Holt McDougal Florida Larson Algebra 1 - RJS SOLUTIONSTeachers using HOLT McDOUGAL FLORIDA Larson...

Holt McDougal Florida

Larson Algebra 1

Notetaking Teacher’s GuideVolume 2: Chapters 7–12

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iiiAlgebra 1

Notetaking Guide

ContentsChapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176

Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218

Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250

Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281

Chapter 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303

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7.1

Your Notes

Solve Linear Systems by GraphingGoal p Graph and solve systems of linear equations.

VOCABULARY

Systems of linear equations

Solution of a system of linear equations

Consistent independent system

SOLVING A LINEAR SYSTEM USING THE GRAPH- AND-CHECK METHOD

Step 1 both equations in the same coordinate plane. For ease of graphing, you may want to write each equation in .

Step 2 Estimate the coordinates of the .

Step 3 the coordinates algebraically by substituting into each equation of the original linear system.

176 Lesson 7.1 • Algebra 1 Notetaking Guide Copyright © Holt McDougal. All rights resvered.

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7.1

Your Notes

Solve Linear Systems by GraphingGoal p Graph and solve systems of linear equations.

VOCABULARY

Systems of linear equations A system of linear equations consists of two or more linear equations in the same variables.

Solution of a system of linear equations A solution of a system of linear equations in two variables is an ordered pair that satisfies each equation in the system.

Consistent independent system A linear system that has exactly one solution

SOLVING A LINEAR SYSTEM USING THE GRAPH- AND-CHECK METHOD

Step 1 Graph both equations in the same coordinate plane. For ease of graphing, you may want to write each equation in slope-intercept form .

Step 2 Estimate the coordinates of the point of intersection .

Step 3 Check the coordinates algebraically by substituting into each equation of the original linear system.



Copyright © Holt McDougal. All rights resvered. Lesson 7.1 • Algebra 1 Notetaking Guide 177

Your Notes

Solve the linear system: 3x 1 y 5 9 Equation 1

x 2 y 5 21 Equation 2

Solution

1. both equations.

x

y

2

6

10

22222

6 10

2. Estimate the point of intersection. The two lines

appear to intersect at ( , ).

3. Check whether ( , ) is a solution by substituting

for x and for y in each of the original

equations.

Equation 1 Equation 2

3x 1 y 5 9 x 2 y 5 21

0 9 0 21

5 9 ✓ 5 21 ✓

Because ( , ) is a solution of each equation in the

linear system, it is a .

Example 1 Use the graph-and-check method

To ease graphing, write each equation in slope intercept form.


Copyright © Holt McDougal. All rights resvered. Lesson 7.1 • Algebra 1 Notetaking Guide 177

Your Notes

Solve the linear system: 3x 1 y 5 9 Equation 1

x 2 y 5 21 Equation 2

Solution

1. Graph both equations.

x

y

2

6

10

22222

6 10

3x 1 y 5 9

x 2 y 5 21

2. Estimate the point of intersection. The two lines

appear to intersect at ( 2 , 3 ).

3. Check whether ( 2 , 3 ) is a solution by substituting

2 for x and 3 for y in each of the original

equations.


3x 1 y 5 9 x 2 y 5 21

3(2) 1 3 0 9 2 2 3 0 21

9 5 9 ✓ 21 5 21 ✓

Because ( 2 , 3 ) is a solution of each equation in the

linear system, it is a solution of the linear system.

Example 1 Use the graph-and-check method



Your Notes

Homework

1. 2y 1 4x 5 12 2. 4x 1 2y 5 6

2x 2 y 5 210 3x 2 3y 5 9

x

y

1

3

5

7

9

1212325 3

x

y

1

3

5

12123 3 521

23

25

3. 2y 5 6x 1 8 4. y 5 4x 1 4

4x 1 y 5 23 2y 5 23x 2 14

x

y

1

3

5

212325 3121

23

x

y

1

3

21232527 3121

23

25

Checkpoint Solve the linear system by graphing.



Your Notes

Homework

1. 2y 1 4x 5 12 2. 4x 1 2y 5 6

2x 2 y 5 210 3x 2 3y 5 9

x

y

1

3

5

7

9

1212325 3

2y 1 4x 5 12

2x 2 y 5 210

x

y

1

3

5

12123 3 5

4x 1 2y 5 6 3x 2 3y 5 9

21

23

25

(21, 8) (2, 21)

3. 2y 5 6x 1 8 4. y 5 4x 1 4

4x 1 y 5 23 2y 5 23x 2 14

x

y

1

3

5

212325 31

4x 1 y 5 23 2y 5 6x 1 8

21

23

x

y

1

3

21232527 31

y 5 4x 1 4

2y 5 23x 2 14

21

23

25

(21, 1) (22, 24)

Checkpoint Solve the linear system by graphing.



7.2

Copyright © Holt McDougal. All rights reserved. Lesson 7.2 • Algebra 1 Notetaking Guide 179

Solve Linear Systemsby SubstitutionGoal p Solve systems of linear equations by substitution.

SOLVING A LINEAR SYSTEM USING THE SUBSTITUTION METHOD

Step 1 one of the equations for one of its

variables. When possible, solve for a variable

that has a coefficient of or .

Step 2 the expression from Step 1 into the

other equation and solve for the other variable.

Step 3 the value from Step 2 into the

revised equation from Step 1 and solve.

Your Notes

Solve the linear system: x 5 22y 1 2 Equation 1

3x 1 y 5 16 Equation 2

1. for x. Equation 1 is already solved for x.

2. Substitute for x in Equation 2 and solve

for y.

3x 1 y 5 16 Write Equation 2.

3( ) 1 y 5 16 Substitute

for x.

1 y 5 16 Distributive property

5 16 Simplify.

5 Subtract from each side.

y 5 Divide each side by.

3. Substitute for y in the original Equation 1 to

find the value of x.

x 5 22y 1 2 5 22( ) 1 2 5 4 1 2 5

The solution is ( , ).

Example 1 Use the substitution method

Remember to check your solution in each of the original equations.


7.2


Solve Linear Systemsby SubstitutionGoal p Solve systems of linear equations by substitution.

SOLVING A LINEAR SYSTEM USING THE SUBSTITUTION METHOD

Step 1 Solve one of the equations for one of its

variables. When possible, solve for a variable

that has a coefficient of 1 or 21 .

Step 2 Substitute the expression from Step 1 into the

other equation and solve for the other variable.

Step 3 Substitute the value from Step 2 into the

revised equation from Step 1 and solve.

Your Notes



1. Solve for x. Equation 1 is already solved for x.

2. Substitute 22y 1 2 for x in Equation 2 and solve

for y.

3x 1 y 5 16 Write Equation 2.

3( 22y 1 2 ) 1 y 5 16 Substitute 22y 1 2

for x.

26y 1 6 1 y 5 16 Distributive property

25y 1 6 5 16 Simplify.

25y 5 10 Subtract 6 from each side.

y 5 22 Divide each side by 25 .

3. Substitute 22 for y in the original Equation 1 to

find the value of x.

x 5 22y 1 2 5 22( 22 ) 1 2 5 4 1 2 5 6

The solution is ( 6 , 22 ).


Remember to check your solution in each of the original equations.


Your Notes

Homework

Solve the linear system: 4x 2 2y 5 14 Equation 1


Solution 1. Solve Equation 2 for y.

2x 1 y 5 23 Write original Equation 2.

y 5 Revised Equation 2

2. Substitute for y in Equation 1 and solve for x.

4x 2 2y 5 14 Write Equation 1.

4x 2 2( ) 5 14 Substitute for y.

4x 1 5 14 Distributive property

5 14 Simplify.


x 5 Divide each side by .

3. Substitute for x in the revised Equation 2 to find the value of y.

y 5 5 5 5



1. 5x 2 4y 5 21 2. x 1 y 5 5

y 5 6x 1 5 7x 2 9y 5 3

Checkpoint Solve the linear system using the substitution method.

180 Lesson 7.2 • Algebra 1 Notetaking Guide Copyright © Holt McDougal All rights reserved


Your Notes

Homework



Solution 1. Solve Equation 2 for y.

2x 1 y 5 23 Write original Equation 2.

y 5 22x 2 3 Revised Equation 2

2. Substitute 22x 2 3 for y in Equation 1 and solve for x.


4x 2 2( 22x 2 3 ) 5 14 Substitute 22x 2 3 for y.

4x 1 4x 1 6 5 14 Distributive property

8x 1 6 5 14 Simplify.

8x 5 8 Subtract 6 from each side.

x 5 1 Divide each side by 8 .

3. Substitute 1 for x in the revised Equation 2 to find the value of y.

y 5 22x 2 3 5 22(1) 2 3 5 22 2 3 5 25



1. 5x 2 4y 5 21 2. x 1 y 5 5

y 5 6x 1 5 7x 2 9y 5 3

(21, 21) (3, 2)

Checkpoint Solve the linear system using the substitution method.

180 Lesson 7.2 • Algebra 1 Notetaking Guide Copyright © Holt McDougal All rights reserved


Solve Linear Systems byAdding or SubtractingGoal p Solve linear systems using elimination.

SOLVING A LINEAR SYSTEM USING THE ELIMINATION METHOD

Step 1 the equations to one variable.

Step 2 the resulting equation for the other variable.

Step 3 Substitute in either original equation to .

Your Notes


4x 2 5y 5 214 Equation 2

Solution

1. the equations to x 1 5y 5 9

eliminate one variable. 4x 2 5y 5 214

5

2. Solve for x. x 5

3. Substitute for x in either equation and .

x 1 5y 5 9 Write Equation 1.

1 5y 5 9 Substitute for x.

y 5 Solve for y.


Example 1 Use addition to eliminate a variable

Make sure to check your solution by substituting it into each of the original equations.


7.3


Solve Linear Systems byAdding or SubtractingGoal p Solve linear systems using elimination.

SOLVING A LINEAR SYSTEM USING THE ELIMINATION METHOD

Step 1 Add or subtract the equations to eliminate one variable.

Step 2 Solve the resulting equation for the other variable.

Step 3 Substitute in either original equation to find the value of the eliminated variable .

Your Notes



Solution

1. Add the equations to x 1 5y 5 9

eliminate one variable. 4x 2 5y 5 214

5x 5 25

2. Solve for x. x 5 21

3. Substitute 21 for x in either equation and solve for y .

x 1 5y 5 9 Write Equation 1.

21 1 5y 5 9 Substitute 21 for x.

y 5 2 Solve for y.


Example 1 Use addition to eliminate a variable

Make sure to check your solution by substituting it into each of the original equations.


7.3


Your Notes



Solution

1. the equations 3x 2 4y 5 2

to eliminate one variable. 3x 1 2y 5 26

5

2. Solve for y. y 5

3. Substitute for y in either equation and .



x 5 Solve for x.


Example 2 Use subtraction to eliminate a variable

1. 28x 1 3y 5 12 2. x 1 6y 5 13

8x 2 9y 5 12 22x 1 6y 5 28

Checkpoint Solve the linear system.

182 Lesson 7.3 • Algebra 1 Notetaking Guide Copyright © Holt McDougal All rights reserved.


Your Notes



Solution

1. Subtract the equations 3x 2 4y 5 2

to eliminate one variable. 3x 1 2y 5 26

26y 5 224

2. Solve for y. y 5 4

3. Substitute 4 for y in either equation and solve for x.


3x 1 2( 4 ) 5 26 Substitute 4 for y.

x 5 6 Solve for x.


Example 2 Use subtraction to eliminate a variable

1. 28x 1 3y 5 12 2. x 1 6y 5 13

8x 2 9y 5 12 22x 1 6y 5 28

(23, 24) (7, 1)


182 Lesson 7.3 • Algebra 1 Notetaking Guide Copyright © Holt McDougal All rights reserved.


Your Notes


y 5 6x 2 32 Equation 2

Solution

1. Equation 2 so that the like terms are arranged in columns.

6x 1 7y 5 16 6x 1 7y 5 16

y 5 6x 2 32

2. the equations. 5

3. Solve for y. y 5

4. Substitute for y in either equation and .



x 5 .


Example 3 Arrange like terms

3. 4x 2 5y 5 5 4. 7y 5 4 2 2x

5y 5 x 1 10 2x 1 y 5 28



Homework


Your Notes


y 5 6x 2 32 Equation 2

Solution

1. Rewrite Equation 2 so that the like terms are arranged in columns.

6x 1 7y 5 16 6x 1 7y 5 16

y 5 6x 2 32 26x 1 y 5 232

2. Add the equations. 8y 5 216


4. Substitute 22 for y in either equation and solve for x .


6x 1 7( 22 ) 5 16 Substitute 22 for y.

x 5 5 Solve for x .


Example 3 Arrange like terms

3. 4x 2 5y 5 5 4. 7y 5 4 2 2x

5y 5 x 1 10 2x 1 y 5 28

(5, 3) (25, 2)



Homework


7.4 Solve Linear Systemsby Multiplying FirstGoal p Solve linear systems by multiplying first.

Your Notes

184 Lesson 7.4 • Algebra 1 Notetaking Guide Copyright © Holt McDougal. All rights reserved.



Solution

1. Multiply Equation 1 by so that the coefficients of y are .

3x 2 3y 5 21 3

8x 1 6y 5 214 8x 1 6y 5 214

2. Add the equations. 5

3. Solve for x. x 5

4. Substitute for x in either of the original equations and .


3( ) 2 3y 5 21 Substitute for x.

y 5 Solve for y.


CHECK Substitute for x and for y in the original equations.


3x 2 3y 5 21 8x 1 6y 5 214

3( ) 2 3( ) 0 21 8( ) 1 6( ) 0 214

5 21 ✓ 5 214 ✓

Example 1 Multiply one equation, then add


7.4 Solve Linear Systemsby Multiplying FirstGoal p Solve linear systems by multiplying first.

Your Notes




Solution

1. Multiply Equation 1 by 2 so that the coefficients of y are opposites .

3x 2 3y 5 21 32 6x 2 6y 5 42

8x 1 6y 5 214 8x 1 6y 5 214

2. Add the equations. 14x 5 28

3. Solve for x. x 5 2

4. Substitute 2 for x in either of the original equations and solve for y.


3( 2 ) 2 3y 5 21 Substitute 2 for x.

y 5 25 Solve for y.


CHECK Substitute 2 for x and 25 for y in the original equations.


3x 2 3y 5 21 8x 1 6y 5 214

3( 2 ) 2 3( 25 ) 0 21 8( 2 ) 1 6( 25 ) 0 214

21 5 21 ✓ 214 5 214 ✓

Example 1 Multiply one equation, then add


Your Notes

Solve the linear system: 3y 5 22x 1 17 Equation 1


Solution 1. Arrange the equations so that like terms are in

columns.

2x 1 3y 5 17 Rewrite Equation 1.


2. Multiply Equation 1 by and Equation 2 by so

that the coefficient of x in each equation is the

of 2 and 3, or .

2x 1 3y 5 17 3 x 1 y 5

3x 1 5y 5 27 3 x 1 y 5

3. the equations. 5

4. Solve for y. y 5

5. Substitute for y in either of the original

equations and solve for x.


3x 1 5( ) 5 27 Substitute for x.

x 5 Solve for x.


Example 2 Multiply both equations, then subtract

Homework

1. 7x 1 2y 5 26 2. 5y 5 9x 2 8

10x 2 5y 5 210 220x 1 10y 5 210

Checkpoint Solve the linear system using elimination.



Your Notes

Solve the linear system: 3y 5 22x 1 17 Equation 1


Solution 1. Arrange the equations so that like terms are in

columns.

2x 1 3y 5 17 Rewrite Equation 1.


2. Multiply Equation 1 by 3 and Equation 2 by 2 so

that the coefficient of x in each equation is the least common multiple of 2 and 3, or 6 .

2x 1 3y 5 17 33 6 x 1 9 y 5 51

3x 1 5y 5 27 32 6 x 1 10 y 5 54

3. Subtract the equations. 21y 5 23


5. Substitute 3 for y in either of the original

equations and solve for x.


3x 1 5( 3 ) 5 27 Substitute 3 for x.

x 5 4 Solve for x.


Example 2 Multiply both equations, then subtract

Homework

1. 7x 1 2y 5 26 2. 5y 5 9x 2 8

10x 2 5y 5 210 220x 1 10y 5 210

(2, 6) (23, 27)

Checkpoint Solve the linear system using elimination.



7.5 Solve Special Types ofLinear SystemsGoal p Identify the number of solutions of a linear

system.

VOCABULARY

Inconsistent system

Consistent dependent system

Show that the linear system has no solution.



SolutionMethod 1 Graphing

x

y

1

3

1212321

23

25

3

Graph the linear system.The lines are because they have the same slope but different y-intercepts. Parallel lines do , so thesystem has .

Method 2 Elimination

Subtract the equations. 22x 1 y 5 1 22x 1 y 5 23 5

The variables are and you are left with a regardless of the values of x and y. This tells you that the system has .

Example 1 A linear system with no solutions


Your Notes



7.5 Solve Special Types ofLinear SystemsGoal p Identify the number of solutions of a linear

system.

VOCABULARY

Inconsistent system A linear system with no solutions

Consistent dependent system A linear system with infinitely many solutions

Show that the linear system has no solution.




x

y

1

3

1212321

23

25

3

22x 1 y 5 23

22x 1 y 5 1Graph the linear system.The lines are parallel because they have the same slope but different y-intercepts. Parallel lines do not intersect , so thesystem has no solution .

Method 2 Elimination

Subtract the equations. 22x 1 y 5 1 22x 1 y 5 23 0 5 4

The variables are eliminated and you are left with a false statement regardless of the values of x and y. This tells you that the system has no solution .

Example 1 A linear system with no solutions


Your Notes



Your Notes

Show that the linear system has infinitely many solutions.

x 1 3y 5 23 Equation 1



x

y

1

3

1212321

23

25

3

Graph the linear system.

The equations represent the

, so any point

on the line is a solution.

So, the linear system has

.

Method 2 Substitution

x 5 Solve Equation 1 for x.


3( ) 1 9y 5 29 Substitute

for x.

1 9y 5 29 Distributive property

5 29 Simplify.

The variables are and you are left with a

statement that is regardless of the values of

x and y. This tells you that the system has

.

Example 2 A linear system with infinitely many solutions



Your Notes

Show that the linear system has infinitely many solutions.

x 1 3y 5 23 Equation 1



x

y

1

3

1212321

23

25

3

3x 1 9y 5 29

x 1 3y 5 23Graph the linear system.

The equations represent the

same line , so any point

on the line is a solution.

So, the linear system has

infinitely many solutions .

Method 2 Substitution

x 5 23y 2 3 Solve Equation 1 for x.


3( 23y 2 3 ) 1 9y 5 29 Substitute 23y 2 3 for x.

29y 2 9 1 9y 5 29 Distributive property

29 5 29 Simplify.

The variables are eliminated and you are left with a

statement that is true regardless of the values of

x and y. This tells you that the system has infinitely many solutions .

Example 2 A linear system with infinitely many solutions



Your Notes

Homework

1. y 5 2x 2 7 2. 2y 5 8x 1 4

4x 2 2y 5 14 24x 1 y 5 4

Checkpoint Tell whether the linear system has no solution or infinitely many solutions.

NUMBER OF SOLUTIONS OF A LINEAR SYSTEM

One solution

x

y

The lines .

The lines have

slopes.

No solution

x

y

The lines are . The lines have the same slope and y-intercepts.

Infinitely manysolutions

x

y

The lines .

The lines have the same slope and the

.



Your Notes

Homework

1. y 5 2x 2 7 2. 2y 5 8x 1 4

4x 2 2y 5 14 24x 1 y 5 4

infinitely many no solutionsolutions

Checkpoint Tell whether the linear system has no solution or infinitely many solutions.

NUMBER OF SOLUTIONS OF A LINEAR SYSTEM

One solution

x

y

The lines intersect . The lines have different slopes.

No solution

x

y

The lines are parallel . The lines have the same slope and different y-intercepts.

Infinitely manysolutions

x

y

The lines coincide . The lines have the same slope and the same y-intercept.



Your Notes

7.6 Solve Linear Systems ofLinear InequalitiesGoal p Solve systems of linear inequalities in two

variables.

VOCABULARY

System of linear inequalities

Solution of a system of linear inequalities

Graph of a system of linear inequalities

GRAPHING A SYSTEM OF LINEAR INEQUALITIES

Step 1 each inequality.

Step 2 Find the of the graphs. The graph of the system is this intersection.



Your Notes

7.6 Solve Linear Systems ofLinear InequalitiesGoal p Solve systems of linear inequalities in two

variables.

VOCABULARY

System of linear inequalities A system of linear inequalities in two variables consists of two or more linear inequalities in the same variables.

Solution of a system of linear inequalities An ordered pair that is a solution of each inequality in the system

Graph of a system of linear inequalities The graph of all solutions of the system

GRAPHING A SYSTEM OF LINEAR INEQUALITIES

Step 1 Graph each inequality.

Step 2 Find the intersection of the graphs. The graph of the system is this intersection.



Your Notes

Graph the system of inequalities.

y > 1 Inequality 1

x ≤ 4 Inequality 2

3y < 6x 2 6 Inequality 3

SolutionGraph all three inequalities in the same coordinate plane. The graph of the system is the shown.

The region is the line

x

y

1

3

5

12121

3 5

y 5 1.

The region is of the line x 5 4.

The region is the line3y 5 6x 2 6.

Example 1 Graph a system of three linear inequalities

1. x 1 y ≤ 5

x

y

1

3

5

1212321

3 5

y < x 1 3

2. x > 22

2226 2 6

2

22

26

6

x

y

y ≤ 4

3x 1 4y ≤ 24

Checkpoint Graph the system of linear equations.



Your Notes

Graph the system of inequalities.

y > 1 Inequality 1

x ≤ 4 Inequality 2

3y < 6x 2 6 Inequality 3

SolutionGraph all three inequalities in the same coordinate plane. The graph of the system is the triangular region shown.

The region is above the line

x

y

1

3

5

12121

3 5

y 5 1.

The region is on and to the left of the line x 5 4.

The region is below the line3y 5 6x 2 6.

Example 1 Graph a system of three linear inequalities

1. x 1 y ≤ 5

x

y

1

3

5

1212321

3 5

y < x 1 3

2. x > 22

2226 2 6

2

22

26

6

x

y

y ≤ 4

3x 1 4y ≤ 24

Checkpoint Graph the system of linear equations.



Your Notes

Homework

Write a system of inequalities for the shaded region.

SolutionInequality 1 One boundary

x

y

1

3

12123

23

25

3

line for the shaded region is . Because the shaded region is the

line, the inequality is .

Inequality 2 Another boundaryline for the shaded region hasa slope of and a y-intercept of . So, its equation is . Because the shaded region is the line, the inequality is .

The system of inequalities for the shaded region is:

Inequality 1

Inequality 2

Example 2 Write a system of linear inequalities


3.

x

y

1

3

5

125 3

4.

21 1 3 5

1

21

3

7

x

y

Checkpoint Write a system of inequalities that defines the shaded region.


Your Notes

Homework

Write a system of inequalities for the shaded region.

SolutionInequality 1 One boundary

x

y

1

3

12123

23

25

3

line for the shaded region is y 5 22 . Because the shaded region is above the dashed line, the inequality is y > 22 .

Inequality 2 Another boundaryline for the shaded region hasa slope of 1 and a y-intercept of 21 . So, its equation is y 5 x 2 1. Because the shaded region is below the solid line, the inequality is y ≤ x 2 1.

The system of inequalities for the shaded region is:

y > 22 Inequality 1

y ≤ x 2 1 Inequality 2

Example 2 Write a system of linear inequalities


3.

x

y

1

3

5

125 3

4.

21 1 3 5

1

21

3

7

x

y

y ≤ 2x 1 5 y < 2 5 } 6 x 1 5

x < 21 x ≥ 2

Checkpoint Write a system of inequalities that defines the shaded region.


Words to ReviewGive an example of the vocabulary word.

System of linear

equations

Consistent independent

system

Dependent system

Solution of a system of

linear inequalities


linear equations

Inconsistent system

System of linear

inequalities

Graph of a system of

linear inequalities

Review your notes and Chapter 7 by using the Chapter Review on pages 490–493 of your textbook.

192 Words to Review • Algebra 1 Notetaking Guide Copyright © Holt McDougal. All rights reserved.



System of linear

equations

3x 1 y 5 9

x 2 y 5 21

Consistent independent

system

3x 1 y 5 9

x 2 y 5 21

Dependent system

x 1 3y 5 23

3x 1 9y 5 29


linear inequalities

A solution of the system

y > 1

x ≤ 4

3y < 6x 2 6

is (3, 2).


linear equations

The solution of the system

3x 1 y 5 9

x 2 y 5 21 is (2, 3).

Inconsistent system

22x 1 y 5 1

22x 1 y 5 23

System of linear

inequalities

y > 1

x ≤ 4

3y < 6x 2 6

Graph of a system of

linear inequalities

x

y

3

5

2121

3 5




VOCABULARY

Order of magnitude


Your Notes

PRODUCT OF POWERS PROPERTY

Let a be a real number, and let m and n be positive integers.

Words: To multiply powers having the same base, .

Algebra: am p an 5 a

Example: 56 p 53 5 5 5 5

Simplify the expression.

a. 22 p 23 5 2

5 2

b. w9 p w2 p w7 5 w

5 w

c. 44 p 4 5 44 p 4

5 4

5 4

d. (26)(26)6 5 (26) p (26)6

5 (26)

5 (26)

Example 1 Use the product of powers property

When simplifying powers with numerical bases only, write your answers using exponents.

Apply Exponent PropertiesInvolving ProductsGoal p Use properties of exponents involving products.

8.1


VOCABULARY

Order of magnitude The order of magnitude of a quantity is the power of 10 nearest the quantity.


