] � [ x 1 � x 2 y 1 � y 2 ]

(a � bi ) � (c � di ) � (a � c) � (b � d )i

x

y

A + B

O

6

8

21

21 4 6 73 5 8

4

7

3

5 → →

B→

A→ x

y

21

21 43 5

43

5

A - B→ →

B→

A→

-4-5 -2-3 -1

-4-5

-2-3

-1

If _

› a � (1, 2) and

_ › b � (3, 5), find the vector sum

_ › a �

_ › b using matrix and complex

addition.

a. [ 1 2

] � [ 3 5

] � [ 1 � 3 2 � 5

] � [ 4 7

] b. (1 � 2i) � (3 � 5i) � (1 � 3) � (2 � 5)i � 4 � 7i

If _

› u � (6, 2) and

_ › v � (3, 2), fi nd the vector difference

_ › u �

_ › v using matrix and

complex subtraction.

c. [ 6 2 ] � [ 3

2 ] � [ 6 � 3

2 � 2 ] � [ 3

0 ] d. (6 � 2i) � (3 � 2i) � (6 � 3) � (2 � 2)i

� 3 � 0i

� 3

PracticeComplete the steps to add or subtraction the vectors.

5. If _ › a � (2, 4) and

_ › b � (5, 3), find

_ › a �

_ › b 6. If

_ › a � (6, 2) and

_ › b � (1, 1), find

_ › a �

_ › b ,

using matrix addition. using complex subtraction.

[ 2 4

] � [ 5 3 ] � [ 2 � 5

4 � 3

] � [ 7 __ 7

] (6 � 2i ) � (1 � 1i ) � (6 � 1 ) � (2 � 1)i

� 5 � 1i

Add or subtract the vectors as indicated.

7. If _ › a � (1, 7) and

_ › b � (9, 6), find

_ › a �

_ › b 8. If

_ › u � (2, 8) and

_ › v � (1, 6), find

_ › u �

_ › v

using complex addition. using matrix subtraction.

10 � 13i [ 2 8

] � [ 1 6

] � [ 1 2

]

99

Name Date Class


The graph of g(x) is the graph of f(x) � 1 __ x

translated 2 units right and 3 units down.

You have solved rational equations. Now you will graph rational functions.

A rational function is a function that can be written as the ratio of two polynomials.

One example of a rational function can be written in the form f (x) � a _____ x � h

� k.

This function has a vertical asymptote at x � h and a horizontal asymptote at y � k. The domain is {x | x � h} and the range is {y|y � k}.

Graph g(x) � 1 _____ x � 2

� 3.

Step 1: Identify h and k from the equation: h � 2 and k � �3

There is a vertical asymptote at x � 2.

There is a horizontal asymptote at y � �3.

Step 2: Draw the asymptotes as dotted lines.

Step 3: Sketch the graph.

PracticeComplete the steps to graph the rational function.

1. g(x) � 1 _____ x � 1

� 2

h � �1 and k � �2

vertical asymptote at x � �1

horizontal asymptote at y � �2

Graph the rational function.

2. g(x) � 1 _____ x � 1

� 3

100

x

y

42

42 86 10

86

-8-10 -4-6 -2

-8

-4-6

-2

-10

10

x

y

42

4 86 10

86

-8-10 -4-6 -2

-8

-4-6

-2

-10

2

10

x

y

42

4 86 10

86

-8-10 -4-6 -2

-8

-4-6

-2

-10

10

2

ReteachingGraphing Rational Functions I


Reteachingcontinued

A rational function can be written in the form f (x) � p(x)

____ q(x)

, for

polynomials p(x) and q(x) � 0. The zeros of f (x) occur where p(x) � 0. The vertical asymptote(s) of f (x) occur where q(x) � 0.

Graph f(x) � x 2 � 2x � 8 ___________ x � 1

.

Step 1 Find the zeros.

Factor the numerator: x 2 � 2x � 8 � (x � 2)(x � 4).

The zeros occur at x � �2 and x � 4.

Step 2 Find the vertical asymptote(s).

x � 1 � 0 at x � 1

Step 3 Graph.

Plot the zeros at (�2, 0) and (4, 0).

Draw the vertical asymptote at x � 1.

Make a table of values and plot the points.

x �8 0 2 10

y �8 8 �8 8

Practice Complete the steps to graph the rational function.

3. f(x) � x 2 � x � 12 __________ x � 2

Zeros:

x 2 � x � 12 � (x � 4)(x � 3) � 0

x � 4 and x � �3

Vertical Asymptote:

x � 2 � 0 at x � 2

Graph the rational function.

4. f(x) � x 2 � x � 6 _________ x � 1

100

x y

�8 �6

�2 1.5

7 6

x

y

42

4 86 10

86

-8-10 -4-6 -2

-8

-4-6

-2

-10

10

2

x

y

42

4 86 10

86

-8-10 -4-6 -2

-8

-4-6

-2

-10

10

2

x

y

42

4 86 10

86

-8-10 -4-6 -2

-8

-4-6

-2

-10

10

2

Name Date Class


INV

10

You have plotted points on the polar coordinate plane. Now you will graph and analyze polar equations.

A polar coordinate system in a plane is formed by a fixed point, O called the pole, and a ray, called the polar axis, whose endpoint is O.

Polar Coordinates

Every point P in the polar coordinate system has an ordered pair of polar coordinates (r, �), defined as follows:

r is the directed distance from O to P.

� is the directed angle measure from the polar axis to _

OP .

One of the simplest polar equations is r � 1. Its graph is the circle of radius one, with center at the origin.

1. This equation does not include the variable �. How is this equation similar to the Cartesian equation y � 1?

In each case the variable is equal to a constant. 2. How do the equation and the graph conform to the definition

of a circle?

The radius of the circle is constant.

One of the simplest polar equations that uses the variable �is r � �. Its graph is a spiral.

3. How does the value of r change as the value of � increase?

The value of r increases as � increases.

To graph polar equations with your graphing calculator, click on the MODE key and select the Pol option. Once in this mode, clicking the Y� key allows you to input an equation in r � f(�) form. Use the interval � � [0, 3�] for your graph.

4. Use your graphing calculator graph r � c� for different values of the constant c.

a. For what values of c is the spiral narrower than the graph of the equation r � �? (Hint: Try c � 1, 0 � c � 1, and c � 0)

0 � c � 1 b. For what values of c is the spiral wider than the graph of the equation

r � �? (Hint: Test the intervals given in part a.)

c � 1

ReteachingGraphing Polar Models

r = 1π

4θ =

π

24

2

2 4

-2

-2-4

-4

Oπ 2π

3π2

x

π

2y8

4

4 8

-4

-4-8

-8

Oπ 2π

3π2

Reteachingcontinued

INV

10


For equations of the form r � a� � b, the graph is also a spiral. The graphs pictured are r � � and r � � � 2.

5. Describe the role of a and b in the graph of r � a� � b. (Hint: Think about what you found in question 4 and translations.)

The coefficient a determines the curvature of the spiral and b is the radial translation of the graph.

To draw a graph that does not spiral outwards, you must use trigonometric functions, which cycle through the periodic nature of �. The graph of r � sin(�) is pictured.

6. Describe the graph. How is it a periodic graph? A circle of radius

0.5 centered at � 0.5, � __ 2 � . For increasing values of

�, the values of r through the same values.

Trigonometric equations of the form r � a sin(b�) � c can result in graphs of different shapes depending on the values of a, b, and c. The graph of r � sin(�) � 1 is a heart-shaped curve known as a cardioid. The graph of r � 3 sin(2�) is called a rose curve.

7. Use a graphing calculator to graph r � a sin(b�) � c for different values of a, b, and c, and generalize to answer the following questions.

a. If a � 0, b � 1, and c �0, what shape will the graph be? circle b. If a � 0, b � 1, and c � 0, what shape will the graph be and where will it

be located? Circle; below the horizontal axis c. What shape is the graph if a � 0, b is an integer, and c � 0? rose curve d. If a � 0, c � 0, and b is an odd integer the graph will be a rose curve.

How many petals will the rose curve have? There will be b petals.

π

28

4

4 8

-4

-4-8

-8

Oπ 2π

3π2

r = θ + 2r = θ

r = 0.5

π

24

2

2 4

-2

-2-4

-4

Oπ 2π

3π2

π

20.5, =

( )

r = sin(θ) + 1

π

24

2

2 4

-2

-2-4

-4

Oπ 2π

3π2

r = 3sin(2θ)

π

24

2

2 4

-2

-2-4

-4

Oπ 2π

3π2

Name Date Class


You can use the degree and the leading coefficient of a function to identify that function’s end behavior. The end behavior describes what happens to the function as its x-values become very small, approach negative infinity (→ ��) or very large, approach positive infinity (→ ��).

Describe each polynomial function.

This is a graph of a polynomial function with an even-number degree and a leading coefficient � 0.

as x → ��, P(x) → ��

as x → ��, P(x) → ��

This is a graph of a polynomial function with an even-number degree and a leading coefficient � 0.

as x → ��, P(x) → ��

as x → ��, P(x) → ��

This is a graph of a polynomial function with an odd-number degree and a leading coefficient � 0.

as x → ��, P(x) → ��

as x → ��, P(x) → ��

This is a graph of a polynomial function with anodd-number degree and a leading coefficient � 0.

as x → ��, P(x) → ��

as x → ��, P(x) → ��

PracticeComplete the steps to describe the function.

1. P (x) � 2 x 4 � x 3 � 3x � 1

The degree of this function is 4 or even.

The leading coefficient is 2, which is � 0.

As x → ��, P(x) → ��.

As x → ��, P(x) → ��.

2. f (x) � � x 3 � 5 x 2 � 2x

The degree of this function is 3 or odd.

The leading coefficient is �1, which is � 0.

As x → ��, f(x) → ��.

As x → ��, f(x) → ��.

Identify the leading coefficient and degree, and describe the end behavior of each function.

3. P(x) � x 2 � 4x � 1

leading coefficient � 1, deg � 1; as x, → ��, P(x) → ��, as x, → ��, P(x) → ��

4. f (x) � � 3x 3 � 5 x � 1

leading coefficient � �3, deg � 3; as x → ��, f(x) → ��, as x → ��, f(x) → ��

101ReteachingMaking Graphs of Polynomial Functions

x

y4

2

2 4-4 -2

-4

-2

Ox

y4

2

2 4-4 -2

-4

-2

O

x

y4

2

2 4-4 -2

-4

-2

O x

y4

2

2 4-4 -2

-4

-2

O


Reteachingcontinued

Graph the polynomial function.

f (x) � x 3 � 3 x 2 � x � 3.

Step 1: Identify the real zeros of the

equation x 3 � 3x 2 � x � 3 � 0.

According to the Rational Root Theorem, p � �3 and q � 1. The possible rational roots are �1 and �3.

Step 2: Use synthetic division to test the roots until you find a zero.

Test x � 1. 1 1 3 −1 −3 1 4 3 By the

1 4 3 0

results, you can see that x � 1 is a zero, so f (x) � (x � 1) ( x 2 � 4x � 3).

