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Course 15 Numbers and Their Properties

Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mafah_k.

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Module: Rules for Exponents and Radicals

Objective: To practice applying rules for exponents when the exponents are rational numbers

Name: _______________________ Date: __________________

Fill in the blanks.

When multiplying expressions with exponents, you should add the

exponents if the ___bases______________ are the same.

The term 2x 3 is the same as (2x) 3 . _______false_________ (true, false)

Any quantity raised to the zero power is equal to ____1______________.

A non-zero number raised to a negative exponent is equal to 1

over the number raised to a(n) positive exponent.

Any quantity raised to the one-half power is equal to the

____square__________ ____root___________ of the quantity.

Problem Set: Simplify each expression. 1. (2x) 4 = 416x 2. 25 2/1 = 5

3. (r 3 ) 5 = 15r 4. (4n) 1− = n41

5. (9a 3 ) (4a5) = 836a 6. (5t) 0 = 1

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mafah_k.

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7. 6

3

432bb = 3

8b

8. 4

4

28 = 44 or 256

9. zyxzyx

23

46

321 = 237 yx 10. ( ) 3572 −cba = 15216

1cba

11. ( )4232 yx = 81216 yx 12. 2

62

3

62

−

−−

−

yxyx = 14

22

49ybx

13. 8/94/38/1 •• xxx = 2x 14. ( ) ( ) 123252 2•4 −cbabca = 98ac

15. 235

623

155

zyxzyx−−

−−

= 8

52

3zyx 16. ( )

( ) 2/1410

3/169

368

yxyx = 23

1x

Reflection: Explain the difference between 4x3 and (4x)3.

4x3 is equal to 4 • x • x • x, while (4x)3 is equal to 4x • 4x • 4x which is the same

as 4 • 4 • 4 • x • x • x • or 64 • x • x • x.


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mafbh_k.

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Module: Rationalizing the Denominator in Rational Expressions

Objective: To practice rationalizing the denominator in rational expressions

Name: _______________________ Date: __________________

Fill in the blanks.

12x is the same as the square root of x.

Any nonzero number raised to the zero power is equal to one .

mnb is the same as taking the thn root of mb .

Rationalizing the denominator involves taking a rational expression that

has a radical in the denominator and converting it to an

equivalent expression that does not have a radical in the

denominator.

To rationalize the denominator, multiply the numerator and denominator

by a factor that changes the denominator from a(n) irrational

number to a(n) rational number.

Problem Set: Simplify the following.

1. 1416 2.

238 3. 4 1081a b

22224 ••• 3 28 2 59a b

2 3 2•2•2•2•2•2 3 33 2•2

4

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mafbh_k.

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4. 6 93 64x y 5. 14 77 128x y 6. 6 4 24 81a b c

( )1

6 9 364x y ( ) ( )1 1 1

14 77 7 7128 x y ( ) ( ) ( )1 1 1 1

6 4 24 4 4 481 a b c

( ) ( )1 1 1

6 93 3 364 x y 22x y 21

23

3 cba •••

2 34x y 33b a c

Rationalize the denominator.

7. 17

8. 33

9. 6 23

77

77•

71

= 3333

33•

33

== 62366

33•

326

==

10. 23x +

11. 4 2

5 3

2x yx y

12. 15x −

23x +

•332

33

++

=++

xx

xx 2 2

xy xy= • 2xy xy

xyxy= 1

5x −• 5 5

55x x

xx+ +

=−+

13. 3 4 2

15a b

14. 2 44

416a b

15. 2

2 64

6481aba b

2 2 13 3 3

1 4 2 2 2 13 3 3 3 3 3

1 5*5 5

a b

a b a b

21

21

21

42

41 •

2

4

16

4

a

a

baba= 3

42

3

3

3 38

38

abba

abab

abab =•

abba

5253 2

4 22a aba ba

= = 2

3

8 83 3abab b

= =

Reflection: Explain in your own words what you are doing when you “rationalize the denominator”. Rationalizing the denominator is taking a rational expression that has a radical in

its denominator and converting it to an equivalent expression that is without a

radical in its denominator.

•


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mafch_k.

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Module: Applying Rules for Exponents and Radicals

Objective: To practice simplifying rational expressions with exponents and radicals

Name: _______________________ Date: __________________

Simplifying ____rational____ ___expressions___ often requires the use

of the rules for exponents and radicals.

Simplifying rational expressions with radicals generally includes

rationalizing the ___denominator___.

Rationalize the denominator when there is an ____irrational____ number

in the denominator.

Problem Set: Simplify the expressions. Rationalize the denominator if necessary.

1. 27

27

27

68

66•

62

baba

baba

27

2821521

6••12

baba=

33

63•2 2242/1772/1 abbaa ==

−−

2. 3 246

1yx

3/13/23/2

3/13/23/2

3/23/43/1 66•

61

yxyx

yx=

yxyx

yxyx

2

3 2

2

3/13/23/2

636

66 ==

3. 3/1

3/1

3/63/53/13 65•

271

271

xx

yxyx=

22

3/1

3 yxx=

22

3

3 yxx=

4. yy

yy

y 12127

1212•

127 =

yy

123•47

=

yy

yy

637

12314

==

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mafch_k.

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5. 3232•

325

++

−

32)32(5

−+=

35251

)32(5 −−=−+=

6. 14102

8622

75

43

)8(4

84

−−−

−−−−

−=

− vu

vuvuvu

1664 2216vu=

22164 vu=

7.

zyx

zyx

zyx

zyx63

63

63

269

2

2•2

8

zyxzyx

63

2/32/122/122/1

216=

zx32=

8. 4/14/3

4/14/3

4/34/54/14 35•

161

161

yxyx

yxyx=

yxyx2

4/14/3

2=

yxyx

2

4 3

2=

9. 5 155243

1yx

5/155/55/12431

yx=

331xy

=

10. zz

zz

z 322•169

3232•

329 =

zz

32236=

zz

829=

Reflection: Explain in your own words what rationalizing the denominator involves.

Answers will vary.__________________________________________________

________________________________________________________________

________________________________________________________________

________________________________________________________________

________________________________________________________________

________________________________________________________________


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mafdh_k.

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Module: Scientific Notation

Objective: To practice representing values using scientific notation

Name: _______________________ Date: __________________

Fill in the blanks.

In scientific notation, numbers are represented as the product of a number

and a power of 10 .

The number 600,000 is written in scientific notation as 6•10 5 /6•10 4 .

In scientific notation, •10nm , m must be greater than or equal to 1 and

less than 10 .

When dividing numbers written in scientific notation, apply the quotient

rule for exponents. That means the exponents are subtracted .

Problem Set: Express the following numbers in scientific notation.

1. 700 7 • 10 2 2. 14,000 1.4 • 10 4 3. 417,000,000 4.17 • 10 8 4. .00374 3.74 • 10 3− Translate the following numbers from scientific notation.

5. 8 • 10 7− 0.0000008 6. 2.26 • 10 15 2,260,000,000,000,000 7. 3.1 • 10 10 31,000,000,000 8. 8.1 • 10 2 810

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mafdh_k.

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Indicate whether the following numbers are represented in proper scientific notation by writing “correct” or “incorrect” after each number.

9. –4 • 10 7 incorrect 10. 7.1 • 10 3 correct

11. 9 • 10 6− correct 12. 78.94 • 10 7 incorrect Solve these problems. Express you answer in scientific notation.

13. (4.73 • 10 4 ) + (2.11 • 10 3 )= 4.941 • 10 4

14. =11

10

10•1.510•7.5 5.0 • 10 1−

15. (9.10 • 10 3 ) • (3.2 • 10 t ) = 2.912 • 10 8

16. (9.26 • 10 2 ) – (8.12 • 10 2 ) = 1.14 • 10 2 Extension Activity: Pretend you are an algebra instructor teaching a lesson on adding, subtracting, multiplying, or dividing numbers in scientific notation. Make up a story problem that uses one of the above operations. Using another sheet of paper, write a lesson that shows your students how to solve this problem. Reflection: Name some professions that use scientific notation to represent numbers.

physicist, epidemiologist, biologist, mathematician, statistician

How does scientific notation help people in these jobs?

It makes calculations manageable and also provides a standard so that numbers

can be quickly identified.


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mafeh_k.

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Module: Simplifying Algebraic Expressions

Objective: To practice simplifying algebraic expressions by collecting like terms and following grouping symbols

Name: _______________________ Date: __________________

Circle the like terms in this expression. 15x2yz – 3x2y + x2yz – xy2z +

31 x2yz – 14xyz

Describe what makes the others unlike terms. _different variables or the

same variables to different powers. ______________________________ Fill in the blank. When you collect and combine like terms,

• what part(s) of the terms are changed?___coefficient______ (coefficient,

variables, exponents) • what part(s) of the terms are unchanged?_variables, exponents_______

(coefficient, variables, exponents)

• what kind of change occurs?_addition or subtraction_______(addition or

subtraction, multiplication) When removing an expression from parentheses or other grouping

symbol, it's important to note the ___+ or – sign___ in front of the grouping

symbol, as well as in front of the term inside the group.

In the example below, write the operator (+ / –) that belongs in the blanks.

c – (a2 + 3a – b) = c – a2 – 3a + b

Problem Set: Simplify the expressions. 1. a2 – ab2 + 5a2b + 2ab2 – b2 2. 3x2 + 4x – x + 7 – 5x2 + 13 + 9x2 +2x a2 + ab2 + 5a2b – b2 7x2 + 5x + 20

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mafeh_k.

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3. 8b2 – 2b + 17 + 5b 4. -30x + y + z + 3x + y + 2z 8b2 + 3b + 17 -27x + 2y + 3z 5. 2m – 4mn2 + 5m2n + 2mn2 – n2 6. 2a2 + 3a – a + 12 – 4a2 – 22 + 7a 2m2 – 2mn2 + 5m2n – n2 5a2 + 2a – 10 7. 3b2 – [-4b – (b2 – 4b)] – [1 + (4b – 3)] 8. 6y – {-3y + [2y – (4y + 12)]} 3b2 – (-4b – b2 + 4b) – (1 + 4b – 3) 6y – (-3y + (2y – 4y + 12) 3b2 + 4b + b2 – 4b – 1 – 4b + 3 6y – (-3y + 2y – 4y + 12) 4b2 – 4b + 2 6y – (-5y – 12) 6y + 5y + 12 11y + 12 9. -8a2 – [6a – (7a2 + 2a)] – [5 –(-3a + 7)] 10. x + {-3x2 – [2x2 + 2x – (3x – x2)]} -8a2 – (6a – 7a2 – 2a) – (5 + 3a + 7) x + (-3x2 – (2x2 + 2x – 3x – x2)

-8a2 – (4a – 7a2) – (3a – 2) x + (-3x2 – (3x2 – x) -8a2 – 4a + 7a2 – 3a + 2 x + (-3x2 – 3x2 + x) -a2 – 7a + 2 x – 6x2 + x

-6x2 + 2x

Reflection: Describe to a friend or classmate how to keep track of the positive and negative signs as you simplify algebraic expressions. Include any tricks you’ve found helpful when this is especially difficult to do correctly. Answers will vary. Example: I rewrite each line as I undo parentheses. I think

to myself “minus negative x2 = +x2”. I double check each step._______________

________________________________________________________________

________________________________________________________________

List 2 to 3 properties of numbers that you can remember and find examples of when you use each one as you simplify an algebraic expression. Answers will vary. Example: Distributive_5 – (-3a + 7) = 5 – (-3a) – 7________

_____________________________________________= 5 + 3a – 7__________

________________________Associative____________= 3a – 2_____________

________________________________________________________________


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document maffh_k.

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Module: Multiplying Algebraic Expressions

Objective: To practice multiplying algebraic expressions, including a shortcut approach to multiplying binomials

Name: _______________________ Date: __________________

Fill in the blanks. To multiply binomials, such as 13m2 • 2mn2o8, do the following:

• multiply the _constants (or coefficients) Ex: 13 • 2 = __26________

• multiply the __variables__________ by ___adding_____ the exponents.

Ex: m2mn2o8 = m3n2o8___

Final answer: 13m2 • 2mn2o8 = ___26m3n2o8_______

To multiply a monomial times a binomial, use the __distributive_______

property and multiply the monomial by each term of the binomial.

Example: -5(2x + y) = __(-5)__ 2x + _(-5)___ y

To multiply a trinomial times a trinomial, multiply each term of one

trinomial by __other trinomial____________________________.

Example: (2a + 3b – c)(a2 – b + c)

= 2a (a2 – b + c) + 3b (a2 – b + c) + -c (a2 – b + c)

A binomial squared follows this pattern: (a + b)2 = ___a2_____ + __2ab____ + __b2______ 1st term middle term last term The sum and difference of two terms follows this pattern:

(a + b)(a – b) = ___a2________ + ab – ab = 0 + ___-b2___ 1st term middle term last term = __a2___ – __b2___ simplified form To multiply this type of a binomial times a binomial, follow this pattern:

(x + 3)(x + 4) = ____•____ + ____+____ + ____• ___ 1st term middle term last term (Fill in the boxes.)

x x 4x 3x 3 4

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document maffh_k.

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Problem Set: Find the products. 1. 3a • 2a4b2c 2. -

81 x • 32x2y4

6a5b2c -4x3y4 3. 5x4(3x3 – 8) 4. -2x2(3x2 – 5)

15x7–40x4 -6x4+10x2 5. (a – 2b + c)(2a – 2b – c) 6. (2x – 3)(2x + 5) 2a2–6ab + ac + 4b2 + c2 4x2 + 4x – 15 7. (a2 + 2)(a2 – 4) 8. (3x + 2)2 a4 – 2a2 – 8 9x2 + 12x + 4 9. (2x – 5)2 10. (2x – 3)(2x + 3) 4x2 – 20x + 25 4x2 – 9 Extension: Write down two sample patterns from the beginning of this activity, in equation form, and determine a value for each variable. Substitute those values into both sides of the problem (the equation) and verify that it comes out to a true statement. Example: (a + b)2 = a2 + 2ab + b2, when a = 5, b = 3

(5 + 3)2 = 52 + 2 • 5 • 3 + 32 82 = 25 + 30 + 9 64 = 64

Answers will vary.


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2 Document mafgh_k.

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Module: Factoring Algebraic Expressions

Objective: To practice factoring algebraic expressions

Name: _______________________ Date: __________________

Fill in the blanks with one of the words in parentheses. Some polynomials can be factored completely by finding the

greatest common factor: 12ac2 + 2ac – 4c = 2c(6ac + a – 2)

(lowest, greatest) x2 + (a + b)x + ab = (x – a)(x – b)

This pattern shows how to factor another type of polynomial. The sum of

a and b must equal the coefficient of the middle term, while the product

of a and b must equal the last term. (difference, sum, product, quotient) If the sign of the third term is positive, the signs of the factors a

and b will be the same . (opposites, the same) In this case, the sign

of the middle term determines whether both factors are positive or

negative. (first, middle, third) The first two steps in factoring polynomials in the form ax2 + bx + c are

1) to find the product (factors, product, sum) of a and c and

2) to find factors of this value whose sum equals b . (a, b, c) In the

final step, use the distributive property. (associative, distributive)

Problem Set: Look carefully at each polynomial to determine its form, then factor it. 1. 7ax2 + 14ax + 7a 2. y2 – 17y – 60 3. 3c2d + 6cd2 7a(x2 + 2x + 1) (y – 20)(y + 3) 3cd(c + 2d) 7a(x + 1)(x + 1) 4. 10 x2y2 + 15xy2 – 5x2y – 60xy 5. x2 + 10x + 16 6. 2y2 + 11y + 14 5xy(2xy + 3y – x – 12) (x + 8)(x + 2) (2y + 7)(y + 2)

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2 Document mafgh_k.

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7. y2 + 34y + 120 8. 88x2 – 13x + 0.25 9. g2 – g – 20 (y + 4)(y + 30) (11x – 0.25)(8x – 1) (g – 5)(g + 4) 10. c2 – 41c + 40 11. 3x2 – 11x + 6 12. 8x2 – 26x + 6 (c – 40)(c – 1) (3x – 2)(x – 3) 2(4x2 – 13x + 3) 2(4x – 1)(x – 3) 13. 72c2 + 15c + 0.5 14. 21k2 – 2k – 3 15. g2 – 5g – 24 (12c + 0.5)(6c + 1) (3k + 1)(7k – 3) (g – 8)(g + 3) Extension Activity: Here are some trinomials. Plug the values of a, b and c into the expression below to see if each one is factorable. acb 4–2 If the resulting value is an integer, the trinomial is factorable.