Your Notes

PRODUCT OF POWERS PROPERTY


Words: To multiply powers having the same base, add the exponents .

Algebra: am p an 5 a m 1 n

Example: 56 p 53 5 5 6 1 3 5 5 9


a. 22 p 23 5 2 2 1 3

5 2 5

b. w9 p w2 p w7 5 w 9 1 2 1 7

5 w 18

c. 44 p 4 5 44 p 4 1

5 4 4 1 1

5 4 5

d. (26)(26)6 5 (26) 1 p (26)6

5 (26) 1 1 6

5 (26) 7

Example 1 Use the product of powers property


Apply Exponent PropertiesInvolving ProductsGoal p Use properties of exponents involving products.

8.1


Your NotesPOWER OF A POWER PROPERTY


Words: To find a power of a power, .

Algebra: (am)n 5 a

Example: (34)2 5 3 5 3


a. (52)3 5 5 5 5

b. (n7)2 5 n 5 n

c. [(23)5]3 5 (23)

5 (23)

d. [(z 2 4)2]5 5 (z 2 4)

5 (z 2 4)

Example 2 Use the power of a power property

POWER OF A PRODUCT PROPERTY

Let a and b be real numbers, and let m be a positive integer.

Words: To find a power of a product, find the .

Algebra: (ab)m 5

Example: (23 p 17)5 5


a. (4 p 16)7 5

b. (23rs)2 5 ( )2 5 ( )2 p 2 p 2

5

c. 2(3rs)2 5 2( )2 5 2( 2 p 2 p 2) 5

Example 3 Use the power of a product property

When simplifying powers with numerical and variable bases, evaluate the numerical power.



Your NotesPOWER OF A POWER PROPERTY


Words: To find a power of a power, multiply exponents .

Algebra: (am)n 5 a mn

Example: (34)2 5 3 4 p 2 5 3 8


a. (52)3 5 5 2 p 3 5 5 6

b. (n7)2 5 n 7 p 2 5 n 14

c. [(23)5]3 5 (23) 5 p 3

5 (23) 15

d. [(z 2 4)2]5 5 (z 2 4) 2 p 5

5 (z 2 4) 10

Example 2 Use the power of a power property

POWER OF A PRODUCT PROPERTY

Let a and b be real numbers, and let m be a positive integer.

Words: To find a power of a product, find the power of each factor and multiply .

Algebra: (ab)m 5 ambm

Example: (23 p 17)5 5 235 p 175


a. (4 p 16)7 5 47 p 167

b. (23rs)2 5 ( 23 p r p s )2 5 ( 23 )2 p r 2 p s 2

5 9r2s2

c. 2(3rs)2 5 2( 3 p r p s )2 5 2( 3 2 p r 2 p s 2) 5 29r2s2

Example 3 Use the power of a product property





Your Notes

1. (27)8(27)5 2. k3 p k p k2 3. (p3)4

4. [(q 1 8)2]6 5. (8cd)2 6. 2(5z)3

Checkpoint Simplify the expression.

Simplify x2 p (3x3y)3.

Solution

x2 p (3x3y)3 5 property

5 property

5 property

Example 4 Use all three properties

7. (2x5)4 8. (3y3)4 p y5


Homework



Your Notes

1. (27)8(27)5 2. k3 p k p k2 3. (p3)4

(27)13 k6 p12

4. [(q 1 8)2]6 5. (8cd)2 6. 2(5z)3

(q 1 8)12 64c2d2 2125z3


Simplify x2 p (3x3y)3.

Solution

x2 p (3x3y)3 5 x2 p 33 p (x3)3 p y3 Power of a product property

5 x2 p 27 p x9 p y3 Power of a power property

5 27x11y3 Product of powers property

Example 4 Use all three properties

7. (2x5)4 8. (3y3)4 p y5

16x20 81y17


Homework


8.2 Apply Exponent PropertiesInvolving QuotientsGoal p Use properties of exponents involving quotients.

QUOTIENT OF POWERS PROPERTY

Let a be a nonzero real number, and let m and n be positive integers such that m > n.

Words: To divide powers having the same base, the exponents.

Algebra: am }

an 5 a , a Þ 0

Example: 47 }

42 5 4 5 4


a. 612 }

65 5 6 5 6

b. (22)7

} (22)4

5 (22) 5 (22)

c. 42 p 48

} 44 5 4}

44

5 4

5

d. 1 } y9 p y12 5

y12 }

y9

5 y

5

Example 1 Use the quotient of powers property


Your Notes



8.2 Apply Exponent PropertiesInvolving QuotientsGoal p Use properties of exponents involving quotients.

QUOTIENT OF POWERS PROPERTY

Let a be a nonzero real number, and let m and n be positive integers such that m > n.

Words: To divide powers having the same base, subtract the exponents.

Algebra: am }

an 5 a m 2 n , a Þ 0

Example: 47 }

42 5 4 7 2 2 5 4 5


a. 612 }

65 5 6 12 2 5 5 6 7

b. (22)7

} (22)4

5 (22) 7 2 4 5 (22) 3

c. 42 p 48

} 44 5 4

10 }

44

5 4 10 2 4

5 46

d. 1 } y9 p y12 5

y12 }

y9

5 y 12 2 9

5 y3

Example 1 Use the quotient of powers property


Your Notes



Your NotesPOWER OF A QUOTIENT PROPERTY

Let a and b be real numbers with b Þ 0, and let m be a positive integer.

Words: To find a power of a quotient, find the power of the and the power of the and divide.

Algebra: 1 a } b 2 m 5 , b Þ 0 Example: 1 4 } 7 2 3 5 ____ ____

Example 2 Use the power of a quotient property


a. 1 r } s 2 5 5

____

b. 1 2 4 } w 2 3 5 1 2 3 5 5 5

____ ______ ______ ______


1. (28)8

} (28)5

2. 35 p 34

} 33

3. 1 2 r } 3 2 2 4. 1 5 } t 2

4




Your NotesPOWER OF A QUOTIENT PROPERTY

Let a and b be real numbers with b Þ 0, and let m be a positive integer.

Words: To find a power of a quotient, find the power of the numerator and the power of the denominator and divide.

Algebra: 1 a } b 2 m 5 am }

bm , b Þ 0 Example: 1 4 } 7 2 3 5 43 }

73 ____ ____

Example 2 Use the power of a quotient property


a. 1 r } s 2 5 5 r

5 }

s5

____

b. 1 2 4 } w 2 3 5 1 24 } w 2 3 5 (24)3 }

w3 5 264 }

w3 5 2 64 } w3

____ ______ ______ ______


1. (28)8

} (28)5

2. 35 p 34

} 33

(28)3 36

3. 1 2 r } 3 2 2 4. 1 5 } t 2

4

r2 } 9 625 }

t4




Your Notes

Homework

Example 3 Use properties of exponents

Simplify 1 2y7 }

y5 2 3.

Solution

1 2y7 }

y5 2 3 5 property

5 property

5 property

5 property

5. 1 7y3z } y 2 2 6. 2s4

} t p 1 2t } s 2 3

7. 1 6m3n2

} 3mn 2 3 8. 4a }

b2 p 1 2a2b3

} a 2 4




Your Notes

Homework


Simplify 1 2y7 }

y5 2 3.

Solution

1 2y7 }

y5 2 3 5 (2y7)3

} (y5)3

Power of a quotient property

5 23 p (y7)3

} (y5)3

Power of a product property

5 8y21

} y15

Power of a power property

5 8y6 Quotient of powers property

5. 1 7y3z } y 2 2 6. 2s4

} t p 1 2t } s 2 3

49y4z2 16st2

7. 1 6m3n2

} 3mn 2 3 8. 4a }

b2 p 1 2a2b3

} a 2 4

8m6n3 64a5b10




8.3 Define and Use Zero andNegative ExponentsGoal p Use zero and negative exponents.

DEFINITION OF ZERO AND NEGATIVE EXPONENTS

Words Algebra Example

a to the zero power a0 5 , a Þ 0 50 5 is 1.

a2n is the reciprocal a2n 5 , a Þ 0 221 5 of an. ___ ____

an is the reciprocal an 5 , a Þ 0 2 5 of a2n. ____ ____

Example 1 Use definition of zero and negative exponents

Evaluate the expression.

a. 223 5 Definition of ____

5 Evaluate exponent. ___

b. (210)0 5 Definition of

c. 1 1 } 4 2 23

5 Definition of

____

5 Evaluate exponent.

___

5 Simplify.

d. 027 5 a2n is defined only for a number a.


Your Notes


8.3 Define and Use Zero andNegative ExponentsGoal p Use zero and negative exponents.

DEFINITION OF ZERO AND NEGATIVE EXPONENTS

Words Algebra Example

a to the zero power a0 5 1 , a Þ 0 50 5 1 is 1.

a2n is the reciprocal a2n 5 1 } an , a Þ 0 221 5 1 }

2

of an. ___ ____

an is the reciprocal an 5 1 } a2n , a Þ 0 2 5 1 }

221 of a2n. ____ ____

Example 1 Use definition of zero and negative exponents


a. 223 5 1 } 23

Definition of negative ____ exponents

5 1 } 8 Evaluate exponent.

___

b. (210)0 5 1 Definition of zero exponent

c. 1 1 } 4 2 23

5 1 } 1 1 } 4 2 3

Definition of negative

____ exponents

5 1 } 1 } 64

Evaluate exponent.

___

5 64 Simplify.

d. 027 5 undefined a2n is defined only for a nonzero number a.


Your Notes


Your NotesPROPERTIES OF EXPONENTS

Let a and b be real numbers, and let m and n be integers.

am p an 5 a property

(am)n 5 a property

(ab)m 5 property

am }

an 5 a , a Þ 0 property

1 a } b 2 m 5 , b Þ 0 property ____


a. (25)4 p (25)24 5 Product of powersproperty

5 exponents.

5 Definition of

b. (522)22 5 property

5 exponents.

5 Evaluate power.

c. 1 } 422 5 Definition of

5 Evaluate power.

d. 32 }

321 5 property

5 exponents.

5 Evaluate power.

Example 2 Evaluate exponential expressions



Your NotesPROPERTIES OF EXPONENTS

Let a and b be real numbers, and let m and n be integers.

am p an 5 a m 1 n Product of powers property

(am)n 5 a mn Power of a power property

(ab)m 5 ambm Power of a product property

am }

an 5 a m 2 n , a Þ 0 Quotient of powers property

1 a } b 2 m 5 am }

bm , b Þ 0 Power of a quotient property ____


a. (25)4 p (25)24 5 (25)4 1 (24) Product of powersproperty

5 50 Add exponents.

5 1 Definition of zero exponent

b. (522)22 5 522 p (22) Power of a power property

5 54 Multiply exponents.

5 625 Evaluate power.

c. 1 } 422 5 42 Definition of

negative exponents


d. 32 }

321 5 32 2 (21) Quotient of powers property

5 33 Subtract exponents.


Example 2 Evaluate exponential expressions



Your Notes

Homework

1. 1 1 } 8 2 21 2. 1 } 322

3. 621 } 6 4. (521)2

Checkpoint Evaluate the expression.


Simplify the expression 2w23x } (2wx)2

. Write your answer using

only positive exponents.

Solution

2w23x } (2wx)2

5 Definition of negative exponents ___________

5 property ___________

5 property ___________

5 property ___________

5. 6fg24

} 2f2g

6. (3yz2)22




Your Notes

Homework

1. 1 1 } 8 2 21 2. 1 } 322

8 9

3. 621 } 6 4. (521)2

1 } 36 1 } 25



Simplify the expression 2w23x } (2wx)2

. Write your answer using

only positive exponents.

Solution

2w23x } (2wx)2

5 2x } w3(2wx)2

Definition of negative exponents ___________

5 2x } w3(4w2x2)

Power of a product property ___________

5 2x } 4w5x2

Product of powers property ___________

5 1 } 2w5x

Quotient of powers property ___________

5. 6fg24

} 2f2g

6. (3yz2)22

3 } fg5 1 }

9y2z4




Define and Use Fractional Exponents

Goal p Use fractional exponents.

a. 25 1 } 2

5 b. 362 1 } 2

5

5 5

5

c. 4 5 } 2

5 d. 492 3 } 2

5

5 5

5 5

5 5

5 5

5

Example 1 Evaluate expressions involving square roots

1. 16 3 } 2 2. 64

2 1 } 2

3. 1442 3 } 2

4. 25 5 } 2


202 8.3 Focus On Operations • Algebra 1 Notetaking Guide Copyright © Holt McDougal. All rights reserved.

Your Notes

Focus On OperationsUse after Lesson 8.3


Define and Use Fractional Exponents

Goal p Use fractional exponents.

a. 25 1 } 2

5 Ï}

25 b. 362 1 } 2

5 1 }

36 1 } 2

5 5 5 1 }

Ï}

36

5 1 } 6

c. 4 5 } 2

5 4 1 1 } 2 2 p 5

d. 492 3 } 2

5 49 1 1 } 2 2 p 1 23 2

5 1 4 1 } 2 2 5 5 1 49

1 } 2 2 23

5 1 Ï}

4 2 5 5 1 Ï}

49 2 23

5 25 5 723

5 32 5 1 } 73

5 1 } 343

Example 1 Evaluate expressions involving square roots

1. 16 3 } 2 64 2. 64

2 1 } 2 1 }

8

3. 1442 3 } 2

1 } 1728

4. 25 5 } 2 3125



Your Notes



Your Notes

Homework

5. 64 2 } 3 6. 125

2 4 } 3

7. 1 1 } 4 2 2 3 } 2

1 1 } 4 2

5 } 2 8. 1 27 2

5

} 3 1 27 2

2 4 } 3



a. 152 1 } 2 p 15

5 b. 4

4 } 3 p 4 } 4

1 } 3 5

5 5

5 5

5 5

5

Copyright © Holt McDougal. All rights reserved. 8.3 Focus On Operations • Algebra 1 Notetaking Guide 203

a. 64 1 } 3 5 b. 216

2 1 } 3 5

5 5

5 5

c. 8 4 } 3 5 d. 27

2 2 } 3 5

5 5

5 5

5 5

5 5

5

Example 2 Evaluate expressions involving cube roots

5 } 2

LAH_A1_11_FL_NTG_Ch08_193-217.indd 203LAH_A1_11_FL_NTG_Ch08_193-217.indd 203 11/14/09 1:39:12 AM11/14/09 1:39:12 AM

Your Notes

Homework

5. 64 2 } 3 16 6. 125

2 4 } 3 1 }

625

7. 1 1 } 4 2 2 3 } 2

1 1 } 4 2

5 } 2 1 }

4 8. 1 27 2

5

} 3 1 27 2

2 4 } 3 3



a. 152 1 } 2 p 15

5 15 1 2 1 } 2 2 1 1 5 } 2 2 b. 4

4 } 3 p 4 } 4

1 } 3 5 4

1 4 } 3 2 1 1

}

4 1 } 3

5 15 4 } 2 5 4

1 7 } 3 2 }

4 1 } 3

5 152 5 4 1 7 } 3 2 2 1 1 }

3 2

5 225 5 42

5 16


a. 64 1 } 3 5

3 Ï}

64 b. 2162 1 } 3

5 1 }

216 1 } 3

5 3 Ï}

43 5 1 }

3 Ï}

216

5 4 5 1 } 6

c. 8 4 } 3 5 8 1

1 } 3 2 p 4 d. 27

2 2 } 3 5 27

1 1 } 3 2 p 1 22 2

5 1 8 1 } 3 2 4 5 1 27

1 } 3 2 22

5 1 3 Ï}

8 2 4 5 1 3 Ï}

27 2 22

5 24 5 322

516 5 1 }

32

5 1 } 9

Example 2 Evaluate expressions involving cube roots

5 } 2


8.4 Use Scientific NotationGoal p Read and write numbers in scientific notation.

VOCABULARY

Scientific notation Your Notes

SCIENTIFIC NOTATION

A number is written in scientific notation when it is of the form where 1 ≤ c < 10 and n is an integer.

Number Standard form Scientific notation

Sixteen million

Two hundredths

a. 7,820,000 5 3 10 Move decimal point places to the . Exponent is .

b. 0.00401 5 3 10 Move decimal point places to the . Exponent is .

Example 1 Write numbers in scientific notation

a. 3.89 3 109 5 Exponent is . Move decimal point

places to the .

b. 9.097 3 1025 5 Exponent is . Move decimal point places to the

.

Example 2 Write numbers in standard form



8.4 Use Scientific NotationGoal p Read and write numbers in scientific notation.

VOCABULARY

Scientific notation A number is written in scientific notation when it is of the form c 3 10n where 1 ≤ c < 10 and n is an integer.

Your Notes

SCIENTIFIC NOTATION

A number is written in scientific notation when it is of the form c 3 10n where 1 ≤ c < 10 and n is an integer.

Number Standard form Scientific notation

Sixteen million 16,000,000 1.6 3 107

Two hundredths 0.02 2 3 1022

a. 7,820,000 5 7.82 3 10 6 Move decimal point 6 places to the left . Exponent is 6 .

b. 0.00401 5 4.01 3 10 23 Move decimal point 3 places to the right . Exponent is 23 .

Example 1 Write numbers in scientific notation

a. 3.89 3 109 5 3,899,000,000 Exponent is 9 . Move decimal point 9 places to the right .

b. 9.097 3 1025 5 0.00009097 Exponent is 25 . Move decimal point 5 places to the left .

Example 2 Write numbers in standard form



Your Notes


1. Write the number 0.0899 in scientific notation. Then write the number 6.0001 3 107 in standard form.

Checkpoint Complete the following exercise.

Order 3.2 3 1024, 0.0004, and 2.8 3 1025 from least to greatest.

SolutionStep 1 Write each number in scientific notation, if

necessary.

0.0004 5

Step 2 Order the numbers. First order the numbers with different powers of 10. Then order the numbers with the same power of 10.

Because 1025 1024, you know that is less than both and

. Because 3.2 4, you know that is less than .

So, < < .

Step 3 Write the original numbers in order from least to greatest.

Example 3 Order numbers in scientific notation

2. Order 225,000, 1,740,000, and 1.75 3 105 from least to greatest.



Your Notes


1. Write the number 0.0899 in scientific notation. Then write the number 6.0001 3 107 in standard form.

8.99 3 1022; 60,001,000


Order 3.2 3 1024, 0.0004, and 2.8 3 1025 from least to greatest.

SolutionStep 1 Write each number in scientific notation, if

necessary.

0.0004 5 4 3 1024

Step 2 Order the numbers. First order the numbers with different powers of 10. Then order the numbers with the same power of 10.

Because 1025 < 1024, you know that 2.8 3 1025 is less than both 3.2 3 1024 and 4 3 1024 . Because 3.2 < 4, you know that 3.2 3 1024 is less than 4 3 1024 .

So, 2.8 3 1025 < 3.2 3 1024 < 4 3 1024 .

Step 3 Write the original numbers in order from least to greatest.

2.8 3 1025; 3.2 3 1024; 0.0004

Example 3 Order numbers in scientific notation

2. Order 225,000, 1,740,000, and 1.75 3 105 from least to greatest.

1.75 3 105; 225,000; 1,740,000



Your Notes

Evaluate the expression. Write your answer in scientific notation.

a. (5.6 3 1024)(1.4 3 1025) 5 (5.6 p 1.4) 3 (1024 p 1025) Commutative property

and associative property

5 3 Product of powers property

b. (3.2 3 102)3

5 3 Power of a product property

5 3 Power of a power property

5 ( ) 3 Write in scientific notation.

5 3 ( ) Associative property

5 Product of powers property

c. 3.5 3 1023 }}

1.75 3 1025

5 3.5 } 1.75 3 1023 }

1025 Product rule for fractions

5 3 Quotient of powers property

Example 4 Compute with numbers in scientific notation

Homework

3. (2.01 3 1027)2 4. 4.8 3 1024 }

6 3 1024

Checkpoint



Your Notes

Evaluate the expression. Write your answer in scientific notation.

a. (5.6 3 1024)(1.4 3 1025) 5 (5.6 p 1.4) 3 (1024 p 1025) Commutative property

and associative property

5 7.84 3 1029 Product of powers property

b. (3.2 3 102)3

5 3.23 3 (102)3 Power of a product property

5 32.768 3 106 Power of a power property

5 ( 3.2768 3 101 ) 3 106 Write 32.768 in scientific notation.

5 3.2768 3 ( 101 3 106 ) Associative property

5 3.2768 3 107 Product of powers property

c. 3.5 3 1023 }}

1.75 3 1025

5 3.5 } 1.75 3 1023 }

1025 Product rule for fractions

5 2 3 102 Quotient of powers property

Example 4 Compute with numbers in scientific notation

Homework

3. (2.01 3 1027)2 4. 4.8 3 1024 }

6 3 1024

4.0401 3 10214 8 3 1021

Checkpoint



Your Notes

8.5 Write and Graph ExponentialGrowth FunctionsGoal p Write and graph exponential growth models.

VOCABULARY

Exponential function

Exponential growth

Compound interest

Write a rule for the function.

x 22 21 0 1 2

y 2 } 9 2 }

3 2 6 18

SolutionStep 1 Tell whether the function is exponential. Here the

y-values are multiplied by for each increase

of 1 in x, so the table represents an exponential

function of the form where .

Step 2 Find the value of a by finding the value of y when

x 5 0. When x 5 0, y 5 5 5 .

The value of y when x 5 0 is , so .

Step 3 Write the function rule. A rule for the function is

y 5 .

Example 1 Write a function rule



Your Notes

8.5 Write and Graph ExponentialGrowth FunctionsGoal p Write and graph exponential growth models.

VOCABULARY

Exponential function A function of the form y 5 abx where a Þ 0, b > 0, and b Þ 1

Exponential growth A quantity that increases by the same percent over equal time periods

Compound interest Interest earned on both an initial investment and on previously earned interest

Write a rule for the function.

x 22 21 0 1 2

y 2 } 9 2 }

3 2 6 18

SolutionStep 1 Tell whether the function is exponential. Here the

y-values are multiplied by 3 for each increase

of 1 in x, so the table represents an exponential

function of the form y 5 abx where b 5 3 .

Step 2 Find the value of a by finding the value of y when

x 5 0. When x 5 0, y 5 ab0 5 a p 1 5 a .

The value of y when x 5 0 is 2 , so a 5 2.

Step 3 Write the function rule. A rule for the function is

y 5 2 p 3x .

Example 1 Write a function rule



Your Notes

Graph the function y 5 3x. Identify its domain and range.

SolutionStep 1 Make a table by choosing a

x

y

1

3

5

7

9

12123 3

few values for x and finding the values of y. The domain is .

x 22 21 0 1 2

y

Step 2 Plot the points.

Step 3 Draw a smooth curve through the points. From either the table or the graph, you can see that the range is .

Example 2 Graph an exponential function

Graph y 5 2 p 3x. Compare the graph with the graph of y 5 3x.

SolutionTo graph each function, make x y 5 3x y 5 2 p 3x

22

21

0

1

2

a table of values, plot the points, and draw a smooth curve through the points.

x

y

2

6

10

14

18

12123 3

Because the y-values for y 5 2 p 3x are the corresponding y-values for y 5 3x, the graph of y 5 2 p 3x is a of the graph of y 5 3x.

Example 3 Compare graphs of exponential functions



Your Notes

Graph the function y 5 3x. Identify its domain and range.

SolutionStep 1 Make a table by choosing a

x

y

1

3

5

7

9

12123 3

y 5 3x

(2, 9)

(1, 3)

(0, 1)( )1922,

( )1321,

few values for x and finding the values of y. The domain is all real numbers .

x 22 21 0 1 2

y 1 } 9 1 }

3

1 3 9


Step 3 Draw a smooth curve through the points. From either the table or the graph, you can see that the range is all positive real numbers .


Graph y 5 2 p 3x. Compare the graph with the graph of y 5 3x.

SolutionTo graph each function, make x y 5 3x y 5 2 p 3x

22 1 } 9 2 }

9

21 1 } 3 2 }

3

0 1 2

1 3 6

2 9 18

a table of values, plot the points, and draw a smooth curve through the points.

x

y

2

6

10

14

18

12123 3

y 5 3xy 5 2 ? 3x

Because the y-values for y 5 2 p 3x are 2 times the corresponding y-values for y 5 3x, the graph of y 5 2 p 3x is a vertical stretch of the graph of y 5 3x.




Your Notes

1. Write a rule for the function.

x 22 21 0 1 2

y 2 1 } 16 2 1 } 4 21 24 216

2. Graph y 5 4x. Identify its

x

y

2

6

10

14

12123 322

domain and range.

3. Graph y 5 22 p 3x. Compare

x

y

3

9

12123 323

29

215

the graph with the graph of y 5 3x.

Checkpoint Complete the following exercises.



Your Notes

1. Write a rule for the function.

x 22 21 0 1 2

y 2 1 } 16 2 1 } 4 21 24 216

y 5 21 p 4x

2. Graph y 5 4x. Identify its

x

y

2

6

10

14

12123 3

y 5 4x

22

(0, 1)

(1, 4)

(2, 16)

( )11622,

( )1421,

domain and range.

The domain is all real numbers. The range is all positive real numbers.

3. Graph y 5 22 p 3x. Compare

x

y

3

9

12123 3

y 5 3x

y 5 22 ? 3x23

29

215

the graph with the graph of y 5 3x.

The graph of y 5 22 p 3x is a vertical stretch and a reflection in the x-axis of the graph of y 5 3x.




Your Notes

Homework

EXPONENTIAL GROWTH MODEL

y 5 a(1 1 r)t

a is the . r is the .

1 1 r is the . t is the .

Investment You put $250 in a savings account that earns 4% annual interest compounded yearly. You do not make any deposits or withdrawals. How much will your investment be worth in 10 years?