Completely factoring the second term gives you (x � 3)(x � 1). So, the real zeros and x-intercepts are 1, �3, and �1.

Step 3: Make a table of ordered pairs that includes the x-intercepts, the x-values between the x-intercepts, and the y-intercept.

x �3 �2 �1 0 1

y 0 3 0 �3 0

Step 4: Determine the end behavior. The degree is odd and the leading coefficient is � 0, so

as x → ��, f (x) → ��

and as x → ��, f (x) → ��.

Step 5: Plot the points and sketch the graph.

x

y321

21 3-3-1-2-1

-3-2

O

Practice

5. f (x) � x 3 � 3 x 2 � 6x � 8

Step 1: Use the Rational Root Theorem,

p = 8 and q = 1 . The possible rational

roots are �1, �2, �4, and �8.

Step 2: Use synthetic division to find the

real zeros. x = 1 is a zero. The other real

zeros and x-intercepts are �2 and 4.

Step 3: Make a table of points including the x-intercepts, x-values between the x-intercepts, and the y-intercept.

Step 4: Determine the end behavior.

The degree is odd; the leading

coefficient is � 0, so

as x → ��, f (x) → �� and

as x → ��, f (x) → ��.

Step 5: Plot the points and sketch the graph.

x �2 �1 0 1 2 3 4

y �8 �10100 8 0 0

x

y

68

1012

42

21 3 4 5 6-3-4-5-6-2-2-1

-6-8-10-12

-4

O

101

Name Date Class


You have learned about logarithms and their properties. Now you will apply those properties to solve equations and inequalities that contain logarithms.

If lo g b x � lo g b y, then x � y.

b lo g b x � x and lo g b b x � x Inverse Properties of Logarithms and Exponents

lo g b mn � lo g b m � lo g b n Product Property of Logarithms

lo g b m __ n � lo g b m � lo g b n Quotient Property of Logarithms

lo g b a p � p lo g b a Power Property of Logarithms

Solve lo g 3 � x � 1 � 2 � .lo g 3 � x � 1 � � 2 3 2 � x � 1 Definition of a logarithm

x � 3 2 � 1 Subtract 1 from both sides. x � 9 � 1 Simplify.

x � 8

Solve lo g 2 x 4 � 8.

lo g 2 x 4 � 84lo g 2 x � 8 Power Property of Logarithms lo g 2 x � 2 Divide both sides by 4. x � 2 2 Definition of a logarithm x � 4

Solve log x � log � x � 2 � � log 8. log x � log � x � 2 � � log 8 log [ x � x � 2 � ] � log 8 Product Property of Logarithms x(x � 2) � 8 If lo g b x � lo g b y, then x � y. x 2 � 2x � 8 Distributive Property of Multiplication x 2 � 2x � 8 � 0 Substract 8 from both sides. � x � 4 � � x � 2 � � 0 Factor. x � 4 � 0 or x � 2 � 0 If ab � 0, then a � 0 or b � 0 or both a and b � 0. x � 4 or x � � 2A logarithmic function is undefined for negative numbers. Therefore, the only solution is x �4.

Practice 1. log x � log � x � 2 � � log 3

log [ x � x � 2 � ] � log 3 product Property of Logarithms.

x � x � 2 � � 3 If lo g b x � lo g b y, then x � y.

x 2 � 2x � 3 Distributive Property of Multiplication

x 2 � 2x � 3 � 0 Substract 3 from both sides.

� x � 3 � � x � 1 � � 0 Factor.

x � 3 � 0 or x � 1 � 0 Zero Product Property

x � � 3 or x � 1 The only solution is x � 1.

Solve each logarithmic equation.

2. lo g 4 (x � 2) � 3 x � 66 3. log x � log(x � 4) � log32 x � 8

102ReteachingSolving Logarithmic Equations and Inequalities



You can use the properties of logarithms to solve logarithmic inequalities.

Solve log8x � log2 � log10.

log8x � log2 � log10

log � 8x ___ 2

� � log10 Quotient Property of Logarithms

log4x � log10 Simplify.

4x � 10 If lo g b x � lo g b y, then x � y.

x � 5 __ 2 Divide both sides by 4 and Simplify.

A logarithm function is undefined for negative numbers, therefore, 0 � x � 5 __ 2 .


4. log10x � log5 � log12

log � 10x _____

5 � � log12 Quotient Property of Logarithms

log 2x � log12 Simplify.

2x � 12 Divide both sides by 2x.

x � 6 5. The population of a certain city grows at the rate of 2% per year.

Approximately how many years will it take for the population to triple?

Use the equation 3 q 0 � q 0 e 0.02t and solve for t.

3 q 0 � q 0 e 0.02t

3 � e 0.02t Divide both sides by q 0 . ln 3 � ln e 0.02t Take the natural log of both sides.

ln 3 � 0.02 t � ln e Power Property of Logarithms

ln 3 � 0.02 t � 1 ln e � 1

t � ln3

_____ 0.02

Solve for t.

t � 1.099

______ 0.02

Use a calculator to approximate the natural log.

t � 55 years

Solve. 6. log2x � log4 � log7 7. log16x � log8 � log11

x � 14 0 � x � 11 ___ 2

Name Date Class


You have graphed the sine, cosine, and tangent functions. Now you will graph the reciprocals of these functions.

csc(x) � 1 _____ sin(x)

sec(x) � 1 ______ cos(x)

The cosecant function is the reciprocal of the sine function. The secant function is the reciprocal of the cosine function. The graph of a cosecant

function, y � a � csc(bx) has a period of 2� ___

� b � and asymptotwes at �n ___

� b � .

The graph of a secant function, y � a � sec(bx) has a period of

2�

___

� b �

and asymptotes at

�

____

2 � b �

�

�n

___

� b �

where n is an integer.

Graph the function y � csc � x __ 4 � and identify its period and asymptotes.

a � 1 and b � 1 __ 4

The period is 2� ___ � b �

� 2� ___ � 1 __ 4 � � 8�.

The asymptotes occur at �n ___ � b �

� �n ___ � 1 __ 4 � � 4�n.

PracticeComplete the steps to identify the period and asymptotes of the function. Then graph the function.

1. y � 2 sec � x __ 2 �

Period: 2� ___ � b �

� 2� ____ � 1 __ 2 �

� 4�

Asymptotes: � ____ 2 � b �

� �n ___ � b �

� � ____ 2 � 1 __

2 � � �n ____

� 1 __ 2 � � � � 2�n.

Graph the function and identify its period and asymptotes.

2. y � 2 csc(2x)

Period: � Asymptote: �n ___ 2

103ReteachingGraphing Reciprocal Trignometric Functions

x

y4

2

2π 4π 6π 8π

1

3

-2-1-2π-4π

-3-4

O

x

y

45

2

π 2π 3π 4π

1

3

-2-1-π-2π

-3-4

O

x

y4

2

π 2π

1

3

-2-1

-3-4

O3π2

-

π

2π

2-π



The reciprocal of the tangent function is the cotangent function:

cot(x) � 1 _____ tan(x)

. The cotangent function is undefined when tan(x) � 0.

The function y � a � cot(bx) has a period of � ___ � b �

and asymptotes at �n ___ � b �

.

Graph the function y � 3 cot � x � and identify its period and asymptotes.

a � 3 and b � 1

The period is � ___ � 1 �

� �.

The asymptotes occur at �n ___ � 1 �

� �n.

PracticeGraph the function and complete the steps to identify its period and asymptotes.

3. y � 2 cot � 2x �

Period: � ___ � b �

� � ___ � 2 �

� � __ 2

Asymptotes: �n ___ � b �

� �n ___ 2

Graph the function and identify its period and asymptotes.

4. y � 2 cot � x __ 2

�

Period: 2� Asymptotes: 2�n

5. An airplane is flying at an altitude of 25,000 feet. Signals are sent to the airplane from the runway at an angle of elevation x. Let y be the horizontal distance from the runway to the point directly below the airplane. Graph the function y � 25,000 cot(x).

x

y12

63

9

-6-3

-9-12

Oπ 2π3π

2-

π

2π

2-π

x

y8

42

6

-4-2

-6-8

Oπ 2π3π

2-

π

2π

2-π

x

y8

42

6

-4-2

-6-8

Oπ 3π-π

x

y

O

80,000

60,000

40,000

20,000

2π

Name Date Class


You have applied translations to parabolas. Now you will apply transformations to square root and exponential functions.One type of transformation is a translation. A translation shifts a graph up, down, left, or right.vertical translation of y � f(x): y � f(x) � a; If a � 0, the graph is shifted a units up. If a � 0, the graph is shifted a units down.horizontal translation of y � f(x): y � f(x � a); If a � 0, the graph is shifted a units left. If a � 0, the graph is shifted a units right.

The graph of y � �� x is shown. Graph the function y � ��

x � 4.

The equation y � �� x � 4 has the form y � f(x) � a,

which is a vertical translation.

f(x) � �� x and a � 4

This means that the graph of y � �� x is

translated 4 units up.

PracticeComplete the steps to identify the transformation and graph the function.

1. y � 2 x�3

The equation has the form y � f(x � a).

f(x) � 2 x a � �3 The graph of y � 2 x�3 is a horizontal translation of 2 x , 3 units right.

Identify the transformation and graph the function.

2. y � ��

x � 2

horizontal translation of y � �� x

2 units left

104ReteachingFinding Transformations

x

y

y = x + 4

O

6

8

10

2

2-2

+4

4 6 8 10

4

-2

y = x

x

y

O

6

8

10

12

2

2 4 6 8

4

y = 2x

-2-4

x

y

O

2

2 4 6 8 10

4

-2

-2



Another way to transform a function is to reflect it about the x- or y-axis.

y � �f(x) is a reflection of f(x) about the x-axis.

y = f(�x) is a reflection of f(x) about the y-axis.

The graph of y � �� x is shown. Graph the function y � � ��

x.

The equation y � � �� x has the form y � �f(x),

where f(x) � �� x .

This means that the graph of y � �� x is

reflected about the x-axis.

PracticeComplete the steps to graph the function.

3. y � 3 �x

The equation has the form y � f(�x).

f(x) � 3 x The graph of y � 3 �x is a reflection of

y � 3 x about the y-axis.

Graph the function.

4. y � �� x

5. A square painting has an area of x inches. The frame surrounding the painting has a width of 2.5 inches. Graph the function y � ��

x � 5 to show the possible side lengths of the frame.

x

y

O

4

6

2-2 4 6 8 10

2

y = - x

-2

-4

-6

y = x

x

y

O

4

6

8

10

2 4

2

y = 3x

-2

-2

-4

-4

x

y

O

2

2 4 6

4

6

-2

-2

-4

-4-6

x

y

O

2

2 4 6 8 10

4

6

8

-2

-2

10

Name Date Class


You have found arithmetic sequences. Now you will find arithmetic series.

An arithmetic series is the sum of the terms in an arithmetic sequence. When the sequence is infinite, you can find partial sums. Summation notation, or sigma notation, is one way to write this.