Trinomial ax2 + bx + c Value of acb 4–2 Factorable?

YES or NO Factored form (if possible)

3g2 – 9g – 25 75•4–)9( 2 −− = 381 no

5x2 + 2x – 3 15•4–22 − = 64 = 8 yes (5x – 3)(x + 1)

2m2 – 7m + 6 12•4–)7( 2− = 1 = 1 yes (2m – 3)(m – 2)

x2 – x + 19 19•4–)1( 2− = 75− no

Reflection: The expression above comes from the quadratic formula:

a

acbbx2

42 −±−=

The term under the radical sign is called the discriminant. It determines the number and type of solutions a quadratic equation will have. What must be true about the value of the discriminant for a quadratic equation to be factorable? ________________________________________________________________

The discriminant must be a perfect square for the quadratic equation to be_____

factorable.______________________________________________________


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mafhh_k.

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Module: Factoring Sums and Differences of Perfect Cubes

Objective: To practice factoring the sums and differences of cubes

Name: _______________________ Date: __________________

The sum of cubes can be used to factor expressions such as 8x3 + y6.

Write the formula you would use to factor this binomial as the product of a

binomial and a trinomial. ))(( 2233 babababa +−+=+

The difference of cubes can be used to factor expressions such as

27x9 – 64x3. Write the formula you would use to factor this binomial as the

product of a binomial and a trinomial. ))(( 2233 babababa ++−=−

Sometimes you must factor out a ______ common______ factor before

you can apply the formula for perfect cubes.

Problem Set: Factor these binomials as the product of a binomial and a trinomial.

1. 6327 yx + 323 )()3( yx +

)39)(3( 4222 yxyxyx +−+

2. 912 8nm + 3334 )2()( nm +

)42)(2( 634834 nnmmnm +−+

3. 39125 yx − 333 )5( yx −

)525)(5( 2363 yyxxyx ++−

4. 3664 ba + 332 )4( ba +

)416)(4( 2242 bbaaba +−+

5. 66 21627 gf + 3232 )6()3( gf +

)36189)(63( 442422 ggffgf +−+

6. 153 278 yx − 353 )3()2( yx −

)964)(32( 10525 yxyxyx ++−

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mafhh_k.

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7. 1512216 zy − 3534 )()6( zy −

)636)(6( 1054854 zzyyzy ++−

8. 123 cb + 343 )(cb +

))(( 8424 cbcbcb +−+ Factor out a common factor and then apply one of the formulas for perfect cubes.

9. 9927 nrnp − )27( 99 rpn −

)39)(3( 633633 rrpprpn ++−

10. 99 254 yx + )27(2 99 yx +

)39)(3(2 633633 yyxxyx +−+

11. 6232 12816 ywxw +

)8(16 632 yxw +

)42)(2(16 42222 yxyxyxw +−+

12. 93 8124 yx − )278(3 93 yx −

)964)(32(3 6323 yxyxyx ++−

13. 915 19281 acab + )6427(3 915 cba +

)16129)(43(3 6351035 ccbbcba +−+

14. 15595 432 pmnm −

)8(4 1595 pnm −

24)(2(4 10536535 ppnnpnm ++−

Reflection: It can be tricky to keep the formulas for perfect cubes straight. How do you

remember the formula for the sum of perfect cubes and the difference of perfect

cubes? Use an example from above to demonstrate your strategy or thought

process.

Answers will vary. The binomial in the sum of cubes is a sum and the trinomial

has one subtraction sign, whereas the binomial in the difference of cubes is a

difference and the trinomial has only addition.


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mafih_k.

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Module: Factoring or Using the Quadratic Formula

Objective: To practice solving quadratic equations by factoring or using the quadratic formula

Name: _______________________ Date: __________________

Fill in the blanks. One way to find the solution set of a quadratic equation is by factoring and

using the ____zero_____ ____product____ ______rule_____. Another way to find the solution set of a quadratic equation is to use the

___quadratic___ formula. a

acbbx2

42 −±−=

Use this method to decide if a quadratic equation is easy to factor. If b2 – 4ac, the discriminant, is a ____perfect_____ _____square___,

then the equation is factorable. If b2 – 4ac, the discriminant, is not a perfect square, then the equation is

____not___ ___factorable___.

Problem Set: Find the solution set by factoring and using the zero product rule or by using the quadratic formula.

1. x2 – 17x + 72 = 0 (x – 8)(x – 9) = 0 x – 8 = 0 x – 9 = 0 x = 8 or x = 9 {8,9}

2. 2x2 – 5x = 12 (2x + 3)(x – 4) = 0 2x + 3 = 0 x – 4 = 0 2x = -3 or x = 4

23−=x

{ }4,23−

3. 3x2 + 6x + 1 = 0

)3(2)1)(3(466 2 −±−

=x

363 ±−=x

−−

+−

363,

363

4. 4x2 – 5x = 1

)4(2)1)(4(4)5()5( 2 −−−±−−

=x

8415 ±=x

−

+

8415,

8415

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mafih_k.

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5. 7x2 + 9x – 1 = 0

)7(2)1)(7(499 2 −−±−

=x

141099 ±−=x

−−

+−

141099,

141099

6. 2x2 – x – 15 = 0 (2x + 5)(x – 3) = 0 2x + 5 = 0 x – 3 = 0 2x = -5

25−=x or x = 3

{ }3,25−

Set up each problem as a quadratic equation. Solve. 7. The square of a number plus four times the number equals 32. Find the

number(s). Let n = the number

n2 + 4n = 32 n2 + 4n – 32 = 0 (n + 8)(n – 4) = 0 n + 8 = 0 n – 4 = 0

n = -8 or n = 4 {-8,4}

8. Find two consecutive positive integers such that their product added to four times the smaller integer equals 36.

Let n = the smaller positive integer n + 1 = second integer n(n + 1) + 4n = 36 n2 + 5n – 36 = 0 (n + 9)(n – 4) = 0 n + 9 = 0 n – 4 = 0

n = -9 or n = 4 n + 1 = 5

The numbers are 4 and 5.

Reflection: You can find the solution set of a quadratic equation by factoring and using the zero product rule or by using the quadratic formula. Find the solution set of 10x2 + x – 3 = 0 using both methods. Then write the benefits and drawbacks of each method.

{ }21,

53−

Answers will vary. Factoring is fast as long as you can quickly find the correct combination of numbers for the terms. It may be difficult to find the correct combination. Using the quadratic formula works with every quadratic equation, but there is more room for error in calculating the answer.


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mafjh_k.

–1 of 3–

Module: Rational Expressions: Simplify

Objective: To practice simplifying rational expressions by factoring

Name: _______________________ Date: __________________

Complete each statement. A ____rational____ expression is the quotient of two polynomials.

A rational expression is in ___simplest___ _____form_____ if the

numerator and denominator have 1 as their only common factor.

To put a rational expression in simplest form, follow these steps:

1. ____Factor_____ the numerator and denominator.

2. ____Cancel_____, or divide out, common factors.

Removing ____negative___ signs or changing signs of factors can also

help simplify rational expressions.

Problem Set: Simplify each rational expression completely.

1. 2 16

4tt−

=+

( )( )4 44

4t t

tt

− += −

+

2. =−+++

2832811

2

2

xxxx ( )( )

( )( ) 44

4747

−+

=−+++

xx

xxxx

3. 24

6xx=

2 • 2 22 •3 3x x xx

=

4. ( )( )

( )( ) 11

88

8888

−=−−−

=−+−−

=−−

yxyx

yxxy

yxxy

5. 2

24 4b

b b−

=− +

( )( )

2 12 2 2b

b b b−

=− − −

6. 2

13 131

xx−

=−

( )( )( )

13 1 131 1 1x

x x x−

=− + +

KEY



–2 of 3–

7. 29

3abab

= bbbabba 3

13

39

==•

•••••

8. 2 2

2 2

2510 25x y

x xy y−

=− +

( )( )( )( )

5 5 55 5 5

x y x y x yx y x y x y− + +

=− − −

9. 2 2 8

2x xx+ −−

= ( )( ) 42

42+=

−+− x

xxx

10. =++−+

12102422

2

2

xxxx ( )

( )( )( )( )( ) 3

1322122

65222

2

2

+−

=++−+

=++−+

xx

xxxx

xxxx

Determine if each rational expression is in simplest form. If it is not, then write it in simplest form.

11. 7xx +

yes 12. 2

7 71

xx−−

no, 1

7+x

13. 218

6xx

no, x3 14. 5

5+xy yes

Extension Activity: Anne had an algebra test yesterday. One question was:

Simplify the following expression completely and give reasons for each of your steps:

22

22

6644yxxy

−−

Determine if Anne’s solution is correct and if her answer is simplified completely. Explain your findings on a separate piece of paper.

Anne’s solution: 22

22

6644yxxy

−− ( )

( )22

22

64

yxxy

−−

= factor out a common factor

( )( )( )( )yxyx

xyxy+−+−

=64 factor the difference of squares

( )( )( )( )yxyx

xyxy+−+−

=64 cancel common factors

32

= reduce fraction to simplest form



–3 of 3–

Student solutions should address these major points:

Anne’s first and second steps are correct. When she canceled common factors,

she was correct to cancel the (x + y) and the (y + x) terms because they are

equivalent.

However, to cancel (y – x) and (x – y) is not correct. It is necessary to first factor

-1 out of one of the terms. For example, if -1 is factored out of (y – x), then

(y – x) = -1(-y + x) = -1(x – y). Now, the two (x – y) terms can be canceled. And the

problem can be completely simplified.

The completely simplified answer to this test question is 32

− .


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 1, Document mafkh_k.

–1 of 2–

Module: Rational Expressions: Add and Subtract

Objective: To practice adding and subtracting rational expressions

Name: _______________________ Date: __________________

Read each statement and decide whether it is true or false.

The rule for adding rational expressions is the same as the rule for adding

fractions. _____true_______ (true, false)

When adding rational expressions with equal denominators, you simply

add the numerators and the denominators. _____false_____ (true, false)

The rule for subtracting rational expressions is not the same as the rule for

subtracting fractions. ____false_______ (true, false)

When subtracting rational expressions with equal denominators, you

subtract the numerators and place the difference over the common

denominator. _____true_______ (true, false)

When adding or subtracting rational expressions with different

denominators, you should find the least common denominator before

adding or subtracting. ____true________ (true, false)

Problem Set: Add these rational expressions. Simplify if necessary.

=++ 6–32

4–52 x

xxx

)4–(3419

2

2

xxx ++ or

123419

2 −++

xxx

9–511

9–102

22

2

xx

xx +

++ =

3–52

xx +

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 1, Document mafkh_k.

–2 of 2–

365–6–83

2 ++xx

xxx =

15–18–2444

2

2

xxxx +

xxx

xx

2323

2 ++

+ =xxxx234139

2

2

+++

bab 339

+ =abab )(3 2 + or ab

ab 33 2 +

Subtract these rational expressions. Simplify if necessary.

yxyx

yxyx

++

++ 6–23 =

yxyx

+4–2

baba

baba

––– +

+ = 22 –

4baab−

xxxx

+22

1–1–

=))(1–)(1(

1––2

xxxxx

+ or

xxxx−3

2 1––

xyx

xxy

23–

34 =

yy69–8 2

121–

1 2 +++

+ aaa

aa =

11–

+aa

Reflection: Why must the denominators of rational expressions be the same before you add or subtract the rational expressions? Answer will probably mention that adding rational expressions is like adding__

fractions, and just as you can only add like fractions, you can only add like___

rational expressions._____________________________________________

How does factoring help you find the least common denominator for two rational expressions? Answer will probably make the analogy with adding/subtracting fractions, and the

way that LCDs are found by factoring in that situation._____________________

Course15 Numbers and Their Properties

Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 1, Document maflh_k.

–1 of 2–

Module: Rational Expressions: Multiply and Divide

Objective: To practice multiplying and dividing rational expressions

Name: _______________________ Date: __________________

Read each statement about multiplying and dividing. Decide whether it is true or false. When solving a multiplication or division problem with several rational

expressions, use the order of operations. That is, multiply first, then

divide. __false_____________ (true, false)

An answer is in simplified form if the numerator and denominator do not

have any common factors other than 1. __true____________ (true, false)

A simple fraction has one fraction bar. __true_____________ true, false)

A complex fraction must have a fractional numerator and a fractional

denominator. ____false_____________ (true, false)

Problem Set: Express each product in simplest form.

1. yxyx

xyx

–2•

7_2 ++

= 1 2. cba

cab

4•

3

22

= 2

23

12cba

3. 3–82•

415–5

rr

rr ++

= 10 4. 149

•632 +

+yyxy

xy =

73+yy

5. yx

xyy

+224•

623

2

= 2

4xy

KEY

Course15 Numbers and Their Properties

Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 1, Document maflh_k.

–2 of 2–

Express each quotient in simplest form.

1. 2

2

78 yy

yx÷ = 3

3

87yx 2.

aa

aa

+÷

+ 3–3

3–3 = 1

3. xyyx

yxyx

5–55–5

–

22

÷+ = 2)–(

–yxxy 4.

bb

bb 1243

4 +÷

+ = 2

2

)3( +bb

5. 1558–2

312–3

+÷

+ xx

xx =

215

Reflection: Explain how dividing rational expressions is related to multiplying rational expressions. Be sure to tell why you must invert the divisor.

The answers will vary, but it should include the idea that division is the inverse of

multiplication._____________________________________________________

________________________________________________________________

________________________________________________________________

________________________________________________________________

________________________________________________________________

Course 16 Special Equations and Inequalities

Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document magah_k.

–1 of 2–

Module: Evaluating Expressions with Absolute Value

Objective: To practice evaluating simple equations that involve absolute value or distance

Name: _______________________ Date: __________________

Fill in the blanks. Use one of the words in parentheses when choices are given. |x| = ____-x____ if x < 0

|x| = _____x____ if x ≥ 0

The absolute value symbol is treated like _____parentheses________

with regard to the order of operations.

-|-8| = ___-8_____

Absolute value is computed from the ____inner most_______________

symbol to the outermost symbol.

|x| is the distance of x from _____0 or the origin______ on a number line.

A measurement of distance must be ____positive___________.

(negative, inverted, positive, even, odd)

If you are at -5 on a number line, then you are ____5____ units from zero.

The distance between any two points a and b on a number line is given by

the algebraic expression __|a – 6|______________.

Problem Set: Evaluate the expression. 1. |-4 + 2| = ____2_____________ 2. -|1 – 6| = ____-5____________ 3. |5| + |-5| = ___10____________ 4. |3| – |7| = ____-4____________

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document magah_k.

–2 of 2–

5. | |-5| + |4| | = ___9___________ 6. | |-9| – |-3| | = ___6_____________ 7. Find |x – 8| if x = 0. __8_______ 8. Find -|2 + 3x| if x = 1. ___-5_____ 9. Find |-x + 4| if x = 6. ___2______ 10. Find |x – 2| if x = 7. ___5_______ Each pair of points lies on a number line. Find the unit distance between the two points. 11. -3,9 ____12______________ 12. 3,9 ______6______________ 13. -7,-8 ____1______________ 14. -2,2 _____4______________ 15. -1,21 ___22______________ 16. 0,18 ____18______________ Extension Activity: Below is a map of interstate highway 8 stretching from Ligurta to Gila Bend. The distance markers are given in kilometers. The map is not to scale. Use the map to fill in the blanks that follow.

A. Find the distance from Aztec to Ligurta in kilometers. ____32__________ B. Find the distance from Wellton to Sentinel in kilometers. __36_________ C. The equation |-19| = 19 refers to the distance between ___Aztec________

and ___Tacuna__________. D. The equation |9 – 21| = 12 refers to the distance between ___Sentinel___ and __Theba____________. E. The two cities that lie the same distance from Aztec are __Wellton______ and ___Gila Bend_______.


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document magbh_k.