Solution

The initial amount is , the interest rate is , or , and the time period is .

y 5 a(1 1 r)t Write exponential growth model.

5 (1 1 ) Substitute for a, for r, and for t.

5 250( )10 Simplify.

ø Use a calculator.

You will have in 10 years.

Example 4 Solve a compound interest problem

4. In Example 4, suppose the annual interest rate is 5%. How much will your investment be worth in 10 years?




Your Notes

Homework

EXPONENTIAL GROWTH MODEL

y 5 a(1 1 r)t

a is the initial amount . r is the growth rate .

1 1 r is the growth factor . t is the time period .

Investment You put $250 in a savings account that earns 4% annual interest compounded yearly. You do not make any deposits or withdrawals. How much will your investment be worth in 10 years?

Solution

The initial amount is $250 , the interest rate is 4% , or 0.04 , and the time period is 10 years .

y 5 a(1 1 r)t Write exponential growth model.

5 250 (1 1 0.04 ) 10 Substitute 250 for a, 0.04 for r, and 10 for t.

5 250( 1.04 )10 Simplify.

ø 370.06 Use a calculator.

You will have $370.06 in 10 years.

Example 4 Solve a compound interest problem

4. In Example 4, suppose the annual interest rate is 5%. How much will your investment be worth in 10 years?

about $407.22




8.6 Write and Graph ExponentialDecay FunctionsGoal p Write and graph exponential decay functions.

VOCABULARY

Exponential decay

Graph the function y 5 1 1 } 3 2 x and identify its domain and

range.

SolutionStep 1 Make a table of values.

The domain is .

x 22 21 0 1 2

y


x

y

1

3

5

7

9

12123 3

Step 3 Draw a smooth curve through the points. From either the table or the graph, you can see that the range is .



Your Notes


8.6 Write and Graph ExponentialDecay FunctionsGoal p Write and graph exponential decay functions.

VOCABULARY

Exponential decay A quantity that decreases by the same percent over equal time periods

Graph the function y 5 1 1 } 3 2 x and identify its domain and

range.

SolutionStep 1 Make a table of values.

The domain is all real numbers .

x 22 21 0 1 2

y 9 3 1 1 } 3 1 }

9


x

y

1

3

5

7

9

12123 3

(22, 9)

(21, 3)

(0, 1) ( )192,

( )131,

y 5x( )1

3

Step 3 Draw a smooth curve through the points. From either the table or the graph, you can see that the range is all positive real numbers .



Your Notes


Your Notes

Graph y 5 2 p 1 1 } 3 2 x. Compare the graph with the graph

of y 5 1 1 } 3 2 x.

Solution

x y 5 1 1 } 3 2 x y 5 2 p 1 1 } 3 2 x

22

21

0

1

2

x

y

1

3

5

7

9

12123 3

Because the y-values for y 5 2 p 1 1 } 3 2 x are

the corresponding y-values for y 5 1 1 } 3 2 x, the graph of

y 5 2 p 1 1 } 3 2 x is a of the graph of

y 5 1 1 } 3 2 x.


1. Graph y 5 22 p 1 1 } 3 2 x. Compare the graph with the

graph of 1 1 } 3 2 x.

x

y

3

9

12123 323

29

215




Your Notes

Graph y 5 2 p 1 1 } 3 2 x. Compare the graph with the graph

of y 5 1 1 } 3 2 x.

Solution

x y 5 1 1 } 3 2 x y 5 2 p 1 1 } 3 2 x

22 9 18

21 3 6

0 1 2

1 1 } 3 2 }

3

2 1 } 9 2 }

9

x

y

1

3

5

7

9

12123 3

y 5x( )1

3

y 5 2 ? x( )1

3

Because the y-values for y 5 2 p 1 1 } 3 2 x are 2 times

the corresponding y-values for y 5 1 1 } 3 2 x, the graph of

y 5 2 p 1 1 } 3 2 x is a vertical stretch of the graph of

y 5 1 1 } 3 2 x.


1. Graph y 5 22 p 1 1 } 3 2 x. Compare the graph with the

graph of 1 1 } 3 2 x.

x

y

3

9

12123 323

29

215

y 5x( )1

3

y 5 22 ? x( )1

3

The graph of y 5 22 p 1 1 } 3 2 x

is a vertical stretch and a

reflection in the x-axis of

the graph of y 5 1 1 } 3 2 x.




Your Notes

Tell whether the graph represents exponential growth or exponential decay. Then write a rule for the function.

SolutionThe graph represents

x

y

1

3

5

12121

3 5

(1, 1)

(0, 5)

(y 5 abx where 0 < b < 1).

The y-intercept is , so

a 5 . Find the value

of b by using the point

(1, 1) and a 5 .

y 5 abx Write function.

5 p b Substitute.

5 b Solve.

A function rule is .

Example 3 Classify and write rules for functions

EXPONENTIAL GROWTH AND DECAY

Exponential Growth Exponential Decay

y 5 abx, a > 0 y 5 abx, a > 0

and b > 1 and 0 < b < 1

x

y

(0, a)

x

y

(0, a)


EXPONENTIAL DECAY MODEL

y 5 a(1 2 r)t

a is the . r is the .

1 2 r is the . t is the .


Your Notes

Tell whether the graph represents exponential growth or exponential decay. Then write a rule for the function.

SolutionThe graph represents

x

y

1

3

5

12121

3 5

(1, 1)

(0, 5) exponential decay

(y 5 abx where 0 < b < 1).

The y-intercept is 5 , so

a 5 5 . Find the value

of b by using the point

(1, 1) and a 5 5 .

y 5 abx Write function.

1 5 5 p b 1 Substitute.

0.2 5 b Solve.

A function rule is y 5 5(0.2)x .

Example 3 Classify and write rules for functions

EXPONENTIAL GROWTH AND DECAY

Exponential Growth Exponential Decay

y 5 abx, a > 0 y 5 abx, a > 0

and b > 1 and 0 < b < 1

x

y

(0, a)

x

y

(0, a)


EXPONENTIAL DECAY MODEL

y 5 a(1 2 r)t

a is the initial amount . r is the decay rate .

1 2 r is the decay factor . t is the time period .


Your Notes

Homework

Population The population of a city decreased from 1995 to 2003 by 1.5% annually. In 1995 there were about 357,000 people living in the city. Write a function that models the city's population since 1995. Then find the population in 2003.

SolutionLet P be the population of the city (in thousands), and let t be the time (in years) since 1995. The initial value is , and the decay rate is .

P 5 a(1 2 r)t Write exponential decay model.

5 (1 2 )t Substitute for a, and for r.

5 Simplify.

To find the population in 2003, years after 1995, substitute for t.

P 5 Substitute for t.

ø Use a calculator.

The city's population was about in 2003.

Example 4 Use the exponential decay model

2. The graph of an exponential function passes through the points (0, 4) and (1, 10).

x

y

2

6

10

14

1212322

3

Graph the function. Tell whether the graph represents exponential growth or exponential decay. Then write a rule for the function.

3. In Example 4, suppose that the decay rate of the city's population remains the same beyond 2003. What will be the population in 2020?




Your Notes

Homework

Population The population of a city decreased from 1995 to 2003 by 1.5% annually. In 1995 there were about 357,000 people living in the city. Write a function that models the city's population since 1995. Then find the population in 2003.

SolutionLet P be the population of the city (in thousands), and let t be the time (in years) since 1995. The initial value is 357 , and the decay rate is 0.015 .

P 5 a(1 2 r)t Write exponential decay model.

5 357 (1 2 0.015 )t Substitute 357 for a, and 0.015 for r.

5 357(0.985)t Simplify.

To find the population in 2003, 8 years after 1995, substitute 8 for t.

P 5 357(0.985)8 Substitute 8 for t.

ø 316.3 Use a calculator.

The city's population was about 316,300 in 2003.

Example 4 Use the exponential decay model

2. The graph of an exponential function passes through the points (0, 4) and (1, 10).

x

y

2

6

10

14

1212322

3

(1, 10)

(0, 4)

Graph the function. Tell whether the graph represents exponential growth or exponential decay. Then write a rule for the function.

Exponential growth;y 5 4(2.5)x

3. In Example 4, suppose that the decay rate of the city's population remains the same beyond 2003. What will be the population in 2020?

about 244,700




Graph the sequence 243, 81, 27, 9, 3, …. Let each term’s position number be the x-value. The term is the corresponding y-value.

Position, x 1 2

Term, y 243 27

Example 2 Graph a geometric sequence

Copyright © Holt McDougal. All rights reserved. 8.6 Focus On Functions • Algebra 1 Notetaking Guide 215

Relate Geometric Sequences to Exponential FunctionsGoal p Identify, graph, and write geometric sequences.

VOCABULARY

Geometric sequence

Common ratio

Tell whether the sequence 243, 81, 27, 9, 3, ... is arithmetic or geometric. Then write the next term of the sequence.

Solution

The first term is a1 5 . Find the of consecutive terms:

a2

} a1 5 5

a3 } a2 5 5 5 } 5 5 5

The ratios are . The sequence is . The

common ratio is .

The next term of the sequence is a6 5 p 5 .

Example 1 Identify a geometric sequence

The sequence is a function. The domain is the set of positive numbers (all positive whole numbers). The range is the set of terms.

1 2 3 4 5 6

y

x

200

50

100

150

Your Notes

Focus On FunctionsUse after Lesson 8.6


Graph the sequence 243, 81, 27, 9, 3, …. Let each term’s position number be the x-value. The term is the corresponding y-value.

Position, x 1 2 3 4 5 6Term, y 243 81 27 9 3 1

Example 2 Graph a geometric sequence


Relate Geometric Sequences to Exponential FunctionsGoal p Identify, graph, and write geometric sequences.

VOCABULARY

Geometric sequence Sequence where the ratio of any term to the previous term is constant

Common ratio The constant ratio in a geometric sequence

Tell whether the sequence 243, 81, 27, 9, 3, ... is arithmetic or geometric. Then write the next term of the sequence.

Solution

The first term is a1 5 243. Find the ratios of consecutive terms:

a2

} a1 5 81

} 243 5 1 } 3 a3

} a2 5 27

} 81 5 1 } 3 a4

} a3 5

9 }

27 5 1 } 3

a5 } a4 5

3 } 9 5

1 } 3

The ratios are constant. The sequence is geometric. The

common ratio is 1 } 3 .

The next term of the sequence is a6 5 a5 p 1 } 3 5 1.

Example 1 Identify a geometric sequence

The sequence is a function. The domain is the set of positive numbers (all positive whole numbers). The range is the set of terms.

1 2 3 4 5 6

y

x

200

50

100

150

Your Notes

Focus On FunctionsUse after Lesson 8.6


Homework

Your NotesGENERAL RULE OF A GEOMETRIC SEQUENCE

The nth term of a with first a1 and common r is given by: an 5 .

Write a rule for the nth term of the geometric sequence 243, 81, 27, 9, 3, …. Then find a10 .

Solution

Write the general rule an 5 . Then substitute

values for and to give an 5 p . For the

tenth term, n 5 , so

a10 5 p 5 .

Example 3 Write a rule for a geometric sequence

1. Tell whether the sequence is arithmetic or geometric. Write the next term in the sequence.

7, 221, 63, 2189, …

2. Graph the sequence 7, 221, 63, 2189, …

3. Write a rule for the nth term of the geometric sequence and find a8. 3, 33, 363, 3993, …


1 2 3 4 5

400

600

200

�200

y

x

216 8.6 Focus On Functions • Algebra 1 Notetaking Guide Copyright © Holt McDougal. All rights reserved.


Homework

Your NotesGENERAL RULE OF A GEOMETRIC SEQUENCE

The nth term of a geometric sequence with first term a1 and common ratio r is given by: an 5 a1r n21.

Write a rule for the nth term of the geometric sequence 243, 81, 27, 9, 3, …. Then find a10 .

Solution

Write the general rule an 5 a1r n21 . Then substitute

values for a1 and r to give an 5 243 p 1 1 } 3 2 n 21 . For the

tenth term, n 5 10, so

a10 5 243 p 1 1 } 3 2 10 21 5 1 } 81 .

Example 3 Write a rule for a geometric sequence

1. Tell whether the sequence is arithmetic or geometric. Write the next term in the sequence.

7, 221, 63, 2189, …

Geometric; 567

2. Graph the sequence 7, 221, 63, 2189, …

3. Write a rule for the nth term of the geometric sequence and find a8. 3, 33, 363, 3993, …

an 5 3 p 11n21; 58,461,513


1 2 3 4 5

400

600

200

�200

y

x





Order of magnitude


Compound interest

Geometric sequence

Scientific notation

Exponential growth

Exponential decay

Common ratio

Copyright © Holt McDougal. All rights reserved. Words to Review • Algebra 1 Notetaking Guide 217




Order of magnitude

The order of magnitude of 96,000,000 is 108.


y 5 3 p 2x

Compound interest

Interest earned on both an initial investment and previously earned interest

Geometric sequence

2, 10, 50, 250, 1250,…

Scientific notation

3.7 3 1025

Exponential growth

y 5 16(2.5)t

Exponential decay

y 5 2.85(0.05)t

Common ratio

3 for the sequence 4, 12, 36, 108,…



9.1

Your NotesVOCABULARY

Monomial

Degree of a monomial

Polynomial

Degree of a polynomial

Leading coefficient

Binomial

Trinomial

Goal p Add and subtract polynomials.


Add and Subtract Polynomials

Write 7 1 2x4 2 4x so that the exponents decrease from left to right. Identify the degree and leading coefficient of the polynomial.

SolutionConsider the degree of each of the polynomial's terms.

Degree is . Degree is . Degree is .

7 1 2x4 2 4x

The polynomial can be written as . The greatest degree is , so the degree of the polynomial is , and the leading coefficient is .

Example 1 Rewrite a polynomial

LAH_A1_11_FL_NTG_Ch09.indd 218LAH_A1_11_FL_NTG_Ch09.indd 218 3/2/09 4:27:58 PM3/2/09 4:27:58 PM

9.1

Your NotesVOCABULARY

Monomial A number, a variable, or the product of a number and one or more variables with whole number exponents.

Degree of a monomial The sum of the exponents of the variables in a monomial

Polynomial A monomial or a sum of monomials

Degree of a polynomial The greatest degree of the terms in a polynomial

Leading coefficient The coefficient of the first term in a polynomial that is written with the exponents of a variable decreasing from left to right

Binomial A polynomial with two terms

Trinomial A polynomial with three terms

Goal p Add and subtract polynomials.


Add and Subtract Polynomials

Write 7 1 2x4 2 4x so that the exponents decrease from left to right. Identify the degree and leading coefficient of the polynomial.

SolutionConsider the degree of each of the polynomial's terms.

Degree is 0 . Degree is 4 . Degree is 1 .

7 1 2x4 2 4x

The polynomial can be written as 2x4 2 4x 1 7 . The greatest degree is 4 , so the degree of the polynomial is 4 , and the leading coefficient is 2 .

Example 1 Rewrite a polynomial


Your Notes

Tell whether the expression is a polynomial. If it is a polynomial, find its degree and classify it by the number of terms. Otherwise, tell why it is not a polynomial.

a.

Expression Is it a polynomial? Classify by degreeand number of terms

26 0 degree monomial

m23 1 4

2h3 1 4h2 Yes

9 2 5x4 1 3x Yes

2w3 1 4w

b.

c.

d.

e.

Example 2 Identify and classify polynomials

Checkpoint Write the polynomial so that the exponents decrease from left to right. Identify the degree and leading coefficient of the polynomial.

1. 5x 1 13 1 8x3

2. 4y4 2 7y5 1 2y

Checkpoint Tell whether the expression is a polynomial. If it is a polynomial, find its degree and classify it by the number of terms. Otherwise, tell why it is not a polynomial.

3. 4x 2 x7 1 5x3 4. v3 1 v22 1 2v



Your Notes

Tell whether the expression is a polynomial. If it is a polynomial, find its degree and classify it by the number of terms. Otherwise, tell why it is not a polynomial.

a.

Expression Is it a polynomial? Classify by degreeand number of terms

26 Yes 0 degree monomial

m23 1 4 No; negativeexponent

2h3 1 4h2 Yes 3rd degreebinomial

9 2 5x4 1 3x Yes 4th degreetrinomial

2w3 1 4w No; variableexponent

b.

c.

d.

e.

Example 2 Identify and classify polynomials

Checkpoint Write the polynomial so that the exponents decrease from left to right. Identify the degree and leading coefficient of the polynomial.

1. 5x 1 13 1 8x3

8x3 1 5x 1 13; 3; 8

2. 4y4 2 7y5 1 2y

27y5 1 4y4 1 2y; 5; 27

Checkpoint Tell whether the expression is a polynomial. If it is a polynomial, find its degree and classify it by the number of terms. Otherwise, tell why it is not a polynomial.

3. 4x 2 x7 1 5x3 4. v3 1 v22 1 2v

polynomial; 7; not a polynomial;trinomial negative exponent




Homework

Your Notes

Find the sum (a) (4x3 1 x2 2 5) 1 (7x 1 x3 2 3x2) and (b) (x2 1 x 1 8) 1 (x2 2 x 2 1).

Solutiona. Vertical format: Align like 4x3 1 x2 2 5

terms in vertical columns. 1 x3 2 3x2 1 7x

b. Horizontal format: Use the associative and

commutative properties to group like terms. then

simplify

(x2 1 x 1 8) 1 (x2 2 x 2 1)

5 ( ) 1 ( ) 1 ( )

5

Example 3 Add polynomials

Find the difference (a) (4z2 2 3) 2 (22z2 1 5z 2 1) and (b) (3x2 1 6x 2 4) 2 (x2 2 x 2 7).

Solutiona. ( 4z2 2 3) 4z2 2 3

2 (22z2 1 5z 2 1) 2z2 5z 1

b. (3x2 1 6x 2 4) 2 (x2 2 x 2 7)

5 3x2 1 6x 2 4

5

5

Example 4 Subtract polynomials

If a particular power of the variable appears in one polynomial but not the other, leave a space in that column, or write the term with a coefficient of 0.

Remember to multiply each term in the polynomial by 21 when you write the subtraction as addition.

5. (3x4 2 2x2 2 1) 1 (5x3 2 x2 1 9x4)

6. (3t2 2 5t 1 t4) 2 (11t4 2 3t2)

Checkpoint Find the sum or difference.



Homework

Your Notes

Find the sum (a) (4x3 1 x2 2 5) 1 (7x 1 x3 2 3x2) and (b) (x2 1 x 1 8) 1 (x2 2 x 2 1).

Solutiona. Vertical format: Align like 4x3 1 x2 2 5

terms in vertical columns. 1 x3 2 3x2 1 7x

5x3 2 2x2 1 7x 2 5

b. Horizontal format: Use the associative and

commutative properties to group like terms. then

simplify

(x2 1 x 1 8) 1 (x2 2 x 2 1)

5 ( x2 1 x2 ) 1 ( x 2 x ) 1 ( 8 2 1 )

5 2x2 1 7

Example 3 Add polynomials

Find the difference (a) (4z2 2 3) 2 (22z2 1 5z 2 1) and (b) (3x2 1 6x 2 4) 2 (x2 2 x 2 7).

Solutiona. ( 4z2 2 3) 4z2 2 3

2 (22z2 1 5z 2 1) 1 2z2 2 5z 1 1

6z2 2 5z 2 2

b. (3x2 1 6x 2 4) 2 (x2 2 x 2 7)

5 3x2 1 6x 2 4 2x2 1 x 1 7

5 (3x2 2 x2) 1 (6x 1 x ) 1 (24 1 7)

5 2x2 1 7x 1 3

Example 4 Subtract polynomials

If a particular power of the variable appears in one polynomial but not the other, leave a space in that column, or write the term with a coefficient of 0.

Remember to multiply each term in the polynomial by 21 when you write the subtraction as addition.

5. (3x4 2 2x2 2 1) 1 (5x3 2 x2 1 9x4)

12x4 1 5x3 2 3x2 2 1

6. (3t2 2 5t 1 t4) 2 (11t4 2 3t2)

210t4 1 6t2 2 5t



Your Notes

9.2 Multiply PolynomialsGoal p Multiply polynomials.

Find the product 3x3(2x3 2 x2 2 7x 2 3).

Solution

3x3(2x3 2 x2 2 7x 2 3) 5 3x3( ) 2 3x3( ) 2 3x3( ) 2 3x3( )

5 2 2 2

Example 1 Multiply a monomial and a polynomial


Find the product.a. (a2 2 6a 2 3)(2a 2 5) b. (3b2 2 2b 1 5)(5b 2 6)

Solutiona. Vertical format: a2 2 6a 2 3 Write the product 3 2a 2 5 in vertical format.

2 a2 1 a 1 Multiply by .

a3 2 a2 2 a Multiply by .

Add products.

b. Horizontal format:

(3b2 2 2b 1 5)(5b 2 6) 5 (5b 2 6) 2 (5b 2 6) 1 (5b 2 6)

5

5

Example 2 Multiply polynomials vertically and horizontally

Remember that the terms of (2a 2 5) are 2a and 25. They are not 2a and 5.


Your Notes

9.2 Multiply PolynomialsGoal p Multiply polynomials.

Find the product 3x3(2x3 2 x2 2 7x 2 3).

Solution

3x3(2x3 2 x2 2 7x 2 3) 5 3x3( 2x3 ) 2 3x3( x2 ) 2 3x3( 7x ) 2 3x3( 3 )

5 6x6 2 3x5 2 21x4 2 9x3

Example 1 Multiply a monomial and a polynomial


Find the product.a. (a2 2 6a 2 3)(2a 2 5) b. (3b2 2 2b 1 5)(5b 2 6)

Solutiona. Vertical format: a2 2 6a 2 3 Write the product 3 2a 2 5 in vertical format.

2 5 a2 1 30 a 1 15 Multiply by 25 .

2 a3 2 12 a2 2 6 a Multiply by 2a .

2a3 2 17 a2 1 24 a 1 15 Add products.

b. Horizontal format:

(3b2 2 2b 1 5)(5b 2 6) 5 3b2 (5b 2 6) 2 2b (5b 2 6) 1 5 (5b 2 6)

5 15b3 2 18b2 2 10b2 1 12b 1 25b 2 30

5 15b3 2 28b2 1 37b 2 30

Example 2 Multiply polynomials vertically and horizontally

Remember that the terms of (2a 2 5) are 2a and 25. They are not 2a and 5.


Your Notes

1. 2x2(x3 2 5x2 1 3x 2 7)

2. (a2 1 5a 2 4)(2a 1 3)

Checkpoint Find the product.


Find the product (2c 1 7)(c 2 9).

Solution

(2c 1 7)(c 2 9)

5 2c( ) 1 2c( ) 1 7( ) 1 7( )

5

5

Example 3 Multiply binomials using the FOIL pattern

3. Find the product (m 1 3)(5m 2 4).



Your Notes

1. 2x2(x3 2 5x2 1 3x 2 7)

2x5 2 10x4 1 6x3 2 14x2

2. (a2 1 5a 2 4)(2a 1 3)

2a3 1 13a2 1 7a 2 12



Find the product (2c 1 7)(c 2 9).

Solution

(2c 1 7)(c 2 9)

5 2c( c ) 1 2c( 29 ) 1 7( c ) 1 7( 29 )

5 2c2 1 (218c) 1 7c 1 (263)

5 2c2 2 11c 2 63

Example 3 Multiply binomials using the FOIL pattern

3. Find the product (m 1 3)(5m 2 4).

5m2 1 11m 2 12



Your Notes


Area The dimensions of a rectangle are x 1 4 and x 1 5. Write an expression that represents the area of the rectangle.

SolutionArea 5 length p width Formula for area

of a rectangle

5 ( )( ) Substitute for length and width.

5 Multiply binomials.

5 Combine like terms.

CHECK Use a graphing calculator to check your answer. Graphy1 5 andy2 5 in the same viewing window. The graphs

, so the product of x 1 4 and x 1 5 is .

Example 4 Multiply polynomials to find an area

4. The dimensions of a rectangle are x 1 3 and x 1 11. Write an expression that represents the area of the rectangle.



Your Notes


Area The dimensions of a rectangle are x 1 4 and x 1 5. Write an expression that represents the area of the rectangle.

SolutionArea 5 length p width Formula for area

of a rectangle

5 ( x 1 4 )( x 1 5 ) Substitute for length and width.

5 x2 1 5x 1 4x 1 20 Multiply binomials.

5 x2 1 9x 1 20 Combine like terms.

CHECK Use a graphing calculator to check your answer. Graphy1 5 (x 1 4)(x 1 5) andy2 5 (x2 1 9x 1 20) in the same viewing window. The graphs coincide , so the product of x 1 4 and x 1 5 is x2 1 9x 1 20 .

Example 4 Multiply polynomials to find an area

4. The dimensions of a rectangle are x 1 3 and x 1 11. Write an expression that represents the area of the rectangle.

x2 1 14x 1 33



Your Notes

Homework


Walkway You are making a x ft

18 ft

25 ftx ft

a walkway around part of your swimming pool. The dimensions of the swimming pool and walkway are shown in the diagram.

• Write a polynomial that represents the area of the swimming pool.

• What is the area of the swimming pool if the walkway is 2 feet wide?

SolutionStep 1 Write a polynomial using the formula for the area

of a rectangle. The length is . The width is .

Area 5 p

5

5

5

Step 2 Substitute for x and evaluate. Area 5 5

The area of the swimming pool is .

Example 5 Solve a multi-step problem

5. Swimming Pool Your neighbor

22 ft

30 ft

x ft

x ft

has a walkway around his entire pool as shown in the diagram. The width of the walkway is the same on every side. Write a polynomial that represents the area of the pool. What is the area of the pool if the walkway is 3 feet wide?