� k � 1

n

a k � a 1 � a 2 � a 3 �...� a n

This notation means to find the sum of the values in the sequence, replacing k with each number from 1 to n.

Find the sum � k � 1

5 k � 4 .

Replace k with each of the values from 1 to 5 and find the sum.

� k � 1

5

k � 4 � (1 � 4) � (2 � 4) � (3 � 4) � (4 � 4) � (5 � 4)

� 5 � 6 � 7 � 8 � 9 � 35

PracticeComplete the steps to fi nd the sum.

1. � k � 1

6

3k � 3(1) � 3(2) � 3(3) � 3(4) � 3(5) � 3(6)

� 3 � 6 � 9 � 12 � 15 � 18 � 63

Find the sum.

2. � k � 1

4

k � 2 3. � k � 1

5

5k

2 75

4. � k � 1

5

�2k 5. � k � 1

7

k � 3

�30 49

6. � k � 1

6

6 � k 7. � k � 1

5

2k � 3

15 45

105ReteachingFinding Arithmetic Series


Reteachingcontinued

If you know the fi rst and nth term in an arithmetic series, you can fi nd the sum of the fi rst n terms, S n , with a formula.

S n � n � a 1 � a n _______

2 �

Find S15 for the series 2 � 4 � 6 � 8 � . . .

The first term in the series is 2, so and the common difference between each term is 2.

Use a n � a 1 � (n � 1)d, where d is the common difference, to find a 15 .

a 15 � 2 � (15 � 1)(2) � 2 � 28 � 30, so a 15 � 30.

Use the formula, with n � 15, a 1 � 2, and a 15 � 30.

S 15 � 15 � 2 � 30 ______ 2 �

� 15 � 32 ___ 2 �

� 15(16)

� 240

PracticeComplete the steps to find �

k � 1

20 5k � 4 .

8. n � 20

a 1 � 5(1) � 4 � 1 a 20 � 5(20) � 4 � 96

S 20 � 20 � 1 � 96 __________

2 �

� 970

Find the partial sum.

9. S 12 ; 3 � 4 � 5 � 6 � . . . 10. S 16 ; 5 � 12 � 19 � 26 � . . .

102 920

11. � k � 1

25

3k � 1 12. � k � 1

18

6k � 3

1000 972

13. A swimmer swims laps each day for 12 days. The first day, the swimmer swims 10 laps. Each day after, the swimmer does 2 additional laps. How many total laps will the swimmer have done at the end of the 12 days?

252 laps

105

Name Date Class


You have found real roots of polynomial equations. Now you will use the Fundamental Theorem of Algebra to find all roots of polynomial equations.

A polynomial equation of degree n has exactly n roots. These may include real numbers, complex numbers, and repeats. For example:

x 2 � 1 → roots: 1, �1 x 2 � �4 → roots: 2i, �2i x 2 � 0 → roots: 0, 0

Solve x 3 � x 2 � x � 1 � 0 by finding all the roots.

The polynomial has degree 3, so the equation has 3 roots.

By the Rational Roots Theorem, possible roots are x � �1. Test with synthetic division.

1 1 �1 1 �1

1 0 1 1 is a root.

1 0 1 0

Therefore, (x � 1) is a factor. Rewrite the equation as (x � 1) ( x 2 � 1) � 0

The equation x 2 � 1 � 0 has roots i and �i.

So, the 3 solutions of x 3 � x 2 � x � 1� 0 are 1, i and �i.

PracticeComplete the steps to solve x 3 � 2x � 4 � 0.

1. The equation has 3 roots. �2 1 0 �2 4Possible roots: �1, �2, and �4 �2 4 �4 �2 is a root. 1 �2 2 0

(x � 2)( x 2 � 2x � 2) � 0

Use the quadratic formula for the second factor.

2 � ��

( �2 ) 2

� 4( 1 )( 2 ) ___________________________

2( 1 )

� 2 � ��

�4 __________

2

� 1 � i The roots are 2 , 1 � i , and 1 � i .

Solve each polynomial equation.

2. x 3 � 3 x 2 � x � 3 � 0 3. x 3 � 2x 2 � x � 2 � 0

�3, i, �i 2, i, �i

4. x 3 � 3x 2 � 4x � 12 � 0 5. x 3 � x 2 � 9x � 9 � 0

3, 2i, �2i 1, 3i, �3i

ReteachingUsing the Fundamental Theorem of Algebra 106


Reteachingcontinued

If you know roots of a polynomial, you can write a polynomial with those roots by finding the product of the factors. When a polynomial has a complex root a � bi, the conjugate a � bi is also a root. This is also true of irrational roots, a � �

� b and a � �

� b must both be roots of a

polynomial if one of them is.

Write the simplest polynomial function with zeros 3i and ��

5 .

If 3i is a root, then �3i must be a root, so (x � 3i ) and (x � 3i ) will both be factors.

If ��

5 is a root, then � ��

5 is a root, so (x � ��

5 ) and (x � ��

5 ) will both be factors.

Write the function as a product of the factors, then multiply.

P(x) � (x � 3i )(x � 3i )(x � ��

5 )(x � ��

5 )

� [ x 2 � x(3i ) � x(3i ) � (3i ) 2 ] [ x 2 � ��

5 x � ��

5 x � ( ��

5 ) 2 ] � ( x 2 � 9)( x 2 � 5)

� x 4 � 5x 2 � 9x 2 � 45

� x 4 � 4 x 2 � 45

PracticeComplete the steps to write the simplest polynomial function with zeros 1 � 2i and �

� 3 .

6. The roots of the polynomial are 1 � 2i, 1 � 2i, �� 3 , and � �

� 3 .

P(x) � [ x � (1 � 2i) ] [ x � (1 � 2i) ] � x � ��

3 � � x � ��

3 � � [ x 2 � x(1 � 2i) � x(1 � 2i) � (1 � 2i)(1 � 2i ] [ x 2 � x �

� 3 � x �

� 3 � � ��

3 � 2 ]

� ( x 2 � 2x � 5) ( x 2 � 3)

� x 4 � 2x 3 � 2x 2 � 6x � 15

Write the simplest polynomial function with the given zeros.

7. i, ��

6 8. 3i, � ��

8

x 4 � 5x 2 � 6 x 4 � x 2 � 72

9. 1 � i, ��

2 10. 2i, 3 � ��

5

x 4 � 2x 3 � 4x � 4 x 4 � 6x 3 � 8 x 2 � 24x � 16

106

Name Date Class


You have graphed rational functions which have vertical asymptotes. Now you will learn about rational functions with slant asymptotes and find all of the asymptotes and holes in a rational function.

A rational function has a slant asymptote when the degree of the numerator is 1 greater than the degree of the denominator.

Find the equation of the slant asymptote for f(x) � x 3 � x 2 � 4x � 7 _______________ x 2 � x � 3

.

Step 1: Check that the function has a slant asymptote.

Degree of numerator � 3. Degree of denominator � 2. The degree of the numerator is 1 greater than the degree of the denominator.

Step 2: Divide the numerator by the denominator.

x � 2

x 2 � x � 3 � �

x 3 � x 2 � 4x � 7

__ �( x 3 � x 2 � 3x)

�2 x 2 � x � 7

__ �(�2 x 2 � 2x � 6)

x � 1 f(x) � x 3 � x 2 � 4x � 7 _______________

x 2 � x � 3 � x � 2 � x � 1 __________

x 2 � x � 3

Step 3: The equation of the slant asymptote is the linear term of the quotient.

y � x � 2

PracticeComplete steps to find the equation of the slant asymptote for the rational function.

1. f(x) � 2 x 3 � 11 x 2 � x � 1 _________________ x 2 � 5x � 4

Degree of numerator: 3 Degree of denominator: 2

2x � 1 Divide: x 2 � 5x � 4 �

� 2x 3 � 11x 2 � x � 1

� ( 2 x 3 � 10 x 2 � 8x )

� x 2 � 9x � 1 f(x) � 2x � 1 � 4x � 3

___________ x 2 � 5x � 4

�( �x 2 � 5x � 4)

4x � 3 Slant asymptote: y � 2x � 1

Find the equation of the slant asymptote for each rational function.

2. 3 x 3 � 8 x 2 � 4x � 8 _________________ 3 x 2 � 2x � 5

3. 2 x 3 � 5 x 2 � 12x � 2 __________________ x 2 � 2x � 5

y � x � 2 y � 2x � 1

107

Use polynomial division.

ReteachingGraphing Rational Functions II


Reteachingcontinued

When graphing a rational function, first find all asymptotes and holes.

Identify all asymptotes and holes for f(x) � x 4 � 16 ________________ x 3 � 5 x 2 � 2x � 8

.

Step 1: Check the function for a slant asymptote. The degree of the numerator is 1 greater than the degree of the denominator. It has a slant asymptote.

Step 2: Factor the numerator and the denominator.

Numerator: x 4 � 16 � ( x 2 � 4)( x 2 � 4) � (x � 2)(x � 2)( x 2 � 4)

Denominator: Factors of 1 are �1.

Factors of 8 are �1, �2, �4, �8.

p __ q � { �1, �2, �4, �8 }

Try dividing by each root using synthetic division until you find a factor.

1 1 �5 2 8 �1 1 �5 2 8 1 �4 �2 �1 6 �8 1 �4 �2 6 ✕ 1 �6 8 0 ✓

f(x) � (x � 2)(x � 2)( x 2 � 4)

___________________ (x � 1)( x 2 � 6x � 8)

� (x � 2)(x � 2)( x 2 � 4)

___________________ (x � 1)( x � 4) (x � 2)

� x 3 � 2 x 2 � 4x � 8 ________________ x 2 � 3x � 4

Step 3: Find the slant asymptote.

x � 5

x 2 � 3x � 4 � �

x 3 � 2 x 2 � 4x � 8

__ �( x 3 � 3 x 2 � 4x)

5 x 2 � 8x � 8

__ �(5 x 2 �15x � 20)

23x � 28

PracticeComplete the steps to identify all asymptotes and holes for the rational function.

4. f (x) � x 4 � 16 ________________ x 3 � 2x 2 � 5x � 6

f (x) � (x � 2)(x � 2)( x 2 � 4)

_________________________

(x � 1)( x 2 � x � 6 )

� (x � 2)(x � 2)( x 2 � 4)

________________________

(x � 1)( x � 3 )( x � 2 )

x � 2

x 2 � 4x � 3 � ��

x 3 � 2 x 2 � 4x � 8

107

q � �1, p � �1, �2, �3, �6, p __ q � { �1, �2, �4, �6 }

Slant asymptote: y � x � 5

Vertical asymptotes: x � �1 and x � 4

One factor is (x � 1).

Hole: x � 2

Values of p

Values of q

Hole: x � � 2; Vertical

asymptotes: x � 1 and x � 3

Slant asymptote: y � x � 2

Name Date Class


You have learned about trigonometric functions and their reciprocals. Now you will use these functions to write fundamental trigonometric identities.