–1 of 2–

Module: Absolute Value, Inequalities, and Interval Notation

Objective: To practice using interval notation to describe solution sets for equations with absolute values and inequalities

Name: _______________________ Date: __________________

Fill in the blanks.

A(n) ____open____________ interval is a set of real numbers that

contains neither endpoint but still contains all points in between.

A(n) ___half-open_________ interval is a set of real numbers that

includes only one endpoint.

A(n) ___unbounded_________ interval is a set of real numbers that

approaches infinity.

A(n) _____closed__________ interval is a set of real numbers that

contains both endpoints and all points in between.

|x| < 5 if and only if ___-5_____ < x < ____5_____.

x ≤ -9 or x ≥ 9 if and only if ___| x |___ ≥ 9.

For any real numbers a and b on a number line, the midpoint between

a and b is defined by this algebraic expression: ___a + ___________.

The distance from the midpoint of interval [a,b] to either endpoint is

defined by this algebraic expression: ____________________.

Problem Set: Identify the interval as open, closed, half-open, or unbounded. 1. [4,9) ____half open_______ 2. (-1,10) _____open___________

KEY

(b - a) 2

|a-b| 2


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document magbh_k.

–2 of 2–

3. x < 3 ___unbounded_______ 4. -4 ≤ x < 4 ____half open_____ 5. -∞ < x < ∞ __unbounded_______ 6. [6,7] ___closed______________ Graph the following intervals on the number line. You will need to mark your own points as needed. 7. (2,∞)

8. -3 ≤ x ≤ 6

9. { x| |x – 4| < 4 }

Find the midpoint of the interval. Some answers may be in the form of a decimal or fraction. 10. (-3,9) __3___________ 11. [2,13) ___7.5___________ 12. [-6,0] __-3___________ 13. -11 < x < -2 ___-6.5______ Express the following inequalities in interval notation. 14. { x| |x – 4| ≤ 5 } __[-1,9]______ 15. { x| |x| < 3 } ____(-3,3)________ 16. { x| |x – 2| < 4 } __(-2,6)______ 17. { x| |x + 7| ≤ 2 } __[-9,-5]_______ Reflection: Unbounded intervals approach infinity in either the positive direction, the

negative direction, or both. Define infinity (∞) using your own words. Answers will vary, but references to something that forever decreases or______

increases is the main idea. _________________________________________

_______________________________________________________________

Can infinity (∞) ever be reached? Circle one. Yes No

2

-3 6

8


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document magch_k.

–1 of 2–

Module: Graphing Linear Inequalities in 1 Variable

Objective: To practice graphing the solution sets to inequalities in 1 variable

Name: _______________________ Date: __________________

Fill in the blanks.

If you multiply or divide both sides of an inequality by a negative number,

the inequality sign will ___switch___________ direction.

A(n) _____union_________________ of two or more inequalities means

that the solutions are graphed together on a single number line.

A(n) ___intersection_____________ of two or more inequalities means

that only solutions common to all the inequalities are graphed.

When solving inequalities, the word “or” is used interchangeably with the

symbol ___U_______.

When solving inequalities, the word “and” is used interchangeably with the

symbol I .

If the union or intersection of two or more inequalities yields no solution,

then we say that the solution set is a(n) ____empty_____________ set.

Problem Set: Solve the inequality for x. Express your answer as an inequality.

1. -2x + 5 < 3 2. 3 + 5 < 2x 3. 9x – 4 ≥ x + 12

x > 1 4 < x x ≥ 2

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document magch_k.

–2 of 2–

Solve for all real numbers x. Graph the solution set on the number line. You will need to mark your own points as needed. Some problems may have no solution. 4. x > 3 or x < -2 5. -2x + 5 ≥ 2x + 13 I 3x < 12 -2 ≥ x I x < 4

6. 4x + 4 < 19 – x U 2x – 1 < 9 7. x + 6 ≤ 28 + 12x and 36 > 6x 5x < 15 -22 ≤ 11x I 6 > x x < 3 U x < 5 -2 ≤ x

8. x > 6 I 4x + 7 ≤ -13 9. x + 8 ≤ 3 or 2x – 2 ≥ -27 – 3x x ≤ -5 x ≤ -5 x ≥ -5

Extension Activity: You are studying for your final exam of the semester. Up to this point, you have received 4 exam scores of 98%, 85%, 82%, and 100%. To receive a grade of A in the class, you need an average exam score between 90% and 100% for all 5 exams including the final. Find the range of scores that you can get on the final exam to receive a grade of A in the class. Write your answer as an inequality. 98 + 85 + 82 + 100 + final > 500 (.9) = 450 100 ≥ final > 85

No solutions (empty set) All real numbers are solutions

-2 3 -2

5 -2 6


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document magdh_k.

–1 of 2–

Module: Graphing with Restrictions on the Variable

Objective: To practice graphing the solution sets to absolute value inequalities in one variable

Name: _______________________ Date: __________________

Fill in the blanks. |x| > 7 if and only if x < __-7__ or x > __7__.

-9 ≤ x ≤ 9 if and only if |x| ≤ __9__.

The solution set to |x| < -5 is a(n) ____ _empty______ set.

The solution set to |x| > -5 is ______all_______ real numbers.

|x – 3| < 6 if and only if -6 < ___x – 3____< 6.

|x + 4| > 8 if and only if ____x + 4____< -8 or ____x + 4____> 8.

Problem Set: Solve for x. Write your answer as an inequality. 1. |x – 1| > 4 2. |x + 3| ≤ 3 3. |x – 9| > -8 x > 5 -6 ≤ x ≤ 0 all real numbers or x < -3 4. |6 – x| ≥ -1 5. |x + 2| ≤ -3 6. |4 + x| < 12 all real numbers empty set -16 < x < 8

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document magdh_k.

–2 of 2–

Solve for x. Graph the solution set on the number line. You will need to mark your own points as needed. Some problems may have no solution. 7. |4x + 6| > 10 8. |3x – 9| ≤ 12 x >1 or x < -4 -1 ≤ x ≤ 7

9. |9x + 14| ≥ -4 10. |5x – 15| < 25 all real numbers -2 < x < 8

11. |7 + x – 2| ≥ 11 12. |5 – 3x + 8x| < -8 x ≥ 6 or x ≤ -16 no solution (empty set)

13. |6x + 9| ≤ 1 + 14 14. |-7x + 5x + 18| > 2 + 4 -4 ≤ x ≤ 1 x > 12 or x < 6

Reflection: Explain in your own words why the solution to |x| < -b is an empty set. When b is positive, the solution is an empty set because the absolute value on

the left will always be positive, thus it will never be negative or smaller than a

negative number.

Explain in your own words why the solution to |x| > -b is all real numbers. When b is positive, the solution is all real number because the absolute value on

the left will always be positive. Therefore it will always be greater than a negative

number.


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mageh_k.

–1 of 2–

Module: Graphing Solution Sets of Associated Inequalities

Objective: To practice graphing the solution sets to quadratic inequalities in one variable and to other unions of solution sets

Name: _______________________ Date: __________________

Fill in the blanks.

An inequality of the form ax2 + bx + c < 0 is called a ___quadratic____

inequality.

For all real numbers a and b, a • b < 0 if a < 0 and ______b > 0_______ or

a > 0 and ______b < 0_______ .

For all real numbers a and b, a • b > 0 if ______a > 0_______ and b > 0 or

______a < 0_______ and b < 0.

The intersection of x ≥ -3 and x ≥ 2 is 2≥x .

The union of x ≥ -3 and x ≥ 2 is 3-≥x .

The intersection of x < 5 and x > 9 yields a(n) ______empty_______ set.

If you multiply or divide both sides of an inequality by x in a problem, you

must split the problem into two parts. For part 1, assume that

______x > 0________. For part 2, assume that ______x < 0_______.

Problem Set: Solving a quadratic inequality requires four basic steps. Write the correct number of the step in the blank provided. Step __2__ Factor the left side of the inequality.

Step __1__ Write the inequality in standard form.

Step __4__ Check your sign assumptions by solving for the variable.

Step __3__ Assume a sign, positive or negative, for each factor in the pair.

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mageh_k.

–2 of 2–

Solve for all real numbers x. Graph the solution set on the number line. You will need to mark your own points as needed. 1. x2 – 2x – 15 > 0 2. x2 + 6x ≥ 0 (x – 5)(x + 3) > 0 x(x + 6) ≥ 0 x – 5 > 0 and x + 3 > 0 or x – 5 < 0 and x + 3 < 0 x ≥ 0 and x + 6 ≥ 0 or x ≤ 0 and x + 6 ≤ 0 x > 5 and x > -3 or x < 5 and x < -3 x ≥ 0 and x ≥ -6 or x ≤ 0 and x ≤ -6 x > 5 or x < -3 x ≥ 0 or x ≤ -6

3. x2 – 5x ≤ -4 4. x2 + 1 > -2x x2 – 5x + 4 ≤ 0 x2 + 2x + 1 > 0 (x – 4)(x – 1) ≤ 0 (x + 1)(x + 1) > 0 x – 4 ≤ 0 and x – 1 ≥ 0 or x – 4 ≥ 0 and x – 1 ≤ 0 x + 1 > 0 or x + 1 < 0 x ≤ 4 and x ≥ 1 or x ≥ 4 and x ≤ 1 x > -1 or x < -1 1 ≤ x ≤ 4 or no solution

5. x +

x14 < -9 6. x >

x24 – 5

x2 + 14 < -9x x2 > 24 – 5x x2 + 9x + 14 < 0 x2 + 5x – 24 > 0 (x + 2)(x + 7) < 0 (x + 8)(x – 3) > 0 if x ≥ 0 if x > 0 x + 2 < 0 and x + 7 > 0 or x + 2 > 0 and x + 7 < 0 x + 8 > 0 and x – 3 > 0 or x + 8 < 0 and x – 3 < 0 x < -2 and x > -7 or x > -2 and x < -7 x > -8 and x > 3 or x < -8 and x < 3 Since x > 0 → no solution x > 3 or x < -8 Since x > 0 → x > 3 if x < 0 x + 2 > 0 and x + 7 > 0 or x + 2 < 0 and x + 7 < 0 if x < 0 x > -2 and x > -7 or x < -2 and x < -7 x + 8 < 0 and x – 3 > 0 or x + 8 > 0 and x – 3 < 0 x > -2 or x < -7 x < -8 and x > 3 or x > -8 and x < 3 Since x < 0 → x < -7 U – 2 < x < 0 -8 < x < 3 Since x < 0 → -8 < x < 0

Extension Activity: Explain in words or numbers why the quadratic inequality x2 + 2x + 1 < 0 has no solution (empty set). (x + 1)(x + 1) < 0; x + 1 < 0 and x + 1 > 0; x < -1 and x > -1 → no solution

-3 5 -6

4 1 -1

-7 -2 -8 3

Course 17 Coordinates and Curves

Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mahah_k.

–1 of 2–

Module: Calculating the Slope of a Line

Objective: To practice calculating the slope of a line from a graph or from two points

Name: _______________________ Date: __________________

Fill in the blanks.

The slope of a line can be determined by finding the rise over

the run .

Lines that move from upper left to lower right have a negative

slope.

Lines that move from lower left to upper right have a positive

slope.

A line with a slope of zero is horizontal .

A line with an undefined slope is vertical .

Problem Set: Determine the slope of the line that contains each pair of points. 1. (4,2) and (10,14) 2. (-3,-5) and (1,8) 3. (2,-3) and (-4,-7)

14 2 1210 4 6

m −= = =

− 2 8 5 13

1 3 4m − −= =

− − 7 3 4 2

4 2 6 3m − − − −= = =

− − −

4. (3,6) and (3,-10) 5. (0,-8) and (-4,-8) 6. (17, 12) and (3,2)

10 6 163 3 0

m − − −= = =

−no

slope 8 8 0 0

4 0 4m − − −= = =

− − − 2 12 10 5

3 17 14 7m − −= = =

− −

KEY


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–2 of 2–

7. (-2.3,.4) and (5.7,-1.6) 8. (4,1.8) and (6.2,10) 9. 2 7,3 8

and 1 1,9 2

−

1.6 .4 2 15.7 2.3 8 4

m − − −= = = −

− − 10 1.8 8.2

6.2 4 2.2m −= =

−

1 7 32 8 81 2 79 3 9

m

−−

= =−

− −

41415

11 115

= = 2756

=

Use the graph to find two additional points that lie on the line that has the given point and the given slope.

10. (1,3); m = 12

11. ( ) 2-2,-4 ; = 5

m

(3,4) (-1,2) (-3,1) (-5,0) (3,-2) (8,0) Extension Activity: The points (1,5), (6,20), and (k,41) all lie on the same line. Find k.

20 5 15 36 1 5

m −= = =

− so . . . 41 20 3

6k−

=−

21 36k=

− 6 7k − = so 13k =

Reflection: Using the idea that slope = rise

run, explain why vertical lines have no slope.

Any vertical line will have a “rise” that is represented by some quantity. The

“run”, however, will always be zero. Since any number divided by zero is

undefined, vertical lines have no slope.

6

2 4

-6

y

x-2

2

6-4-6-2

-4

6

2 4

-6

y

x-2

2

6-4-6-2

-4


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–1 of 3–

Module: Point-Slope and Slope-Intercept Forms of Equations

Objective: To practice how to write the point-slope and the slope-intercept forms of linear equations

Name: _______________________ Date: __________________

Fill in the blanks. Use one of the words in parentheses when choices are given. Given the slope m of a line and a point (a,b) on the line, the

point-slope (point-slope, slope-intercept) form of the

equation of a line is y – b = m(x – a).

The slope-intercept (point-slope, slope-intercept) form of the

equation of a line is y = mx + b where m is the slope

and b is the y-intercept .

All vertical (horizontal, vertical) lines are defined by

equations in the form x = a where a is a real number.

Problem Set: On a separate piece of paper, write the equation of the line in point-slope form that satisfies the given conditions.

1. The line passes through (2,-4) and has a slope of -21 . y + 4 = -

21 (x – 2)

2. The line passes through (-3,-8) and has a slope of 34 . y + 8 =

34 (x + 3)

3. The line passes through (6,12) and has a slope of -53 . y – 12 = -

53 (x – 6)

4. The line passes through (-5,-7) and has slope of 6. y + 7 = 6(x + 5)

KEY



–2 of 3–

On a separate piece of paper, write the equation of the line in slope-intercept form that satisfies the given conditions.

1. The line passes through the points (1,-8) and (-2,4). y = -4x – 4

2. The line passes through the points (3,5) and (7,11). y = 23 x +

21

3. The line passes through the points (-9,-2) and (-5,6). y = 2x + 16 4. The line passes through the points (-2,-4) and (0,8). y = 6x + 8

Determine if the equation is true or false. On a separate piece of paper, show work to support each answer. 1. True False The slope of a line through the points (4,2) and (6,3) is 2.

The slope of a line through the points (4,2) and (6,3) is 21 .

2. True A line represented by the equation x = -2 is a horizontal line. A line represented by the equation x = -2 is a vertical line. 3. True A line with slope of -3 that passes through (-1,4) has this

equation written in point-slope form: y + 4 = -3(x – 1). A line with slope of -3 that passes through (-1,4) has this equation written in point-slope form: y – 4 = -3(x + 1). 4. False An equation in point-slope form can be rewritten in slope-

intercept form. Examples will vary. 5. False It is possible to determine the equation of a line in slope-

intercept form simply by knowing two points that are on the line.

Examples will vary. Extension Activity: Triangle ABC has vertices at A (7,6), B (1,1), and C (7,-4). Triangle EFG has vertices at E (4,2), F (-2,-3), and G (4,-8). Sketch each triangle. Write the slope-intercept form for the equation of the line that represents each side of each triangle. With a partner, state 3 to 5 observations about the information you have compiled. (Hint: You may want to discuss the positions of the triangles, the slopes of the sides of each triangle, the size of the triangles, the lengths of the sides of the triangles, etc.)

False

False

False

True

True



–3 of 3–

A

B

C

E

F

G

y

x

AB: y = 65 x +

61 BC: y = -

65 x +

611 AC: x = 7

EF: y = 65 x –

34 FG: y = -

65 x –

314 EG: x = 4

Observations: (will vary) • ABC and EFG are the same size • EFG is ABC shifted left 3 units and down 4 units.