Your Notes

Homework


Walkway You are making a x ft

18 ft

25 ftx ft

a walkway around part of your swimming pool. The dimensions of the swimming pool and walkway are shown in the diagram.

• Write a polynomial that represents the area of the swimming pool.

• What is the area of the swimming pool if the walkway is 2 feet wide?

SolutionStep 1 Write a polynomial using the formula for the area

of a rectangle. The length is 25 2 x . The width is 18 2 x .

Area 5 length p width

5 (25 2 x)(18 2 x)

5 450 2 25x 2 18x 1 x2

5 450 2 43x 1 x2

Step 2 Substitute 2 for x and evaluate. Area 5 450 2 43(2) 1 (2)2 5 368

The area of the swimming pool is 368 square feet .


5. Swimming Pool Your neighbor

22 ft

30 ft

x ft

x ft

has a walkway around his entire pool as shown in the diagram. The width of the walkway is the same on every side. Write a polynomial that represents the area of the pool. What is the area of the pool if the walkway is 3 feet wide?

660 2 104x 1 4x2; 384 square feet



Your Notes


9.3 Find Special Products ofPolynomialsGoal p Use special product patterns to multiply

polynomials.

SQUARE OF A BINOMIAL PATTERN

Algebra

(a 1 b)2 5 a2 1 b2

(a 2 b)2 5 a2 1 b2

Example

(x 1 4)2 5 x2 1 16

(3x 2 2)2 5 9x2 1 4

Find the product.

Solution

a. (4x 1 3)2 5 (4x)2 1 32

5 16x2 1 9

b. (3x 2 5y)2 5 (3x)2 1 (5y)2

5 9x2 1 25y2

Example 1 Use the square of a binomial pattern

1. (x 1 9)2

2. (2x 2 7)2

3. (5r 1 s)2


When you use special product patterns, remember that a and b can be numbers, variables, or variable expressions.


Your Notes


9.3 Find Special Products ofPolynomialsGoal p Use special product patterns to multiply

polynomials.

SQUARE OF A BINOMIAL PATTERN

Algebra

(a 1 b)2 5 a2 1 2ab 1 b2

(a 2 b)2 5 a2 2 2ab 1 b2

Example

(x 1 4)2 5 x2 1 8x 1 16

(3x 2 2)2 5 9x2 2 12x 1 4

Find the product.

Solution

a. (4x 1 3)2 5 (4x)2 1 2(4x)(3) 1 32

5 16x2 1 24x 1 9

b. (3x 2 5y)2 5 (3x)2 2 2(3x)(5y) 1 (5y)2

5 9x2 2 30xy 1 25y2

Example 1 Use the square of a binomial pattern

1. (x 1 9)2

x2 1 18x 1 81

2. (2x 2 7)2

4x2 2 28x 1 49

3. (5r 1 s)2

25r2 1 10rs 1 s2


When you use special product patterns, remember that a and b can be numbers, variables, or variable expressions.


Your NotesSUM AND DIFFERENCE PATTERN

Algebra(a 1 b)(a 2 b) 5 2 2 2

Example(x 1 4)(x 2 4) 5 2 2

Find the product.

Solution

a. (n 1 3)(n 2 3) 5 2 2 2 Sum and difference pattern

5 2 2 Simplify.

b. (4x 1 y)(4x 2 y) 5 2 2 2 Sum and difference pattern

5 2 2 2 Simplify.

Example 2 Use the sum and difference pattern


Use special products to find the product 17 p 23.

Solution

Notice that 17 is 3 less than while 23 is 3 more than .

17 p 23 5 ( 2 3)( 1 3) Write as product.

5 Sum and difference pattern

5 Evaluate powers.

5 Simplify.

Example 3 Use special products and mental math


Your NotesSUM AND DIFFERENCE PATTERN

Algebra(a 1 b)(a 2 b) 5 a 2 2 b 2

Example(x 1 4)(x 2 4) 5 x 2 2 16

Find the product.

Solution

a. (n 1 3)(n 2 3) 5 n 2 2 3 2 Sum and difference pattern

5 n 2 2 9 Simplify.

b. (4x 1 y)(4x 2 y) 5 (4x) 2 2 y 2 Sum and difference pattern

5 16x 2 2 y 2 Simplify.

Example 2 Use the sum and difference pattern


Use special products to find the product 17 p 23.

Solution

Notice that 17 is 3 less than 20 while 23 is 3 more than 20 .

17 p 23 5 ( 20 2 3)( 20 1 3) Write as product.

5 202 2 32 Sum and difference pattern

5 400 2 9 Evaluate powers.

5 391 Simplify.

Example 3 Use special products and mental math


Your Notes


4. Find the product (z 1 6)(z 2 6).

5. Find the product (4x 1 3)(4x 2 3).

6. Find the product (x 1 5y)(x 2 5y).

7. Describe how you can use special products to find 392.



Your Notes


4. Find the product (z 1 6)(z 2 6).

z2 2 36

5. Find the product (4x 1 3)(4x 2 3).

16x2 2 9

6. Find the product (x 1 5y)(x 2 5y).

x2 2 25y2

7. Describe how you can use special products to find 392.

Use the square of a binomial pattern to find the product (40 2 1)2.



Your Notes


Homework

Eye Color An offspring's eye color is determined by a combination of two genes, one inherited from each parent. Each parent has two color genes, and the offspring has an equal chance of inheriting either one.The gene B is for brown eyes,

BB

ParentBb

ParentBb

Bb

B b

Bb

B

b bb

and the gene b is for blue eyes. Any gene combination with a B results in brown eyes. Suppose each parent has the same gene combination Bb. The Punnett square shows the possible gene combinations of the offspring and the resulting eye color.• What percent of the possible gene combinations of the

offspring result in blue eyes?• Show how you could use a polynomial to model the

possible gene combinations of the offspring.

Solution

Step 1 Notice that the Punnett square shows that out of 4, or of the possible gene combinations result in blue eyes.

Step 2 Model the gene from each parent with . The possible gene of the offspring

can be modeled by . Notice that this product also represents the area of the Punnett square.

5

5

The coefficients show that of the possible gene combinations will result in blue eyes.


8. Eye Color Look back at Example 4. What percent of the possible gene combinations of the offspring result in brown eyes?



Your Notes


Homework

Eye Color An offspring's eye color is determined by a combination of two genes, one inherited from each parent. Each parent has two color genes, and the offspring has an equal chance of inheriting either one.The gene B is for brown eyes,

BB

ParentBb

ParentBb

Bb

B b

Bb

B

b bb

and the gene b is for blue eyes. Any gene combination with a B results in brown eyes. Suppose each parent has the same gene combination Bb. The Punnett square shows the possible gene combinations of the offspring and the resulting eye color.• What percent of the possible gene combinations of the

offspring result in blue eyes?• Show how you could use a polynomial to model the

possible gene combinations of the offspring.

Solution

Step 1 Notice that the Punnett square shows that 1 out of 4, or 25% of the possible gene combinations result in blue eyes.

Step 2 Model the gene from each parent with 0.5B 1 0.5b . The possible gene of the offspring can be modeled by (0.5B 1 0.5b)2 . Notice that this product also represents the area of the Punnett square.

(0.5B 1 0.5b)2

5 (0.5B)2 1 2(0.5B)(0.5b) 1 (0.5b)2

5 0.25B2 1 0.5Bb 1 0.25b2

The coefficients show that 25% of the possible gene combinations will result in blue eyes.


8. Eye Color Look back at Example 4. What percent of the possible gene combinations of the offspring result in brown eyes?

75%



Your Notes

9.4 Solve Polynomial Equationsin Factored FormGoal p Solve polynomial equations.

VOCABULARY

Roots

Vertical motion model

ZERO-PRODUCT PROPERTY

Let a and b be real numbers. If ab 5 0, then 5 0 or 5 0.

Solve (x 2 5)(x 1 4) 5 0.

Solution (x 2 5)(x 1 4) 5 0 Write original

equation.

5 0 or 5 0 property

x 5 or x 5 Solve for x.

The solutions of the equation are .

CHECK Substitute each solution into the original equation to check.

( 2 5)( 1 4) 0 0 ( 2 5)( 1 4) 0 0

0 0 0 0

5 0 5 0

Example 1 Use the zero-product property



Your Notes

9.4 Solve Polynomial Equationsin Factored FormGoal p Solve polynomial equations.

VOCABULARY

Roots The solutions of an equation in which one side is zero and the other side is a product of polynomial factors

Vertical motion model A model that describes the height of a projectile

ZERO-PRODUCT PROPERTY

Let a and b be real numbers. If ab 5 0, then a 5 0 or b 5 0.

Solve (x 2 5)(x 1 4) 5 0.

Solution (x 2 5)(x 1 4) 5 0 Write original

equation.

x 2 5 5 0 or x 2 4 5 0 Zero-product property

x 5 5 or x 5 24 Solve for x.

The solutions of the equation are 5 and 24 .

CHECK Substitute each solution into the original equation to check.

( 5 2 5)( 5 1 4) 0 0 ( 24 2 5)( 24 1 4) 0 0

0 p 9 0 0 29 p 0 0 0

0 5 0 0 5 0

Example 1 Use the zero-product property



Your Notes


Factor out the greatest common monomial factor.

a. 16x 1 40y b. 6x2 1 30x3

Solution

a. The GCF of 16 and 40 is . The variables x and y have . So, the greatest common monomial factor of the terms is .

16x 1 40y 5

b. The GCF of 6 and 30 is . The GCF of x2 and x3 is . So, the greatest common monomial factor of the

terms is .

6x2 1 30x3 5

Example 2 Find the greatest common monomial factor

Solve the equation.

a. 3x2 1 15x 5 0 Original equation

5 0 Factor left side.

5 0 or 5 0 Zero-product property



b. 9b2 5 24b Original equation

5 0 Subtract from each side.



b 5 or b 5 Solve for b.


Example 3 Solve an equation by factoring

To use the zero-product property, you must write the equation so that one side is 0. For this reason, must be subtracted from each side of the equation.


Your Notes


Factor out the greatest common monomial factor.

a. 16x 1 40y b. 6x2 1 30x3

Solution

a. The GCF of 16 and 40 is 8 . The variables x and y have no common factor . So, the greatest common monomial factor of the terms is 8 .

16x 1 40y 5 8(2x 1 5y)

b. The GCF of 6 and 30 is 6 . The GCF of x2 and x3 is x2 . So, the greatest common monomial factor of the terms is 6x2 .

6x2 1 30x3 5 6x2(1 1 5x)

Example 2 Find the greatest common monomial factor

Solve the equation.

a. 3x2 1 15x 5 0 Original equation

3x(x 1 5) 5 0 Factor left side.

3x 5 0 or x 1 5 5 0 Zero-product property



b. 9b2 5 24b Original equation

9b2 2 24b 5 0 Subtract 24b from each side.

3b(3b 2 8) 5 0 Factor left side.

3b 5 0 or 3b 2 8 5 0 Zero-product property

b 5 0 or b 5 8 } 3 Solve for b.

The solutions of the equation are 0 and 8 } 3 .

Example 3 Solve an equation by factoring

To use the zero-product property, you must write the equation so that one side is 0. For this reason, 24b must be subtracted from each side of the equation.


Your Notes


1. (x 1 6)(x 2 3) 5 0

2. (x 2 8)(x 2 5) 5 0

Checkpoint Solve the equation.

VERTICAL MOTION MODEL

The height h (in feet) of a projectile can be modeled by

h 5 216t2 1 vt 1 s

where t is the (in seconds) the object has been in the air, v is the (in feet per second), and s is the (in feet).

The vertical motion model takes into account the effect of gravity but ignores other, less significant, factors such as air resistance.

3. 10x2 2 24y2

4. 3t6 1 8t4

Checkpoint Factor out the greatest common monomial factor.


Your Notes


1. (x 1 6)(x 2 3) 5 0

26, 3

2. (x 2 8)(x 2 5) 5 0

8, 5


VERTICAL MOTION MODEL

The height h (in feet) of a projectile can be modeled by

h 5 216t2 1 vt 1 s

where t is the time (in seconds) the object has been in the air, v is the initial vertical velocity (in feet per second), and s is the initial height (in feet).

The vertical motion model takes into account the effect of gravity but ignores other, less significant, factors such as air resistance.

3. 10x2 2 24y2

2(5x2 2 12y2)

4. 3t6 1 8t4

t4(3t2 1 8)

Checkpoint Factor out the greatest common monomial factor.


Your Notes

Fountain A fountain sprays water into the air with an initial vertical velocity of 20 feet per second. After how many seconds does it land on the ground?

SolutionStep 1 Write a model for the water's height above ground.

h 5 216t2 1 vt 1 s Vertical motion model

h 5 216t2 1 t 1 v 5 and s 5

h 5 216t2 1 Simplify.

Step 2 Substitute for h. When the water lands, its height above the ground is feet. Solve for t.

5 216t2 1 Substitute for h.

5 Factor right side.

or Zero-product property

or Solve for t.

The water lands on the ground seconds after it is sprayed.


5. Solve d2 2 7d 5 0. 6. Solve 8b2 5 2b.

7. What If? In Example 4, suppose the initial vertical velocity is 18 feet per second. After how many seconds does the water land on the ground?


Homework

The solution t 5 0 means that before the water is sprayed, its height above the ground is 0 feet.



Your Notes

Fountain A fountain sprays water into the air with an initial vertical velocity of 20 feet per second. After how many seconds does it land on the ground?

SolutionStep 1 Write a model for the water's height above ground.


h 5 216t2 1 20 t 1 0 v 5 20 and s 5 0

h 5 216t2 1 20t Simplify.

Step 2 Substitute 0 for h. When the water lands, its height above the ground is 0 feet. Solve for t.

0 5 216t2 1 20t Substitute 0 for h.

0 5 4t(24t 1 5) Factor right side.

4t 5 0 or 24t 1 5 5 0 Zero-product property

t 5 0 or t 5 1.25 Solve for t.

The water lands on the ground 1.25 seconds after it is sprayed.


5. Solve d2 2 7d 5 0. 6. Solve 8b2 5 2b.

0, 7 0, 1 } 4

7. What If? In Example 4, suppose the initial vertical velocity is 18 feet per second. After how many seconds does the water land on the ground?

1.125 seconds


Homework

The solution t 5 0 means that before the water is sprayed, its height above the ground is 0 feet.



Your Notes

9.5 Factor x2 1 bx 1 cGoal p Factor trinomials of the form x2 1 bx 1 c.

FACTORING x2 1 bx 1 c

Algebra

x2 1 bx 1 c 5 (x 1 p)(x 1 q) provided 5 band 5 c.

Example

x2 1 6x 1 5 5 ( )( ) because 5 6and 5 5.

Factor x2 1 10x 1 16.

Solution

Find two factors of whose sum is . Make an organized list.

Factors of Sum of factors

16, 16 1 5

8, 8 1 5

4, 4 1 5

The factors 8 and have a sum of , so they are the correct values of p and q.

x2 1 10x 1 16 5 (x 1 8)( )

CHECK (x 1 8)( ) 5 Multiply.

5 Simplify.

Example 1 Factor when b and c are positive



Your Notes

9.5 Factor x2 1 bx 1 cGoal p Factor trinomials of the form x2 1 bx 1 c.

FACTORING x2 1 bx 1 c

Algebra

x2 1 bx 1 c 5 (x 1 p)(x 1 q) provided p 1 q 5 band pq 5 c.

Example

x2 1 6x 1 5 5 ( x 1 5 )( x 1 1 ) because 5 1 1 5 6and 5 p 1 5 5.

Factor x2 1 10x 1 16.

Solution

Find two positive factors of 16 whose sum is 10 . Make an organized list.

Factors of 16 Sum of factors

16, 1 16 1 1 5 17

8, 2 8 1 2 5 10

4, 4 4 1 4 5 8

The factors 8 and 2 have a sum of 10 , so they are the correct values of p and q.

x2 1 10x 1 16 5 (x 1 8)( x 1 2 )

CHECK (x 1 8)( x 1 2 ) 5 x2 1 2x 1 8x 1 16 Multiply.

5 x2 1 10x 1 16 Simplify.

Example 1 Factor when b and c are positive



Your Notes

Factor a2 2 5a 1 6.

SolutionBecause b is negative and c is positive, p and q must .


1 ( ) 5

1 ( ) 5

a2 2 5a 1 6 5 ( )( )

Example 2 Factor when b is negative and c is positive

Factor y2 1 3y 2 10.

Solution

Because c is negative, p and q must .


210, 210 1 5

10, 10 1 5

25, 25 1 5

5, 5 1 5

y2 1 3y 2 10 5 ( )( )

Example 3 Factor when b is positive and c is negative


1. x 1 7x 1 12 2. x 1 9x 1 8

Checkpoint Factor the trinomial.


Your Notes

Factor a2 2 5a 1 6.

SolutionBecause b is negative and c is positive, p and q must both be negative .


26, 21 26 1 ( 21 ) 5 27

23, 22 23 1 ( 22 ) 5 25

a2 2 5a 1 6 5 ( a 2 3 )( a 2 2 )


Factor y2 1 3y 2 10.

Solution

Because c is negative, p and q must have different signs .


210, 1 210 1 1 5 29

10, 21 10 1 (21) 5 9

25, 2 25 1 2 5 23

5, 22 5 1 (22) 5 3

y2 1 3y 2 10 5 ( y 1 5 )( y 2 2 )



1. x 1 7x 1 12 2. x 1 9x 1 8

(x 1 4)(x 1 3) (x 1 8)(x 1 1)



Your Notes

Solve the equation x2 1 7x 5 18.

x2 1 7x 5 18 Write original equation.

x2 1 7x 2 5 0 Subtract from each side.



or Solve for x.


Example 4 Solve a polynomial equation


3. x 1 12x 1 27 4. x2 2 9x 1 20

5. y2 1 4y 2 21 6. z2 1 2z 2 24



Your Notes

Solve the equation x2 1 7x 5 18.


x2 1 7x 2 18 5 0 Subtract 18 from each side.

(x 1 9)(x 2 2) 5 0 Factor left side.






3. x 1 12x 1 27 4. x2 2 9x 1 20

(x 1 9)(x 1 3) (x 2 5)(x 2 4)

5. y2 1 4y 2 21 6. z2 1 2z 2 24

(y 1 7)(y 2 3) (z 1 6)(z 2 4)



Your Notes


7. Solve the equation s2 2 12s 5 13.

8. What If? In Example 5, suppose the area of the bandage is 27 square centimeters. What is the width?


Dimensions The bandage shown

w cm

w cm

3 cm 3 cmhas an area of 16 square centimeters. Find the width of the bandage.

SolutionStep 1 Write an equation using the fact that the area of

the bandage is 16 square centimeters.

A 5 l p w Formula for area

5 p w Substitute values.

0 5 Simplify.

Step 2 Solve the equation for w.

0 5 Write equation.

0 5 Factor right side.


or Solve for w.

The bandage cannot have a negative width, so the width is .


Homework


Your Notes


7. Solve the equation s2 2 12s 5 13.

21, 13

8. What If? In Example 5, suppose the area of the bandage is 27 square centimeters. What is the width?

3 centimeters


Dimensions The bandage shown

w cm

w cm

3 cm 3 cmhas an area of 16 square centimeters. Find the width of the bandage.

SolutionStep 1 Write an equation using the fact that the area of

the bandage is 16 square centimeters.

A 5 l p w Formula for area

16 5 (3 1 w 1 3) p w Substitute values.

0 5 w2 1 6w 2 16 Simplify.

Step 2 Solve the equation for w.

0 5 w2 1 6w 2 16 Write equation.

0 5 (w 1 8)(w 2 2) Factor right side.

w 1 8 5 0 or w 2 2 5 0 Zero-product property

w 5 28 or w 5 2 Solve for w.

The bandage cannot have a negative width, so the width is 2 centimeters .


Homework


Your Notes

9.6 Factor ax2 1 bx 1 cGoal p Factor trinomials of the form ax2 1 bx 1 c.

Factor 2x2 2 11x 1 5.

SolutionBecause b is negative and c is positive, both factors of c must be . You must consider the of the factors of 5, because the x-terms of the possible factorizations are different.

Factors of 2

Factors of 5

Possible factorization

Middle term whenmultiplied

1, 2 21, (x 2 1)(2x ) 2 2x 5

1, 2 25, (x 2 5)(2x ) 2 10x 5

2x2 2 11x 1 5 5 (x 2 )(2x )


Factor 5n2 1 2n 2 3.

SolutionBecause b is positive and c is negative, the factors of c have .

Factors of 5

Factors of 23



1, 5 1, (n 1 1)(5n )

1, 5 21, (n 2 1)(5n )

1, 5 3, (n 1 3)(5n )

1, 5 23, (n 2 3)(5n )

5n2 1 2n 2 3 5 (n )(5n )




Your Notes

9.6 Factor ax2 1 bx 1 cGoal p Factor trinomials of the form ax2 1 bx 1 c.

Factor 2x2 2 11x 1 5.

SolutionBecause b is negative and c is positive, both factors of c must be negative . You must consider the order of the factors of 5, because the x-terms of the possible factorizations are different.

Factors of 2

Factors of 5



1, 2 21, 25 (x 2 1)(2x 2 5 ) 25x 2 2x 5 27x

1, 2 25, 21 (x 2 5)(2x 2 1 ) 2x 2 10x 5 211x

2x2 2 11x 1 5 5 (x 2 5 )(2x 2 1 )


Factor 5n2 1 2n 2 3.

SolutionBecause b is positive and c is negative, the factors of c have different signs .

Factors of 5

Factors of 23



1, 5 1, 23 (n 1 1)(5n 2 3 ) 23n 1 5n 5 2n

1, 5 21, 3 (n 2 1)(5n 1 3 ) 3n 2 5n 5 22n

1, 5 3, 21 (n 1 3)(5n 2 1 ) 2n 1 15n 5 14n

1, 5 23, 1 (n 2 3)(5n 1 1 ) n 2 15n 5 214n

5n2 1 2n 2 3 5 (n 1 1 )(5n 2 3 )




Your Notes


Factor 24x2 1 4x 1 3.

Solution

Step 1 Factor from each term of the trinomial.

24x2 1 4x 1 3 5 ( )

Step 2 Factor the trinomial . Because b and c are both , the factors of c must have .

Factors of 4

Factors of 23



1, 4 1, (x 1 1)(4x )

1, 4 3, (x 1 3)(4x )

1, 4 21, (x 2 1)(4x )

1, 4 23, (x 2 3)(4x )

2, 2 1, (2x 1 1)(2x )

2, 2 21, (2x 2 1)(2x )

24x2 1 4x 1 3 5

Example 3 Factor when a is negative

1. 3x2 2 5x 1 2 2. 2m2 1 m 2 21


Remember to include the that you factored out in Step 1.

3. Factor 22y2 2 11y 2 5.



Your Notes


Factor 24x2 1 4x 1 3.

Solution

Step 1 Factor 21 from each term of the trinomial.

24x2 1 4x 1 3 5 2 ( 4x2 2 4x 2 3 )

Step 2 Factor the trinomial 4x2 2 4x 2 3 . Because b and c are both negative , the factors of c must have different signs .

Factors of 4

Factors of 23



1, 4 1, 23 (x 1 1)(4x 2 3 ) 23x 1 4x 5 x

1, 4 3, 21 (x 1 3)(4x 2 1 ) 2x 1 12x 5 11x

1, 4 21, 3 (x 2 1)(4x 1 3 ) 3x 2 4x 5 2x

1, 4 23, 1 (x 2 3)(4x 1 1 ) x 2 12x 5 211x

2, 2 1, 23 (2x 1 1)(2x 2 3 ) 26x 1 2x 5 24x

2, 2 21, 3 (2x 2 1)(2x 1 3 ) 6x 2 2x 5 4x

24x2 1 4x 1 3 5 2(2x 1 1)(2x 2 3)

Example 3 Factor when a is negative

1. 3x2 2 5x 1 2 2. 2m2 1 m 2 21

(x 2 1)(3x 2 2) (m 2 3)(2m 1 7)


Remember to include the 21 that you factored out in Step 1.

3. Factor 22y2 2 11y 2 5.

2(2y 1 1)(y 1 5)



Your Notes

4. What If? In Example 4, suppose another athlete hits the tennis ball with an initial vertical velocity of 20 feet per second from a height of 6 feet. After how many seconds does the ball hit the ground?



Homework

Tennis An athlete hits a tennis ball at an initial height of 8 feet and with an initial vertical velocity of 62 feet per second.

a. Write an equation that gives the height (in feet) of the ball as a function of the time (in seconds) since it left the racket.

b. After how many seconds does the ball hit the ground?

Solution

a. Use the to write an equation for the height h (in feet) of the ball.

h 5 216t2 1 vt 1 s

h 5 216t2 1 t 1 v 5 and s 5

b. To find the number of seconds that pass before the ball lands, find the value of t for which the height of the ball is . Substitute for h and solve the equation for t.

5 216t2 1 t 1 Substitute for h.

5 ( ) Factor out .

5 ( )( ) Factor the trinomial.


or Solve for t.

A negative solution does not make sense in this situation. The tennis ball hits the ground after .

Example 4 Write and solve a polynomial equation


Your Notes

4. What If? In Example 4, suppose another athlete hits the tennis ball with an initial vertical velocity of 20 feet per second from a height of 6 feet. After how many seconds does the ball hit the ground?

1.5 seconds



Homework

Tennis An athlete hits a tennis ball at an initial height of 8 feet and with an initial vertical velocity of 62 feet per second.

a. Write an equation that gives the height (in feet) of the ball as a function of the time (in seconds) since it left the racket.

b. After how many seconds does the ball hit the ground?

Solution

a. Use the vertical motion model to write an equation for the height h (in feet) of the ball.


h 5 216t2 1 62 t 1 8 v 5 62 and s 5 8

b. To find the number of seconds that pass before the ball lands, find the value of t for which the height of the ball is 0 . Substitute 0 for h and solve the equation for t.