From the Pythagorean Theorem: x 2 � y 2 � 1

From the definition of sine and cosine:

cos(�) � adjacent

__________ hypotenuse

� x __ 1 � x sin(�) �

opposite __________

hypotenuse �

y __

1 � y

Substitute cos(�) sin(�) x 2 � y 2 � 1 → cos 2 (�) � sin 2 (�) � 1

For more trigonometric identities, begin with this identity:

cos 2 (�) � sin 2 (�) � 1 cos 2 (�) � sin 2 (�) � 1

cos 2 (�)

______ cos 2 (�)

� sin 2 (�)

______ cos 2 (�)

� 1 ______ cos 2 (�)

cos 2 (�)

______ sin 2 (�)

� sin 2 (�)

______ sin 2 (�)

� 1 ______ sin 2 (�)

1 � tan 2 (�) � sec 2 (�) cot 2 (�) � 1 � csc 2 (�)

Show that cos 2 (�) � sin 2 (�) � 1 for � � 50�.

cos 2 (50�) � sin 2 (50�) � 0.4132 � 0.5868 � 1

Show that 1 � tan 2 (�) � sec 2 (�) for � � 50�.

1 � tan 2 (50�) � 1 � 1.4203 � 2.4203 sec 2 (50�) � 1 ________ cos 2 (50�)

� 2.4203

PracticeComplete the steps to show that the trigonometric identity is true.

1. cot 2 (�) � 1 � csc 2 (�), for � � 50�

1 � cot 2 (50�) � 1 � 1 ___________ tan 2 (50�)

� 1 � 1 ________ 1.4203

� 1.7041

csc 2 (50�) � 1 __________ sin 2 (50�)

� 1 ________ 0.5868

� 1.7041

Show that each trigonometric identity is true for � � 37�.

2. cos 2 (�) � sin 2 (�) � 1 cos 2 (37�) � sin 2 (37�) � 0.6378 � 0.3622 � 1

3. 1 � tan 2 (�) � sec 2 (�) 1 � tan 2 (37�) � 1 � 0.5678 � 1.5678

sec 2 (37�) � 1 ________ cos 2 (37�)

� 1 ______ 0.6378

� 1.5678

4. cot 2 (�) � 1 � csc 2 (�) cot 2 (37�) � 1 � 1 _________ tan 2 (37�)

� 1 � 1 _______ 0.5678

�1 � 2.7610

csc 2 (37�) � 1 _________ sin 2 (37�)

� 1 _______ 0.3622

� 2.7610

108Reteaching Using Fundamental Trigonometric Identities

unit circle (x, y)

x

yθ

1



Graphing can be used to show that a trigonometric identity is true.

Show graphically that the trigonometric identity sin 2 (�) � cos 2 (�) � 1 is true.

Step 1: Graph y 1 � sin 2 (x) and y 2 � cos 2 (x) on the same coordinate plane.

Step 2: Notice that both graphs are symmetric about the line y � 0.5. Add this line to the graph.

Step 3: Draw a vertical line x � k anywhere on the graph. This line intersects each graph the same distance from the line y � 0.5.

So, cos 2 (k) � 0.5 � 0.5 � sin 2 (k). Rewrite this equation:

cos 2 (k) � 0.5 � 0.5 � sin 2 (k)

cos 2 (k) � 0.5 �0.5 � 0.5 � sin 2 (k) �0.5

cos 2 (k) � sin 2 (k) � 1 � sin 2 (k) � sin 2 (k)

cos 2 (k) � sin 2 (k) � 1

The result would be the same for any vertical line, so the identity is true.

PracticeComplete the steps to show graphically that the trigonometric identity is true.

5. cot 2 (�) � 1 � csc 2 (�)

Graph y 1 � cot 2 (x) and y 2 � csc 2 (x) .

A vertical line intersects the graphs at points 1 unit(s) apart.

The graph of y 2 is the graph of y 1 shifted up 1 unit(s).

Therefore, cot 2 (�) � 1 � csc 2 (�).

Show graphically that the trigonometric identity is true.

6. 1 � tan 2 (�) � sec 2 (�)

Graph y 1 � tan 2 (x) and y 2 � sec 2 (x).

A vertical line intersects the graphs at points 1 unit apart. The graph of y 2 is the graph of y 1 shifted up 1 unit.

Therefore, 1 � tan 2 (�) � sec 2 (�).

Both functions complete a full cycle in 2�, so draw the graphs over the range �� to �.

x

y

π-π

-0.5

Oπ

2-

2π

1

0.5

y1 = sin2(x)y2 = cos2(x)

x

y

π-π

-0.5

Oπ

2-

2π

y1 = sin2(x)

y= 0.5

x = ky2 = cos2(x)

1

0.5

x

y

O3π4

π

2π

4

3

2

1

y2 = csc2(x)

y1 = cot2(x)

x

y

Oπ

2π

2-π

4-π

2

3

2

1

y2 = sec2(x)y1 = tan2(x)

Name Date Class


Positive coefficient of x 2 :hyperbola is horizontal and vertices lie on x-axis.

Positive coefficient of y 2 :hyperbola is vertical and vertices lie on y-axis.

You have graphed parabolas, circles, and ellipses. Now you will graph hyperbolas.

The basic hyperbola is centered at the origin.

Horizontal Opening Vertical Opening

x 2 ___ a 2

� y 2

___ b 2

� 1

Vertices: (a, 0) and (�a, 0)

Co-vertices: (0, b) and (0, �b)

y 2

___ a 2

� x 2 ___ b 2

� 1

Vertices: (0, a) and (0, �a)

Co-vertices: (b, 0) and (�b, 0)

Asymptotes: y � � b __ a

x Asymptotes: y � � a __ b x

Graph x 2 ___ 9

� y 2

___ 4 � 1.

Step 1: Determine the direction in which the hyperbola opens. Since the coefficient of x 2 is positive, the hyperbola opens horizontally.

Step 2: Find and plot the vertices and co-vertices.

a 2 � 9, so a � 3. Vertices: (3, 0) and (�3, 0)

b 2 � 4, so b � 2. Co-vertices: (0, 2) and (0, �2)

Step 3: Draw a rectangle that has the vertices and co-vertices as the midpoints to each side.

Step 3: Draw and extend the diagonals of the rectangle. These are the asymptotes. Write their equations.

y � � b __ a x → y � � 2 __ 3 x

PracticeComplete the steps to graph the hyperbola.

1. y 2

___ 16

� x 2 __ 9 � 1

Graph opens vertically a � 4 b � 3

Vertices: (0, 4), (0, �4) Co-vertices: (3, 0), (�3, 0)

Asymptotes: y � � 4 __ 3 x

Graph each hyperbola.

2. y 2

__ 4

� x 2 ___ 25

� 1 3. x 2 __ 9 �

y 2 ___

25 � 1

ReteachingMaking Graphs and Using Equations of Hyperbolas 109

x

y8

4

4 8

-4

-4-8

-8

x

y8

4

4 8

-4

-4-8

-8

y8

4

84-4-8

-8

-4

y8

4

84-4-8

-8

-4


Reteachingcontinued

A hyperbola does not need to be centered at the origin.

Horizontal Opening Vertical Opening

(x � h) 2

_______ a 2

� (y � k) 2

_______ b 2

� 1

Center: (h, k)

(y � k) 2

_______ a 2

� (x � h) 2

_______ b 2

� 1

Center: (h, k)

Slope of asymptotes: � b __ a Slope of asymptotes � a __ b

Graph (x � 1) 2

_______ 9 �

(y � 3) 2 _______

4 � 1.

Step 1: Since the coefficient of x 2 is positive, the hyperbola opens horizontally.

Step 2: Find the center: (h, k) � (1, �3)

Step 3: Find and plot the vertices and co-vertices.

a 2 � 9, so a � 3. Vertices: Start at the center and move left 3 and right 3.

b 2 � 4, so b � 2. Co-vertices: Start at the center and move up 2 and down 2.

Step 4: Draw a rectangle that has the vertices and co-vertices as the midpoints to each side.

Step 5: Draw and extend the diagonals of the rectangle. These are the asymptotes.

Write their equations. m � � b __ a � � 2 __ 3 , ( x 1 , y 1 ) � (1, �3).

y � y 1 � m(x � x 1 ) y � y 1 � m(x � x 1 )

y � (�3) � 2 __ 3 (x � 1) y � (�3) � � 2 __

3 (x � 1)

y � � 2 __ 3 x � 11 ___

3 y � � 2 __

3 x � 7 __

3

Step 6: Draw the hyperbola.

PracticeComplete the steps to graph the hyperbola.

4. (y � 2) 2

_______ 16

� (x � 1) 2

_______ 9 � 1 Graph opens Vertically

Center: (�1, 2) a � 4 b � 3

Vertices: (�1, 6), (�1, �2) Co-vertices: (2, 2), (�4, 2)

Asymptotes: y � � 4 __ 3 x + 10 ___

3 and y � � 4 __

3 x + 2 __

3

Graph each hyperbola.

5. (y � 4) 2

_______ 4

� (x � 3) 2

_______ 25

� 1

y8

4

84-4-8

-8

-4

109

To find the equations of the asymptotes, use the point-slope formula with this slope and (h, k) as the point.

y8

4

8-4-8

-8

y8

4

84-4-8

-8

-4

Name Date Class


You have evaluated logarithmic expressions and solved logarithmic equations and inequalities. Now you will graph logarithmic functions.

Some transformations of a logarithmic function can stretch the graph of the function. Transformed logarithmic functions that include stretches have the form f (x) � a lo g b (cx).

Graph and compare y � 3 log(x) y � 6 log(x).Step 1: Graph y � 3 log(x).

Step 2: Graph y � 6 log(x).

Step 3: Compare the graphs. The graph of y � 6log(x) is taller than the graph of y � 3log(x).

Graph and compare y � log(3x) and y � log(6x).

Step 1: Graph y � log(3x).

Step 2: Graph y � log(6x).

Step 3: Compare the graphs. The graph of y � log(6x) is taller than the graph of y � log(3x).

PracticeComplete the steps to graph and compare the logarithmic functions.

1. y � 0.5 log(x) and y � log (x) 2. y � log (0.5x) and y � log(x)

The graph of The graph of

y � log(x) is taller y � log(x) is taller than the graph of than the graph of y � 0.5log(x). y � log(0.5x).

Graph and compare the logarithmic functions.

3. y � 0.25log(x) and y � 2log(x) 4. y � log(0.25x) and y � 2log(x)

The graph of The graph of

y � 2log(x) is taller y � log(2x) is taller than the graph of than the graph of y � 0.25log(x). y � log(0.25x).

ReteachingGraphing Logarithmic Functions 110

x

y

3

1

2

42 6 8O

y = log(6x)

y = log(3x)

x

y

3

1

2

42 6 8O

y = 6log(x)

y = 3log(x)

x

y

1

2

42 6 8O

y = log(x)

y = 0.5log(x)

x

y

1

2

42 6 8O

y = 2log(x)

y = 0.25log(x)x

y

1

2

42 6 8O

y = log(2x)

y = log(0.25x)

x

y

2

1

2 4 6 8

-1

O

y = log (0.5x)

y = log (x)


Reteachingcontinued

The graph of a logarithmic function can also be translated vertically and horizontally. Transformed logarithmic functions that include translations have the form f (x) � a lo g b (cx � d) � e.