• Each triangle has one side which is vertical, one side with a slope of 65 , and

one side with a slope of -65 .

• AB || EF, BC || FG, AC || EG.


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–1 of 2–

Module: Equation of a Line Given a Point and Parallel Line

Objective: Finding the equation of a line through a point and parallel to a given line

Name: _______________________ Date: __________________

Fill in the blanks.

If two lines are parallel, they will have the same __slope______________.

If the equation of a line is given in the form bmxy += , you can quickly

determine the ____slope_________ and the__ y-intercept___________.

The equation ( )243 −=− xy is in ___point-slope_____ form.

Name a point that ( )243 −=− xy passes through. _(2, 3)______________

(other possible answers).

All lines parallel to the line ( )243 −=− xy will have a slope of ___4_____.

Problem Set: Determine the equation that satisfies each of the following. 1. passes through (5,2) and is 2. passes through (-3,4) and is

parallel to 132

+= xy . parallel to ( )247 +=+ xy .

( )5322 −+− xy or

34

32

−= xy ( )344 +=− xy or 164 += xy

3. passes through (-2,-6) and is 4. passes through (-6,10) and is

parallel to 2054 =+ yx . parallel to 1121

+= xy .

454

+−= xy ( )62110 +=− xy or 13

21

+= xy

( )2546 +−=+ xy or

538

54

−−+ xy

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 1, Document mahch_k.

–2 of 2–

5. passes through (0,4) and is 6. passes through (3,6) and is

parallel to 155 =+ yx . parallel to ( )2323 −=+ xy .

351

+−= xy , ( )0514 −−=− xy , ( )3

326 −=− xy or 4

32

+= xy

xy514 −=− or 4

51

+−= xy

7. passes through (4,5) and is 8. passes through (-1,5) and is

parallel to 7=x . parallel to .1+y

4=x 5=y

Extension Activity: Are any of the following lines parallel? If so, name those that are parallel. Explain.

12104 =− yx 1352

++ xy ( )64.1 +=+ xy

56

52

−= xy 57

52

+= xy

Explanation: All three lines are parallel because they each have a slope of . All three____

lines are distinct. Reflection: If two lines have different slopes, describe at least one thing that is certain to be true. If two lines have different slopes, they will certainly intersect at exactly one point.

25


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–1 of 2–

Module: Equation of a Line Given a Point and Perpendicular Line

Objective: To practice finding the equation of a line through a point and parallel to a given line

Name: _______________________ Date: __________________

Fill in the blanks.

Two lines are perpendicular if the7 intersect and form a _____right______

angle.

If two lines are perpendicular, the product of their respective slopes is

equal to _______-1___________.

Another way of determining if two lines are perpendicular is to chec if one

slope is the ______negative_______ ______reciprocal_____ of the other.

The negative reciprocal of ba is

ab

−

If a line has a slope of 41

− , a line perpendicular to it will have a slope of

________4___________.

Problem Set: Determine an equation that satisfies each of the following. 1. Find the equation of a line 2. Find the equation of a line that

that contains (2, 7) and is that contains (-4, 5) and is

perpendicular to 132

−−= xy . perpendicular to .8=+ yx

( )2237 −=− xy or 4

23

+= xy ( )415 +=− xy or 9+= xy

KEY

.


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–2 of 2–

3. Find the equation of a line 4. Find the equation of a line

that contains (3, 4) and is passes through (-2, -4) and is

perpendicular to ( ).1218 −−=− xy perpendicular to .75 += xy

( )324 −=+ xy or 102 −= xy ( )2514 +−=+ xy or

522

51

−−= xy

5. Find the equation of a line 6. Find the equation of the

that passes through (14, -6) line that passes through

and is perpendicular to (-3, -7) and is perpendicular

2173 =− yx . to 3+x . (Think)

( )14376 −−=+ xy or

380

37

+−= xy 7−=y

Determine if the following pair of equations are parallel, perpendicular, or neither.

7. ;632 =+ yx ( )4326 +−=+ xy 8. 3

71

+= xy ; 57 =+ yx

parallel perpendicular

9. 2484 =− yx ; 62 −=+− yx 10. 1243 =+ yx ; ( )4431 −=+ xy

neither (same line) neither Extension Activity: Find the value of k so that the line passing through (-3,5) and (5,k) is perpendicular to y + 2x – 4.

Any line perpendicular to 42 −= xy will have a slope of 21

− . So:

21

)3(–55 −=−−k or

21

85

−=−k

45 −=−k

1=k


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–1 of 2–

Module: Perpendicular Bisector of a Line Segment

Objective: To practice finding the equation of the perpendicular bisector of a line segment

Name: _______________________ Date: __________________

Fill in the blanks.

If two lines are perpendicular, their slopes will be ______negative______

____reciprocals______.

A perpendicular bisector of a line segment is not only perpendicular to the

segment but passes through the ______midpoint_______ of the segment.

To find the midpoint of a segment whose endpoints are ( )11 , yx , and ( )22 , yx ,

the x-coordinate of the midpoint is found by taking 221 xx +

and

the y-coordinate of the midpoint is found by taking 221 yy +

After finding a segment’s ______midpoint______ and _____slope______,

it is usually easiest to put the equation of the perpendicular bisector in

____point-slope______ form.

It is possible to find the perpendicular bisector of any line just as you can find

the perpendicular bisector of line segments. (true, false)

KEY

.

.


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–2 of 2–

Problem Set: Find the slope of the line segment with the given endpoints. Then give the slope of any line perpendicular to the segment. 1. (4,-6) and (3,7) 2. (-8,2) and (6,-4)

( ) 13113

43)6(7

1 −=−

=−−−=m ( ) 7

3146

)8(624

1 −=−=−−−−=m

( )131

2 =m ( )37

2 =m

3. (2,2) and (1,-9) 4. (-2,10) and (4,-3)

( ) 11111

2129

1 =−−

=−−−

=m ( ) 613

)2(4103

1−=

−−−−=m

( )111

2 −=m ( )136

2 =m

Determine the midpoint of the following line segments with the given endpoints. 5. (-8,10) and (-2,2) 6. (-4,-6) and (4,4)

( )6,52210,2

)2(8 −=

+−+− ( )1,0

246,

244

−=

+−+−

7. (-6,-8) and (-4,-2) 8. (10,12) and (3,3)

( )5,52)2(8,2

)4(6 −−=

−+−−+− ( )5.7,5.6

215,

213

2312,

2310

=

=

++

Determine the equation of the perpendicular bisector of the line segment with the following endpoints. 9. (2,4) and (8,10) 10. (-4,-2) and (16,8)

( )517 −=− xy ( )6213 −=− xy

11. (-3,2) and (5,-4) 12. (2,8) and (6,0)

( )1431 −−=+ xy ( )424 −−=− xy

13. (5,7) and (-1,-3) 14. (0,6) and (-8,0)

( )2352 −=− xy ( )4

433 +=− xy

15. (1,1) and (11,13) 16. (-10,6) and (-2,-2)

( )6567 −=− xy ( )612 +−=− xy


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–1 of 2–

Module: Distance between 2 Points

Objective: To practice finding the distance between 2 points

Name: _______________________ Date: __________________

Fill in the blanks.

The distance formula is derived from the Pythagorean theorem, which is

2 2 2c a b= + .

It is often helpful when finding the distance between two points to create a

right (left, right) triangle.

Given two points (x1,y1) and (x2,y2), the distance between them is

d = √ 212

212 )()( yyxx −+−

Problem Set: Find the distance between the following points. Round your answer to the nearest hundredth. 1. (1,2) and (5,7) 2. (-3,2) and (-7,10)

( ) ( )2 25 1 7 2d = − + − ( ) ( )2 27 3 10 2d = − − − + −

2 24 5= + ( )2 24 8= − +

2516 += 16 64= + 41 6.40= ≈ 80 8.94= ≈ 3. (1,-4) and (3,-13) 4. (-2,-4) and (-8,-1)

( ) ( )2 23 1 13 4d = − + − − − ( ) ( )2 28 2 1 4d = − − − + − − −

( ) ( )2 22 9= + − ( ) ( )2 26 3= − +

4 81= + 36 9= + 85 9.22= ≈ 45 6.71= ≈

KEY


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–2 of 2–

5. (-1,2) and (5,-6) 6. (-5,-4) and (2,9)

( ) ( )2 25 1 6 2d = − − + − − ( ) ( )2 22 5 9 4d = − − + − −

( )226 8= + − 2 27 13= +

36 64= + 49 169= + 100= 7614218 .≈= = 10 7. (0,6) and (11,0) 8. (4,3) and (0,-4)

( ) ( )2 211 0 0 6d = − + − ( ) ( )2 20 4 4 3d = − + − −

( )2211 6= + − ( ) ( )2 24 7= − + −

121 36= + 16 49= + 157 12.53= ≈ 65 8.06= ≈ 9. (2,8) and (-5,8) 10. (-9,-2) and (5,7)

( ) ( )2 25 2 8 8d = − − + − ( ) ( )2 25 9 7 2d = − − + − −

( )2 27 0= − + 2 214 9= +

49= 196 81= + = 7 277 16.64= ≈ Extension Activity: The distance between (6,2) and (11,y) is 13. Find y.

22 )2()611(13 −+−= y 22 )2(513 −+= y 2)2(2513 −+= y

( )2169 25 2y= + −

( )2144 2y= −

( )2144 2y= − 12 = |y-2| so: y −2 = 12 y − 2 = −12 y = 14 or y = -10

5

5

y

x-5

-5

10

10

-10

-10


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mahgh_k.

–1 of 3–

Module: Distance between a Point and a Line

Objective: To practice finding the distance between a line and a point not on the line

Name: _______________________ Date: __________________

Fill in the blanks. Use one of the words in parentheses when choices are given. The distance from a line to a point not on the line is always represented by

the perpendicular (parallel, perpendicular) distance between them.

If two lines are perpendicular, their slopes will be negative

(positive, negative) reciprocals of each other.

Given two equations of perpendicular lines, one can find the point of

intersection by solving a system (system, group) of equations.

Once the point of intersection is found for two perpendicular lines, use the

distance formula (distance formula, distance rule)

to find the distance between the original point and the point of intersection.

The distance between two points is always positive .

(negative, positive)

Problem Set: Find the distance from the given point to the given line. Round your answer to the nearest hundredth.

1. (-1,4) and y = x –1 2. (4,4) and y + 1 = - )4–(21 x

( )4 1 1y x− = − + ( )4 2 4y x− = − 3y x= − + - equation of ⊥ 2 4y x= − - equation of ⊥

1y x= − 1 12

y x= − +

KEY



–2 of 3–

3y x= − + 12 4 12

x x− = − +

2 2y = 4 8 2x x− = − +

1y = ⇒ 2x = 5 10x = 2x = ⇒ 0y = Distance between ( )1,4− and (2,1) Distance between (4,4) and (2,0)

( ) ( )2 22 1 1 4d = − − + − ( ) ( )2 22 4 0 4d = − + −

( )223 3= + − ( ) ( )2 22 4= − + −

18 4.24= ≈ 4 16 20 4.47= + = ≈

3. (-3,1) and y + 4 = )2–(31 x 4. (6,1) and -2x + 4 = 4

( )1 3 3y x− = − + ( )11 62

y x− = − −

3 8y x= − − - equation of ⊥ 1 42

y x= − + - equation of ⊥

1 243 3

y x= − 2 4x y− + =

3 14y x= − 1 42

y x= − +

( )3 3 8 14x x− − = − 2 8y x= − + 9 24 14x x− − = − ( )2 2 4 8x x+ = − + 10 10x− = 4 8 8x x+ = − + 1x = − ⇒ 5y = − 5 0x =

0x = ⇒ 4y = Distance between (-3,1) and (-1,-5) Distance between (6,1) and (0,4)

( ) ( )2 21 3 5 1d = − − − + − − ( ) ( )2 20 6 4 1d = − + −

( )222 6= + − ( ) ( )2 26 3= − +

40 6.32= ≈ 36 9= + 45 6.71= ≈



–3 of 3–

5. (-3,-1) and 3x + 4y = 12 6. (-7,-2) and y = 7x – 3

( )41 33

y x+ = + ( )12 77

y x+ = − +

4 33

y x= + 1 37

y x= − −

3 4 12x y+ = 7 3y x= −

4 33

y x= + 1 37

y x= − −

43 4 3 123

x x + + =

1 3 7 37x x− − = −

163 12 123

x x+ + = 21 49 21x x− − = −

9 16 36 36x x+ + = 50 0x− = 25 0x = 0=x 3−=y

0 3x y= =

Distance between (-3,-1) and (0,3) Distance between (-7,-2) and (0,-3)

( ) ( )2 26 3 3 1d = − − + − − ( ) ( )2 20 7 3 2d = − − + − − −

= 2 23 4= ( )227 1= + −

9 16= + 49 1= + = 5 50 7.07= ≈

Reflection: When describing the distance between a line and a point not on the line, why do we use the perpendicular distance? There are many distances between a point and a line. We need to make the

assumption that it is the shortest distance that we want to find. The shortest

distance will always be found along the perpendicular.


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–1 of 3–

Module: Distance and Circles

Objective: To practice finding the radius and center of a circle given an equation and points on the circle

Name: _______________________ Date: __________________

Fill in the blanks.

The standard form of the equation of a circle allows you to identify the

radius and the center of a circle.

The process of completing the square can change an

equation of the form ax2 + ay2 + bx + cy + d = 0 into the standard form

of the equation of a circle, which is __(x – h)2 + (y – k)2 = r2___.

Problem Set: Find the radius and center of the circles defined by the following equations. Show your work. 1. x2 + y2 – 2x – 4y + 1 = 0 2. 2x2 + 2y2 + 4x – 12y – 30 = 0

)2,1(,24

4)2–()1–(51)44–()12–(

1)4–()2–(

22

22

22

===

=+

+−=+++

−=+

centerr

yxyyxx

yyxx

30)6–(2)2(2 22 =++ yyxx 18230)96–(2)12(2 22 ++=++++ yyxx

250

2)3(2

2)1(2 22

=+

++ yx

25)3–()1( 22 =++ yx )3,1(,525 −=== centerr

3. x2 + y2 – 4x – 5 = 0

5)()4–( 22 =+ yxx 45)()44–( 22 +=++ yxx

9)2–( 22 =+ yx )0,2(,39 === centerr

4. 25x2 + 25y2 + 150x + 50y + 246 = 0 246)2(25)6(25 22 −=+++ yyxx

25225246)12(25)96(25 22 ++−=+++++ yyxx

254

25)1(25

25)3(25 22

=+

++ yx

254)1()3( 22 =+++ yx

)1,3(,52

254

−−=== centerr

KEY



–2 of 3–

Now find the radius and center of the circles that pass through the given points. Round your answers to the nearest hundredth where necessary. Show your work.