0 5 216t2 1 62 t 1 8 Substitute 0 for h.

0 5 22 ( 8t2 2 31t 2 4 ) Factor out 22 .

0 5 22 ( 8t 1 1 )( t 2 4 ) Factor the trinomial.

8t 1 1 5 0 or t 2 4 5 0 Zero-product property

t 5 2 1 } 8 or t 5 4 Solve for t.

A negative solution does not make sense in this situation. The tennis ball hits the ground after 4 seconds .

Example 4 Write and solve a polynomial equation


Your Notes

9.7 Factor Special ProductsGoal p Factor special products.

VOCABULARY

Perfect square trinomial

DIFFERENCE OF TWO SQUARES PATTERN

Algebra

a2 2 b2 5 (a 1 b)( )

Example

9x2 2 4 5 (3x)2 2 22 5 ( )( )


Factor the polynomial.

a. z2 2 81 5 z2 2 2

5 (z 1 )(z 2 )

b. 16x2 2 9 5 ( )2 2 2

5 ( 1 )( 2 )

c. a2 2 25b2 5 a2 2 ( )2

5 (a 1 )(a 2 )

d. 4 2 16n2 5 ( 2 ) 5 [( )2 2 ( )2] 5 ( 1 )( 2 )

Example 1 Factor the differences of two squares

1. x2 2 100 2. 49y2 2 25

3. c2 2 9d2 4. 45 2 80m2

Checkpoint Factor the polynomial.


Your Notes

9.7 Factor Special ProductsGoal p Factor special products.

VOCABULARY

Perfect square trinomial A trinomial of the form a2 1 2ab 1 b2 or a2 2 2ab 1 b2

DIFFERENCE OF TWO SQUARES PATTERN

Algebra

a2 2 b2 5 (a 1 b)( a 2 b )

Example

9x2 2 4 5 (3x)2 2 22 5 ( 3x 1 2 )( 3x 2 2 )



a. z2 2 81 5 z2 2 9 2

5 (z 1 9 )(z 2 9 )

b. 16x2 2 9 5 ( 4x )2 2 3 2

5 ( 4x 1 3 )( 4x 2 3 )

c. a2 2 25b2 5 a2 2 ( 5b )2

5 (a 1 5b )(a 2 5b )

d. 4 2 16n2 5 4 ( 1 2 4n2 ) 5 4 [( 1 )2 2 ( 2n )2] 5 4 ( 1 1 2n )( 1 2 2n )

Example 1 Factor the differences of two squares

1. x2 2 100 2. 49y2 2 25

(x 1 10)(x 2 10) (7y 1 5)(7y 2 5)

3. c2 2 9d2 4. 45 2 80m2

(c 1 3d)(c 2 3d) 5(3 1 4m)(3 2 4m)



Your Notes


PERFECT SQUARE TRINOMIAL PATTERN

Algebra

a2 1 2ab 1 b2 5 ( )2

a2 2 2ab 1 b2 5 ( )2

Example

x2 1 8x 1 16 5 x2 1 2(x p 4) 1 42 5 ( )2

x2 2 6x 1 9 5 x2 2 2(x p 3) 1 32 5 ( )2


a. x2 2 16x 1 64 5 x2 2 2( ) 2 2

5 ( )2

b. 4y2 2 12y 1 9 5 ( )2 2 2( ) 1 2

5 ( )2

c. 9s2 1 6st 1 t2 5 ( )2 1 2( ) 1 2

5 ( )2

d. 23z2 1 24z 2 48 5 (z2 2 8z 1 16) 5 [z2 2 2( ) 1 2] 5 ( )2

Example 2 Factor perfect square trinomials

5. x2 1 14x 1 49 6. 9y2 2 6y 1 1

7. 16x2 2 40xy 1 25y2 8. 25r2 2 20r 2 20



Your Notes


PERFECT SQUARE TRINOMIAL PATTERN

Algebra

a2 1 2ab 1 b2 5 ( a 1 b )2

a2 2 2ab 1 b2 5 ( a 2 b )2

Example

x2 1 8x 1 16 5 x2 1 2(x p 4) 1 42 5 ( x 1 4 )2

x2 2 6x 1 9 5 x2 2 2(x p 3) 1 32 5 ( x 2 3 )2


a. x2 2 16x 1 64 5 x2 2 2( x p 8 ) 2 8 2

5 ( x 2 8 )2

b. 4y2 2 12y 1 9 5 ( 2y )2 2 2( 2y p 3 ) 1 3 2

5 ( 2y 2 3 )2

c. 9s2 1 6st 1 t2 5 ( 3s )2 1 2( 3s p t ) 1 t 2

5 ( 3s 1 t )2

d. 23z2 1 24z 2 48 5 23 (z2 2 8z 1 16) 5 23 [z2 2 2( z p 4 ) 1 4 2] 5 23 ( z 2 4 )2

Example 2 Factor perfect square trinomials

5. x2 1 14x 1 49 6. 9y2 2 6y 1 1

(x 1 7)2 (3y 2 1)2

7. 16x2 2 40xy 1 25y2 8. 25r2 2 20r 2 20

(4x 2 5y)2 25(r 1 2)2



Your Notes


Falling Object A brick falls off of a building from a height of 144 feet. After how many seconds does the brick land on the ground?

SolutionUse the vertical motion model. The brick fell, so its initial vertical velocity is . Find the value of time t (in seconds) for which the height h (in feet) is .

h 5 Vertical motion model

5 Substitute values.

5 ( ) Factor out .

5 ( )( ) Difference of two squares


or Solve for t.

The brick lands on the ground after it falls.

Example 4 Solve a vertical motion problem

Solve the equation x2 1 x 1 1 } 4 5 0.

x2 1 x 1 1 } 4 5 0 Write original equation.

5 0 Multiply each side by .

5 0 Write left side as a2 1 2ab 1 b2.

5 0 Perfect square trinomial pattern

5 0 Zero-product property

x 5 Solve for x.


This equation has two identical solutions, because it has two identical factors.


Your Notes


Falling Object A brick falls off of a building from a height of 144 feet. After how many seconds does the brick land on the ground?

SolutionUse the vertical motion model. The brick fell, so its initial vertical velocity is 0 . Find the value of time t (in seconds) for which the height h (in feet) is 0 .


0 5 216t2 1 0t 1 144 Substitute values.

0 5 216 ( t2 2 9 ) Factor out 216 .

0 5 216 ( t 1 3 )( t 2 3 ) Difference of two squares

t 1 3 5 0 or t 2 3 5 0 Zero-product property

t 5 23 or t 5 3 Solve for t.

The brick lands on the ground 3 seconds after it falls.

Example 4 Solve a vertical motion problem

Solve the equation x2 1 x 1 1 } 4 5 0.

x2 1 x 1 1 } 4 5 0 Write original equation.

4x2 1 4x 1 1 5 0 Multiply each side by 4 .

(2x )2 1 2(2x p 1) 1 12 5 0 Write left side as a2 1 2ab 1 b2.

(2x 1 1)2 5 0 Perfect square trinomial pattern

2x 1 1 5 0 Zero-product property

x 5 2 1 } 2 Solve for x.


This equation has two identical solutions, because it has two identical factors.


Your Notes


12. What If? In Example 4, suppose the brick falls from

a height of 225

} 4 feet. After how many seconds does

the brick land on the ground?


Homework

9. m2 2 8m 1 16 5 0

10. w2 1 16w 1 64 5 0

11. t2 2 121 5 0



Your Notes


12. What If? In Example 4, suppose the brick falls from

a height of 225

} 4 feet. After how many seconds does

the brick land on the ground?

15 } 8 seconds


Homework

9. m2 2 8m 1 16 5 0

4

10. w2 1 16w 1 64 5 0

28

11. t2 2 121 5 0

611



Your Notes

9.8 Factor Polynomials CompletelyGoal p Factor polynomials completely.

VOCABULARY

Factor by grouping

Factor completely

Factor the expression.

a. 3x(x 1 2) 2 2(x 1 2) b. y2(y 2 4) 1 3(4 2 y)

Solution

a. 3x(x 1 2) 2 2(x 1 2) 5 (x 1 2)( )

b. The binomials y 2 4 and 4 2 y are . Factor from 4 2 y to obtain a common binomial factor.

y2(y 2 4) 1 3(4 2 y) 5 y2(y 2 4)

5 (y 2 4)

Example 1 Factor out a common binomial


a. y3 1 7y2 1 2y 1 14 b. y2 1 2y 1 yx 1 2x

Solution

a. y3 1 7y2 1 2y 1 14 5 ( ) 1 ( ) 5 ( ) 1 ( ) 5 ( )( )

b. y2 1 2y 1 yx 1 2x 5 ( ) 1 ( )

5 ( ) 1 ( )

5 ( )( )

Example 2 Factor by grouping


Remember that you can check a factorization by multiplying the factors.


Your Notes

9.8 Factor Polynomials CompletelyGoal p Factor polynomials completely.

VOCABULARY

Factor by grouping Factoring a polynomial with four terms by factoring a common monomial from a pair of terms, then looking for a common binomial term

Factor completely Factoring a polynomial until it is written as a product of unfactorable polynomials with integer coefficients


a. 3x(x 1 2) 2 2(x 1 2) b. y2(y 2 4) 1 3(4 2 y)

Solution

a. 3x(x 1 2) 2 2(x 1 2) 5 (x 1 2)( 3x 2 2 )

b. The binomials y 2 4 and 4 2 y are opposites . Factor 21 from 4 2 y to obtain a common binomial factor.

y2(y 2 4) 1 3(4 2 y) 5 y2(y 2 4) 23(y 2 4)

5 (y 2 4) (y2 2 3)

Example 1 Factor out a common binomial


a. y3 1 7y2 1 2y 1 14 b. y2 1 2y 1 yx 1 2x

Solution

a. y3 1 7y2 1 2y 1 14 5 ( y3 1 7y2 ) 1 ( 2y 1 14 ) 5 y2 ( y 1 7 ) 1 2 ( y 1 7 ) 5 ( y 1 7 )( y2 1 2 )

b. y2 1 2y 1 yx 1 2x 5 ( y2 1 2y ) 1 ( yx 1 2x )

5 y ( y 1 2 ) 1 x ( y 1 2 )

5 ( y 1 2 )( y 1 x )



Remember that you can check a factorization by multiplying the factors.


Your Notes

Factor x3 2 12 1 3x 2 4x2.

SolutionThe terms x3 and 212 have no common factor. Use the to rearrange the terms so that you can group terms with a common factor.

x3 2 12 1 3x 2 4x2 5

5

5

5


1. 5z(z 2 6) 1 4(z 2 6) 2. 2y2(y 2 1) 1 7(1 2 y)

3. x3 2 4x2 1 5x 2 20 4. n3 1 48 1 6n 1 8n2

Checkpoint Factor the expression.

GUIDELINES FOR FACTORING POLYNOMIALS COMPLETELY

To factor a polynomial completely, you should try each of these steps.

1. Factor out the common monomial factor.

2. Look for a difference of two squares or a .

3. Factor a trinomial of the form ax2 1 bx 1 c into a product of factors.

4. Factor a polynomial with four terms by .



Your Notes

Factor x3 2 12 1 3x 2 4x2.

SolutionThe terms x3 and 212 have no common factor. Use the commutative property to rearrange the terms so that you can group terms with a common factor.

x3 2 12 1 3x 2 4x2 5 x3 2 4x2 1 3x 2 12

5 (x3 2 4x2) 1 (3x 2 12)

5 x2(x 2 4) 1 3(x 2 4)

5 (x 2 4)(x2 1 3)


1. 5z(z 2 6) 1 4(z 2 6) 2. 2y2(y 2 1) 1 7(1 2 y)

(z 2 6)(5z 1 4) (y 2 1)(2y2 2 7)

3. x3 2 4x2 1 5x 2 20 4. n3 1 48 1 6n 1 8n2

(x 2 4)(x2 1 5) (n 1 8)(n2 1 6)

Checkpoint Factor the expression.

GUIDELINES FOR FACTORING POLYNOMIALS COMPLETELY

To factor a polynomial completely, you should try each of these steps.

1. Factor out the greatest common monomial factor.

2. Look for a difference of two squares or a perfect square trinomial .

3. Factor a trinomial of the form ax2 1 bx 1 c into a product of binomial factors.

4. Factor a polynomial with four terms by grouping .



Your Notes

Factor the polynomial completely.

a. x2 1 3x 2 1

b. 3r3 2 21r2 1 30r

c. 9d4 2 4d2

Solutiona. The terms of the polynomial have no common

monomial factor. Also, there are no factors of that have a sum of . This polynomial be factored.

b. 3r3 2 21r2 1 30r 5

5

c. 9d4 2 4d2 5

5

Example 4 Factor completely

Solve 5x3 2 25x2 5 230x.

Solution

5x3 2 25x2 5 230x Write original equation.

5x3 2 25x2 30x 5 0 30x to each side.

5 0 Factor out .

5 0 Factor trinomial.

or or Zero-product property

x 5 x 5 x 5 Solve for x.



Remember that you can check your answers by substituting each solution for x in the original equation.


Your Notes

Factor the polynomial completely.

a. x2 1 3x 2 1

b. 3r3 2 21r2 1 30r

c. 9d4 2 4d2

Solutiona. The terms of the polynomial have no common

monomial factor. Also, there are no factors of 21 that have a sum of 3 . This polynomial cannot be factored.

b. 3r3 2 21r2 1 30r 5 3r(r2 2 7r 1 10)

5 3r(r 2 2)(r 2 5)

c. 9d4 2 4d2 5 d 2(9d 2 2 4)

5 d 2(3d 2 2)(3d 1 2)

Example 4 Factor completely

Solve 5x3 2 25x2 5 230x.

Solution

5x3 2 25x2 5 230x Write original equation.

5x3 2 25x2 1 30x 5 0 Add 30x to each side.

5x (x2 2 5x 1 6) 5 0 Factor out 5x .

5x (x 2 3)(x 2 2) 5 0 Factor trinomial.

5x 5 0 or x 2 3 5 0 or x 2 2 5 0 Zero-product property

x 5 0 x 5 3 x 5 2 Solve for x.



Remember that you can check your answers by substituting each solution for x in the original equation.


Your Notes

Volume A crate in the shape of a rectangular prism has a volume of 180 cubic feet. The crate has a width of w feet, a length of (9 2 w) feet, and a height of (w 1 4) feet. The length is more than half the width. Find the crate's length, width, and height.

SolutionStep 1 Write and solve an equation for w.

Volume 5 p p

5

0 5

0 5

0 5

0 5

0 5

5 0 or 5 0 or 5 0

w 5 w 5 w 5

Step 2 Choose the solution that is the correct value for w. Disregard , because the width cannot be .

You know that the length is more than half the width. Test the solutions in the length expression.

Length 5 5 or Length 5 5 .

The solution gives a length of feet, which is more than half the width.

Step 3 Find the height.

Height 5 5 5 .

The width is , the length is , and the height is .




Your Notes

Volume A crate in the shape of a rectangular prism has a volume of 180 cubic feet. The crate has a width of w feet, a length of (9 2 w) feet, and a height of (w 1 4) feet. The length is more than half the width. Find the crate's length, width, and height.

SolutionStep 1 Write and solve an equation for w.

Volume 5 length p width p height

180 5 (9 2 w)(w)(w 1 4)

0 5 2w3 1 5w2 1 36w 2 180

0 5 2w2(w 2 5) 1 36(w 2 5)

0 5 (w 2 5)(2w2 1 36)

0 5 21(w 2 5)(w2 2 36)

0 5 21(w 2 5)(w 2 6)(w 1 6)

w 2 5 5 0 or w 2 6 5 0 or w 1 6 5 0

w 5 5 w 5 6 w 5 26

Step 2 Choose the solution that is the correct value for w. Disregard 26 , because the width cannot be negative .

You know that the length is more than half the width. Test the solutions 5 and 6 in the length expression.

Length 5 9 2 5 5 4 or Length 5 9 2 6 5 3 .

The solution 5 gives a length of 4 feet, which is more than half the width.

Step 3 Find the height.

Height 5 w 1 4 5 5 1 4 5 9 .

The width is 5 feet , the length is 4 feet , and the height is 9 feet .




Your Notes


5. 22x3 1 6x2 1 108x

6. 12y4 2 75y2


Homework

7. Solve 2x3 1 2x2 5 40x.

8. What If? A box in the shape of a rectangular prism has a volume of 180 cubic feet. The box has a length of x feet, a width of (x 1 9) feet, and a height of (x 2 4) feet. Find the dimensions of the box.



Your Notes


5. 22x3 1 6x2 1 108x

22x(x 1 6)(x 2 9)

6. 12y4 2 75y2

3y2(2y 2 5)(2y 1 5)


Homework

7. Solve 2x3 1 2x2 5 40x.

0, 25, 4

8. What If? A box in the shape of a rectangular prism has a volume of 180 cubic feet. The box has a length of x feet, a width of (x 1 9) feet, and a height of (x 2 4) feet. Find the dimensions of the box.

6 feet long by 15 feet wide by 2 feet high




Monomial

Polynomial

Leading coefficient

Trinomial


Factor by grouping



Binomial

Roots


Factor completely





Monomial

29x5

Polynomial

2x3 2 3x2 1 21

Leading coefficient

The leading coefficient of 2x3 2 3x2 1 21 is 2.

Trinomial

x2 1 4x 2 5


h 5 216t2 1 vt 1 s

Factor by grouping

x3 2 6x2 1 3x 2 18

5 x2(x 2 6) 1 3(x 2 6)

5 (x 2 6)(x2 1 3)


The degree of 29x5 is 5.


The degree of 2x3 2 3x2 1 21 is 3.

Binomial

3x 1 7

Roots

The roots of x2 1 4x 2 5 are 25 and 1.


x2 2 12x 1 36

Factor completely

2x2 2 8

5 2(x2 2 4) 5 2(x 2 2)(x 1 2)




10.1 Graph y 5 ax2 1 cGoal p Graph simple quadratic functions.

VOCABULARY

Quadratic function

Parabola

Parent quadratic function

Vertex

Axis of Symmetry

PARENT QUADRATIC FUNCTION

The most basic quadratic function in the family of quadratic functions, called the

, is y 5 x2. The graph is shown below.

The line that passes through

x

y

1

3

5

12121

23 3

(0, 0)

x 5 0y 5 x2

the vertex and divides the parabola into two symmetric parts is called the

. The axis of symmetry for the graph of y 5 x2 is the y-axis, .

The lowest or highest point on the parabola is the . The vertex of the graph of y 5 x2 is ( , ).


Your Notes


10.1 Graph y 5 ax2 1 cGoal p Graph simple quadratic functions.

VOCABULARY

Quadratic function A non-linear function that can be written in the standard form y 5 ax2 1 bx 1 c where a Þ 0

Parabola A U-shaped graph of a quadratic function

Parent quadratic function y 5 x2

Vertex The lowest or highest point on a parabola

Axis of Symmetry The line that passes through the vertex and divides the parabola into two symmetric parts

PARENT QUADRATIC FUNCTION

The most basic quadratic function in the family of quadratic functions, called the parent quadratic function , is y 5 x2. The graph is shown below.

The line that passes through

x

y

1

3

5

12121

23 3

(0, 0)

x 5 0y 5 x2

the vertex and divides the parabola into two symmetric parts is called the axis of symmetry . The axis of symmetry for the graph of y 5 x2 is the y-axis, x 5 0 .

The lowest or highest point on the parabola is the vertex . The vertex of the graph of y 5 x2 is ( 0 , 0 ).


Your Notes



1. y 5 25x2

x

y3

121

29

23

215

221

23 3

Checkpoint Graph the function. Compare the graph with the graph of y 5 x2.

Graph y 5 1 } 2 x2. Compare the graph with the graph of

y 5 x2.

SolutionStep 1 Make a table of values

x

y

1

3

5

7

12123 3

for y 5 1 } 2 x2.

x 24 22 0 2 4

y

Step 2 the points from the table.

Step 3 Draw a through the points.

Step 4 Compare the graphs of y 5 1 } 2 x2 and y 5 x2. Both

graphs have the same vertex, ( , ), and axis of symmetry, . However, the graph of

y 5 1 } 2 x2 is than the graph of y 5 x2. This

is because the graph of y 5 1 } 2 x2 is a vertical

1 by a factor of 2 of the graph of

y 5 x2.

Example 1 Graph y 5 ax2 where ⏐a⏐< 1Your Notes



1. y 5 25x2

y 5 25x2 is narrower and x

y3

121

29

23

215

221

23 3

y 5 x2

y 5 25x2

opens down because it is a vertical stretch (by a factor of 5) and a reflection in the x-axis of the graph y 5 x2.


Graph y 5 1 } 2 x2. Compare the graph with the graph of

y 5 x2.

SolutionStep 1 Make a table of values

x

y

1

3

5

7

12123 3

y 5 x2

y 5 x2 12

for y 5 1 } 2 x2.

x 24 22 0 2 4

y 8 2 0 2 8

Step 2 Plot the points from the table.

Step 3 Draw a smooth curve through the points.

Step 4 Compare the graphs of y 5 1 } 2 x2 and y 5 x2. Both

graphs have the same vertex, ( 0 , 0 ), and axis of symmetry, x 5 0 . However, the graph of

y 5 1 } 2 x2 is wider than the graph of y 5 x2. This

is because the graph of y 5 1 } 2 x2 is a vertical

shrink 1 by a factor of 1 } 2 2 of the graph of

y 5 x2.

Example 1 Graph y 5 ax2 where ⏐a⏐< 1Your Notes


Graph y 5 x2 2 2. Compare the graph with the graph of y 5 x2.

Step 1 Make a table of values for

x

y

1

3

12121

23

23 3

y 5 x2 2 2.

x 22 21 0 1 2

y



Step 4 Compare the graphs of y 5 x2 2 2 and y 5 x2. Both graphs open and have the same axis of symmetry, . However, the vertex of the

graph of y 5 x2 2 2, ( , ), is different thanthe vertex of the graph of y 5 x2, ( , ), because the graph of y 5 x2 2 2 is a

(of units ) of the graph of y 5 x2.

Example 2 Graph y 5 x2 1 c

Graph y 5 23x2 1 3. Compare the graph with the graph of y 5 x2.

Step 1 Make a table of values for x

y2

12122

26

210

23 3y 5 23x2 1 3.

x 22 21 0 1 2

y



Step 4 Compare the graphs. Both graphs have the same axis of symmetry. However, the graph of y 5 23x2 1 3 is and has a vertex than the graph of y 5 x2 because the graph of y 5 23x2 1 3 is a and a of the graph of y 5 x2.

Example 3 Graph y 5 ax2 1 c


Your Notes


Graph y 5 x2 2 2. Compare the graph with the graph of y 5 x2.

Step 1 Make a table of values for

x

y

1

3

12121

23

23 3

y 5 x2 2 2

y 5 x2

y 5 x2 2 2.

x 22 21 0 1 2

y 2 21 22 21 2



Step 4 Compare the graphs of y 5 x2 2 2 and y 5 x2. Both graphs open up and have the same axis of symmetry, x 5 0 . However, the vertex of the

graph of y 5 x2 2 2, ( 0 , 22 ), is different thanthe vertex of the graph of y 5 x2, ( 0 , 0 ), because the graph of y 5 x2 2 2 is a vertical translation (of 2 units down ) of the graph

of y 5 x2.

Example 2 Graph y 5 x2 1 c

Graph y 5 23x2 1 3. Compare the graph with the graph of y 5 x2.

Step 1 Make a table of values for x

y2

12122

26

210

23 3

y 5 23x2 1 3

y 5 x2

y 5 23x2 1 3.

x 22 21 0 1 2

y 29 0 3 0 29



Step 4 Compare the graphs. Both graphs have the same axis of symmetry. However, the graph of y 5 23x2 1 3 is narrower and has a higher vertex than the graph of y 5 x2 because the graph of y 5 23x2 1 3 is a vertical stretch and a vertical translation of the graph of y 5 x2.

Example 3 Graph y 5 ax2 1 c


Your Notes


Compared with the graph of y 5 ax2, a > 0y 5 x2, the graph of y 5 ax2 is:

x

y

a 5 10 , a , 1

a . 1

• a vertical if a > 1,

• a vertical if 0 < a < 1.

Compared with the graph of y 5 ax2, a < 0y 5 x2, the graph of y 5 ax2 is:

x

y

a 5 2121 , a , 0

a , 21• a vertical and a

in the x-axis ifa < 21,

• a vertical and a in the x-axis if

21 < a < 0.

Compared with the graph of y 5 x2, y 5 x2 1 cthe graph of y 5 x2 1 c is:

x

y

c 5 0c , 0

c . 0

• an vertical translation if c > 0,

• a vertical translation if c < 0.

Homework

2. y 5 1 } 4 x2 2 6 x

y

1

12121

23

25

23 3



Your Notes


Compared with the graph of y 5 ax2, a > 0y 5 x2, the graph of y 5 ax2 is:

x

y

a 5 10 , a , 1

a . 1

• a vertical stretch if a > 1,

• a vertical shrink if 0 < a < 1.

Compared with the graph of y 5 ax2, a < 0y 5 x2, the graph of y 5 ax2 is:

x

y

a 5 2121 , a , 0

a , 21• a vertical stretch and a

reflection in the x-axis ifa < 21,

• a vertical shrink and a reflection in the x-axis if21 < a < 0.

Compared with the graph of y 5 x2, y 5 x2 1 cthe graph of y 5 x2 1 c is:

x

y

c 5 0c , 0

c . 0

• an upward vertical translation if c > 0,

• a downward vertical translation if c < 0.