Horizontal Translation Vertically Translation

Shift � d � unites right for d � 0.

Shift � d � unites left for d � 0.

Shift � e � units up for e � 0.

Shift � e � units down for e � 0.

Graph the logarithmic functions.

y � ln(x) y � ln(x � 1) y � ln(x � 3)

Step 1: Graph y � ln(x).

Step 2: Graph y � ln(x � 1). d � �1 and � d � � � �1 � � 1, so the graph is the same as the graph of y � ln(x), but shifted 1 unit left.

Step 3: Graph y � ln(x � 3). d � 3 and � d � � � 3 � � 3, so the graph is the same as the graph of y � ln(x), but shifted 3 units right.


y � ln(x) y � ln(x) � 1 y � ln(x) � 3Step 1: Graph y � ln(x).

Step 2: Graph y � ln(x) � 1. e � 1 and � e � � � 1 � � 1, so the graph is the same as the graph of y � ln(x), but shifted 1 unit up.

Step 3: Graph y � ln(x) � 3. e � �3 and � e � � � �3 � � 3, so the graph is the same as the graph of y � ln(x), but shifted 3 units down.

PracticeComplete the steps to graph the logarithmic functions.

5. y � ln(x) y � ln(x � 1) y � ln(x � 2)

y � ln(x � 1): shift the graph of y � ln(x) right 1 unit(s)

y � ln(x � 2): shift the graph of y � ln(x) left 2 unit(s)


6. y � ln(x) y � ln(x � 2) y � ln(x � 3) 7. y � ln(x) y � ln(x) � 2 y � ln(x) � 3

110

x

y4

2

42 6

-2

-2

-4

O

y = ln(x + 1)

y = ln(x - 3)

y = ln(x)

x

y4

2

42 6

-2

-2

-4

O

y = ln(x) + 1

y = ln(x) - 3

y = ln(x)

x

y4

2

42 6

-2

-2

-4

O

y = ln(x + 2)

y = ln(x - 1)

y = ln(x)

x

y4

2

2 4

-2

-4 -2

-4

Oy = ln(x + 3)

y = ln(x - 2)

y = ln(x)

x

y4

2

42 6

-2

-2

-4

O

y = ln(x) + 3

y = ln(x) - 2

y = ln(x)

Name Date Class


INV

11

You have graphed polar coordinates and complex numbers. Now you will use De Moivre’s Theorem.

All complex numbers can be graphed on a complex plane. A complex number of the form a � bi can be graphed with a as the x-coordinate and b as the y-coordinate.

Graph the complex number 4 � 3i.

x

y

2

4

6

2 4 6O

When the number is graphed, a right triangle is formed with the angle, �, and the horizontal and vertical components of a � bi.

x

a + bi

y

2

4

6

2 4 6Oθ

To calculate � based on the values of a and b, use tan �1 � b __ a � . Also,

make note of the quadrant in which the complex number is found.

Plot the following complex numbers. Then find the length of the hypotenuse and measure of the angle � formed with the positive x-axis. Round answers to the nearest hundredth.

1. 4 � 4i 2. �7 � 10i 3. 3 � (�9i)

hypotenuse:

5.66 hypotenuse:

12.21 hypotenuse:

9.49 Since z is in

Quadrant I,

� � tan �1 b __ a

� 45�

Since z is in

Quadrant II,

� � 180� � tan �1 b __ a

� 124.99�

Since z is in Quadrant IV,

� � 360� � tan �1 b __ a

� 288.43�

ReteachingUsing De Moivre’s Theorem

When graphing complex numbers, the horizontal axis is the real axis and the vertical axis is the imaginary axis.

To calculate the length of the hypotenuse formed, use the Pythagorean Theorem. For a complex number a � bi, the

hypotenuse will equal ��

a 2 � b 2 .

x

y

8

4

4 8 12

-4

-4-8

-8

O

12

-12

-12

4 + 4i

3 + (-9i)

-7 + 10i

Reteachingcontinued

INV

11


Another way of writing and graphing complex numbers is to use trigonometric ratios. If a complex number is defined as z � a � bi, then the horizontal and vertical coordinates of the complex number can be calculated using the following equations.

� z � � � ��

a 2 � b 2 � length of hypotenuse

x � � z � cos(�) horizontal coordinate

y � � z � sin(�) vertical coordinate

An equivalent form of z � a � bi using trigonometric ratios is z � � z � cos(�) � i � z � sin(�).

Complete the steps to convert the following complex numbers of the form a � bi to trigonometric form.

4. 3 � 6i 5. �1 � 4i

� z � � ��

3 2 � 6 2 � � tan �1 � 6 __ 3 � � z � � �

� (�1 ) 2 � 4 2 � � 180� � tan �1 � 4 ___

�1 �

� ��

9 � 36 � tan �1 (2) � ��

1 � 16 � 180� � tan �1 (�4)

� ��

45 � 3 ��

5 � 63.43� � ��

17 � 104.04�

z � 3 ��

5 cos(63 .43�) � i 3 ��

5 sin (63.43�) z � ��

17 cos(104.04�) � i ��

17 sin(104.04�)

One reason for converting to trigonometric form is De Moivre’s Theorem, which states,

( � z � cos(�) � i � z � sin(�) ) n � � z � n cos(n�) � i � z � n sin(n�)

According to the theorem, a complex number in trigonometric form, when raised to any integer power, remains a binomial and is therefore easier to calculate than if it were left in a � bi form.

Complete the steps to evaluate the following complex number using De Moivre’s Theorem. Write the answer in trigonometric form.

6. (�5 � 6i ) 2

(�5 � 6i ) � ��

61 cos(129.81�) � i ��

61 sin(129.81�)

(�5 � 6i ) 2 � ( ��

61 ) 2 cos(2�129.81�) � i( ��

61 ) 2 sin(2�129.81�)

� 61cos(259.62�) � i 61sin(259.62�)

Evaluate the following complex numbers using De Moivre’s Theorem. Write the answer in trigonometric form.

7. (3 � 4i ) 3 8. (10 � 10i ) 4

125 cos(159.39�) � i 125 40,000 cos(1260�) � i 40,000 sin(159.39�) sin(1260�)

x

z = a + bi

y = |z|cosθ

x = |z|sinθ

y

2

4

6

2 4 6Oθ

|z|

Name Date Class


You have learned how to graph a polynomial function. Now you will learn how to explain shifts in the graph of polynomial functions using transformations.

Vertical Translation

If f (x) is a polynomial function, g(x) � f(x) � k is a vertical translation of f(x).

For f(x) � x 3 � 2, write the rule for the function and sketch its graph.

a. g(x) � f(x) � 5

Vertical translation 5 units down.

g(x) � f(x) � 5

g(x) � x 3 � 2 � 5

g(x) � x 3 � 3

Horizontal Translation

If f(x) is a polynomial function, g(x) � f(x � h) is a horizontal translation of f(x).

For f(x) � x 3 � 2, write the rule for the function and sketch its graph.

a. g(x) � f (x � 4)

Horizontal translation 4 units left

g(x) � f (x � (�4))

g(x) � (x � 4) 3 � 2


1. Given f (x) � x 3 � 2, write the rule for

g(x) � f(x) � 1. Sketch the graphs of f and g.

Shift the graph of f (x) 1 unit up.

g(x) � x 3 � 3

Solve.

2. Given f (x) � x 3 � 2, write the rule for

g(x) � f(x � 3). Sketch the graphs of f and g.

g(x) � (x � 3) 3 � 2

111ReteachingTransforming Polynomial Functions

To graph g (x), shift the graph of f (x) down 5 units.

x

y4

2

2 4

-2

-2-4

-4

f(x) = x3+ 2

g(x) = x3- 3

x

y4

2

-2

-4 -2-6

-4f(x) = x3

+ 2

g(x) = (x + 4)3 + 2To graph g(x), shift the graph of f(x) left 4 units.

x

y

4

6

8

2

2 4-2-4

-2

x

y

4

6

8

2

2 4-2-4

-2


Reteachingcontinued

Vertical Stretch or Compression Horizontal Stretch or Compression

If f(x) is a polynomial function, g(x) � af(x) is a vertical stretch—or “compression”—of f(x).

If f(x) is a polynomial function, g(x) � f � 1 __ b x �

is a horizontal stretch—or “compression”—of f(x).

Let f(x) � 2 x 4 � 6 x 2 � 4 and g(x) � 1 __ 2 f(x).

Graph f and g on the same coordinate plane. Describe g as a transformation of f.Solution:

g(x) � 1 __ 2 f(x)

g(x) � 1 __ 2 (2 x 4 � 6 x 2 � 4)

g(x) � x 4 � 3 x 2 � 2

g(x) is a vertical compression of f(x).

Let f(x) � 2 x 4 � 6 x 2 � 4 and g(x) � f � 1 __ 2 x � .

Graph f and g on the same coordinate plane. Describe g as a transformation of f.Solution:

g(x) � f � 1 __ 2 x �

g(x) � 2 � 1 __ 2

x � 4 � 6 � 1 __ 2 x � 2 � 4

g(x) � 1 __ 8 x 4 � 3 __

2 x 2 � 4

g(x) is a horizontal stretch of f(x).


3. Let f(x) � 2 x 4 � 6 x 2 � 4 and g(x) � 2f(x).

Graph f and g on the same coordinate plane. Describe g as a transformation of f. g(x) � 2f(x)

g(x) � 2(2 x 4 � 6 x 2 � 4)

g(x) � 4 x 4 � 12 x 2 � 8 g(x) is a vertical stretch of f(x).

4. Let f(x) � 2 x 4 � 6 x 2 � 4 and g(x) � f(2x).

Graph f and g on the same coordinate plane. Describe g as a transformation of f.

horizontal compression;

g(x) � 32 x 4 � 24 x 2 � 4

111

x

y

4

6

8

2

2 4-2-4

-2

f(x) = 2x4- 6x2

+ 4

g(x) = x4- 3x2

+ 2

x

y

4

6

8

2

2 4-2-4

-2

f(x) = 2x4- 6x2

+ 4

g(x) = 18 x4

-

32 x2

+ 4- -

x

y

8

12

16

4

2 4-2-4

-4

x

y

8

12

16

4

2 4-2-4

-4

Name Date Class


You have learned how to use fundamental trigonometric identities. Now you will learn how to use angle sum and difference identities to find the exact value of some trigonometric expressions.

Sum Identities Difference Identities

sin(A � B) � sin A cos B � cos A sin B

cos(A � B) � cos A cos B � sin A sin B

tan(A � B) � tan A � tan B _____________ 1 � tan A tan B




Find the exact value of cos 105�.

cos 105� � cos(60� � 45�) Think: 60� � 45� � 105�.

� cos 60� cos 45� � sin 60� sin 45� Apply cos(A � B) identity.