5. (1,0), (2,1), and (0,-2) 6. (-2,3), (0,-3), and (4,-3)

(1 – h)2 + (0 – k)2 = r 2 (2 – h)2 + (1 – k)2 = r 2

(0 – h)2 + (-2 – k)2 = r 2

expand and simplify (1 – 2h + h2) + k2 = r 2

(4 – 4h + h2) + (1 – 2k + k2) = r 2

(h2) + (4 + 4k + k2) = r 2

which leads to two equations: 2h + 4k – 4 = 0 2h + 4k + 3 = 0

(-2 – h)2 + (3 – k)2 = r 2 (0 – h)2 + (-3 – k)2 = r 2

(4 – h)2 + (-3 – k)2 = r 2

expand and simplify (4 + 4h + h2) + (9 – 6k + k2) = r 2

(h2) + (9 + 6k + k2) = r 2 (16 + 8h + h2) + (9 + 6k + k2) = r2

use addition method to eliminate r and you get : 8h = 16, -4h + 12k = 4 , so h = 2 k = 1 r = 2 47.45 =

7. (3,2), (-1,-2), and (-5,2) 8. (-3,0), (0,3), and (3,0)

(3 – h)2 + (2 – k)2 = r2 (-1 – h)2 + (-2 – k)2 = r2 (-5 – h)2 + (2 – k)2 = r2

expand and simplify (9 – 6h + h2) + (4 – 4k + k2) = r2 (1 + 2h + h2) + (4 + 4k + k2) = r2 (25 + 10h + h2) + (4 – 4k + k2) = r2

use addition method to eliminate r and you get : 16h = -16, 8h + 8k = 8, so h = -1 k = 2 r = 4

(-3 – h)2 + (0 – k)2 = r2 (0 – h)2 + (3– k)2 = r2 (3 – h)2 + (0 – k)2 = r2

expand and simplify (9 + 6h + h2) + (k2) = r2 (h2) + (9 – 6k + k2) = r 2

(9 – 6h + h2) + (k2) = r2

use addition method to eliminate r and you get : 12h = 0, 6k = 0, so h = 0 k = 0 r = 3



–3 of 3–

9. (1,1), (0,2), and (-1,1) 10. (-2,0), (0,2), and (2,0)

(1 – h)2 + (1 – k)2 = r 2 (0 – h)2 + (2 – k)2 = r 2 (-1 – h)2 + (1 – k)2 = r 2

expand and simplify (1 – 2h + h2) + (1 – 2k + k2) = r2 (h2) + (4 – 4k + k2) = r2

(1 + 2h + h2) + (1 – 2k + k2) = r2

use addition method to eliminate r and you get : 4h = 0, 2h – 2k = -2, so h = 0 k = 1 r = 1

(-2 – h)2 + (0 – k)2 = r 2 (0 – h)2 + (2 – k)2 = r 2 (2 – h)2 + (0 – k)2 = r 2 expand and simplify (4 + 4h + h2) + k2 = r2

h2 + (4 – 4k + k2) = r2

(4 – 4h + h2) + k2 = r 2

use addition method to eliminate r and you get : h = 0 k = 0 r = 2


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mahih_k.

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Module: Parabola and Its Intercepts

Objective: To practice finding the x- and y-intercepts of a parabola

Name: _______________________ Date: __________________

Complete each statement. ____Parabolas ______have equations of the form y = ax2 + bx + c where

a, b, and c are constants and a ≠ 0.

Parabolas are graphs of _____quadratic______ equations. They are

smooth, symmetric, u-shaped curves when graphed.

A(n) ____x-intercept_____ of a graph is any point at which the graph

crosses the x-axis. It is the value of ______x_____________ when y = 0.

Two methods used to calculate the x-intercepts of an equation are

____factoring_______ and the ___quadratic________

____formula________. Using a graphing calculator is another common

way to find intercepts.

The ____y-intercept_____ of a graph is the point at which the graph

crosses the y-axis. It is the value of ______y_____________ when x = 0.

To find the y-intercept, let ______x = 0_________ and solve the new

equation for _________y_________.

Problem Set: Find the x-intercepts and y-intercept for each parabola. Write your answers in (x,y) form. Round, if necessary, to the nearest hundredth. 1. y = x2 – 6x – 7 x-int: __(-1,0)___ __(7,0)____ y-int: _(0,-7)____

0 = x2 – 6x – 7 y = 02 – 6 (0) – 7 = (x – 7) (x + 1) y = -7 x – 7 = 0 x +1 = 0 x = 7 x = -1

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mahih_k.

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2. y = 2x2 + 4x – 1 x-int: _(-2.22,0)__ _(0.22,0)__ y-int: _(0,-1)____ a = 2 b = 4 c = -1

x = )2(2

)1)(2(4–44 2 −±− x = 22.0

4244 =+− y = 2(0)2 + 4(0) –1

= 4

8164 +±− x = 22.2424–4 −=− = -1

= 4244 ±−

3. y = 2x2 – 4x – 6 x-int: __(3,0) ___ __(-1,0)___ y-int: __(0,-6)____ 0 = 2x2 – 4x – 6 y = 2(0)2 – 4(0) – 6 = 2(x2 – 2x – 3) = -6 = 2(x – 3)(x +1) x –3 = 0 x + 1 = 0 x = 3 x = -1 4. y = 5x2 – 8x – 4 x-int: _(2,0)_____ __(-0.40,0) y-int: _(0,-4)____ a = 5 b = 8 c = -4

y = 5(0)2 – 8(0) – 4

x = )5(2

)4)(5(4–)8()8( 2 −−±− x = 2

1020

10128 ==+

= 10

80648 +± x =52

104

1012–8 −=−=

= 101448 ± = -0.40

= 10128 ±

Extension Activity: Ryan is studying parabolas and intercepts at school. One evening, while he is discussing his classes with his father, he learns some interesting information. His father, who is an engineer, mentions that it is possible to find the equation of a parabola when you know only the x-intercepts. Unfortunately, before Ryan can ask how this is done, his father must leave. So, Ryan needs your help! On the back of this paper, explain how it is possible to determine the equation of a parabola by knowing only the x-intercepts. (Hint: Pick two x-intercepts. Then, think about using the factoring method, which is used to find x-intercepts, in reverse.) Example: Suppose x-int are (-4,0) (5,0) x = -4 x = 5 x + 4 = 0 x – 5 = 0 (x + 4) (x – 5) = 0 x2 – x – 20 = 0 y = x2 – x – 20


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mahjh_k.

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Module: Parabola and Its Vertex

Objective: To practice how to find the vertex of a parabola Name: _______________________ Date: __________________

Complete each statement. The _____vertex_________ of a parabola is the point at which the

parabola reaches a maximum or minimum value.

A parabola that opens downward has a vertex that is called a

__maximum__________, or highest point. A parabola that opens upward

has a vertex that is called a ____minimum_________, or lowest point.

The axis of symmetry____ divides the graph into two pieces that are

mirror images of each other.

The equation of a parabola written in standard form is ax2 + bx + c = 0,

where a, b, and c are constants and a ≠ 0. To graph a parabola, it is often

helpful to rewrite the equation in the form y – k = a(x – h)2, where (h,k) is

the ____vertex_________ and x = h is the _axis of symmetry___. The

method of __completing the square __ is used to rewrite the equation into

the new form.

Problem Set: For each parabola:

a. State the direction the parabola opens: up or down. b. Find the vertex. State if it is a maximum or minimum. c. Find the equation for the axis of symmetry.

Additional paper may be necessary. 1. y = -3x2 + 6x + 10 2. y = 2x2 – 12x + 13

a. down a. up b. (1, 13) max b. (3, -5) min c. x = 1 c. x = 3

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mahjh_k.

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3. y = 5x2 + 20x + 14 4. 9221 2 ++−= xxy

a. up a. down b. (-2, -6) min b. (2, 11) max c. x = -2 c. x = 2 For each parabola:

a. State the direction the parabola opens: up or down. b. Find the vertex. State if it is a maximum or minimum. c. Find the equation of the axis of symmetry. d. Find the x-intercepts and y-intercept. e. Graph the parabola.

Additional paper may be necessary. 1. y = x2 – 6x + 4 2. y = -x2 + 4x + 1 a. up a. down b. (3, -5) min b. (2, 5) max c. 3 = x c. x = 2 d. y = 4 d. y = 1 x = 3+ 5 , 3 - 5 x = 2 ± 5 e. e.

Extension Activity: A football travels in a parabolic path when it is kicked. Suppose a player kicks a 36-yard punt. The path of the football can be modeled by the equation y = -0.042x2 + 1.5x +1 where x is the horizontal distance the football travels and y is the height of the ball, and x and y are measured in yards. Answer the following questions on another sheet of paper.

a. What is the maximum height of the football’s path? (Hint: Find the vertex.)

Answer: about 14.4 yards

b. The player was at the 23-yard line when the football was kicked.

Approximately, which yard line did the ball cross when it was at its

maximum height? Answer: 41 yards


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mahkh_k.

–1 of 3–

Module: Ellipse

Objective: To practice finding the center, vertices, and foci of an ellipse

Name: _______________________ Date: __________________

Complete each statement. An ellipse is the set of all points in a plane such that the sum

of their distances from two fixed points is a constant. It looks somewhat

like a “stretched out” circle.

The foci of the ellipse are the two fixed points located inside the

ellipse. Just one of these points is called a focus of the ellipse.

The very middle of the ellipse is called the center . It is the

midpoint of the line connecting the two foci.

Two vertices of an ellipse are located along the major

located along the minor axis, which makes up the width of the

ellipse.

Keep this helpful information in mind when working with ellipses. The equation of an ellipse can be written in either of the following forms:

( ) ( ) 1––2

2

2

2

=+b

kya

hx (horizontal major axis)

( ) ( ) 1––2

2

2

2

=+a

kyb

hx (vertical major axis)

In both cases, (h,k) is the center, and c2 = a2 – b2.

Distance between foci = 2c

Length of the major axis = 2a

Length of the minor axis = 2b

KEY



–2 of 3–

Problem Set: On the back of this paper, find the center, major axis vertices, minor axis vertices, and foci of each ellipse. Round, if necessary, to the nearest hundredth. Then, graph the ellipse.

1. ( ) ( ) 193

161– 22

=+

+yx 2. ( ) ( ) 1

95

41 22

=+

+− yx

center: (1,-3) center: (1,-5) minor axis vertices: (1,0) (1,-6) minor axis vertices:

(-1,-5)(3,-5) major axis vertices: (-3,-3)(5,-3) major axis vertices: (1,-2)(1,-8) foci: (-1.65,-3)(3.64,-3) foci: (1,-2.76)(1,-7.24)

y

x

1)

y

x

2)

3. ( ) ( ) 116

125

2 22

=+

++ yx 4. 1

4925

22

=+yx

center: (-2,-1) center: (0,0) minor axis vertices: (-2,3)(-2,-5) minor axis vertices:

(-5,0)(5,0) major axis vertices: (-7,-1)(3,-1) major axis vertices: (0,-7)(0,7) foci: (-5,-1)(1,-1) foci: (0,4.90)(0,-4.90)

y

x

3)

y

x

4)



–3 of 3–

5. ( ) 18136

1 22

=++ yx 6. ( ) 1

252

64

22

=−

+yx

center: (-1,0) center: ((0,2) minor axis vertices: (-7,0)(5,0) minor axis vertices: (0,7)(0,-3) major axis vertices: (-1,9)(-1,-9) major axis vertices: (-8,2)(8,2) foci: (-1,6.71)(-1,-6,71) foci: (-6.24,2)(6.24,2)

y

x

5)

y

x

6)

Extension Activity: If you know the center, the length of the major axis, and the length of the minor axis, you can find the equation of an ellipse. On the back of this paper, use the given information about each ellipse to write its equation. 1. center: (-4,-2) 2. center: (5,3)

length of major axis: 12 (horizontal) length of major axis: 14 ⇒ a=6 (vertical) ⇒ a=7 length of minor axis: 6 ⇒ b=3 length of minor axis: 8 ⇒ b=4 ( ) ( )2 24 2

136 9x y+ +

+ = ( ) ( )2 25 31

16 49x y− −

+ =

Reflection: The tutorial mentioned “whispering” galleries. On the back of this paper, discuss the unique characteristics of “whispering” galleries. In an elliptic room, the sound waves originating at one focus are reflected off the walls to converge at the other focus. The length of each sound path is the same.


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mahlh_k.

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Module: Hyperbola

Objective: To practice finding the center, vertices, and foci of a hyperbola

Name: _______________________ Date: __________________

Complete each statement. A ____hyperbola_______ is defined as the set of all points in a plane such that

the difference from two fixed points is a constant.

The two fixed points of a hyperbola are called the _______foci_________.

The line drawn through the foci and the center of a hyperbola intersects the

hyperbola in two points called the ____vertices_________.

The equation of an ellipse involves addition, whereas the equation of a

___hyperbola________ involves ___subtraction_________.

____Asymptotes_____ are the two lines on a graph that the hyperbola

approaches, but never touches. They help to define its shape.

Keep this helpful information in mind when working with hyperbolas. The equation of an ellipse can be written in one of the following forms:

( ) ( ) 1–––2

2

2

2

=bky

ahx (opens left and right)

( ) ( ) 1–––2

2

2

2

=bhx

aky (opens up and down)

For both cases, (h,k) is the center and c2 = a2 + b2.

The values of a and b help determine where the fundamental rectangle is

located, so that the hyperbola can be graphed.

The foci are located c units from the center of the hyperbola.

The asymptotes of a hyperbola have one of the following forms:

( )hxabky –– ±= (when the hyperbola opens left and right)

( )hxbaky –– ±= (when the hyperbola opens up and down)

KEY



–2 of 3–

Problem Set: On the back of this paper, find the center, direction of opening, vertices, equations of the asymptotes, and foci. Round, if necessary, to the nearest hundredth. Then, graph the hyperbola. Be sure to draw the fundamental rectangle and asymptotes in your sketch.

1. ( ) ( ) 141

93 22

=+− yx – 2. ( ) ( ) 1

361

163 22

=++ xy –

center: (3,-1) center: (-1,-3) direction of opening: left and right direction of opening: up and down vertices: (0,-1) (6,-1) vertices: (-1,1) (-1,-7) equations of asymptotes: equations of asymptotes:

)3–(321 xy ±=+ 2)1(

323 +±=+ xy

foci: (-0.61,-1) (6.61,-1) foci: (-1,4.21) (-1,-10.21)

y

x

1)

y

x

2)

3. ( ) ( ) 116

392 22

=−yx –– 4. ( ) 1

361

49

22

=+xy –

center: (2,3) center: (-1,0) direction of opening: left and right direction of opening: up and down vertices: (-1,3) (5,3) vertices: (-1,7) (-1,-7) equations of asymptotes: equations of asymptotes:

)2–(343– xy ±= )1(

67

+±= xy

foci: (-3,3) (7,3) foci: (-1,9.22) (-1,-9.22)



–3 of 3–

3. (continued) 4. (continued)

y

x

3)

y

x

4)

5. 141

64

22

=)–(– xy 6. 1

2516

22

=yx –

center: (1,0) center: (0,0) direction of opening: up and down direction of opening: left and right vertices: (1,8) (1,-8 vertices: (-4,0) (4,0) equations of asymptotes: equations of asymptotes:

)1–(4 xy ±= xy45

±=

foci: (1,8.25) (1,-8.25) foci: (-6.40,0) (6.40,0)

y

x

5)

y

x

6)


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mahmh_k.

–1 of 3–

Module: Equations of Ellipses and Hyperbolas

Objective: To practice how to determine whether an equation represents an ellipse or a hyperbola

Name: _______________________ Date: __________________

Keep this information in mind when working with ellipses and hyperbolas. The equation of an ellipse can be written in either of the following forms:

( ) ( ) 12

2

2

2

=+b

kya

hx –– (horizontal major axis)

( ) ( ) 12

2

2

2

=+a

kyb

hx –– (vertical major axis)

In both cases, the center is located at (h,k) and c2 = a2 – b2. The equation of a hyperbola can be written in either of the following forms:

( ) ( ) 12

2

2

2

=b

kya

hx ––– (opens left and right)

( ) ( ) 12

2

2

2

=b

hxa

ky ––– (opens up and down)

In both cases the center is located at (h,k) and c2 = a2 + b2. The asymptotes of a hyperbola have one of the following forms:

( )hxabky –– ±= (when the hyperbola opens left and right)

( )hxbaky −±=– (when the hyperbola opens up and down)

The general form of the equation for ellipses and hyperbolas is ax2 + by2 + cx + dy + e = 0

where a, b, c, d, and e are constants and a ≠ 0 and b ≠ 0.

Complete the following statements. In the general form above, if a and b have the same sign, then the graph of

the equation is a(n) ___ellipse__________.

In the general form above, if a and b have different signs, then the graph of

the equation is a(n) __hyperbola________.

An equation in general form can be rewritten in one of the more useful

graphing forms using the method of _completing the square________.

KEY



–2 of 3–

Problem Set: For each equation, use a separate sheet of paper to:

a. Determine the shape of the graph. b. Determine the center and vertices. c. Determine the foci. Round, if necessary, to the nearest

hundredth. d. Write each equation in a form useful for graphing. e. Graph each equation.