Homework

2. y 5 1 } 4 x2 2 6 x

y

1

12121

23

25

23 3

y 5 x2

y 5 x2 2 614

y 5 1 } 4 x2 2 6 is a vertical

shrink, and downward vertical translation of the graph y 5 x2.



Your Notes


10.2 Graph y 5 ax2 1 bx 1 cGoal p Graph general quadratic functions.

VOCABULARY

Minimum value

Maximum value

PROPERTIES OF THE GRAPH OF A QUADRATIC FUNCTION

The graph of y 5 ax2 1 bx 1 c is a parabola that:

• opens if a > 0 and opens if a < 0.

• is narrower than the graph of y 5 x2 if⏐a⏐ 1 and wider if⏐a⏐ 1.

• has an axis of symmetry of

x

y

(0, c)

x 5 2b2a

x 5 .

• has a vertex with an

x-coordinate of .

• has a y-intercept of . So, the point ( , ) is on the parabola.

Your Notes



10.2 Graph y 5 ax2 1 bx 1 cGoal p Graph general quadratic functions.

VOCABULARY

Minimum value When a > 0, the minimum value of the function y 5 ax2 1 bx 1 c is the y-coordinate of the vertex.

Maximum value When a < 0, the maximum value of the function y 5 ax2 1 bx 1 c is the y-coordinate of the vertex.

PROPERTIES OF THE GRAPH OF A QUADRATIC FUNCTION

The graph of y 5 ax2 1 bx 1 c is a parabola that:

• opens up if a > 0 and opens down if a < 0.

• is narrower than the graph of y 5 x2 if⏐a⏐ > 1 and wider if⏐a⏐ < 1.

• has an axis of symmetry of

x

y

(0, c)

x 5 2b2a

x 5 2 b } 2a .

• has a vertex with an

x-coordinate of 2 b } 2a .

• has a y-intercept of c . So, the point ( 0 , c ) is on the parabola.

Your Notes



1. Graph the function

x

y

3

9

15

21

12123

23 3

y 5 4x2 1 8x 1 3. Label the vertex and axis of symmetry.


Graph y 5 2x2 1 4x 2 1.

Step 1 Determine whether the parabola opens up or down. Because a 0, the parabola opens .

Step 2 Find and draw the axis of

x

y

1

3

12121

23

3 5

symmetry:

x 5 2 b } 2a 5 5 .

Step 3 Find and plot the vertex. The x-coordinate of the vertex is , or .

To find the y-coordinate, substitute for x in the function and simplify.

y 5 2( )2 1 4( ) 2 1 5 3

So, the vertex is ( , ).

Step 4 Plot two points. Choose x 1 0

y two x-values less than the x-coordinate of the vertex. Then find the corresponding y-values.

Step 5 the points plotted in Step 4 in the axis of symmetry.

Step 6 Draw a through the plotted points.

Example 1 Graph y 5 ax2 1 bx 1 c


Your Notes


1. Graph the function

x

y

3

9

15

21

12123

23 3

(21, 21)

x 5 21

y 5 4x2 1 8x 1 3

y 5 4x2 1 8x 1 3. Label the vertex and axis of symmetry.


Graph y 5 2x2 1 4x 2 1.

Step 1 Determine whether the parabola opens up or down. Because a < 0, the parabola opens down .

Step 2 Find and draw the axis of

x

y

1

3

12121

23

3 5

(2, 3)

(3, 2)(1, 2)

(0, 21)(4, 21)

x 5 2

symmetry:

x 5 2 b } 2a 5 2 4 } 2(21) 5 2 .

Step 3 Find and plot the vertex. The x-coordinate of the vertex is 2 b } 2a , or 2 .

To find the y-coordinate, substitute 2 for x in the function and simplify.

y 5 2( 2 )2 1 4( 2 ) 2 1 5 3

So, the vertex is ( 2 , 3 ).

Step 4 Plot two points. Choose x 1 0

y 2 21 two x-values less than the x-coordinate of the vertex. Then find the corresponding y-values.

Step 5 Reflect the points plotted in Step 4 in the axis of symmetry.

Step 6 Draw a parabola through the plotted points.

Example 1 Graph y 5 ax2 1 bx 1 c


Your Notes


MINIMUM AND MAXIMUM VALUES

For y 5 ax2 1 bx 1 c, the y-coordinate of the vertex is the value of the function if a 0 and the value of the function if a 0.

2. Tell whether the function f(x) 5 2 1 } 2 x2 1 6x 1 8 has

a minimum value or a maximum value. Then find the minimum or maximum value.


Tell whether the function f(x) 5 5x2 2 20x 1 17 has a minimum value or a maximum value. Then find the minimum or maximum value.

Solution

Because a 5 and , the parabola opens and the function has a value. To find the value, find the .

x 5 2 b } 2a 5 5 The x-coordinate is

2 b } 2a .

f( ) 5 5( )2 2 20( ) 1 17 Substitute for x.

5 Simplify.

The value of the function is .

Example 2 Find the minimum or maximum value

Homework


The domain of the function is all real numbers and the range is f(x)

Your Notes


MINIMUM AND MAXIMUM VALUES

For y 5 ax2 1 bx 1 c, the y-coordinate of the vertex is the minimum value of the function if a > 0 and the maximum value of the function if a < 0.

2. Tell whether the function f(x) 5 2 1 } 2 x2 1 6x 1 8 has

a minimum value or a maximum value. Then find the minimum or maximum value.

The maximum value of the function is f(6) 5 26.


Tell whether the function f(x) 5 5x2 2 20x 1 17 has a minimum value or a maximum value. Then find the minimum or maximum value.

Solution

Because a 5 5 and 5 > 0 , the parabola opens up and the function has a minimum value. To find the minimum value, find the vertex .

x 5 2 b } 2a 5 2 220 } 2(5) 5 2 The x-coordinate is

2 b } 2a .

f( 2 ) 5 5( 2 )2 2 20( 2 ) 1 17 Substitute 2 for x.

5 23 Simplify.

The minimum value of the function is f (2) 5 23 .

Example 2 Find the minimum or maximum value

Homework


The domain of the function is all real numbers and the range is f(x) $ –3.

Your Notes


Graph Quadratic Functions in Intercept Form

VOCABULARY

Intercept form

GRAPH OF INTERCEPT FORM y 5 a(x 2 p)(x 2 q)

p The x-intercepts are and .

p The axis of symmetry is halfway

between (p, 0) and (q, 0). So, the

axis of symmetry is x 5 .

p The parabola opens if

a > 0 and opens if a < 0.

Goal p Graph quadratic equations in intercept form.


x

yx � p�q

(q,0)(p,0)

2

y

x21 3�5 �3 �1

2

�9

�7

�5

�3

�1

Focus On

Functions

Use after Lesson 10.2

Your Notes

Graph y 5 (x 2 2)(x 1 4).

Step 1 Identify and plot the x-intercepts. They occur

at ( , 0) and ( , 0).

Step 2 Find and draw the axis

of symmetry.

x 5 p 1 q

} 2 5 }

2 5 21

Step 3 Find and plot the vertex.

Its x-coordinate is .

Its y-coordinate is

y 5 ( 2 2)( 1 4)

5 The vertex is .

Step 4 Draw a parabola through the vertex and

x-intercepts.

Example 1 Graph a quadratic equation in intercept form

Notice that the x-intercepts of the graph are also the zeros of the function.0 5 (x – 2)(x 1 4)x 2 2 5 0 or x 1 4 5 0x 5 or x 5


Graph Quadratic Functions in Intercept Form

VOCABULARY

Intercept form A quadratic equation written as y 5 a(x 2 p)(x 2 q), where a Þ 0.

GRAPH OF INTERCEPT FORM y 5 a(x 2 p)(x 2 q)

p The x-intercepts are p and q.

p The axis of symmetry is halfway

between (p, 0) and (q, 0). So, the

axis of symmetry is x 5 p + q

} 2 .

p The parabola opens up if

a > 0 and opens down if a < 0.

Goal p Graph quadratic equations in intercept form.


x

yx � p�q

(q,0)(p,0)

2

(�4, 0)

y

x ��1

x21 3�5 �3 �1

2

(2, 0)

(�1, �9)y � (x�2)(x�4)

�9

�7

�5

�3

�1

Focus On

Functions


Your Notes

Graph y 5 (x 2 2)(x 1 4).

Step 1 Identify and plot the x-intercepts. They occur

at ( 2 , 0) and (24 , 0).

Step 2 Find and draw the axis

of symmetry.

x 5 p 1 q

} 2 5

2 – 4 }

2 5 21

Step 3 Find and plot the vertex.

Its x-coordinate is –1.

Its y-coordinate is

y 5 ( –1 2 2)( –1 1 4)

5 –9 The vertex is (–1, –9) .

Step 4 Draw a parabola through the vertex and

x-intercepts.

Example 1 Graph a quadratic equation in intercept form

Notice that the x-intercepts of the graph are also the zeros of the function.0 5 (x – 2)(x 1 4)x 2 2 5 0 or x 1 4 5 0x 5 2 or x 5 24


Graph y 5 23x2 1 3.

Step 1 Rewrite the quadratic function in intercept form.

y 5 Write original function.

5 Factor out common factor.

5 Difference of two squares

Step 2 Identify and plot x-intercepts at ( , 0) and ( , 0).

Step 3 Find and draw the axis of symmetry.

x 5 p 1 q

} 2 5 } 2 5

Step 4 Find and plot the vertex. The x-coordinate of the vertex is . The

y-coordinate is

y 5 23( )2 1 3 5

The vertex is ( ).

Step 5 Draw a parabola through the vertex and x-intercepts.

Example 2 Graph a quadratic function

The domain of the function is all real numbers, and the range is .


y

x1

1

1. 1 } 2 (x 1 1)(x 2 7) 2. 2x2 2 8x 2 12

Checkpoint Graph the quadratic function. Label the vertex, axis of symmetry, and x-intercepts. Identify the domain and range of the function.

y

x

11

y

x

1

Homework

Your Notes


Graph y 5 23x2 1 3.

Step 1 Rewrite the quadratic function in intercept form.

y 5 –3x2 + 3 Write original function.

5 –3(x2 – 1) Factor out common factor.

5 –3(x + 1)(x – 1) Difference of two squares

Step 2 Identify and plot x-intercepts at (–1, 0) and ( 1 , 0).

Step 3 Find and draw the axis of symmetry.

x 5 p 1 q

} 2 5 –1 + 1

} 2 5 0

Step 4 Find and plot the vertex. The x-coordinate of the vertex is 0 . The

y-coordinate is

y 5 23( 0 )2 1 3 5 3

The vertex is ( 0, 3 ).

Step 5 Draw a parabola through the vertex and x-intercepts.


The domain of the function is all real numbers, and the range is y # 3 .


y

x1

1(�1, 0) (1, 0)

(0, 3)

1. 1 } 2 (x 1 1)(x 2 7) 2. 2x2 2 8x 2 12

domain: all real numbers domain: all real numbersrange: y $ –8 range: y # 4

Checkpoint Graph the quadratic function. Label the vertex, axis of symmetry, and x-intercepts. Identify the domain and range of the function.

y

x

11 (7, 0)(�1, 0)

(3, �8)

x � 3

y

x

1(�6, 0)

(�4, 4)

(�2, 0)

x � �4

Homework

Your Notes


10.3 Solve Quadratic Equations by GraphingGoal p Solve quadratic equations by graphing.

VOCABULARY

Quadratic equation

Solve 2x2 1 2x 5 28 by graphing.

Step 1 Write the equation in .

2x2 1 2x 5 28 Write original equation.

2x2 1 2x 1 8 5 Add to each side.

Step 2 Graph the function y 5 2x2 1 2x 1 8. The x-intercepts are and .

x

y

2

6

10

14

12123 3

The solutions of the equation 2x2 1 2x 5 28 are and .

CHECK You can check and in the original equation.

2x2 1 2x 5 28 2x2 1 2x 5 28

2( )2 1 2( ) 0 28 2( )2 1 2( ) 0 28

5 5

Example 1 Solve a quadratic equation having two solutions

Your Notes



10.3 Solve Quadratic Equations by GraphingGoal p Solve quadratic equations by graphing.

VOCABULARY

Quadratic equation An equation that can be written in standard form as ax2 1 bx 1 c where a Þ 0

Solve 2x2 1 2x 5 28 by graphing.

Step 1 Write the equation in standard form .

2x2 1 2x 5 28 Write original equation.

2x2 1 2x 1 8 5 0 Add 8 to each side.

Step 2 Graph the function y 5 2x2 1 2x 1 8. The x-intercepts are 22 and 4 .

x

y

2

6

10

14

12123 3

y 5 2x2 1 2x 1 8

422

The solutions of the equation 2x2 1 2x 5 28 are 22 and 4 .

CHECK You can check 22 and 4 in the original equation.

2x2 1 2x 5 28 2x2 1 2x 5 28

2( 22 )2 1 2( 22 ) 0 28 2( 4 )2 1 2( 4 ) 0 28

28 5 28 28 5 28

Example 1 Solve a quadratic equation having two solutions

Your Notes



Solve x2 2 4x 5 24 by graphing.

x

y

1

3

5

7

12123 3

Step 1 Write the equation in standard form.


x2 2 4x 1 4 5 Add to each side.

Step 2 the function y 5 x2 2 4x 1 4. The x-intercept is .

The solution of the equation x2 2 4x 5 24 is .

Example 2 Solve a quadratic equation having one solution

Solve x2 1 8 5 2x by graphing.

x

y

2

6

10

14

12123 3


x2 1 8 5 2x Write original equation.

Subtract from each side.

Step 2 the function y 5 . The graph has x-intercepts.

The equation x2 1 8 5 2x has .

Example 3 Solve a quadratic equation having no solution

1. Solve x2 2 6 5 25x by

x

y

3

9

12123

29

2325

graphing.



Your Notes


Solve x2 2 4x 5 24 by graphing.

x

y

1

3

5

7

12123 3

2

y 5 x2 2 4x 1 4



x2 2 4x 1 4 5 0 Add 4 to each side.

Step 2 Graph the function y 5 x2 2 4x 1 4. The x-intercept is 2 .

The solution of the equation x2 2 4x 5 24 is 2 .

Example 2 Solve a quadratic equation having one solution

Solve x2 1 8 5 2x by graphing.

x

y

2

6

10

14

12123 3

y 5 x2 2 2x 1 8


x2 1 8 5 2x Write original equation.

x2 2 2x 1 8 5 0 Subtract 2x from each side.

Step 2 Graph the function y 5 x2 2 2x 1 8 . The graph has no x-intercepts.

The equation x2 1 8 5 2x has no solution .

Example 3 Solve a quadratic equation having no solution

1. Solve x2 2 6 5 25x by

x

y

3

9

12123

29

2325

y 5 x2 1 5x 2 6

26 1

graphing.

The solutions of the equation are 26 and 1.



Your Notes


Homework

NUMBER OF SOLUTIONS OF A QUADRATIC EQUATION

A quadratic equation has two solutions if the graph of its related function has .A quadratic equation has one solution if the graph of its related function has .A quadratic equation has no solution if the graph of its related function has .

Find the zeros of f(x) 5 2x2 2 8x 2 7.

Graph the function

x

y

3

9

22223

29

26 6

f(x) 5 2x2 2 8x 2 7. Thex-intercepts are and .

The zeros of the function are and .

CHECK Substitute and in the original function.

f( ) 5 2( )2 2 8( ) 2 7 5

f( ) 5 2( )2 2 8( ) 2 7 5

Example 4 Find the zeros of a quadratic function

RELATING SOLUTIONS OF EQUATIONS, x-INTERCEPTS OF GRAPHS, AND ZEROS OF FUNCTIONS

Solutions of an EquationThe solutions of the equation x2 2 11x 1 18 are and .

x-Intercepts of a Graph 22 6 10 14

2

22

26

210

214

x

y

2 9

y 5 x2 2 11x1 18

The x-intercepts of the graph of y 5 x2 2 11x 1 18 occur where y 5 , so the x-intercepts are

and , as shown.

Zeros of a FunctionThe zeros of the function f(x) 5 x2 2 11x 1 18 are the values of x for which f(x) 5 , so the zeros are and .


Your Notes


Homework

NUMBER OF SOLUTIONS OF A QUADRATIC EQUATION

A quadratic equation has two solutions if the graph of its related function has two x-intercepts .A quadratic equation has one solution if the graph of its related function has one x-intercept .A quadratic equation has no solution if the graph of its related function has no x-intercepts .

Find the zeros of f(x) 5 2x2 2 8x 2 7.

Graph the function

x

y

3

9

22223

29

26 6

y 5 2x2 2 8x 2 7

27 21

f(x) 5 2x2 2 8x 2 7. Thex-intercepts are 27 and 21 .

The zeros of the function are 27 and 21 .

CHECK Substitute 27 and 21 in the original function.

f( 27 ) 5 2( 27 )2 2 8( 27 ) 2 7 5 0

f( 21 ) 5 2( 21 )2 2 8( 21 ) 2 7 5 0

Example 4 Find the zeros of a quadratic function

RELATING SOLUTIONS OF EQUATIONS, x-INTERCEPTS OF GRAPHS, AND ZEROS OF FUNCTIONS

Solutions of an EquationThe solutions of the equation x2 2 11x 1 18 are 2 and 9 .

x-Intercepts of a Graph 22 6 10 14

2

22

26

210

214

x

y

2 9

y 5 x2 2 11x1 18

The x-intercepts of the graph of y 5 x2 2 11x 1 18 occur where y 5 0 , so the x-intercepts are 2 and 9 , as shown.

Zeros of a FunctionThe zeros of the function f(x) 5 x2 2 11x 1 18 are the values of x for which f(x) 5 0 , so the zeros are 2 and 9 .


Your Notes


10.4 Use Square Roots to SolveQuadratic EquationsGoal p Solve a quadratic equation by finding square roots.

SOLVING x2 5 d BY TAKING SQUARE ROOTS

• If d > 0, then x2 5 d has solutions: .

• If d 5 0, then x2 5 d has solution: .

• If d < 0, then x2 5 d has solution.

Your Notes

Solve the equation.

a. z2 2 5 5 4 b. r2 1 7 5 4 c. 25k2 5 9

Solutiona. z2 2 5 5 4 Write original equation.

z2 5 Add to each side.

z 5 Take square roots of each side.

z 5 Simplify. The solutions are and .

b. r2 1 7 5 4 Write original equation.

r2 5 Subtract from each side.

Negative real numbers do not have real . So, there is .

c. 25k2 5 9 Write original equation.

k2 5 Divide each side by .

k 5 Take square roots of each side.

k 5 Simplify. The solutions are

and .

Example 1 Solve quadratic equations



10.4 Use Square Roots to SolveQuadratic EquationsGoal p Solve a quadratic equation by finding square roots.

SOLVING x2 5 d BY TAKING SQUARE ROOTS

• If d > 0, then x2 5 d has two solutions: x 5 6 Ï}

d .

• If d 5 0, then x2 5 d has one solution: x 5 0 .

• If d < 0, then x2 5 d has no solution.

Your Notes

Solve the equation.

a. z2 2 5 5 4 b. r2 1 7 5 4 c. 25k2 5 9

Solutiona. z2 2 5 5 4 Write original equation.

z2 5 9 Add 5 to each side.

z 5 6 Ï}

9 Take square roots of each side.

z 5 63 Simplify. The solutions are 23 and 3 .

b. r2 1 7 5 4 Write original equation.

r2 5 23 Subtract 7 from each side.

Negative real numbers do not have real square roots . So, there is no solution .

c. 25k2 5 9 Write original equation.

k2 5 9 } 25

Divide each side by 25 .

k 5 6 Ï}

9 } 25 Take square roots of each side.

k 5 6 3 } 5 Simplify. The solutions are 2 3 } 5

and 3 } 5 .

Example 1 Solve quadratic equations



1. 3x2 5 108 2. t2 1 17 5 17 3. 81p2 5 4


Solve 4x2 1 3 5 23. Round the solutions to the nearest hundredth.

Solution4x2 1 3 5 23 Write original equation.

4x2 5 Subtract from each side.

x2 5 Divide each side by .

x 5 Take square roots of each side.

x ø Use a calculator. Round to the nearest hundredth.

The solutions are about and .

Example 2 Approximate solutions of a quadratic equation

Checkpoint Solve the equation. Round the solutions to the nearest hundredth.

4. 2x2 2 7 5 9 5. 6g2 1 1 5 19


Your Notes


1. 3x2 5 108 2. t2 1 17 5 17 3. 81p2 5 4

x 5 66 t 5 0 p 5 6 2 } 9


Solve 4x2 1 3 5 23. Round the solutions to the nearest hundredth.

Solution4x2 1 3 5 23 Write original equation.

4x2 5 20 Subtract 3 from each side.

x2 5 5 Divide each side by 4 .

x 5 6 Ï}


x ø 62.24 Use a calculator. Round to the nearest hundredth.

The solutions are about 22.24 and 2.24 .

Example 2 Approximate solutions of a quadratic equation

Checkpoint Solve the equation. Round the solutions to the nearest hundredth.

4. 2x2 2 7 5 9 5. 6g2 1 1 5 19

x ø 62.83 g ø 61.73


Your Notes


Solve 5(x 1 1)2 5 30. Round the solutions to the nearest hundredth.

Solution5(x 1 1)2 5 30 Write original equation.

(x 1 1)2 5 Divide each side by .

x 1 1 5 Take square roots of each side.

x 5 Subtract from each side.

The solutions are ø and

ø .

CHECK To check the solutions,

ZeroX=-3.5 Y=0

first write the equation so that as follows:

5(x 1 1)2 2 30 5 0. Then graph the related function y 5 5(x 1 1)2 2 30. The x-intercepts appear to be about and about . So, each solution checks.

Example 3 Solve a quadratic equation

Homework

Checkpoint Solve the equation. Round the solutions to the nearest hundredth, if necessary.

6. 3(m 2 4)2 5 12 7. 4(a 2 3)2 5 32


Your Notes


Solve 5(x 1 1)2 5 30. Round the solutions to the nearest hundredth.

Solution5(x 1 1)2 5 30 Write original equation.

(x 1 1)2 5 6 Divide each side by 5 .

x 1 1 5 6 Ï}


x 5 21 6 Ï}

6 Subtract 1 from each side.

The solutions are 21 1 Ï}

6 ø 1.45 and

21 2 Ï}

6 ø 23.45 .

CHECK To check the solutions,

ZeroX=-3.5 Y=0

first write the equation so that 0 is on one side as follows:

5(x 1 1)2 2 30 5 0. Then graph the related function y 5 5(x 1 1)2 2 30. The x-intercepts appear to be about 1.5 and about 23.5 . So, each solution checks.


Homework

Checkpoint Solve the equation. Round the solutions to the nearest hundredth, if necessary.

6. 3(m 2 4)2 5 12 7. 4(a 2 3)2 5 32

m 5 6 and m 5 2 a ø 5.83 and a ø 0.17


Your Notes


10.5 Solve Quadratic Equationsby Completing the SquareGoal p Solve quadratic equations by completing

the square.

VOCABULARY

Completing the square

COMPLETING THE SQUARE

Words To complete the square for the expression x2 1 bx, add the of the term bx.

Algebra x2 1 bx 1 1 b } 2 2 2 5 1 x 1 b } 2 2 2

Find the value of c that makes the expression x2 2 5x 1 c a perfect square trinomial. Then write the expression as the square of the binomial.

SolutionStep 1 Find the value of c. For the expression to be a

perfect square trinomial, c needs to be the square of half the coefficient of the term bx.

c 5 1 2 22

5 Find the square of half the coefficient of bx.

Step 2 Write the expression as a perfect square trinomial. Then write the expression as the square of a binomial.

x2 2 5x 1 c 5 x2 2 5x 1 Substitute for c.

5 2

Square of a binomial

Example 1 Complete the square

Your Notes



10.5 Solve Quadratic Equationsby Completing the SquareGoal p Solve quadratic equations by completing

the square.

VOCABULARY

Completing the square To complete the square, a constant c is added to an expression of the form x2 1 bx to form a perfect square trinomial.

COMPLETING THE SQUARE

Words To complete the square for the expression x2 1 bx, add the square of half the coefficient of the term bx.

Algebra x2 1 bx 1 1 b } 2 2 2 5 1 x 1 b } 2 2 2

Find the value of c that makes the expression x2 2 5x 1 c a perfect square trinomial. Then write the expression as the square of the binomial.

SolutionStep 1 Find the value of c. For the expression to be a

perfect square trinomial, c needs to be the square of half the coefficient of the term bx.

c 5 1 2

25 22

5 25 } 4 Find the square of half the

coefficient of bx.

Step 2 Write the expression as a perfect square trinomial. Then write the expression as the square of a binomial.

x2 2 5x 1 c 5 x2 2 5x 1 25 } 4 Substitute 25 }

4 for c.

5 1 x 2 5 } 2 2 2

Square of a binomial

Example 1 Complete the square

Your Notes



Checkpoint Find the value of c that makes the expression a perfect square trinomial. Then write the expression as the square of a binomial.

1. x2 1 7x 1 c 2. x2 2 6x 1 c

Solve t2 1 6t 5 25 by completing the square.

Solution

t2 1 6t 5 25 Write original equation.

t2 1 6t 1 5 25 1 Add 1 2 2, or , to

each side.

5 25 1 Write left side as the square of a binomial.

5 Simplify the right side.


t 5 Subtract from each side.

The solutions of the equation are and .



Your Notes


Checkpoint Find the value of c that makes the expression a perfect square trinomial. Then write the expression as the square of a binomial.

1. x2 1 7x 1 c 2. x2 2 6x 1 c

c 5 49 } 4 ; 1 x 1 7 }

2 2 2 c 5 9; (x 2 3)2

Solve t2 1 6t 5 25 by completing the square.