� 1 __ 2 � �

� 2 ___

2 � �

� 3 ___

2 � �

� 2 ___

2 Evaluate.

� ��

2 ___ 4 � �

� 6 ___

4 � �

� 2 � �

� 6 ________

4 Simplify.

Find the exact value of sin � � ___ 12

� . sin � � ___

12 � � sin � � __

3 � � __

4 � Think: � __

3 � � __

4 � � ___

12 .

� sin � __ 3 cos � __

4 � cos � __

3 sin � __

4 Apply sin(A � B) identity.

� ��

3 ___ 2 � �

� 2 ___

2 � 1 __

2 � �

� 2 ___

2 Evaluate.

� ��

6 ___ 4 � �

� 2 ___

4 � �

� 6 � �

� 2 ________

4 Simplify.

PracticeComplete the steps to find the exact value of each expression.

1. cos(�15�) � cos(30� � 45�) 2. sin � 7� ___ 12

� � sin � � __ 3 � � __

4 �

� cos 30� cos 45� � sin 30� sin 45� � sin � __ 3 cos � __

4 � cos � __

3 sin � __

4

� ��

3 ___ 2 � �

� 2 ___

2 � 1 __

2 � �

� 2 ___

2

� ��

3 ___ 2 � �

� 2 ___

2 � 1 __

2 � �

� 2 ___

2

� ��

6 � ��

2 _________ 4

� ��

6 � ��

2 _________ 4

Find the exact value of each expression.

3. cos 75� ��

6 � ��

2 _______ 4 4. sin � 5� ___

12 � �

� 6 � �

� 2 _________

4

112ReteachingUsing Sum and Difference Identities


Reteachingcontinued

You can use angle addition and subtraction identities to prove identities. Use an identity to make one side of the equation the same as the other side.

Sum Identities Difference Identities






tan(A � B) � tan A �tan B _____________ 1 � tan A tan B

Prove the identity tan(� � x) � tan x.

tan(� � x) � tan � � tan x _____________ 1 � tan � tan x

Apply the tan(A � B) identity to the left side of the equation.

� 0 � tan x __________ 1 � (0)tan x

Evaluate. Think: tan � � 0.

� tan x Simplify.

Prove the identity sin(� � x) � sin x.

sin(� � x) � sin � cos x � cos � sin x Apply the sin(A � B) identity to the left side of the equation.

� (0)cos x � (�1)sin x Evaluate. Think: sin � � 0 and cos � � �1.

� sin x Simplify.

PracticeComplete the steps to prove the identity.

5. sin � � __ 2 � x � � cos x

sin � � __ 2 � x � � sin � � __

2 � cos x � cos � � __

2 � sin x Reason: Apply the sin(A � B) identity.

� (1)cos x � (0)sin x Reason: sin � � __ 2 � � 1 and cos � � __

2 � � 0.

� cos x Reason: Simplify.

Prove the identity.

6. cos(� � x) � �cos x

cos(� � x) � cos(�)cos x � sin � sin x Reason: Apply the cos(A � B) identity.

� (�1)cos x � (0)sin x Reason: cos � � �1 and sin � � 0. � �cos x Reason: Simplify.

112

Name Date Class


You have learned how to find a given term of a geometric series. Now you will learn how to find the indicated sum of a geometric series.

Sum of a Finite Geometric Series

The sum of the first n terms of a geometric series, S n , is a 1 � 1 � r n ______ 1 � r

� .

Find S 9 for the geometric series 3 � (�6) � 12 � (�24) �. . .


r � �6 ___ 3 � �2


a 1 � 3

Step 3: Use the formula with r � �2 to find S 9 .

S n � a 1 � 1 � r n ______ 1 � r

� Write the rule.

S 9 � 3 � 1 � (�2) 9 _________

1 � (�2) � Substitute 9 for n, �2 for r, and 3 for a 1 .

S 9 � 3 � 1 � (�512) __________

3 � Simplify.

S 9 � 3(171) � 513

The sum of the first 9 terms of the geometric series is 513.

PracticeComplete the steps to find S 8 for the geometric series.

1. 5 � (�10) � 20 � (�40) �. . . 2. 2 � 6 � 18 � 54 �. . .

a 1 � 5 n � 8 a 1 � 2 n � 8 r � 20 � (�10) � �2 r � 3

S 8 � 5 � 1 � ( �2 ) 8

___________

1 � ( �2 ) � � �425 S 8 � 2 � 1 � 3

8 ________

1 � 3 � � 6560

Find S 7 for each geometric series.

3. 1 � 3 � 9 � 27 �. . . 4. 4 � (�8) � 16 � (�32) �. . .

1093 172

113ReteachingUsing Geometric Series


Reteachingcontinued

Sum of an Infinite Geometric Series

The sum of an infinite geometric series, S, is a 1 _____

1 � r , where r is the common ratio and

0 � � r � � 1.

Find the sum of the geometric series 900 � 180 � 36 � 36 ___ 5 � . . .


r � 180 ____ 900

� 1 __ 5


a 1 � 900

Step 3: Use the formula to find the sum.

S � � a 1 _____

1 � r � Write the rule.

S � � 900 _______ 1 � � 1 __

5 � � Substitute 1 __

5 for r, and 900 for a 1 .

S � � 900 ____ 4 __ 5 � Simplify.

S � 1125

The sum of the series is 1125.

PracticeComplete the steps to find the sum of the geometric series.

5. 10 � (�5) � � 5 __ 2

� � � � 5 __ 4 � , . . . 6. 1 � 1 __

3 � 1 __

9 � 1 ___

27 � . . .

a 1 � 10 r � �5

_____ 10

� � 1 __ 2 a 1 � 1 r � 1 __

3 � 1 � 1 __

3

S � 10 _________ 1 � � � 1 __

2 �

� 10 ___

3 __ 2

� 6 2 __

3 S �

1 _______

1 � � 1 __ 3 �

� 1 __

2 __ 3 � 3 __

2

Find the sum of the geometric series.

7. 180 � (�90) � 45 � � � 45 ___ 2

� , . . . 8. 1 � 1 __ 2 � 1 __

4 � 1 __

8 �. . .

120 2 __ 3

113

Name Date Class


You have solved equations of circles and ellipses. Now you will identify conic sections by their equations.

Standard Forms for Conic Sections with Center (h, k)

Circle (x � h) 2 � (y � k) 2 � r 2

Ellipse (x � h) 2

_______ a 2

� (y � k) 2

_______ b 2

� 1 (x � h) 2

_______ b 2

� (y � k) 2

_______ a 2

� 1

Hyperbola (x � h) 2

_______ a 2

� (y � k) 2

_______ b 2

� 1 (y � k) 2

_______ a 2

� (x � h) 2

_______ b 2

� 1

Parabola x � h � 1 ___ 4p

(y � k) 2 y � k � 1 ___ 4p

(x � h) 2

Identify the conic section that each equation represents.

a. (x � 3) 2

_______ 8 2

� (y � 7) 2

_______ 6 2

� 1 b. (x � 5) 2 � (y � 2) 2 � 9 2

The equation shows a sum of squared terms, The equation shows a sum of squaredthe coefficients of x 2 and y 2 will NOT be equal, terms, the coefficients of x 2 and y 2 will beand the right side is equal to 1. equal, and the right side is NOT equal to 1.This is an Ellipse. This is a circle.

c. (y � 1) 2

_______ 4 2

� (x � 9) 2

_______ 7 2

� 1 d. x � 6 � 1 ___ 24

(y � 5) 2

The equation shows a difference of squared The equation only has one term with aterms. squared variable.

This is a hyperbola. This is a parabola.

PracticeComplete the steps to identify the conic section each equation represents.

1. (x � 4) 2 � y 2 � 3 2 2. (x � 6) 2

_______ 2 2

� (y � 4) 2

_______ 10 2

� 1

This equation shows a sum of squared terms. This equation shows a sum of squared terms.

The coeffi cients of x 2 and y 2 will be equal. The coeffi cients of x 2 and y 2 will not be equal.

The right side is not equal to 1. The right side is equal to 1.

This is a circle. This is an ellipse.

Identify the conic section each equation represents.

3. y � 5 � � 1 __ 4

(x � 7) 2 parabola 4. (x � 1) 2

_______ 2 2

� (y � 1) 2

_______ 7 2

� 1 hyperbola

5. (y � 7) 2

_______ 6 2

� (x � 9) 2

_______ 1 2

� 1 ellipse 6. x � 8 � 1 ___ 16

(y � 3) 2 parabola

114ReteachingIdentifying Conic Sections


Reteachingcontinued

The general form of a conic section is A x 2 � Bxy � C y 2 � Dx � Ey � F � 0, where A, B, and C are not all equal to zero.

Conic Section Coeffi cientsCircle B 2 � 4AC � 0, B � 0, and A � CEllipse B 2 � 4AC � 0 and either B � 0 or A � C

Hyperbola B 2 � 4AC � 0Parabola B 2 � 4AC � 0

To identify the conic section from the coeffi cients: • Identify the values A, B, and C.• Substitute into B 2 � 4AC and evaluate the expression.• Compare to 0 to classify the conic section. If

B 2 � 4AC � 0,compare A, B, and C.Identify each conic section.

a. x 2 � y 2 � 4y � 5 � 0 b. 9 x 2 � 18x � 4 y 2 � 27 � 0

A � 1, B � 0, C � 1 A � 9, B � 0, C � 4

B 2 � 4AC � 0 � 4(1)(1) � �4 B 2 � 4AC � 0 � 4(9)(4) � �144

� 4 � 0

B � 0 and A � C � 1 B � 0 and A � C

The conic section is a circle. The conic section is an ellipse.

c. 4 x 2 � 8x � y 2 � 12 � 0 d. 2 y 2 � 2y � 8x � 9 � 0

A � 4, B � 0, C � �1 A � 0, B � 0, C � 2

B 2 � 4AC � 0 � 4(4)(�1) � 16 B 2 � 4AC � 0 � 4(0)(2) � 0

16 � 0 0 � 0

The conic section is a hyperbola. The conic section is a parabola.

Practice Complete the steps to identify each conic section.

7. 16 x 2 � y 2 � 4y � 68 � 0 8. x 2 � 6y � y 2 � 5 � 0

A � 16 , B � 0 , C � �1 A � 1 , B � 0 , C � 1

B 2 � 4AC � 0 � 4(16 )(�1) � 64 B 2 � 4AC � 0 � 4(1 )(1) � �4

64 � 0 �4 � 0,

The conic section is a hyperbola. B � 0 and A � C � 1

The conic section is a circle.

Identify each conic section.

9. 2 x 2 � y � 3 � 0 parabola 10. x 2 � 9 y 2 � 18y � 8 � 0 ellipse

114

A is the coeffi cient of x 2,B is the coeffi cient of xy,C is the coeffi cient of y 2.

Name Date Class


You have used trigonometric functions. Now you will use double-angle and half-angle identities.

Double-Angle Identities

sin 2� � 2 sin� cos�cos 2� � cos 2 � � sin 2 �

tan 2� � 2 tan� _________ 1 � tan 2 �

cos 2� � 2 cos 2 � � 1cos 2� � 1 � 2 sin 2 �

Find sin 2� and cos 2� if cos � � � 2 __ 3 and 180� � � � 270�.