1. 4x2 – 9y2 + 32x + 54y – 161 = 0 2. 4x2 + 25y2 + 8x – 150y + 129 = 0

a) hyperbola – opens left and right a) ellipse – horizontal major axis

b) center: (-4,3) b) center: (-1,3)

vertices: (-10,3) (2,3) vertices: major axis (-6,3) (4,3)

vertices: minor axis (-1,1) (-1,5)

c) (-11.21,3) (3.21,3) c) (-5.58,3) (3.58,3)

d) 116)3–(–

36)4( 22

=+ yx d) 1

4)3–(

25)1( 22

=++ yx

3. 49x2 + 36y2 – 98x – 288y – 1139 = 0 4. -16x2 + 9y2 + 32x – 36y – 124 = 0

a) ellipse – vertical major axis a) hyperbola – opens up and down

b) center: (1,4) b) center: (1,2)

vertices: major axis (1,-3) (1,11) vertices: (1,-2) (1,6)

vertices: minor axis (-5,4) (7,4)

c) (1, 0.39) (1,7.61) c) (1,7) (1,-3)

d) 149)4–(

36)1–( 22

=+yx d) 19

1162 22

=)–(–)–( xy

5. 4x2 + y2 – 48x – 4y + 48 = 0 6. 9x2 – y2 – 72x + 8y + 119 = 0 a) ellipse – vertical major axis a) hyperbola – opens left and right

b) center: (6,2) b) center: (4,4)

vertices: major axis (6,12) (6,-8) vertices: (3,4) (5,4)

vertices: minor axis (1,2) (11,2)

c) )75–2,6( ( )752,6 +− c) (0.84,4) (7.16,4)

d) 1100

)2–(25)6–( 22

=+yx d) 1

9)4–(–)4–(2

2 =yx



–3 of 3–

Extension Activity: When an equation in the form ax2 + bxy + cy2 + dx + ey + f = 0 is a conic section, you may use the discriminant, b2 – 4ac, and the following guidelines to determine the conic’s shape.

• If b2 – 4ac < 0, the equation graphs an ellipse or a circle. (For circles, a = c.)

• If b2 – 4ac = 0, the equation graphs a parabola. • If b2 – 4ac > 0, the equation graphs a hyperbola.

Using this information, determine the shape of the conic sections below. 1. x2 + y2 – 4x – 5y + 4 = 0 2. -x2 + 4y2 + 2x – 8y + 22 = 0 a = 1 b = 0 c = 1 a = 1 b = 0 c = 4 b2 – 4ac = 02 – 4(1)(1) b2 – 4ac = 02 – 4(-1)(4) = 0 – 4 = 0 + 16 = -4 < 0 circle = 16 > 0 hyperbola because a = c 3. 2x2 + 5x + 3 = 0 4. 4x2 + 25y2 + 24x – 50y – 100 = 0 a = 2 b = 0 c = 0 a = 4 b = 0 c = 25 b2 – 4ac = 02 – 4(2)(0) b2 – 4ac = 02 – 4(4)(25) = 0 – 0 = 0 – 400 = 0 = 0 parabola = -400 < 0 ellipse because a ≠ c

Course 18 Functions and Their Graphs

Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 1, Document maiah_k.

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Module: Defining a Function with Its Rule

Objective: To practice defining a function

Name: _______________________ Date: __________________ Fill in the blanks. A relation where every input leads to exactly one output is called a

function .

Given a set of input values, a function must be able to assign a value to

each input value.

Given some specific input value, the function must not assign more

than one value to it.

If the input 3 leads to an output of 6, and the input 4 leads to the output 6,

then the relation cannot be a function. false (true, false)

Problem Set: Determine whether the following diagrams are functions. If not, explain. 1. inputs outputs 2. inputs outputs 2 4 6 14 4 16 7 16 6 36 8 18 8 64 9 20 function Not a function – input value 7 has more than 1 input value

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 1, Document maiah_k.

–2 of 2–

3. inputs outputs 4. inputs outputs 4 2 1 9 6 2 36 9 3 3 81 3 4 -16 Not a function – input value function -16 does not have an output value Extension Activity: Given the set of inputs {-4,-1,0,1,4}, determine if the following relations are functions. If not, explain. 5. 2 7x + 6. |x| 7. 2 4x +

function function function

8. x 9. 2 3 4x x− + 10. 4x +

not a function function function no real number output value for input value -4 – -1


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 1, Document maibh_k.

–1 of 2–

Module: Finding Values of a Function Using Its Rule

Objective: To practice finding values of a function using its rule

Name: _______________________ Date: __________________

Fill in the blanks.

A relation is a function when every input leads to exactly one

output.

The input values of a function are called the domain of the

function.

The output values of a function are called the range of the

function.

A symbolic way to describe the output of a function with input x

is ( )f x .

A function f can be classified as a one-to-one function if the

same element of the range is assigned to only one element of the domain.

Problem Set: Let f denote the function defined by ( ) 2

4 3f x x= − . Find the following. 1. ( )3f = ( )24 3 3− = 33 2. ( )1f − = ( )24 1 3− − = 1

3. ( )0f = ( )24 0 3− = -3 4. ( )2f = ( )24 2 3− = 5 5. ( )2f − = ( )24 2 3− − = 13 6. ( )f a = ( )24 3a − = 24 3a − 7. ( )6f = ( )24 6 3− = 141 8. ( )2f a = ( )224 3a − = 44 3a − 9. ( )5f − = ( )24 5 3− − = 97 10. ( )2f a = ( )24 2 3a − = 316 2 −a

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 1, Document maibh_k.

–2 of 2–

Extension Activity: Find the domain of the following functions.

11. ( ) 1f xx

= 12. ( ) 3f x x= −

All real numbers except 0 All real numbers greater than or

equal to 3 x – 3 ≥ 0

x ≥ 3

13. ( ) 2 3 7f x x x= − + 14. ( ) 2

33 4

f xx x

=− −

All real numbers 2 3 4 0x x− − =

( )( )4 1 0x x− + =

All real numbers except

x = 4 and x = -1

Determine whether or not the following functions are one-to-one functions. 15. f 16. f 17. f

Domain Range Domain Range Domain Range

1 2 2 10 1

3 6 4 12 2 7

4 8 6 14 3

10

No - ( ) ( )3 10 2f f= = Yes No - ( ) ( ) ( )1 2 3 7f f f= = =

Reflection: Describe the difference between ( )2f a and ( ) 2

f a .

( )2f a means take the input 2a and plug it into the function f . ( ) 2f a means

take the input a and plug it into the function f to get the output ( )f a , then take

( )f a and square it.


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document maich_k.

–1 of 2–

Module: Equations and Graphs of Functions, Part 1

Objective: To practice equations and graphs of functions

Name: _______________________ Date: __________________

Fill in the blanks.

A relation is a function if every input leads to exactly one

output.

You can determine whether the graph of a relation is a function by using

the vertical line test.

You can determine whether a function is a one-to-one

function by using the horizontal line test.

A function is a one-to-one function if the same element of the range is

assigned to only one element of the domain.

If a vertical line intersects a graph in more than one place, the relation

is not a function.

Problem Set: Determine if the following relations are functions. 1 y

x5-5

-5

5

2 y

x5-5

-5

5

3 y

x5-5

-5

5

function function not a function

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document maich_k.

–2 of 2–

4 y

x5-5

-5

5

5 y

x5-5

-5

5

6 y

x5-5

-5

5

function function not a function Determine if the following functions are one-to-one functions.

7 y

x5-5

-5

5

8 y

x5-5

-5

5

not one-to-one one-to-one

9 y

x5-5

-5

5

10 y

x5-5

-5

5

not one-to-one one-to-one


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document maidh_k.

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Module: Equations and Graphs of Functions, Part 2

Objective: To practice graphing a function

Name: _______________________ Date: __________________

Fill in the blanks.

The graph of a function is a visual representation of a function

and its properties.

You can graph any function by graphing all points with x-coordinate “a”

and y-coordinate ƒ (a) .

The domain is the set of all inputs for a function. The range

is the set of all outputs of a function.

To graph a function that is defined in a constant manner throughout the

domain, it may be helpful to create a table of values and then

graph the function.

To graph a function that is defined differently on various intervals of the

domain, graph each part separately and then put the pieces

together.

KEY



–2 of 3–

Problem Set: Match the following functions with its corresponding graph. Make a table if necessary.

5

5-5

-5A y

x

5

5-5

-5

B y

x

5

5-5

-5

C y

x

5

5-5

-5

D y

x

5

5-5

-5

E y

x

5

5-5

-5

F y

x

1. ( ) 1f x x= − + E

2. ( ) 1f x x+ + D

3. ( ) 1 23

f x x= + F

4. ( ) 3f x x= H

5. ( )13f x x= B

6. ( ) 2 3 4f x x x= − + A

7. ( ) 1f xx

= C

8. ( ) 3 4f x x= − − G

5

5-5

-5

G y

x

5

5-5

-5

H y

x

Graph the following functions for all x between –4 and 4 (i.e. –4<x<4). 9. f(x) = −x−3 if –4<x< 4 10. f(x) = −3x−1 if −4<x< 0

f(x) = x2−3 if −1≤ x< 2 f(x) = x if 0≤ x< 4 f(x) = 5 if 2 ≤ x < 4



–3 of 3–

9. (continued) 10. (continued)

5

5-5

-5

y

x

5

5-5

-5

y

x


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Module: Translations and Transformations

Objective: To practice altering functions by translating and transforming the graph

Name: _______________________ Date: __________________

Fill in the blanks. A _ __translation______ is also called a horizontal or vertical shift.

A _____transformation_____ involves a change in the width of a graph.

If you alter the graph of ƒ(x) to form the graph of ƒ(x – 5), you are shifting

the original graph ____right ____. (right, left, up, down) To form the graph of ƒ(x) + c, _____translate_____ (translate, transform)

the graph of ƒ(x) upward if c > 0 or downward if c < 0.

The graph of ƒ(cx) ___ horizontally ___ _____stretches_____ the

graph of ƒ(x) when 0 < c < 1 and it __ horizontally_______

__shrinks__ when c > 1. (horizontally, vertically) (stretches, shrinks) The graph of cƒ(x) __ __vertically_ ___ ____stretches____ the graph of

ƒ(x) when c > 1, while it ____ vertically ____ _____shrinks_____ that

graph when 0 < c < 1. (horizontally, vertically) (stretches, shrinks)

Problem Set: Draw the altered graphs as indicated. 1. Suppose ƒ(x) = x. 2. Suppose ƒ(x) = x3. Draw the graph of ƒ(x + 5). Draw the graph of ƒ(x – 3).

y

x5-5

-5

5

y

x5-5

-5

5

KEY



–2 of 3–

3. Suppose ƒ(x) = x2. 4. Suppose ƒ(x) = x. Draw the graph of ƒ(x) + 4. Draw the graph of ƒ(x) – 3.

y

x5-5

-5

5

y

x5-5

-5

5

5. Suppose ƒ(x) = x3 – 2. 6. Suppose ƒ(x) = x – 2.

Draw the graph of ƒ(21 x). Draw the graph of

32 ƒ(x).

y

x5-5

-5

5

y

x5-5

-5

5



–3 of 3–

7. Suppose ƒ(x) = x2 – 4. 8. Suppose ƒ(x) = x3 – 1. Draw the graph of ƒ(2x). Draw the graph of 3ƒ(x).

5-5

-5

5

y

x

y

x5-5

5

Reflection: Remembering which alterations are which can be difficult. On separate paper, come up with a scheme of your own, such as a table or other organizational system, for keeping track of the changes each of these items produces: translation, transformation, whether the constant is negative or positive, whether the constant is an integer or a fraction, whether the constant alters the x- or the y-value. Answers will vary.


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document maifh_k.

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Module: Functional Values

Objective: To practice computing functional values by translating and transforming a function

Name: _______________________ Date: __________________

Complete each statement. To transform a graph, you _mulitiply_the x-value or the y-value by a

quantity.

To translate a graph, you add (or subtract) quantity to the x-value or the

y-value.

Circle any alteration that changes the x-value. Draw a box around any

alteration that changes the y-value.

ƒ(x) → ƒ(x – c) ƒ(x) → ƒ(cx)

ƒ(x) → ƒ(x) + c ƒ(x) → cƒ(x)

Problem Set: Suppose the point given is on the graph of ƒ(x). Find the corresponding point for the altered graph. 1. (3,4) on graph of ƒ(x) point for 8ƒ(x): __(3,32)__

2. (-2,1) on graph of ƒ(x) point for 21 ƒ(x): (-2,

21 )

3. (21 ,2) on graph of ƒ(x) point for ƒ(2x): (

41 ,2)

4. (-1,-3) on graph of ƒ(x) point for ƒ(x) + 3: __(-1,0)__ 5. (-2,5) on graph of ƒ(x) point for ƒ(x – 4): __(2,5) __ 6. (5,-4) on graph of ƒ(x) point for ƒ(x) + 2: __(5,-2)__

7. (9,3) on graph of ƒ(x) point for ƒ(31 x): __(27,3)___

8. (2,-4) on graph of ƒ(x) point for ƒ(4x): (21 ,-4)

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document maifh_k.

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9. (-3,4) on graph of ƒ(x) point for ƒ(x + 2): __(-5,4)__ 10. (2,-6) on graph of ƒ(x) point for ƒ(x) – 4: __(2,-10)__ 11. (-1,5) on graph of ƒ(x) point for ƒ(x – 5): __(4,5)__ 12. (12,4) on graph of ƒ(x) point for ƒ(2x): __(6,4)__ Extension: Directions: Use a graphing calculator to find the graph of y = ƒ(x) = x2. Sketch and label it, then sketch and label the following alterations: 1. 5ƒ(x) = 5x2

2. ƒ(x – 3) = (x – 3)2 5

y

x-5

-5

1)

2)

5-5

-5

y

x

3. ƒ(x) + 2 = x2 + 2

4. ƒ(21 x) = 4

1 x 2

3)

5-5

-5

y

x

4)

5-5

-5

y

x

Reflection: You've learned that translating the x-value changes the graph horizontally. In your own words, explain why that alteration would have that effect. Answers will vary. When you add or subtract a constant to the x-values, your graph shifts only to the left or right. It doesn’t translate vertically because the y-values do not change.

y=5f(x) y=f(x)

y=f(x) y=f(x-3)

y=f(x)+2 y=f(x)

y=f(x)

y=f21 (x)


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document maigh_k.

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Module: Composite Functions

Objective: To practice finding composite functions, their values, and the simpler functions that make up composite functions

Name: _______________________ Date: __________________

Complete each statement. A composite function is formed of two _____simpler_________ functions.

To find (ƒ º g)(x), follow these steps:

First, evaluate ___g(x)_________.

Then evaluate ____f(x)________ for that output value.

You can work _backwards_____ from a composite function to find the

simpler functions that it is built upon.

Circle one. true or false: (ƒ º g)(x) = (g º ƒ)(x)

Problem Set: Part I. Find the value of the composite function.

Part II. Example 1: ƒ(x) = 2x – 7 and g(x) = (x + 4)2 A. Find the value of the composite function. 5. (ƒ º g)(3) = ___91____ 6. (g º ƒ)(3) = __9_____ (3 + 4)² = 49 2(3) – 7 = -1 2(49) – 7 = 91 (-1 + 4)² = 9 7. (ƒ º g)(16) = ___793__ 8. (g º ƒ)(16) = __841___ (16 + 4)² = 400 2(16) – 7 = 25 2(400) – 7 = 793 (25 + 4)² = 841 9. (ƒ º g)(-6) = __1_____ 10. (g º ƒ)(-6) = ___225__ (-6 + 4)² = 4 2(-6) – 7 = 19 2(4) – 7 = 1 (-19 + 4)² = 225

1. (ƒ º g)(8) = ________2. (ƒ º g)(7) = ________3. (g º ƒ)(3) = ________4. (g º ƒ)(4) = ________

ƒ g 1 5 5 12 6 6 23 7 7 34 8 8 4

5 7

42

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document maigh_k.