Solution

t2 1 6t 5 25 Write original equation.

t2 1 6t 1 32 5 25 1 32 Add 1 6 } 2 2 2, or 32 , to

each side.

(t 1 3)2 5 25 1 32 Write left side as the square of a binomial.

(t 1 3)2 5 4 Simplify the right side.

t 1 3 5 62 Take square roots of each side.

t 5 23 6 2 Subtract 3 from each side.

The solutions of the equation are 23 1 2 5 21 and 23 2 2 5 25 .



Your Notes


Example 3 Solve a quadratic equation in standard form

Checkpoint Solve the equation by completing the square. Round your solutions to the nearest hundredth, if necessary.

3. r2 2 8r 5 9 4. 5s2 1 60s 1 125 5 0

Homework

Solve 4m2 2 16m 1 8 5 0 by completing the square.

Solution

4m2 2 16m 1 8 5 0 Write original equation.

4m2 2 16m 5 Subtract from each side.

m2 2 4m 5 Divide each side by .

m2 2 4m 1 5 22 1 Add 1 22,

or , to each side.

5 Write left side as the square of a binomial.


m 5 Add to each side.

The solutions are ø

and ø .


Your Notes


Example 3 Solve a quadratic equation in standard form

Checkpoint Solve the equation by completing the square. Round your solutions to the nearest hundredth, if necessary.

3. r2 2 8r 5 9 4. 5s2 1 60s 1 125 5 0

r 5 9 and r 5 21 s ø 22.68 and s ø 29.32

Homework

Solve 4m2 2 16m 1 8 5 0 by completing the square.

Solution

4m2 2 16m 1 8 5 0 Write original equation.

4m2 2 16m 5 28 Subtract 8 from each side.

m2 2 4m 5 22 Divide each side by 4 .

m2 2 4m 1 (22)2 5 22 1 (22)2 Add 1 24 } 2 2 2,

or (22)2 , to each side.

(m 2 2)2 5 2 Write left side as the square of a binomial.

m 2 2 5 6 Ï}


m 5 2 6 Ï}

2 Add 2 to each side.


2 ø 3.41

and 2 2 Ï}

2 ø 0.59 .


Your Notes



Graph Quadratic Functions in Vertex FormGoal p Graph quadratic functions in Vertex form.

VOCABULARY

Vertex form

GRAPH VERTEX FORM y 5 a(x 2 h)2 1 k

The graph of y 5 a(x 2 h)2 1 k is the graph of

translated units and units

• The vertex is .

• The axis of symmetry is

• The graph opens up if ,

and opens down if .

Graph y 5 2(x 1 3)2 1 1.

Solution

Step 1 Identify the values: a 5 , h 5 , k 5

The parabola opens because .

Step 2 Draw the axis of symmetry, x 5 .

Step 3 Plot the vertex (h, k) 5 .

Step 4 Plot points. For two points, use x-values

than the x-coordinate of the .

If x 5 24, y 5 2( 1 3)2 1 1 5 0

If x 5 25, y 5 5

Plot (24, ), ( ), and

reflections ( ) and ( , 3).

Step 5 Draw the parabola.

Example 1 Graph a quadratic function in vertex form

x

y

k

h

y � ax2

(0, 0)

(h, k)

Focus On

Functions


y

x

1

1

Your Notes



Graph Quadratic Functions in Vertex FormGoal p Graph quadratic functions in Vertex form.

VOCABULARY

Vertex form The form of a quadratic function, y 5 a(x 2 h)2 1 k, where a Þ 0.

GRAPH VERTEX FORM y 5 a(x 2 h)2 1 k

The graph of y 5 a(x 2 h)2 1 k is the graph of y = ax2

translated h units horizontally and k units

vertically.

• The vertex is (h, k).

• The axis of symmetry is x = h.

• The graph opens up if a > 0,

and opens down if a < 0.

Graph y 5 2(x 1 3)2 1 1.

Solution

Step 1 Identify the values: a 5 21, h 5 23, k 5 1 The parabola opens down because a , 0.

Step 2 Draw the axis of symmetry, x 5 23.

Step 3 Plot the vertex (h, k) 5 (23, 1).

Step 4 Plot four points. For two points, use x-values less than the x-coordinate of the vertex .

If x 5 24, y 5 2(−4 1 3)2 1 1 5 0

If x 5 25, y 5 2(25 1 3)2 1 1 5 −3

Plot (24, 0), (−5, −3), and

reflections (−2, 0 ) and (−1, 3).


Example 1 Graph a quadratic function in vertex form

x

y

k

h

y � ax2

(0, 0)

(h, k)

Focus On

Functions


y

x

1

1(�3, 1)

x � �3

Your Notes



Graph y 5 x2 1 4x 1 7.

Solution

Step 1 Write in form. y 5 x2 1 4x 1 7

Complete the . y 5 (x2 1 4x ) 1 7

Write the binomial. y 5 ( )2 1 7

Subtract. y 5 ( )2 1

Step 2 Identify the values: a 5 , h 5 , k 5

The parabola opens because .

Step 3 Draw the , x 5 .

Step 4 Plot the (h, k) 5 .

Step 5 Plot . For , use x-values

than the x-coordinate of the

If x 5 23, y 5 ( 1 2)2 1 3 5 .

If x 5 24, y 5 5 .

Plot (23, ), ( , ), and

reflections ( , ) and ( , ).



1. Graph the quadratic function

y 5 (x 2 3)2 1 2. Label the vertex

and axis of symmetry.

2. Write the function y 5 x2 2 8x 1 14

in vertex form, then graph the

function. Label the vertex and

axis of symmetry.


y

x1

1

y

x1

1

y

x1

1Homework

Your Notes



Graph y 5 x2 1 4x 1 7.

Solution

Step 1 Write in vertex form. y 5 x2 1 4x 1 7

Complete the square. y 1 4 5 (x2 1 4x 1 4) 1 7

Write the binomial. y 1 4 5 (x 1 2)2 1 7

Subtract. y 5 (x 1 2)2 1 3

Step 2 Identify the values: a 5 1 , h 5 22, k 5 3

The parabola opens up because a > 0.

Step 3 Draw the axis of symmetry, x 5 22.

Step 4 Plot the vertex (h, k) 5 (22, 3).

Step 5 Plot four points. For two points, use x-values

less than the x-coordinate of the vertex.

If x 5 23, y 5 (23 1 2)2 1 3 5 4 .

If x 5 24, y 5 (24 1 2)2 1 3 5 7 .

Plot (23, 4 ), (24, 7), and

reflections (21, 4 ) and ( 0 , 7 ).



1. Graph the quadratic function

y 5 (x 2 3)2 1 2. Label the vertex

and axis of symmetry.

2. Write the function y 5 x2 2 8x 1 14

in vertex form, then graph the

function. Label the vertex and

axis of symmetry.

y = (x – 4)2 – 2


y

x1

1

(�2, 3)

x �� 2

y

x1

1(3, 2)

x � 3

y

x1

1

(4, �2)

x � 4Homework

Your Notes


10.6 Solve Quadratic Equations bythe Quadratic FormulaGoal p Solve quadratic equations using the quadratic

formula.

VOCABULARY

Quadratic formula

THE QUADRATIC FORMULA

The solutions of the quadratic equation

ax2 1 bx 1 c 5 0 are x 5 2b 6 Ï}

b2 2 4ac }} 2a when a Þ 0

and b2 2 4ac ≥ 0.

Solve 2x2 2 5 5 3x.

2x2 2 5 5 3x Write original equation.

Write in standard form.

x 5 2b 6 Ï}

b2 2 4ac }} 2a Quadratic formula

5 2 6 Ï

}}}

2 2 4( )( ) }}}}

2( )

Substitute values in the quadratic

formula: a 5 , b 5 , and c 5 .

5 6 Ï

}

}} Simplify.

5 6

} Simplify the square root.

The solutions are 1 } 5 and 2

} 5 .


Check your solution by graphing the related function and finding the x-intercepts.

Your Notes



10.6 Solve Quadratic Equations bythe Quadratic FormulaGoal p Solve quadratic equations using the quadratic

formula.

VOCABULARY

Quadratic formula The quadratic formula gives the solutions of any quadratic equation in standard form.

THE QUADRATIC FORMULA

The solutions of the quadratic equation

ax2 1 bx 1 c 5 0 are x 5 2b 6 Ï}

b2 2 4ac }} 2a when a Þ 0

and b2 2 4ac ≥ 0.

Solve 2x2 2 5 5 3x.

2x2 2 5 5 3x Write original equation.

2x2 2 3x 2 5 5 0 Write in standard form.

x 5 2b 6 Ï}

b2 2 4ac }} 2a Quadratic formula

5 2 (23) 6 Ï

}}}

(23) 2 2 4( 2 )( 25 ) }}}}

2( 2 )


formula: a 5 2 , b 5 23 , and c 5 25 .

5 3 6 Ï

}

49 }} 4 Simplify.

5 3 6 7 } 4

Simplify the square root.

The solutions are 3 1 7 } 4 5 5 } 2 and 3 2 7 } 4 5 21 .


Check your solution by graphing the related function and finding the x-intercepts.

Your Notes



1. Use the quadratic formula to solve 2x2 1 x 5 3.

2. In Example 2, suppose the net was thrown with an initial velocity of 5 feet per second from a height of 20 feet. How long would it take the net to hit the water?


Crabbing A crabbing net is thrown from a bridge, which is 35 feet above the water. If the net's initial velocity is 10 feet per second, how long will it take the net to hit the water?

Solution

The net's initial velocity is v 5 feet per second and the net's initial height is s 5 feet. The net will hit the water when the height is feet.


5 216t2 1 t 1 Substitute for h, v, and s.

t 5 2 6 Ï

}}}

2 2 4( )( ) }}} 2( )


formula: a 5 , b 5 , and c 5 .

5 6 Ï}

}} Simplify.

The solutions are 1 Ï}

}} ø and

2 Ï}

}} ø . So, the net will hit the water

in about seconds.

Example 2 Use the quadratic formula

Because time cannot be a negative number, disregard the negative solution.


Your Notes


1. Use the quadratic formula to solve 2x2 1 x 5 3.

x 5 1 and x 5 2 3 } 2

2. In Example 2, suppose the net was thrown with an initial velocity of 5 feet per second from a height of 20 feet. How long would it take the net to hit the water?

about 1.29 seconds


Crabbing A crabbing net is thrown from a bridge, which is 35 feet above the water. If the net's initial velocity is 10 feet per second, how long will it take the net to hit the water?

Solution

The net's initial velocity is v 5 10 feet per second and the net's initial height is s 5 35 feet. The net will hit the water when the height is 0 feet.


0 5 216t2 1 10 t 1 35 Substitute for h, v, and s.

t 5 2 10 6 Ï

}}}

10 2 2 4( 216 )( 35 ) }}} 2( 216 )


formula: a 5 216 , b 5 10 , and c 5 35 .

5 210 6 Ï}

2340 }} 232 Simplify.


2340 }} 232 ø 21.20 and

210 2 Ï}

2340 }} 232 ø 1.82 . So, the net will hit the water

in about 1.82 seconds.

Example 2 Use the quadratic formula

Because time cannot be a negative number, disregard the negative solution.


Your Notes


METHODS FOR SOLVING QUADRATIC EQUATIONS

Methods When to Use

Factoring Use when a quadratic equation can be easily.

Graphing Use when solutions are adequate.

Finding Use when solving an equation that cansquare roots be written in the form .

Completing Can be used for any quadratic equationthe square ax2 1 bx 1 c 5 0 but is simplest to apply when and b is an number.

Quadratic Can be used for quadratic equation.formula

Tell what method(s) you would use to solve the quadratic equation. Explain your choice(s).

a. 6x2 2 11x 1 7 5 0 b. 4x2 2 36 5 0

Solution

a. The quadratic equation be factored easily and completing the square would result in

. So, the equation can be solved using the .

b. The quadratic equation can be solved using because the equation can be written in the

form x2 5 d.

Example 3 Choose a solution method

3. Tell what method(s) you would use to solve x2 1 8x 5 9. Explain your choices(s).


Homework


Your Notes


METHODS FOR SOLVING QUADRATIC EQUATIONS

Methods When to Use

Factoring Use when a quadratic equation can be factored easily.

Graphing Use when approximate solutions are adequate.

Finding Use when solving an equation that cansquare roots be written in the form x2 5 d .

Completing Can be used for any quadratic equationthe square ax2 1 bx 1 c 5 0 but is simplest to apply when a 5 1 and b is an even number.

Quadratic Can be used for any quadratic equation.formula

Tell what method(s) you would use to solve the quadratic equation. Explain your choice(s).

a. 6x2 2 11x 1 7 5 0 b. 4x2 2 36 5 0

Solution

a. The quadratic equation cannot be factored easily and completing the square would result in many fractions . So, the equation can be solved using the quadratic formula .

b. The quadratic equation can be solved using square roots because the equation can be written in the form x2 5 d.

Example 3 Choose a solution method

3. Tell what method(s) you would use to solve x2 1 8x 5 9. Explain your choices(s).

factoring: the expression can be factored easily; completing the square: the equation is of the form ax2 1 bx 1 c 5 0 where a 5 1 and b is an even number.


Homework


Your Notes


10.7 Interpret the DiscriminantGoal p Use the value of the discriminant.

VOCABULARY

Discriminant

USING THE DISCRIMINANT OF ax2 1 bx 1 c 5 0

Value of the Number of Graph ofdiscriminant solutions y 5 ax2 1 bx 1 c

b2 2 4ac > 0

x

y

b2 2 4ac 5 0

x

y

b2 2 4ac < 0

x

y


Your Notes


10.7 Interpret the DiscriminantGoal p Use the value of the discriminant.

VOCABULARY

Discriminant In the quadratic formula, the expression b2 2 4ac is called the discriminant of the associated equation ax2 1 bx 1 c 5 0.

USING THE DISCRIMINANT OF ax2 1 bx 1 c 5 0

Value of the Number of Graph ofdiscriminant solutions y 5 ax2 1 bx 1 c

b2 2 4ac > 0 Two solutions

x

y

b2 2 4ac 5 0 One solution

x

y

b2 2 4ac < 0 No solution

x

y


Your Notes


Equation Discriminantax2 1 bx 1 c 5 0 b2 2 4ac

a. x2 2 3x 2 2 5 0 2 2 4( )( ) 5

b. 3x2 1 2 5 0 2 2 4( )( ) 5

c. 2x2 1 8x 1 8 5 0 2 2 4( )( ) 5

Number of solutions

a. b. c.

Example 1 Use the discriminant

Checkpoint Tell whether the equation has two solutions, one solution, or no solution.

1. x2 1 2x 5 1 2. 3x2 1 7x 5 25

3. 5x2 2 6 5 0 4. 2x2 2 9 5 6x

Tell whether the equation 22x2 1 4x 5 2 has two solutions, one solution, or no solution.

Step 1 Write the equation in .

22x2 1 4x 5 2 Write equation.

22x2 1 4x 2 2 5 0 Subtract from each side.

Step 2 Find the value of the .

b2 2 4ac 5 2 2 4( )( ) Substitute for a, for b, and for c.

5 Simplify.

The discriminant is , so the equation has .

Example 2 Find the number of solutions


Your Notes


Equation Discriminantax2 1 bx 1 c 5 0 b2 2 4ac

a. x2 2 3x 2 2 5 0 (23) 2 2 4( 1 )( 22 ) 5 17

b. 3x2 1 2 5 0 0 2 2 4( 3 )( 2 ) 5 224

c. 2x2 1 8x 1 8 5 0 8 2 2 4( 2 )( 8 ) 5 0

Number of solutions

a. Two solutions b. No solution c. One solution

Example 1 Use the discriminant

Checkpoint Tell whether the equation has two solutions, one solution, or no solution.

1. x2 1 2x 5 1 2. 3x2 1 7x 5 25

two solutions no solution

3. 5x2 2 6 5 0 4. 2x2 2 9 5 6x

two solutions one solution

Tell whether the equation 22x2 1 4x 5 2 has two solutions, one solution, or no solution.

Step 1 Write the equation in standard form .

22x2 1 4x 5 2 Write equation.

22x2 1 4x 2 2 5 0 Subtract 2 from each side.

Step 2 Find the value of the discriminant .

b2 2 4ac 5 4 2 2 4( 22 )( 22 ) Substitute 22 for a, 4 for b, and 22 for c.

5 0 Simplify.

The discriminant is 0 , so the equation has one solution .

Example 2 Find the number of solutions


Your Notes


Homework

Find the number of x-intercepts of the graph of y 5 2x2 1 3x 1 4.

Solution

Find the of the equation 0 5 2x2 1 3x 1 4.

b2 2 4ac 5 2 2 4( )( ) Substitute for a, for b, and for c.

5 Simplify.

The discriminant is , so the equation has . This means that the graph of y 5 2x2 1 3x 1 4 has x-intercepts.

CHECK You can use a graphing calculator to check the answer. Notice that the graph of y 5 2x2 1 3x 1 4 has intercepts.

Example 3 Find the number of x-intercepts

Checkpoint Find the number of x-intercepts of the graph of the function.

5. y 5 2x2 1 3x 2 3 6. y 5 x2 2 4x 1 4


Your Notes


Homework

Find the number of x-intercepts of the graph of y 5 2x2 1 3x 1 4.

Solution

Find the number of solutions of the equation 0 5 2x2 1 3x 1 4.

b2 2 4ac 5 3 2 2 4( 21 )( 4 ) Substitute 21 for a, 3 for b, and 4 for c.

5 25 Simplify.

The discriminant is positive , so the equation has two solutions . This means that the graph of y 5 2x2 1 3x 1 4 has two x-intercepts.

CHECK You can use a graphing calculator to check the answer. Notice that the graph of y 5 2x2 1 3x 1 4 has two intercepts.

Example 3 Find the number of x-intercepts

Checkpoint Find the number of x-intercepts of the graph of the function.

5. y 5 2x2 1 3x 2 3 6. y 5 x2 2 4x 1 4

no x-intercepts one x-intercept


Your Notes


10.8 Compare Linear, Exponential,and Quadratic ModelsGoal p Compare linear, exponential, and quadratic models.

LINEAR, EXPONENTIAL, AND QUADRATIC FUNCTIONS

Linear Exponential Quadratic Function Function Function

y 5 y 5 y 5

x

y

x

y

x

y

Your Notes

Use a graph to tell whether the ordered pairs represent a linear function, an exponential function, or a quadratic function.

a. (22, 7), (21, 1), (0, 21), (1, 1), (2, 7)

b. (22, 4), (21, 2), (0, 1), 1 1, 1 } 2 2 , 1 2, 1 } 4 2 c. (22, 5), (21, 3), (0, 1), (1, 21), (2, 23)

Solutiona. b. c.

x

y

2

6

10

12122

23 3

x

y

1

3

5

12121

23 3

x

y

2

6

10

12122

23 3

function function function

Example 1 Choose functions using sets of ordered pairs



10.8 Compare Linear, Exponential,and Quadratic ModelsGoal p Compare linear, exponential, and quadratic models.

LINEAR, EXPONENTIAL, AND QUADRATIC FUNCTIONS

Linear Exponential Quadratic Function Function Function

y 5 mx 1 b y 5 abx y 5 ax2 1 bx 1 c

x

y

x

y

x

y

Your Notes

Use a graph to tell whether the ordered pairs represent a linear function, an exponential function, or a quadratic function.

a. (22, 7), (21, 1), (0, 21), (1, 1), (2, 7)

b. (22, 4), (21, 2), (0, 1), 1 1, 1 } 2 2 , 1 2, 1 } 4 2 c. (22, 5), (21, 3), (0, 1), (1, 21), (2, 23)

Solutiona. b. c.

x

y

2

6

10

12122

23 3

x

y

1

3

5

12121

23 3

x

y

2

6

10

12122

23 3

Quadratic Exponential Linear function function function

Example 1 Choose functions using sets of ordered pairs



Use differences or ratios to tell whether the table of values represents a linear function, an exponential function, or a quadratic function.

a. x 22 21 0 1 2

y 212 28 24 0 4

Differences:

The table of values represents function.

b. x 22 21 0 1 2

y 0.25 0.5 1 2 4

Ratios: 5


Example 2 Identify functions using differences or ratios

1. Tell whether the ordered pairs represent a linear function, an exponential function, or a quadratic function: (22, 21), (21, 1), (0, 3), (1, 5), (2, 7).

2. Tell whether the table of values represents a linear function, an exponential function, or a quadratic function:

x 22 21 0 1 2

y 3 0.75 0 0.75 3



Your Notes


Use differences or ratios to tell whether the table of values represents a linear function, an exponential function, or a quadratic function.

a. x 22 21 0 1 2

y 212 28 24 0 4

Differences: 4 4 4 4

The table of values represents a linear function.

b. x 22 21 0 1 2

y 0.25 0.5 1 2 4

Ratios: 0.25

0.5 5 2 2 2 2

The table of values represents an exponential function.

Example 2 Identify functions using differences or ratios

1. Tell whether the ordered pairs represent a linear function, an exponential function, or a quadratic function: (22, 21), (21, 1), (0, 3), (1, 5), (2, 7).

linear function

2. Tell whether the table of values represents a linear function, an exponential function, or a quadratic function:

x 22 21 0 1 2

y 3 0.75 0 0.75 3

quadratic function



Your Notes


Homework

Tell whether the table of values represents a linear function, an exponential function, or a quadratic function. Then write an equation for the function.

x 22 21 0 1 2

y 32 8 2 0.5 0.125

Step 1 Determine which type of function the values in the table represent.

x 22 21 0 1 2

y 32 8 2 0.5 0.25

Ratios: 5


Step 2 Write an equation for the function. The ratio of successive y-values is , so b 5 . Find the value of a using the coordinates of a point that lies on the graph, such as (0, 2).

y 5 Write equation for function.

5 Substitute for b, for x, and for y.

5 a Solve for a.

The equation is .

Example 3 Write an equation for a function

3. Write an equation for the function in Checkpoint 2.



Your Notes


Homework

Tell whether the table of values represents a linear function, an exponential function, or a quadratic function. Then write an equation for the function.

x 22 21 0 1 2

y 32 8 2 0.5 0.125

Step 1 Determine which type of function the values in the table represent.

x 22 21 0 1 2

y 32 8 2 0.5 0.25

Ratios: 32

8 5 0.25 0.25 0.25 0.25

The table of values represents an exponential function.

Step 2 Write an equation for the exponential function. The ratio of successive y-values is 0.25 , so b 5 0.25 . Find the value of a using the coordinates of a point that lies on the graph, such as (0, 2).

y 5 abx Write equation for exponential function.

2 5 a(0.25)0 Substitute 0.25 for b, 0 for x, and 2 for y.

2 5 a Solve for a.

The equation is y 5 2(0.25)x .

Example 3 Write an equation for a function

3. Write an equation for the function in Checkpoint 2.

y 5 0.75x2



Your Notes


Quadratic function

Parabola

Axis of symmetry

Maximum value


Vertex

Minimum value

Intercept form




Quadratic function

y 5 3x2 2 2x 1 4

Parabola

x

y

1

3

5

12121

23 3

Axis of symmetry

For the parabola y 5 x2, the axis of symmetry is x 5 0.

Maximum value

x

y

1

3

5

12121

23 3

maximum


y 5 x2

Vertex

For the parabola y 5 x2, the vertex is (0, 0).

Minimum value

x

y

1

3

5

1 21 21

23 3

minimum

Intercept form

y 5 2(x 2 2)(x 1 3)





Quadratic equation

Vertex form

Discriminant


Quadratic formula




Quadratic equation

3x2 2 2x 1 4 5 0

Vertex form

y 5 2(x 2 3)2 1 6

Discriminant

b2 2 4ac; For 3x2 2 2x 1 4 5 0, the discriminant is 244.


To complete the square

for x2 1 4x add 1 4 } 2 2 2.

Quadratic formula

x 5 2b 6 Ï}

b2 2 4ac }} 2a



11.1Goal p Graph square root functions.

VOCABULARY

Radical expression

Radical function

Square root function

Parent square root function

Graph Square Root Functions

PARENT FUNCTION FOR SQUARE ROOT FUNCTIONS

The most basic square root

x

y

1

23

3

5

1 3 5

(1, 1)(0, 0)

function in the family of all square root functions, called the

, is y 5 .

The graph of the parent squareroot function is shown.

Your Notes



11.1Goal p Graph square root functions.

VOCABULARY

Radical expression An expression that contains a radical, such as a square root, cube root, or other root

Radical function A radical function includes a radical expression with the independent variable in the radicand.

Square root function A radical function that contains a square root

Parent square root function y 5 Ï}

x

Graph Square Root Functions

PARENT FUNCTION FOR SQUARE ROOT FUNCTIONS

The most basic square root

x

y

1

23

3

5

1 3 5

(1, 1)(0, 0)

function in the family of all square root functions, called the parent square root function , is y 5 Ï

}

x .

The graph of the parent squareroot function is shown.

Your Notes



Graph the function y 5 24 Ï}

x and identify its domain and range. Compare the graph with the graph of y 5 Ï

}

x .

SolutionStep 1 Make a table. Because the

x

y

1

21

23

25

27

1 3 5

square root of a negative number is , x must be nonnegative. So, the domain is .

x 0 1 2 3

y


Step 3 Draw a through the points. From either the table or the graph, you can see the range of the function is .

Step 4 Compare the graph with the graph of y 5 Ï}

x . The graph of y 5 24 Ï

}

x is vertical (by a factor of ) and a of

the graph y 5 Ï}

x .

Example 1 Graph a function of the form y 5 a Ï}

x

Graph the function y 5 Ï}

x 2 2 and identify its domain and range. Compare the graph with the graph of y 5 Ï

}

x .