Step 1: Find sin �.

sin 2 � � 1 � cos 2 � Pythagorean Theorem

sin 2 � � 1 � � � 2 __ 3 � 2 Substitute cos� � � 2 __

3 .

sin 2 � � 5 __ 9 Evaluate.

sin� � � ��

5 ___ 3 Slove for sin�.

Step 2: Find sin 2�.

sin 2� � 2 sin� cos� Write the formula.

sin 2� � 2 � � ��

5 ___ 3 � � � 2 __

3 � Substitute values for sin� and cos�.

sin 2� � 4 ��

5 ____ 9 Simplify.

Step 3: Find cos 2�.

cos 2� � 2 cos 2 � � 1 Choose a formula.

cos 2� � 2 � � 2 __ 3 � 2 � 1 Substitute the value for cos�.

cos 2� � 8 __ 9 � 1 � � 1 __

9 Simplify.

PracticeComplete the steps if each expression if cos � � 1 __

4 and 270� � � �� 360�.

1. sin� Use: sin 2 � � 1 � cos 2 � 2. cos 2� Use: cos 2� � 1 � 2 sin 2 �

sin 2 � � 1 � � 1 ______ 4

� 2 cos 2� � 1 � 2 � � ��

15 _______ 4 �

2

sin 2 � � 1 � 1 ______ 16

� 15 ___ 16

cos 2� � 1 � 2 � 15 ___ 16

� � 1 � 30 ___ 16

� � 14 ___ 16

� � 7 __ 8

sin� � ��

15 ___ 16

� � ��

15 ______ 4

Evaluate the expression if each expression if cos � � 1 __ 4 and

270� � � � 360�.

3. sin 2� Use: sin 2� � 2 sin� cos� � 2 ��

15 _____ 16

115

Notice the interval in which � lies. Cosine and sine are both negative in Qudrant III.

ReteachingFinding Double-Angle and Half-Angle Identities

Use the negative value since � lies in Quadrant III.


Reteachingcontinued

You can use half-angle identities to find the exact value of some trigonometric expressions.

Half-Angle Identities

sin � � __ 2

� � � ��

1 � cos � _________ 2

: (�) for � __ 2 in Quadrants I, II; (�) for � __

2 in Quadrants III, IV

cos � � __ 2

� � � ��

1 � cos � _________ 2 : (�) for � __

2 in Quadrants I, IV; (�) for � __

2 in Quadrants II, III

tan � � __ 2 � � � �

�

1 � cos � _________ 1 � cos �

: (�) for � __ 2 in Quadrants I, III; (�) for � __

2 in Quadrants II, IV

Find the exact value of cos 112.5�.

cos 112.5� � cos 225� ____ 2 Find the angle: � � 225�.

� � ��

1 � cos � _________ 2 Choose the half-angle identity.

� � ��

1 � cos 225� ___________ 2 Substitute � � 225�.

��

1 � � � �

� 2 ___

2 � __________

2 Evaluate.

� � ��

� 2 � ��

2 _______ 2 � � 1 __

2 � Simplify.

� � ��

2 � ��

2 _______ 4 � � �

� 2 � �

� 2 ________

2

PracticeComplete the steps using the half-angle identity to find the exact value of each expression.

4. tan 157.5� � tan 315� ____ 2 5. sin 15� � sin 30� ___

2

tan � 315� ____ 2 � � � �

�

1 � cos 315� ___________ 1 � cos 315�

sin 30� ___ 2 � ��

1 � cos 30� ______________

2

� � ��

1 � �

� 2 ___

2 __________

1 � ��

2 ______ 2

� � �

�

2 � �

� 2 _______

2 _______

2 � ��

2 _______ 2

� ��

1 � �

� 3 ___

2 _______

2 � �

�

2 � �

� 3 ________

2 ________

2

� ��

2 � ��

3 _______ 4

� � ��

2 � ��

2 _______ 2 � �

� 2

Use the half-angle identity to find the exact value of the expression.

6. cos 105� � cos 210� ____ 2 � �

� 2 � �

� 3 _________

2

115

Think: 225� has a reference angle of 45�.

112.5� is in QII where cosine is negative.

cos 225� � �cos45�

� � ��

2 ___ 2

Name Date Class


You have found the line of best fit to model data. Now you will determine the best fit model that is not linear by finding finite differences.

To use finite differences to determine the degree of a polynomial:

• Check that the x-values increase by a constant value.

• Find successive differences of the y-values until the differences are constant.

Finite Differences

Function Type Linear Quadratic Cubic Quartic Quintic

Degree 1 2 3 4 5

Constant Finite Differences First Second Third Fourth Fifth

Use finite differences to find the degree of the polynomial that best describes the data.

x � 3 � 2 � 1 0 1 2

y 78 14 0 0 2 18

First Differences

14 � 78�64

0 � 14�14

0 � 00

2 � 02

18 � 216

Second Differences

�14 � (�64)50

0 � (�14)14

2 � 02

16 � 214

Third Differences

14 � 50�36

2 � 14�12

14 � 212

Fourth Differences

�12 � (�36)24

12 � (� 12)24

A fourth degree polynomial best describes the data.

PracticeComplete the steps to use finite differences to determine the degree of the polynomial that best describes the data. 1.

x �2 �1 0 1 2

y �5 2 3 4 11

First Differences

2 � (�5)7

3 � 21

4 � 31

11 � 47

Second Differences

1 � 7�6

1 � 10

7 � 16

Third Differences

0 � (�6)6

6 � 06

116ReteachingFinding Best Fit Models

The Third differences areconstant.

Therefore a Third degreepolynomial best describes thedata.

The x-values increase by 1.

Fourth differences are constant.

First differences are not constant.

Second differences are not constant.

Third differences are not constant.



Data that does not have constant differences may have constant or near constant ratios. This data is best modeled by an exponential function.

To use finite differences to determine if a function is exponential:

• Check that the x-values increase by a constant value.

• Find the successive differences of the y-values.

• Check the ratios of the first differences. If the ratios are constant, the data is exponential.

For each data set, determine whether y is an exponential function of x. If so, find the constant ratio.a. Data Set 1

x �1 0 1 2 3

y 2.2 3 7 27 127

First Differences 3 � 2.2 � 0.8 7 � 3 � 4 27 � 7 � 20 127 � 27 � 100

Check the ratios of the first differences.

4 ___ 0.8

� 5 20 ___ 4 � 5 100 ____

20 � 5

The data set is exponential, with a constant ratio of 5.

b. Data Set 2

x �1 0 1 2 3

y �4 �1 2 5 8

First Differences �1 � (�4) � 3 2 � (� 1) � 3 5 � 2 � 3 8 � 5 � 3

The first differences are constant. The data set is linear, not exponential.

PracticeComplete the steps to determine whether y is an exponential function of x. If so, find the common ratio.

2. x �1 0 1 2 3

y 0.375 1.5 6 24 96

First Difference 1.5 � 0.375 � 1.125 6 � 1.5 � 4.5 24 � 6 � 18 96 � 24 � 72

Ratios: 4.5 _____ 1.125

� 4 18 ____ 4.5

� 4 72 _____ 18

� 4

The data set is exponential, with a constant ratio of 4.


The first differences are not constant, but seem to increase by a constant factor.


Name Date Class


You have learned how to solve systems of linear equations. Now you will learn how to solve systems that contain at least one nonlinear equation.

Solve { x 2 � y 2 � 25

x � y � �1 using substitution.

The first equation produces a circle, the second equation produces a straight line.

The solutions to a system of equations are the intersections of the two graphs. A straight line and a circle can intersect at 0, 1, or 2 points.

Step 1: Solve the linear equation for either x or y. x � �1 � y

Step 2: Rewrite the first equation, substituting (�1 � y) for x and solve for y.

x 2 � y 2 � 25

(� 1 � y) 2 � y 2 � 25

1 � y � y � y 2 � y 2 � 25

2y 2 � 2y � 1 � 25 2y 2 � 2y � 24 � 0

y 2 � y � 12 � 0(y � 4)(y � 3) � 0

y � �4 or y � 3

Step 3: Rewrite either equation, substituting each value for y and solve for x.y � �4 x � y � �1

x � 4 � �1x � 3

y � 3 x � y � �1x � 3 � �1

x � �4So the solutions to the system are the ordered pairs (3, �4) and (�4, 3).

PracticeComplete the steps to solve the system of equations by substitution.

1. { x 2 � y 2 � 34

x � y � 2

The first equation produces a circle.

The second equation produces a straight line.

There could be 0, 1, or 2 solutions.

Step 1: Solve the linear equation for x.

x � y � 2 Step 2: Rewrite the first equation, substituting y � 2 for x and solve for y.

(y � 2) 2 � y 2 � 34

2 y 2 � 4 y � 30 � 0

y 2 � 2 y � 15 � 0

(y � 3 )(y � 5 ) � 0

y � 3 or y � �5 Step 3: Rewrite the second equation, substituting each value for y and solve for x.

y � 3 x � y � 2

x � 3 � 2

x � 5 y � �5 x � y � 2

x � �5 � 2

x � �3The solutions to the system are the ordered pairs (�3, �5) and (5, 3).

Solve each system of equations by substitution.

2. { x 2 � y 2 � 13

x � y � �1

(�3, 2) and (2, �3)

3. { x 2 � y 2 � 35

x � 6y

(�6, �1) and (6, 1)

117ReteachingSolving Systems of Nonlinear Equations


Reteachingcontinued

Solve the system using elimination.

{ 3x 2 � 8y 2 � 21

� 3x 2 � 12y 2 � �30 Both equations produce ellipses so, there could be 0–4 solutions.

Step 1: Eliminate either the x 2 terms or the y 2 terms. The x 2 terms are opposites, so they will cancel out when the two equations are added together.

3x 2 � 8y 2 � 21

__ �3x 2 � 12y 2 � �30

� 4y 2 � �9 Add the terms of each equation.

y 2 � 9 __ 4

Divide both sides by �4.

y � � 3 __ 2 Take the square root of both sides.

Step 2: Substitute 3 __ 2 and � 3 __

2 for y into either equation, then solve for x. Each y-value

produces two x-values.

y � 3 __ 2 3x 2 � 8 � 3 __

2 � 2 � 21 y � � 3 __

2 3x 2 � 8 � � 3 __

2 � 2 � 21

3x 2 � 8 � 9 __ 4 � � 21 3x 2 � 8 � 9 __

4 � � 21

3x 2 � 18 � 21 3x 2 � 18 � 21

3x 2 � 3 3x 2 � 3

x 2 � 1 x 2 � 1

x � �1 x � �1

Step 3: Write the solutions: � 1, � 3 __ 2 � , � 1, 3 __

2 � , � �1, 3 __

2 � , � �1, � 3 __

2 �

Practice Complete the steps to solve the system of equations by elimination.

4. { 2x 2 � y 2 � �1

4x 2 � y 2 � 25 Eliminate the y 2 term by adding the equations together.