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B. Write the expression for the composite function. 11. (ƒ º g)(x) = __2(x + 4)² – 7__ 12. (g º ƒ)(x) = _(2x – 3)²_____

Example 2: ƒ(x) = (x2 – 8)2 and g(x) = 21 x – 7

A. Find the value of the composite function. 13. (ƒ º g)(4) = __289___ 14. (g º ƒ)(4) = __25____

57–)4)(21( −= [4² – 8]² = 64

[(-5)² – 8]² = 289 257–)64(21

= 15. (ƒ º g)(6) = _64_____ 16. (g º ƒ)(6) = __385___

47–)6)(21( −= (6² – 8)² = 784

[(-4)² – 8]² = 64 3857–)784)(21( =

17. (ƒ º g)(8) = __1_____ 18. (g º ƒ)(8) = __1561__

37–)8)(21( −= (8² – 8)² = 3136

[(-3)² – 8]² = 1 15617–)3136)(21( =

B. Write the expression for the composite function.

19. (ƒ º g)(x) = 22 ]8–)7–21[( x 20. (g º ƒ)(x) = 7–)8–(

21 22x

Part III. Find the missing simpler function that forms the composite function. ƒ(x) = (10x – 3)4 = (h º g)(x) 21. g(x) = __10x – 3_______ h(x) = x4

Reflection: Choose one of the Part II examples and explain in your own words why ƒ º g and g º ƒ are different. f(x) = 2x – 7 + g(x) = (x + 4)² ƒ º g (-6) g º ƒ (-6) (-6 + 4)² = 4 2(-6) – 7 = -19 2(4) – 7 = 1 (-19 + 4)² = 225 To find ƒ º g you first evaluate g(x) and then use that value to find f(x)._________ Whereas in g º ƒ, you first find f(x) and then use that value to find g(x).________ ________________________________________________________________


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document maihh_k.

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Module: Domain Values of Composite Functions

Objective: To practice determining if a value is in the domain of a composite function and defining the restrictions on the domain of such functions

Name: _______________________ Date: __________________

Complete each statement. In the composite function ƒ º g, the range values of __g_______ are the

domain values of __f_______.

You must exclude from the domain of the composite function ƒ º g

• all elements x not in the domain of __g__, and

• all elements x such that g(x) is not in the domain of __f___.

What value(s) of x must be excluded from the domain of ƒ(x) = 52

1+x

?

25− Why? 2(

25− ) = 5 = 0 division by 0

What value(s) of x must be excluded from the domain of g(x) = 3−x ?

x < 3________ Why? (3) – 3 = 0 square root of negative number___

Problem Set:

Example 1: Let ƒ(x) = (2x +3)3 and g(x) = x 3. Circle the values below that are in the domain of ƒ º g. Draw a box around those that are not. -5 3 2 0 -25 13 4 -3 4. Circle the values below that are in the domain of g º ƒ. Draw a box around those that are not. -5 -1 3 2 -4 -25 4

ƒ g 1 1 3 12 2 4 23 3 5 34 4 6 4

1. List the values in the domain of ƒ º g. ________________ 2. List the values in the domain of g º ƒ. ________________

1,2,3,4

3,4

1 2

KEY


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–2 of 2–

5. Write the restricted domain of ƒ º g. x ≥ 0 6. Write the restricted domain of g º ƒ. x ≥ Example 2: Let ƒ(x) = x2 and g(x) =

51−x

7. Circle the values below that are in the domain of ƒ º g. Draw a box around those that are not. -5

21 3 2 4 -4 5 5 0

8. Circle the values below that are in the domain of g º ƒ. Draw a box around those that are not.

-5 -21 2 3 4 - 5 16 5

9. Write the restricted domain of ƒ º g. x ≠ 5 10. Write the restricted domain of g º ƒ. x ≠ ± 5 Reflection: With a graphing calculator, graph a function used in this activity that has a restricted domain. Then describe in your own words how the graph represents or illustrates the restrictions on the domain. __Answers will vary.________________________________________________ ________________________________________________________________ __The answer should address how the restricted domain isn’t part of the graph._

Example: All domain values are ≥

-3 2

-3 2


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document maiih_k.

–1 of 2–

Module: Inverse of a Function

Objective: To practice finding the inverse of a function Name: _______________________ Date: __________________

Complete each statement. Give the notation for the inverse of function ƒ: __ƒ-1_____

A function and its inverse switch __domain____ and __range____ values.

If ƒ and ƒ-1 are inverses, the following statement is true:

ƒ-1(x) = _y__ if and only if ƒ(y) = _x___

These are the steps for finding the inverse of the function ƒ(x) = 41 x + 3:

• Rewrite ƒ(x) as an equation for y: y = 341 +x

• Switch x and y: x = 341 +y

• Solve for y: y = ____4x – 12______

• Substitute ƒ-1(x) for y: ƒ-1(x) = ____4x – 12______

Problem Set: A. Write in arrows to show the inverse of these functions.

B. Find the inverse of these functions. 3. ƒ(x) =

21 x – 3 f -1(x) = 2(x + 3) 4. h(x) =

710+x h-1(x) = 7x – 10

3–21 yx =

710+

=yx

321

+= xy 107 += yx

62 += xy 10–7– xy

ƒ ƒ-1 1 5 5 12 6 6 23 7 7 34 8 8 4

g g-1 1 1 1 12 2 2 23 3 3 34 4 4 4

1. 2.

KEY

2. 1.


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document maiih_k.

–2 of 2–

5. g(x) = 15x + 2 15

2–)(1 xxg =− 6. i(x) = 11

94 −x 49

411)(1 +=− xxi

215 += yx 11

9–4yx =

2–15 xy = 9–4–11 yx

15

2–xy = 9114 += xy

4911 += xy

Extension: A. Use the inverse rule for functions to verify two of your answers above

for two specific values for x.

inverse rule for functions: ƒ-1(x) = y if and only if ƒ(y) = x

3–21)( xxf = )3(2)(1 +=− xxf 215)( += xxg

152–)(1 xxg =−

3–)10(21)10( =f 10)32(2 =+ 2)1(15)1( +=g

152–17)17(1 =−g

2)10( =f 10)2(1 =−f 17)1( =g 1)17(1 =−g

3–)8(21)8( =f 8)31(2)1(1 =+=−f 2)2(15)2( +=g

152–32)32(1 =−g

1)8( =f 8)1(1 =−f 32)2( =g 2)32(1 =−g

B. Check one of your inverse function answers by using this statement

about the composition of inverse functions: ƒ[ƒ-1(x)] = x

3–21)( xxf = )3(2)(1 +=− xxf

xx =+ 3–)]3(2[21

xx =+ 3–)62(21

xx =+ 3–3 xx = Reflection: In your own words, describe how inverse functions are similar to and different from additive or multiplicative inverses.

Answers will vary. The response may include a discussion on how the answers

are related to 1 and 1 versus f(8) = 1 f -1(1) = 8


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document maijh_k.

–1 of 2–

Module: Determining if a Function Has an Inverse

Objective: To practice determining if a function has an inverse from looking at a mapping diagram or a graph

Name: _______________________ Date: __________________

Complete each statement. For a function to have an inverse, it must be one-to-one .

A one-to-one function is one that maps each element of its domain

to a unique element of its range .

It's possible to restrict the domain of a function so that the

function is one-to-one.

The horizontal line test shows whether a graphed

function is one-to-one and therefore has an inverse.

Problem Set: A. Answer the questions below about functions ƒ and g.

1. Does ƒ have an inverse function? ___no_____ 2. What domain value(s) of ƒ determine whether ƒ-1 exists? ____3 and 4_______ 3. If the domain of ƒ is restricted to the set {1,3,4}, does it have an inverse? _no___ 4. If the domain of ƒ is restricted to the set {1,2,4}, does it have an inverse? __yes___ 5. Does g have an inverse function? __yes____ 6. What domain value(s) of g determine whether g-1 exists? ___1,2,3,4__________ 7. If the domain of g is restricted to the set {1,3,4}, does it have an inverse? __yes___ 8. If the domain of g is restricted to the set {1,2,3}, does it have an inverse? ___yes__

ƒ 1 5 2 6 3 7 4 8

g 1 5 2 6 3 7 4 8

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document maijh_k.

–2 of 2–

B. Label the graphs YES or NO to indicate if each does or does not have an inverse function. If it does not, restrict the domain by shading or circling a region of the graph so that it does have an inverse. Answers will vary, but here are some possible ones.

9. yes 10. no

11. no 12. no Reflection: You know that 0 is the additive identity element, so that x + 0 = x. Also, 1 is the multiplicative identity element, so that x • 1 = x. There is also an identity function, ƒ(x) = x, which you may know as the equation y = x. Using the identity function ƒ(x) = x and the function g(x) = (x2 + 3)2, find the composite functions (g º ƒ)(x) and (ƒ º g)(x). Then explain in your own words why ƒ(x) = x is called the identity function. (g º ƒ)(x) = (x2 + 3)2 = g(x) (ƒ º g)(x) = (x2 + 3)2 = g(x) ________________________________________________________________ If f(x) = x, then regardless of g(x) you’re going to end up with g(x).____________ ________________________________________________________________

y

x-2

-2

2

2

y

x-2

-2

2

2

y

x-2

-2

2

2

y

x-2

-2

2

2

Course 18 Functions and their Graphs

Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document maikh_k.

–1 of 2–

Module: Solving Problems with Linear Functions

Objective: To practice how to describe real-world situations as linear functions Name: _______________________ Date: __________________

Complete each statement. ____________________ functions have the form f(x) = ax + b where a can be

thought of as a rate and b is a constant.

It is sometimes necessary to solve a ____________________ of two

equations to find the values for a and b. Then, we can write a function that

represents the given situation in terms of the defined variable.

Problem Set: On a separate piece of paper, solve each word problem using linear functions. (Note: You will use a system of two linear equations to solve problem 5.) 1. Ben just joined Jammin’ Tunes, a CD music club. As a new member, he received 8

CDs free. He must purchase 1 new CD every 4 months to keep his membership. a. Write the function, r(x), which represents the number of CDs received after

x years. Answer: ( ) 3 8r x x+ + b. How many CDs will Ben have in three years if he buys according to the

minimum requirement? Answer: 17 CDs

2. After you finish this course, you decide to write a book entitled I Love Algebra. The publisher gives you $20,000 and promises to pay you $5 per book sold.

a. Write the function, p(x), which represents the amount of money you will be paid from the sale of x books. Answer: ( ) 5 20,000p x x= +

b. How much money will you be paid from the sale of 1,550 books? Answer: $27,750

3. Pride Publishing Company produces cookbooks. The setup cost is $25,000. Once the type is set, the cost of printing each cookbook is $8.

a. Write the function, c(x), which represents the total cost of producing x cookbooks. Answer: ( ) 8 25,000c x x= +

b. How much will it cost to produce 5,000 cookbooks? Answer: $65,000 c. Each cookbook sells for $15. Write the function, r(x), which represents the

revenue made from selling x cookbooks. Answer: ( ) 15r x x= d. How much revenue is made when 5,000 cookbooks are sold?

Answer: $75,000

KEY

Course 18 Functions and their Graphs

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–2 of 2–

e. Write the function, p(x), which represents the profit realized from selling x cookbooks. (Remember: profit = revenue – costs) Answer: ( ) ( ) ( ) ,p x r x c x= − ( ) ( )15 8 25,000p x x x= − + , or ( ) 7 25,000p x x= −

f. How much profit does Pride Publishing Company make when 5,000 cookbooks are sold? Answer: $10,000

4. As manager of Penguin Snow Cone Cart, you must hire 3 full-time employees to

cover your area. Each employee is supplied with a $100 cart and a $30 uniform. Each snow cone costs $0.12 in ice, syrup, and cups to make.

a. Write the function, c(x), which represents the total costs of your business when x snow cones are made. Answer: ( ) 0.12 390c x x= +

b. How much will it cost if 4,200 snow cones are made in one month? Answer: $894

c. Snow cones sell for $1.75 each. Write the function, r(x), which represents the revenue received from selling x snow cones. Answer: ( ) 1.75r x x=

d. How much revenue is received when 4,200 snow cones are sold? Answer: $7,350

e. Write the function, p(x), which represents the profit realized from selling x snow cones. (Remember: profit = revenue – costs) Answer: ( ) ( ) ( ) ,p x r x c x= − ( ) ( )1.75 0.12 390p x x x= − + , or ( ) 1.63 390p x x= −

f. How much profit does Penguin Snow Cone Cart make when 4,200 snow cones are sold? Answer: $6,456

5. Hiking Limited’s revenue is a linear function of the number of orienteering

guidebooks sold, x. If 100 books are sold, the revenue is $800. If 300 books are sold, the revenue is $3,000

a. Write the function, r(x), which represents the revenue from the sale of x orienteering guidebooks. Answer: ( ) 11 300r x x= −

b. Find the revenue when 680 guidebooks are sold. Answer: $7,180 Reflection: This is your time to let your teacher know what you do and do not understand about solving problems with linear functions. Complete the chart below.

Solving Problems with Linear Functions What I know… I still have questions about…


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mailh_k.

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Module: Solving Problems with Quadratic Functions

Objective: To practice how to describe real world situations as quadratic functions

Name: _______________________ Date: __________________

Complete each statement.

____Quadratic_______ functions have the form f(x) = ax2 + bx + c where

a, b, and c are constants.

Functions of this form graph ___parabolas________, which are smooth,

u-shaped curves.

Quadratic functions can be used as ___predictors_______. If you have a

general idea of how two quantities are related and a few actual values

demonstrating the relationship, you can often find their relationship for all

values.

The method used to find this relationship for all values is solving a

______system______ of equations for a, b, and c.

Problem Set: On a separate piece of paper, solve each word problem using a system of equations. 1. It costs a men’s clothing company $200 to produce 20 shirts, $600 to produce

40 shirts, and $1,400 to produce 60 shirts. a. Find the quadratic function, c(x), which represents the cost to produce

x shirts. Answer: ( ) 21 10 2002

c x x x= − +

b. How much does it cost the company to produce 150 shirts? Answer: $9,950

2. A ball is shot from a cannon that is 5 feet off the ground. So, before the ball is

shot, it is at a height of 5 feet. After 1 second, the ball is at a height of 117 feet; after 2 seconds, it is at 197 feet.

a. Find the quadratic function, h(x), which represents the height of the ball as function of time, x. Answer: ( ) 5128162 ++−= xxh

b. What is the height of the ball after 4 seconds? Answer: 261 feet

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document mailh_k.

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3. A missile is launched from the ground. After 1 second, it is 185 feet in the air;

after 2 seconds, it is 340 feet in the air. a. Find the quadratic function, h(x), that represents the distance traveled

by the missile as a function of time, x. Answer: ( ) 215 200h x x x= − + b. What is the height of the missile after 6 seconds in the air?

Answer: 660 feet 4. The time, t, it takes a certain computer program to run depends quadratically

on the size of the number x computed in the program. When x is 100, the program takes 31,000 milliseconds; when x is 200, it takes 121,000 milliseconds; and when x is 500, it takes 751,000 milliseconds.

a. Find the quadratic function, t(x), which represents the time it takes the computer program to run. Answer: ( ) 29 3,600 301,000t x x x= − +

b. How long does it take the computer program to run when the number is 20? Answer: 232,600 milliseconds

5. The area, a, covered by algae on a stagnant pond depends quadratically on

the amount of algae, x. If no algae are present, then no area is covered; when 10 liters are present, 70 square decimeters are covered; and when 20 liters are present, 340 square decimeters are covered.

a. What is the quadratic function, a(x), that fits these data. Answer: ( ) 2 3a x x x= −

b. How much algae are present when 400 square decimeters are covered? Answer: 158,800 liters

6. Shadow Art Supply Company manufactures watercolor kits. They have found

that the cost of producing x kits is a quadratic function. The company also discovered that it costs $100 to produce 20 kits, $200 to produce 40 kits, and $500 to produce 60 kits.

a. Find Shadow Art Supply Company’s cost function, c(x).

Answer: ( ) 21 10 2004

c x x x= − +

b. How much will it cost the company to manufacture 100 watercolor kits? Answer: $1,700

Extension Activity: You used a system of equations to solve the problems above. In 1858, Arthur Cayley found another method to use to solve systems of equations. It involves the use of matrices (singular: matrix). A matrix is a rectangular arrangement of numbers in rows and columns. With a partner, research Arthur Cayley. Investigate his contribution to solving systems of equations by use of matrices. On a separate piece of paper, record your discoveries. Share them in class.

Course 19 Exponential and Logarithmic Functions

Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document majah_k.

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Module: Properties of Exponential Functions

Objective: To investigate the properties of exponential functions

Name: _______________________ Date: __________________

Fill in the blanks. Select a word in parentheses when choices are given. Exponential functions have the form y = ƒ(x) = Cax where C is called a

_constant___.