6x 2 � 24 Divide each side by 6.

x 2 � 4 Take the square root of each side.

x � �2 Substitute 2 and �2 for y into either equation.

x � 2 2 (2 ) 2

� y 2 � �1 x � �2 2 (�2 ) 2

� y 2 � �1

8 � y 2 � �1 8 � y 2 � �1

� y 2 � �9 � y 2 � �9 y 2 � 9 y 2 � 9 y � �3 y � �3

The solutions are (2, 3), (2, �3), (�2, 3),(�2, �3).

117

Name Date Class


You have learned how to use statistics to draw conclusions. Now you will learn that the way in which data is used can be misleading.

Explain why the sampling method used is likely to make the survey results misleading.

A group of students at a high school is trying to convince the administration to give any student, not just seniors, with a valid driver’s license a parking permit for the school’s parking lot. For three consecutive days, the group set up a table during all lunch periods and asked students to take their survey. The survey asked, “Should any student with a valid driver’s license be given a permit to park in the student lot?”

Solution: The sampling method used could cause the survey results to be misleading. The results include only those with an interest in the outcome and who choose to respond. This is not a true representation of the population. It is a voluntary-response sample and not a probability sample. Also, the sample includes only the students at the school. This does not represent the rest of the population that should be included, such as parents or residents near the school. Another reason the results could be misleading is that there is no mention of a way to keep students from taking the survey more than once and thereby influencing the results.

PracticeComplete the steps to help explain why the results of the poll are likely to be misleading.

1. A newspaper took a poll to see how people were likely to vote on a proposal in an upcoming election. There are 5,000 registered voters in the community. Since it would be impossible for them to poll each one, the staff decided to go to the mall on a Monday around lunchtime. They set up near the food court and polled every 10th adult who passed by until they talked to 50 people. Why are their polling results likely to be misleading?

Solution: Is the sample a probability sample, a convenience sample, or a voluntary response sample? convenience sample Is this type of sample representative of the population? No Why or why not? Others in the population who do not shop at the local mall, who did not go to the mall on the day of the poll, or who did not go to the food court were not included. Is the sample size large enough? No Why or why not? Polling 50 from a population of 5,000 is not a large enough sample to ensure unbiased results. Are there other ways the sampling methods used are likely to produce misleading results? Explain. Yes; conducting the poll on a weekday at lunchtime excludes many people at work or school.

Give one reason why the result is likely to be misleading.

2. A community group wants the city to build a new animal shelter. To test community support, the group runs an ad on a local TV station asking viewers to call and tell whether or not they support the new shelter. Possible answer: The result of the ad is a voluntary-response sample and not a probability sample. Only people watching the ad and who feel strongly enough to call the number are included in the sample. Also, people who do not have a television or a telephone are excluded.

118ReteachingRecognizing Misleading Data


Reteachingcontinued

Explain how the data are being misused.

Most of the 25 employees at a manufacturing company earn the following salaries: $8 per hour, $9.50 per hour, $11 per hour, or $15 per hour. The two managers at the company earn $25 per hour. The company is trying to attract more workers and is running an ad that states the average pay rate is between $13 and $14 per hour.

Why is this ad misleading? What would be a better way to represent the salaries?

Solution: The ad is misleading because when the salaries of non-management employees are considered, the average is about $10.88 per hour. When the salaries of the managers are included, the mean rises to $13.70 per hour. The managers’ salaries are much higher than the other salaries and should be considered outliers. Outliers alter the mean and make it less reliable as a measure of central tendency. A better way to represent the salaries that still includes the outliers would be to use the median salary, which is $11 per hour. The company could also report the mean salary calculated without the outliers, but should indicate that the outliers were not included in the calculations.

PracticeComplete the steps to explain how the data are being misused.

3. One study shows that, on average, high school students who play in the school band get better grades than students who do not. From these results, can you conclude that playing in the school band causes students to earn higher grades?

Solution: What is the correlation between the two sets of data? Both data sets are made up of high school students. Can you assume that since there is a correlation between the two data sets there is also a relationship between them? No Why or why not? Playing in a school band does not necessarily cause a student to get better grades. What could be an explanation for the results of the study? Students who play in the school band may be more motivated to attend school and do well in their classes. Students who take time to practice a musical instrument may also take more time to study. On the other hand, some students may use much of their free time practicing and have little time left to study.

Explain how the data are being misused.

4. A shoe store is having a sale. Most shoes have been discounted by 5, 10, 15, or 20%. The price of a few pairs of shoes has been reduced by 50%. The sign in the store window says that the average discount during the sale is 20%. Why is this sign misleading? What would be a better way to represent the discounts?

The average discount stated in the sign includes the few pair of shoes that have been reduced by 50%. This discount is an outlier. A more representative reduction would be the median discount, which is 15%.

118

Name Date Class


You have graphed trigonometric functions and used trigonometric identities. Now you will solve trigonometric equations.

Solving trigonometric equations is similar to solving polynomial equations. The goal is to isolate the trigonometric expression and then apply the inverse function. You may need to use methods such as the Zero Product Property or the quadratic formula.

Solve the equation 2 sin 2 x � sin x � 1 for 0� � x � 360�.

2 sin 2 x � sin x � 1

2 sin 2 x � sin x � 1 � 0 Subtract 1 from each side.

(2sin x � 1)(sin x � 1) � 0 Factor the left side of the equation.

Use the Zero Product Property and apply the inverse sine function to solve for x.

2 sin x � 1 � 0 sin x � 1 � 0

2 sin x � �1 sin x � 1

sin x � � 1 __ 2 x � sin �1 (1)

x � sin �1 � � 1 __ 2 � x � 90�

x � 210� or 330�The solutions are 90°, 210°, and 330°.

PracticeComplete the steps to solve the equation for 0 � x �� 2�.

1. 6 cos 2 x � 1 � 5 cos x

6 cos 2 x � 5 cos x � 1 � 0

Use the quadratic formula.

cos x � 5 � ��

(�5 ) 2 � 4(6)(1) ___________________

2(6)

cos x � 5

� 1 ______________

12

cos x � 1 __ 2

cos x � 1 __ 3

x � cos �1 � 1 __ 2

� x � cos �1 � 1 __ 3

� x �

__ 3

, 5 ___ 3

x � 1.23 , 5.05

119

Use a calculator to help find the solutions.

ReteachingSolving Trigonometric Equations


Reteachingcontinued

When an equation contains more than one trigonometric function, you may be able to use a trigonometric identity to help solve.

Solve the equation �2 cos 2 x � sin x � 1 � 0 for 0° � x � 360.

�2 cos 2 x � sin x � 1 � 0

�2(1 � sin 2 x) � sin x � 1 � 0 Rewrite cos 2 x as 1 � sin 2 x.

�2 � 2 sin 2 x � sin x � 1 � 0 Distribute.

2 sin 2 x � sin x � 1 � 0 Simplify.

(2sin x � 1)(sin x � 1) � 0 Factor the left side of the equation.

Use the Zero Product Property and apply the inverse sine function to solve for x.

2 sin x � 1 � 0

2 sin x � 1 sin x � 1 � 0

sin x � 1 __ 2 x � sin �1 (�1)

x � sin �1 � 1 __ 2 � x � 270�

x � 30�, 150�

The solutions are 30°, 150°, and 270°.

PracticeComplete the steps to solve the equation for 0 � x � 360°.

2. sec 2 x � 2tan x � 4 � 0

1 � tan 2 x � 2tan x � 4 � 0

tan 2 x � 2tan x � 3 � 0

(tan x � 3)(tan x � 1) � 0

tan x � 3 � 0 tan x � 1 � 0

tan x � 3 tan x � �1

x � tan �1 (3) x � tan �1 (�1)

x � 72� and 252� x � 135� and 315�

Solve the equation for 0 � x �� 360�.

3. sin 2 x � cos x � 1 � 0 4. sec 2 x � tan x � 3 � 0

x � 90�, 180�, 270� x � 63�, 135�, 236�, 315�

119

Name Date Class

INV

12

You have found arithmetic and geometric series. Now you will use mathematical induction to prove formulas for series.

Mathematical induction is one type of proof that you can use to show that a formula is true. A proof by mathematical induction consists of three parts.

1. Show that the formula is true for a base case n � 1.

2. Assume that the formula is true for a value k.

3. Show that the formula is true for k � 1.

Use mathematical induction to prove that � k � 1

n

4k � 2n(n � 1) .

Step 1: Show that the formula is true for the base case n � 1.

4(1) � 4 2(1)(1 � 1) � 4 ✔

Step 2: Assume that the formula is true for an integer k.

Step 3: Add the term 4(k � 1) to the formula and try to rewrite in the form 2n(n � 1).

2k(k � 1) � 4(k � 1) � (k � 1)(2k � 4)

� 2(k � 1)((k � 1) � 1)

The statement � k � 1

n

4k � 2n(n � 1) is true for all values of n.

PracticeComplete the steps to prove that �

k � 1

n

�k � �n(n � 1)

_________ 2 .

1. Step 1: Show that the formula is true for the base case n � 1 .

�(1 ) � �1 �1 (1 � 1)

_________ 2 � �1 ✔

Step 2: Assume that the formula is true for an integer k.

Step 3: Show the formula is true for k � 1.

�k(k � 1)

_________ 2 ��(k � 1) �

�k(k � 1) � [ �2(k � 1) ] _______________________

2

� �(k � 1) (k � 2) _______________

2

The statement � k � 1

n

�k � �n(n � 1)

_________ 2 .

k � 1

is true for all n.

ReteachingUsing Mathematical Induction


Reteachingcontinued

INV

12


By recognizing patterns, you can sometimes generate a formula for a series. Try to find a pattern that involves the values of n.

Generate a formula for adding the set of numbers that are multiples of 10. The series can be represented as �

k � 1

n

10k . Write the first few sums

and look for a pattern involving the value of n.

10 � 10 � 5(1)(2)

10 � 20 � 30 � 5(2)(3)

10 � 20 � 30 � 60 � 5(3)(4)

10 � 20 � 30 � 40 � 100 � 5(4)(5)

Each sum is the product of 5, n, and n � 1. The formula is 5n(n�1).

Prove by mathematical induction.

Step 1: The formula is true for the first few sums above.

Step 2: Assume the formula is true for k.

Step 3: Show the formula is true for k � 1.

5k(k � 1) � 10(k � 1) � (k � 1)(5k � 10)

� 5(k � 1)(k � 2)

PracticeGenerate a formula for adding the set of numbers that are multiples of 100.

2. The series can be represented as � k � 1

n

100k .

100 � 100 � 50(1)(2)

100 � 200 � 300 � 50(2)(3)

100 � 200 + 300 � 600 � 50(3)(4)

100 � 200 � 300 � 400 � 1000 � 50(4)(5)

The formula is 50n(n � 1) .

Write the sum as a series and generate a formula for the series.3. numbers that are multiples of �2

Series: � k � 1

n

�2n Formula: �n(n�1)

4. numbers that are multiples of 1 __ 3

Series: � k � 1

n

1 __ 3 n Formula:

n(n � 1) ________

6

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