In the function ƒ(x) = ax, the value of C is __1__.

y = ƒ(x) = 2x. If x = 2, then ƒ(x) = __4__.

If the base, a, of ƒ(x) = ax is __less than____ 1, then ƒ(x) decreases with

increasing values for x. (greater than, equal to, less than)

If ƒ(x) = ax, then ƒ( _3_ + _7_ ) = ƒ(3) • ƒ(7)

If ƒ(x) = ax, then ƒ( _5_ • 2) = [ƒ(5)]2

Write T if the statement is true or F if the statement is false. The three rules in the boxes may help you.

a-n = na1

n

ba

=

nb

na nm

b = mn b )(

2-3 = 81 __T__

21

9 = 3 __T__

2

23

=

29 __F__

The point (0,-1) will always lie on the graph of the function ƒ(x) = ax. __F__

The graph of the function ƒ(x) = ax will always lie above the x-axis. __T__

KEY


Problem Set: Complete the data table for the function and then graph it. Use a calculator as needed. 1. y = ƒ(x) = 2x 2. y = ƒ(x) = (1.5)x

4

6

2 4 6-2-4-6-2

-4

-6

2

4

6

2 4 6-2-4-6-2

-4

-6

2

3. y = ƒ(x) =

x

31

4. y = ƒ(x) =

x

21

4

6

2 4 6-2-4-6-2

-4

-6

2

4

6

2 4 6-2-4-6-2

-4

-6

2

x y 3 2

1

0

-1

-2

-3

x y 3

2

1

0

-1

-3 2

8

4

2

1

.5

.25

.125

.037

.111

.333

1

3

5.20

3.38

2.25

1.5

1

.67

.44

.30

y x 3

2

1

0

-1

-2

-3

.125

.25

.5

1

2

4

8

3

2

1

0

-1

-2

-3

y x

Reflection: Alex and Grace were discussing exponential functions of the form y = ƒ(x) = ax where a is greater than 1. Both agreed that as the value of x increases, ƒ(x) approaches positive infinity. However, as the value of x decreases, they disagreed on the behavior of the function. Alex said that ƒ(x) will equal zero for very small values of x. Grace said that ƒ(x) will approach zero, but never equal it. Who is correct and why? State your reasons below. Grace is correct. When ƒ(x) = ax, there is no value of x that makes ƒ(x) = 0. The

value of ƒ(x) will get smaller and smaller as x gets smaller and smaller. The____

value of x_will approach 0 but never equal it. The student could also__________

reference the graphs in problems 1 and 2._______________________________

Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document majah_k.

–2 of 2–


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document majbh_k.

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Module: Properties of Logarithmic Functions

Objective: To investigate the properties of logarithmic functions

Name: _______________________ Date: __________________

Fill in the blanks.

To solve the equation 4y = 12 for y, one could use a function called a

logarithm .

If a and x are positive numbers and a ≠ 1, then the loga x = y and ay = x

are equivalent equations.

A logarithm with a base of 10 is called a common logarithmic

function.

A logarithm with a base of e is usually expressed as ln x and is called

the natural logarithm .

For all logarithmic functions ƒ(x) = loga x, when a > 1, ƒ(x) will be

increasing and when a < 1, the function will be decreasing .

Problem Set: Solve the following problems. Round to the hundredths when necessary. 1. log 232 = 5 2. log 5125= 3 3. log 430 = 2.45 4. log 6100 = 2.57 5. log 750 = 2.01 6. log 315 = 2.46 7. log 10 23 = 1.36 8. log 990 = 2.05

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document majbh_k.

–2 of 2–

Solve the following common logarithmic functions. Round to the hundredths when necessary. 9. log 10 = 1 10. log 150 = 2.18 11. log 1,000 = 3 12. log 101 = 2.00 Solve the following natural logarithms. Round to the hundredths when necessary. 13. ln 20 = 3.00 14. ln 2.72 = 1.00 15. ln 57 = 4.04 16. ln 11.65 = 2.46 Using the properties of logarithms, find the missing piece in each of the following.

17. log2x – log210 = log 210x 18. log42 =

4log2log

10

10

19. 4 log32 = log3

42 20. log23 + log24 = log2 12

21. log94x = log94 + log9x 22. log10 52 = log10 2 – log 10 5

Extension Activity: For the function ƒ(x) = log3x, what is greater: ƒ(1) + ƒ(6) or ƒ(2) + ƒ(3)? ƒ(1) + ƒ(6) = log3(1) + log3(6) = log3(1 • 6) = log36

ƒ(2) + ƒ(3) = log3(2) + log3(3) = log3(2 • 3) = log36

∴ ƒ(1) + ƒ(6) = ƒ(2) + ƒ(3)


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document majch_k.

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Module: Recognizing Graphs of Types of Functions

Objective: To practice identifying exponential and logarithmic functions by looking at the graphs of these functions

Name: _______________________ Date: __________________

Fill in the blanks.

Equations of the form ( ) xf x a= are called exponential functions.

Equations of the form ( ) log af x x= are called logarithmic

functions.

If a graph passes through quadrants I and II, it cannot be a logarithmic

function because x must be greater than zero.

For functions of the form ( ) xf x a= , the graph will increase if a > 1

and will decrease if a < 1.

For functions of the form ( ) log af x x= , the graph will increase if

a > 1 and will decrease if a < 1.

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document majch_k.

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Problem Set: Match the following functions with their corresponding graphs.

5

5

y

x-5

-5

A

B

5

5

y

x-5

-5

C

5

5

y

x-5

-5

D

5

5

y

x-5

-5

E

5

5

y

x-5

-5

5

5

y

x-5

-5

F

1. ( ) .75xf x = D 2. xxf 2.log)( = F 3. ( ) 2xf x = E 4. ( ) 2logf x x= C 5. ( ) .25xf x = H 6. ( ) 5logf x x= G 7. ( ) 5xf x = B 8. ( ) .5logf x x= A

G

5

5

y

x-5

-5

H

5

y

x-5

-5


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document majdh_k.

–1 of 3–

Module: Solving Problems: Exponential and Logarithmic

Objective: To practice how to solve problems that involve exponential and logarithmic functions

Name: _______________________ Date: __________________

Complete each statement. To solve equations that contain variables in the exponent, take the

_____logarithm______ of each side of the equation. Then, solve for the

variable.

To solve equations that contain logarithms, raise each side as the

_______power________ of the base of the logarithm. Then, solve for the

variable.

Problem Set: On a separate sheet of paper, solve each exponential equation for x. Round your final answer to the nearest thousandth. 1. 4x = 9 2. 3x = 1 3. 2(x + 3) = 13 x log 4 = log 9 x log 3 = log 1 (x + 3) log 2 = log 13

x ≈ 1.585 x = 0 x = 32log13log

−

x ≈ 0.700 4. 5(x + 4) = 7 5. e2x = 6 6. 2e(x – 1) = 15 (x + 4)log 5 = log 7 2x = ln 6 5.71 =−xe x ≈ -2.791 x ≈ 0.896 x – 1 = ln 7.5 x ≈ 3.015 7. 8x + 4 = 15 8. -2e(x – 6) = -4 9. 10(3x + 4) = 12 8x = 11 26 =−xe 3x + 4 = log 12

x log 8 = log 11 x – 6 = ln 2 x = 3

412log −

x ≈ 1.153 x ≈ 6.693 x ≈ -0.974

KEY



–2 of 3–

10. 10-5x – 13 = 40 11. 7(-x + 8) + 3 = 27 12.10(4x – 2) = 34 10-5x = 53 7-x + 8 = 24 4x – 2 = log 34

-5x = log 53 (-x – 8)log 7 = log 24 x = 4

234log +

x ≈ -0.345 -x + 8 = 7log24log x ≈ 0.883

x = 7log24log + 8

x ≈ 6.367 On a separate sheet of paper, solve each logarithmic equation for x. Round your answer to the nearest thousandth. 13. log (x + 3) = 4 14. log (2x – 6) = 7 15. log (5x) = 2 104 = x + 3 107 = 2x − 6 102 = 5x x = 9,997 x = 5,000,003 x = 20 16. ln x = -4 17. ln(x – 1) = 6 18. ln (0.5x + 2) = -5 x = e-4 x − 1 = e6 0.5x + 2 = e-5 x ≈ 0.018 x ≈ 404.429 x ≈ -3.987 On a separate sheet of paper, solve each word problem using the properties of logarithmic and exponential functions. 19. Find the amount of time, in years, that it will take an investment of $1,000 to

double at 8% interest compounded continuously. (Hint: A = Pert)

2000 = 1000e0.08t 2 = e0.08t t ≈ 8.7 yrs

20. Find the amount of time, in years, that it would take for an investment of $500

to grow to $1,000,000 at 12% interest compounded continuously. (Hint:

A = Pert)

1,000,000 = 500e0.12t 2000 = e0.12t t ≈ 63.3 yrs

21. R = log x represents earthquake magnitude on the Richter scale. R is the

rating of the earthquake’s size on the Richter scale and x represents the ratio

of the earthquake intensity to some minimal level of intensity that has been

set previously. Find x when an earthquake measures 4.2 on the Richter

scale.

4.2 = log x x = 104.2 x ≈ 15,848.9



–3 of 3–

Extension Activity: Exponential functions can be applied to the values of cars. Suppose that a new car costs $21,000. After 1 year, the depreciated value of the car is $16,000. Find the value of the car when it is 4 years old by using the exponential function y = Cekt where C = the original cost, t = age (in years) of the car, and k = a constant. Round your answer to the nearest hundred dollars. (Hint: First, find the value of k. Then, find y.) 16,000 = 21,000ek(1)

0.7619 = ek

ln 0.7619 = k k ≈ -0.2719 y = 21,000e-0.2719(4)

y ≈ $7,100


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document majeh_k.

–1 of 2–

Module: Exponential Growth

Objective: To practice how to solve problems that involve exponential growth

Name: _______________________ Date: __________________

When the base of a function is greater than one and the exponent is

positive, it is increasing and referred to as _____exponential____

______growth________.

In many cases, _____population_____ can be modeled by a function in

the form P = Aekt, where P is the population at time t, A is the number of

individuals at time t = 0, and k is a positive constant.

List two applications of exponential growth functions.

___population growth___, ____bacteria growth____

Problem Set: On a separate sheet of paper, answer the following word problems about exponential growth. 1. Of the total amount spent by a school district for 1980 through 2000, the

percent, P, spent on athletics can be modeled by P = 2.92(1.08)t where t = 0 represents 1980.

Find the percentage spent on athletics in 1997. P ≈ 10.8% 2. If 3,000 bacteria, with a growth constant k ≈ 2.8 per hour, are present at the

beginning of an experiment, in how many hours will there be 15,000 bacteria? Round your answer to the nearest hundredth hour.

t ≈ 0.57 hour

3. If 7,200 bacteria, with a growth constant k ≈ 1.8 per hour, are present at the

beginning of an experiment, in how many hours will there be 15,000 bacteria? Round to the nearest hundredth hour.

t ≈ 0.41 hour

KEY


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 2, Document majeh_k.

–2 of 2–

4. In 1970, the population of Houston, Texas was 2,100,000. In 1980, it was 3,200,000. Answer the following questions. a. Find k. Round to the nearest ten-thousandth. k ≈ 0.0421 b. According to this data, what would the population have been in 1995?

Round to the nearest hundred thousand. A ≈ 6,000,000 people c. What will the estimated population be in 2010? Round to the nearest

hundred thousand. A ≈ 11,300,000 people 5. The population of Des Moines, Iowa was 320,000 in 1970. In 1980, it was

360,000. Answer the following questions, rounding your answer to the nearest ten thousand. a. Find k. Round to the nearest ten-thousandth. k ≈ 0.0118 b. What will the estimated population be in 2005? Round to the nearest ten

thousand. A ≈ 480,000 people c. What will the estimated population be in 2010? Round to the nearest ten-

thousand. A ≈ 510,000 people 6. After 10 hours there are 8,262 bacteria present in a culture that started with

6,000 bacteria. a. Find the growth constant, k. Round to the nearest ten-thousandth.

k ≈ 0.0320 b. How many bacteria will be present after 48 hours from starting? Round to

the nearest thousand. A ≈ 28,000 bacteria c. How many bacteria will be present after 72 hours from starting? Round to

the nearest thousand. A ≈ 60,000 bacteria 7. There are 20,000 bacteria present in a culture. Two hours earlier the culture

had only 15,000 bacteria. a. Find the growth constant, k. Round to the nearest ten-thousandth.

k ≈ 0.0320 b. How many bacteria are present after 15 hours? Round to the

nearest thousand. A ≈ 130,000 bacteria Extension Activity: With a partner, use an almanac to find the population of your city in 1980 and in 1990. Use this information to help you predict its estimated population in 2010 and 2015. Repeat this exploration for another city of your choice. Answers will vary.


Permission is granted to instructors to copy and distribute this worksheet for instructional purposes only. Copyright © 2002 PLATO Learning, Inc. All rights reserved. Printed in the United States of America. ® PLATO is a registered trademark of PLATO Learning, Inc. Algebra 2, Part 1, Document majfh_k.

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Module: Exponential Decay

Objective: To practice how to solve problems that involve exponential decay

Name: _______________________ Date: __________________

Complete each statement. When the exponent of an exponential function is negative, the function

___decreasing________.

__Exponential________ ___decay____________ can be modeled by

the equation N = N0e-kt where N0 is initial amount of a substance (at

time t = 0), N is the amount of the substance at time t, k is a positive

constant, and t is time.

__Half-life___________ is the amount of time it takes a substance to

lose half of its mass.

Problem Set: Solve the following problems. Show all of your work on a separate piece of paper. 1. Complete the table by finding the half-life of each reactive substance.

substance initial amount k half-life Caesium-137 (Cs-137)

10 g

0.0231

30 years

Tritium-3 (H-3)

12 g

0.0564

12.3 years

Lead-210 (Pb-210)

15 g

0.0311

22.3 years

Iridium-192 (Ir-192)

20 g

0.0094

74 days

Cobalt-60 (Co-60)

25 g

0.1308

5.3 years

KEY



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2. An archeologist found part of the backbone for a wooly mammoth. It contains 32% of its original amount of C-14, where k = 0.0001. Find the age of the bone. Round to the nearest year.

N = N0e-kt 0.32N0 = N0e-0.0001t

0.32 = e-0.0001t

ln 0.32 = ln e-0.0001t

ln 0.32 = -0.0001t

t ≈ 11,394 years

3. A scientist found the remains of a mummy. The remains contain 62% of its original amount of C-14, where k = 0.0001. Find the age of the bone. Round to the nearest year.

N = N0e-kt 0.62 N = N0e-0.0001t 0.62 = e-0.0001t

ln 0.62 = ln e-0.0001t ln 0.62 = -0.0001t t ≈ 4,780 years

Extension Activity: Exponential decay may be represented in this general form. Suppose y = abx where a is the initial amount, 0 < b < 1, and x is a measure of time. Use this definition to answer the following questions on separate piece of paper. If a basketball is properly inflated, it should rebound after bouncing on the ground to about one-half the height from which it was dropped. This can be represented by the following exponential decay equation: y = abx = initial height • fraction of bounce number of bounces

ln 0.32 0.0001 = t

ln 0.62 -0.0001 = t



–3 of 3–

1. Make a table showing the expected heights in the first 6 bounces after a ball is dropped from a height of 20 feet. (Hint: Let x = 0, 1, 2, 3, 4, 5, and 6. Find the corresponding value of y for each of the x-values.)

____ x y_____ 0 20 1 10 2 5 3 2.5 4 1.25 5 0.625

6 0.3125 2. After which bounce will the ball rebound to a height less than ½ foot?

Answer: after the 6th bounce 3. If the ball is dropped from 40 feet, how many bounces will it take to rebound to a height of less than ½ foot? Answer: after the 7th bounce 4. If the ball is dropped from 20 feet, but rebounds to a height of two-thirds instead of one-half, how many bounces would it take to reach a height of less than ½ foot?

xy )32(2 −= 10 bounces

____ x y_____ 0 20.00 1 13.33 2 8.89 3 5.93 4 3.95 5 2.63

6 1.76 7 1.17 8 0.78

9 0.52 10 0.35

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