SolutionTo graph the function, make a table,

x

y

1

21

23

3

1 53

then plot and connect the points. The domain is .

x 0 1 2 3

y

The range is . The graph

of y 5 Ï}

x 2 2 is a (of units

) of the graph of y 5 Ï}

x .

Example 2 Graph a function of the form y 5 Ï}

x 1 k


Your Notes



x and identify its domain and range. Compare the graph with the graph of y 5 Ï

}

x .

SolutionStep 1 Make a table. Because the

x

y

1

21

23

25

27

1 3 5

y 5 x

y 5 24 x

square root of a negative number is undefined , x must be nonnegative. So, the domain is x ≥ 0 .

x 0 1 2 3

y 0 24 25.7 26.9


Step 3 Draw a smooth curve through the points. From either the table or the graph, you can see the range of the function is y ≤ 0 .

Step 4 Compare the graph with the graph of y 5 Ï}

x . The graph of y 5 24 Ï

}

x is vertical stretch (by a factor of 4 ) and a reflection in the x-axis of

the graph y 5 Ï}

x .


x



}

x .


x

y

1

21

23

3

1 53

y 5 x

y 5 x 2 2

then plot and connect the points. The domain is x ≥ 0 .

x 0 1 2 3

y 22 21 20.6 20.3

The range is y ≥ 22 . The graph

of y 5 Ï}

x 2 2 is a vertical translation (of 2 units

down ) of the graph of y 5 Ï}

x .


x 1 k


Your Notes


1. y 5 0.25 Ï}

x

x

y

0.5

1.5

2.5

1 3 5 7

2. y 5 Ï}

x 1 4

x

y

1

3

5

121 3 5 7

Checkpoint Graph the function and identify its domain and range. Compare the graph with the graph of y 5 Ï

}

x .



}

x .


x

y

1

21

23

3

1 3212325

then plot and connect the points. To find the domain, find the values of x for which the radicand, x 1 5, is . The domain is

.

x 25 24 23 22

y

The range is . The graph of y 5 Ï}

x 1 5 is a (of units to the )

of the graph of y 5 Ï}

x .


x 2 h


Your Notes


1. y 5 0.25 Ï}

x

x

y

0.5

1.5

2.5

1 3 5 7

y 5 0.25 x

y 5 x Domain: x ≥ 0; Range: y ≥ 0The graph is a vertical shrink(by a factor of 0.25) of the

graph y 5 Ï}

x .

2. y 5 Ï}

x 1 4

x

y

1

3

5

121 3 5 7

y 5 x

y 5 x 1 4 Domain: x ≥ 0; Range: y ≥ 4

The graph is a vertical translation (of 4 units up)

of the graph y 5 Ï}

x .

Checkpoint Graph the function and identify its domain and range. Compare the graph with the graph of y 5 Ï

}

x .



}

x .


x

y

1

21

23

3

1 3212325

y 5 x

y 5 x 1 5then plot and connect the points. To find the domain, find the values of x for which the radicand, x 1 5, is nonnegative . The domain is x ≥ 25 .

x 25 24 23 22

y 0 1 1.4 1.7

The range is y ≥ 0 . The graph of y 5 Ï}

x 1 5 is a horizontal translation (of 5 units to the left )

of the graph of y 5 Ï}

x .


x 2 h


Your Notes


Homework

GRAPHS OF SQUARE ROOT FUNCTIONS

To graph a function of the form y 5 a Ï}

x 2 h 1 k, you can follow these steps.

Step 1 Sketch the graph of y 5 a Ï}

x . The graph of y 5 a Ï

}

x starts at the and passes through the point .

Step 2 Shift the graph⏐h⏐units (to the right if h is and to the left if h is ) and⏐k⏐units ( if k is positive and if k is negative).


x 2 1 1 2.

Step 1 Sketch the graph of

x

y

1

3

5

7

1 3 5 7

y 5 3 Ï}

x .

Step 2 Shift the graph⏐h⏐units horizontally and⏐k⏐units

vertically. Notice that h 5 and k 5 . Shiftthe graph and .


x 2 h 1 k

3. Graph the function y 5 Ï}

x 2 3

x

y

1

3

1 3 5 7

and identify its domain and range. Compare the graph withthe graph of y 5 Ï

}

x .

4. Identify the domain and range of the function in Example 4.



Your Notes


Homework

GRAPHS OF SQUARE ROOT FUNCTIONS

To graph a function of the form y 5 a Ï}

x 2 h 1 k, you can follow these steps.

Step 1 Sketch the graph of y 5 a Ï}

x . The graph of y 5 a Ï

}

x starts at the origin and passes through the point (1, a) .

Step 2 Shift the graph⏐h⏐units horizontally (to the right if h is positive and to the left if h is negative ) and⏐k⏐units vertically ( up if k is positive and down if k is negative).


x 2 1 1 2.

Step 1 Sketch the graph of

x

y

1

3

5

7

1 3 5 7

y 5 3 x

y 5 3 x 2 1 1 2

(0, 0)

(1, 2)

y 5 3 Ï}

x .

Step 2 Shift the graph⏐h⏐units horizontally and⏐k⏐units

vertically. Notice that h 5 1 and k 5 2 . Shiftthe graph right 1 unit and up 2 units .


x 2 h 1 k

3. Graph the function y 5 Ï}

x 2 3

x

y

1

3

1 3 5 7

y 5 x

y 5 x 2 3

and identify its domain and range. Compare the graph withthe graph of y 5 Ï

}

x .

Domain: x ≥ 3; Range: y ≥ 0;The graph of y 5 Ï

}

x 2 3 is a horizontal translation (of 3 units to the right) of the graph of y 5 Ï

}

x .

4. Identify the domain and range of the function in Example 4.

Domain: x ≥ 1; Range: y ≥ 2



Your Notes


11.2 Simplify Radical ExpressionsGoal p Simplify radical expressions.

VOCABULARY

Simplest form of a radical expression

Rationalizing the denominator

PRODUCT PROPERTY OF RADICALS

Words The square root of a product equals the of the of the factors.

Algebra Ï}

ab 5 p where a ≥ 0 and b ≥ 0

Example Ï}

9x 5 p 5


Your Notes

Simplify Ï}

12x2 .

Solution

Ï}

12x2 5 Ï}}

p p Factor using perfect square factors.

5 p p of radicals

5 Simplify.

Example 1 Use the product property of radicals


11.2 Simplify Radical ExpressionsGoal p Simplify radical expressions.

VOCABULARY

Simplest form of a radical expression A radical expression is in simplest form if: no perfect square factors other than 1 are in the radicand; no fractions are in the radicand; and no radicals appear in the denominator of a fraction.

Rationalizing the denominator The process of eliminating a radical from an expression's denominator

PRODUCT PROPERTY OF RADICALS

Words The square root of a product equals the product of the square roots of the factors.

Algebra Ï}

ab 5 Ï}

a p Ï}

b where a ≥ 0 and b ≥ 0

Example Ï}

9x 5 Ï}

9 p Ï}

x 5 3 Ï}

x


Your Notes

Simplify Ï}

12x2 .

Solution

Ï}

12x2 5 Ï}}

4 p 3 p x2 Factor using perfect square factors.

5 Ï}

4 p Ï}

3 p Ï}

x2 Product property of radicals

5 2x Ï}

3 Simplify.

Example 1 Use the product property of radicals


QUOTIENT PROPERTY OF RADICALS

Words The square root of a quotient equals the of the of the numerator

and denominator.

Algebra Î}

a } b 5 where a ≥ 0 and b > 0

Example Î}

4 } 9 5 5

a. Î}

11 } 49 5 Quotient property of radicals

5 Simplify.

b. Î}

t2 } 36 5 Quotient property of radicals

5 Simplify.

Example 3 Use the quotient property of radicals

a. Ï}

8 p Ï}

2 5 Ï}

p

5

5

b. Ï}

5x3y p 2 Ï}

x 5 Ï}}

p

5 Ï}

5 p p p

5

Example 2 Multiply radicals


Your Notes


QUOTIENT PROPERTY OF RADICALS

Words The square root of a quotient equals the quotient of the square roots of the numerator and denominator.

Algebra Î}

a } b 5 Ï

}

b

Ï}

a where a ≥ 0 and b > 0

Example Î}

4 } 9 5 Ï

}

9

Ï}

4 5 2 }

3

a. Î}

11 } 49 5 Ï

}

49

Ï}

11 Quotient property of radicals

5 7

Ï}

11 Simplify.

b. Î}

t2 } 36 5

Ï}

36

Ï}

t2 Quotient property of radicals

5 t } 6 Simplify.

Example 3 Use the quotient property of radicals

a. Ï}

8 p Ï}

2 5 Ï}

8 p 2

5 Ï}

16

5 4

b. Ï}

5x3y p 2 Ï}

x 5 2 Ï}}

5x3y p x

5 2 Ï}

5x4y

5 2 p Ï}

5 p Ï}

x4 p Ï}

y

5 2x2 Ï}

5y

Example 2 Multiply radicals


Your Notes


1. Ï}

16z4 2. 4 Ï}

mn p Ï}

5m 3. Î}

15 } 25


a. Ï}

2 } Ï

} 5 5 Ï

}

2 } Ï

} 5 p Multiply by Ï

} 5 }

Ï}

5 .

5 Product property of radicals

5 Simplify.

b. 1 } Ï

} 7r 5 1 }

Ï}

7r p Ï

} 7r }

Ï}

7r Multiply by .



5 Simplify.

Example 4 Rationalize the denominator


Your Notes


1. Ï}

16z4 2. 4 Ï}

mn p Ï}

5m 3. Î}

15 } 25

4z2 4m Ï}

5n Ï}

15 } 5


a. Ï}

2 } Ï

} 5 5 Ï

}

2 } Ï

} 5 p

Ï}

5

Ï}

5 Multiply by Ï

} 5 }

Ï}

5 .

5 Ï

}

25

Ï}


5 5

Ï}

10 Simplify.

b. 1 } Ï

} 7r 5 1 }

Ï}

7r p Ï

} 7r }

Ï}

7r Multiply by Ï

}

7r } Ï

}

7r .

5 Ï}

49r2

Ï}

7r Product property of radicals

5 Ï

}

49 p Ï}

r2

Ï}

7r Product property of radicals

5 Ï}

7r } 7r

Simplify.



Your Notes


a. 7 Ï}

5 2 Ï}

11 1 4 Ï}

5

5 Commutative property

5 Distributive property

5 Simplify.

b. 2 Ï}

2 2 Ï}

18

5 Factor using perfect square factors.


5 Simplify.


5 Simplify.

Example 5 Add and subtract radicals

4. 2 }

Ï}

5y 5. 3 Ï

}

11 1 2 Ï}

44



Your Notes


a. 7 Ï}

5 2 Ï}

11 1 4 Ï}

5

5 7 Ï}

5 1 4 Ï}

5 2 Ï}

11 Commutative property

5 (7 1 4) Ï}

5 2 Ï}


5 11 Ï}

5 2 Ï}

11 Simplify.

b. 2 Ï}

2 2 Ï}

18

5 2 Ï}

2 2 Ï}

9 p 2 Factor using perfect square factors.

5 2 Ï}

2 2 Ï}

9 p Ï}


5 2 Ï}

2 2 3 Ï}

2 Simplify.

5 (2 2 3) Ï}


5 2 Ï}

2 Simplify.

Example 5 Add and subtract radicals

4. 2 }

Ï}

5y 5. 3 Ï

}

11 1 2 Ï}

44

2 Ï

}

5y } 5y 7 Ï

}

11



Your Notes


6. Ï}

7 (2 Ï}

7 1 Ï}

3 )

7. (3 Ï}

5 1 7)2

8. (2 1 Ï}

6 )(8 2 Ï}

6 )


Multiply (4 1 Ï}

3 )(3 2 Ï}

3 ).

Solution(4 1 Ï

}

3 )(3 2 Ï}

3 )

5 1 1 1 Multiply.


5 Simplify.

5 Simplify.

Example 6 Multiply radical expressions

Homework


Your Notes


6. Ï}

7 (2 Ï}

7 1 Ï}

3 ) 14 1 Ï

}

21

7. (3 Ï}

5 1 7)2

94 1 42 Ï}

5

8. (2 1 Ï}

6 )(8 2 Ï}

6 ) 10 1 6 Ï

}

6


Multiply (4 1 Ï}

3 )(3 2 Ï}

3 ).

Solution(4 1 Ï

}

3 )(3 2 Ï}

3 )

5 12 1 4(2 Ï}

3 ) 1 3 Ï}

3 1 Ï}

3 (2 Ï}

3 ) Multiply.

5 12 2 4 Ï}

3 1 3 Ï}

3 2 Ï}


5 12 1 (24 1 3) Ï}

3 2 3 Simplify.

5 9 2 Ï}

3 Simplify.

Example 6 Multiply radical expressions

Homework


Your Notes


a. 3 Ï}

9 p 3 Ï}

23 5 3 Ï}

p

5

5

b. 3

Ï}

5x

} 8 5 }

5

}

Example 1 Use properties of radicals

Operations with Cube RootsGoal p Perform operations with cube roots.

PRODUCT PROPERTY OF CUBE ROOTS

Words The cube root of a product equals the of the .

Algebra 3 Ï}

ab 5 p

QUOTIENT PROPERTY OF CUBE ROOTS

Words The cube root of a quotient equals the of the of the numerator and denominator.

Algebra 3

Ï}

a } b 5 } , b Þ 0

property

of cube roots

Multiply.

Simplify.

property of

cube roots

Simplify.

1. 3 Ï}

1000 p 3 Ï}

10 2. 3

Ï}

264

} 216


Recall that

every real

number a has

exactly one

cube root,

written 3 Ï

} a .

Your Notes


Focus On

Operations


Homework


a. 3 Ï}

9 p 3 Ï}

23 5 3 Ï}

9 p 23

5 3 Ï}

227

5 23

b. 3

Ï}

5x

} 8 5

3 Ï}

5x }

3 Ï}

8

5 3 Ï

}

5x }

2

Example 1 Use properties of radicals

Operations with Cube RootsGoal p Perform operations with cube roots.

PRODUCT PROPERTY OF CUBE ROOTS

Words The cube root of a product equals the product of the cube roots.

Algebra 3 Ï}

ab 5 3 Ï}

a p 3 Ï}

b

QUOTIENT PROPERTY OF CUBE ROOTS

Words The cube root of a quotient equals the quotient of the cube roots of the numerator and denominator.

Algebra 3

Ï}

a } b 5

3 Ï}

a }

3 Ï}

b , b Þ 0

Product property

of cube roots

Multiply.

Simplify.

Quotient property of

cube roots

Simplify.

1. 3 Ï}

1000 p 3 Ï}

10 2. 3

Ï}

264

} 216

10 3 Ï}

10 2 2 } 3


Recall that

every real

number a has

exactly one

cube root,

written 3 Ï

} a .

Your Notes


Focus On

Operations


Homework


3 Ï}

254 2 7 3 Ï}

2

5 2 Factor using a perfect cube.

5 2 property of cube roots

5 Simplify.


5 Simplify.

Example 3 Add and subtract cube roots

7 }

3 Ï}

4 5 7

} 3 Ï}

4 p } Multiply by } .

5 } property of cube roots

5 } Simplify.


3 Ï}

4 1 3 1 3 Ï}

16m 2 5 Distributive property

5 property of cube roots

5 property of cube roots

5 Simplify.

Example 4 Multiply expressions involving cube roots

3. 3 Ï}

5k }

3 Ï}

9 4. 216

3 Ï}

6y 1 20 3 Ï}

6y

5. 9 3 Ï}

256 1 4 3 Ï}

7 6. 1 6 1 3 Ï}

25 2 1 1 2 3 Ï}

5 2



Use the fact that

3 Ï}

21 5 21 to

rewrite 3 Ï}

254

as a cube root

with a positive

radicand.

Homework

Your Notes


3 Ï}

254 2 7 3 Ï}

2

5 3 Ï}

227 • 2 2 7 3 Ï}

2 Factor using a perfect cube.

5 3 Ï}

227 • 3 Ï}

2 2 7 3 Ï}

2 Product property of cube roots

5 23 3 Ï}

2 2 7 3 Ï}

2 Simplify.

5 1 23 2 7 2 3 Ï}


5 210 3 Ï}

2 Simplify.

Example 3 Add and subtract cube roots

7 }

3 Ï}

4 5 7

} 3 Ï}

4 p

3 Ï}

2 }

3 Ï}

2 Multiply by

3 Ï}

2 }

3 Ï}

2 .

5 7

3 Ï}

2 }

3 Ï}

8 Product property of cube roots

5 7

3 Ï}

2 }

2 Simplify.


3 Ï}

4 1 3 1 3 Ï}

16m 2 5 3

3 Ï}

4 1 3 Ï}

4 • 3 Ï}

16m Distributive property

5 3 3 Ï}

4 1 3 Ï}

64m Product property of cube roots

5 3 3 Ï}

4 1 3 Ï}

64 • 3 Ï}

m Product property of cube roots

5 3 3 Ï}

4 1 4 3 Ï}

m Simplify.

Example 4 Multiply expressions involving cube roots

3. 3 Ï}

5k }

3 Ï}

9 4. 216

3 Ï}

6y 1 20 3 Ï}

6y

3 Ï}

15k } 3 4 3 Ï

}

6y

5. 9 3 Ï}

256 1 4 3 Ï}

7 6. 1 6 1 3 Ï}

25 2 1 1 2 3 Ï}

5 2 214

3 Ï}

7 1 2 6 3 Ï}

5 1 3 Ï}

25



Use the fact that

3 Ï}

21 5 21 to

rewrite 3 Ï}

254

as a cube root

with a positive

radicand.

Homework

Your Notes


11.3 Solve Radical EquationsGoal p Solve radical equations.

VOCABULARY

Radical equation

Extraneous solution

SQUARING BOTH SIDES OF AN EQUATION

Words If two expressions are equal, then their squares are .

Algebra If a 5 b, then .

Example If Ï}

x 5 4, then .

Solve 3 Ï}

x 1 1 2 15 5 26.

Solution

3 Ï}

x 1 1 2 15 5 26 Write original equation.

3 Ï}

x 1 1 5 Add to each side.

Ï}

x 1 1 5 Divide each side by .

5 Square each side.

5 Simplify.

x 5 Subtract from each side.

The solution is .

Example 1 Solve a radical equation

Check the solution by substituting it in the original equation.

Your Notes



11.3 Solve Radical EquationsGoal p Solve radical equations.

VOCABULARY

Radical equation An equation that contains a radical expression with a variable in the radicand

Extraneous solution Squaring both sides of the equation a 5 b can result in a solution of a2 5 b2 that is not a solution of the original equation. Such a solution is called an extraneous solution.

SQUARING BOTH SIDES OF AN EQUATION

Words If two expressions are equal, then their squares are equal .

Algebra If a 5 b, then a2 5 b2 .

Example If Ï}

x 5 4, then ( Ï}

x )2 5 42 .

Solve 3 Ï}

x 1 1 2 15 5 26.

Solution

3 Ï}

x 1 1 2 15 5 26 Write original equation.

3 Ï}

x 1 1 5 9 Add 15 to each side.

Ï}

x 1 1 5 3 Divide each side by 3 .

( Ï}

x 1 1 )2 5 32 Square each side.

x 1 1 5 9 Simplify.

x 5 8 Subtract 1 from each side.

The solution is 8 .

Example 1 Solve a radical equation

Check the solution by substituting it in the original equation.

Your Notes



1. Solve Ï}

4x 2 19 2 2 5 5.


Solve Ï}

3x 2 3 5 Ï}

2x 1 8 .

Solution

Ï}

3x 2 3 5 Ï}

2x 1 8 Write original equation.

5 Square each side.

5 Simplify.


x 5 Add to each side.

The solution is .

Example 2 Solve an equation with a radical on both sides

2. Ï}

5x 2 4 5 Ï}

3x 1 20 3. Ï}

13 2 x 5 Ï}

3x 2 15



To solve a radical equation that contains two radical expressions, be sure that each side of the equation has only one radical expression before squaring each side.

Your Notes


1. Solve Ï}

4x 2 19 2 2 5 5.

x 5 17


Solve Ï}

3x 2 3 5 Ï}

2x 1 8 .

Solution

Ï}

3x 2 3 5 Ï}


( Ï}

3x 2 3 )2 5 ( Ï}

2x 1 8 )2 Square each side.

3x 2 3 5 2x 1 8 Simplify.

x 2 3 5 8 Subtract 2x from each side.

x 5 11 Add 3 to each side.

The solution is 11 .

Example 2 Solve an equation with a radical on both sides

2. Ï}

5x 2 4 5 Ï}

3x 1 20 3. Ï}

13 2 x 5 Ï}

3x 2 15

x 5 12 x 5 7



To solve a radical equation that contains two radical expressions, be sure that each side of the equation has only one radical expression before squaring each side.

Your Notes


Solve x 5 Ï}

2x 1 15 .

Solution

x 5 Ï}


5 Square each side.

5 Simplify.

5 0 Write in standard form.

( )( ) 5 0 Factor.

( ) 5 0 or ( ) 5 0

x 5 or x 5

CHECK Check and in the original equation.

x 5 : x 5 :

0 Ï}}

2( ) 1 15 0 Ï}}

2( ) 1 15

5 5 ✓ 23 5 ✗

Because does not check in the original equation, it is an . The only solution to the equation is .

Example 3 Solve an equation with an extraneous solution

Homework

4. Ï}

20 2 x 5 x 5. Ï}

7 1 6x 5 x



Your Notes


Solve x 5 Ï}

2x 1 15 .

Solution

x 5 Ï}


x2 5 ( Ï}

2x 1 15 )2 Square each side.

x2 5 2x 1 15 Simplify.

x2 2 2x 2 15 5 0 Write in standard form.

( x 2 5 )( x 1 3 ) 5 0 Factor.

( x 2 5 ) 5 0 or ( x 1 3 ) 5 0

x 5 5 or x 5 23

CHECK Check 5 and 23 in the original equation.

x 5 5 : x 5 23 :

5 0 Ï}}

2( 5 ) 1 15 23 0 Ï}}

2( 23 ) 1 15

5 5 5 ✓ 23 5 3 ✗

Because 23 does not check in the original equation, it is an extraneous solution . The only solution to the equation is 5 .

Example 3 Solve an equation with an extraneous solution

Homework

4. Ï}

20 2 x 5 x 5. Ï}

7 1 6x 5 x

x 5 4 x 5 7



Your Notes


11.4 Apply the Pythagorean Theorem and its ConverseGoal p Use the Pythagorean theorem and its converse.

VOCABULARY

Hypotenuse

Legs of a right triangle

Pythagorean theorem


THE PYTHAGOREAN THEOREM

Words If a triangle is a right triangle, a

b

cthen the equals

the .

Algebra

The lengths of the legs of a right triangle are a 5 8 and b 5 15. Find c.

Solutionc2 5 a2 1 b2 Pythagorean theorem

c2 5 2 1 2 Substitute for a and for b.

c2 5 Simplify.

c 5 Take positive square root of each side.

The side length of c is .

Example 1 Use the Pythagorean theorem

Your Notes


11.4 Apply the Pythagorean Theorem and its ConverseGoal p Use the Pythagorean theorem and its converse.

VOCABULARY

Hypotenuse The hypotenuse of a right triangle is the side opposite the right angle.

Legs of a right triangle The two sides that form the right angle

Pythagorean theorem If a triangle is a right triangle, then the sum of the squares of the lengths of the legs equals the square of the length of the hypotenuse.


THE PYTHAGOREAN THEOREM

Words If a triangle is a right triangle, a

b

cthen the sum of the squares of the lengths of the legs equals the square of the length of the hypotenuse.

Algebra a2 1 b2 5 c2

The lengths of the legs of a right triangle are a 5 8 and b 5 15. Find c.

Solutionc2 5 a2 1 b2 Pythagorean theorem

c2 5 8 2 1 15 2 Substitute 8 for a and 15 for b.

c2 5 289 Simplify.

c 5 17 Take positive square root of each side.

The side length of c is 17 .


Your Notes


1. The lengths of the legs of a right triangle are a 5 7 and b 5 9. Find c.

2. The length of a leg of a right triangle is a 5 20 and the length of the hypotenuse is c 5 52. Find b.


A right triangle has one leg that is 4 inches longer than the other leg. The hypotenuse is Ï

}

106 inches. Find the unknown lengths.

SolutionSketch a right triangle and label the sides with their lengths. Let x be the length of the shorter leg.

a2 1 b2 5 c2 Pythagorean theorem

2 1 ( )2 5 ( )2 Substitute.

1 5 Simplify.


5 0 Factor.

( ) 5 0 or ( ) 5 0 Zero-product property


Because length is nonnegative, the solution x 5 does not make sense. The legs have lengths of inches and 1 4 5 inches.


Your Notes



1. The lengths of the legs of a right triangle are a 5 7 and b 5 9. Find c.

c 5 Ï}

130

2. The length of a leg of a right triangle is a 5 20 and the length of the hypotenuse is c 5 52. Find b.

b 5 48


A right triangle has one leg that is 4 inches longer than the other leg. The hypotenuse is Ï

}

106 inches. Find the unknown lengths.

SolutionSketch a right triangle and label

x 1 4

x 106

the sides with their lengths. Let x be the length of the shorter leg.

a2 1 b2 5 c2 Pythagorean theorem

x 2 1 ( x 1 4 )2 5 ( Ï}

106 )2 Substitute.

x2 1 x2 1 8x 1 16 5 106 Simplify.

2x2 1 8x 2 90 5 0 Write in standard form.

2(x 1 9)(x 2 5) 5 0 Factor.

( x 1 9 ) 5 0 or ( x 2 5 ) 5 0 Zero-product property


Because length is nonnegative, the solution x 5 29 does not make sense. The legs have lengths of 5 inches and 5 1 4 5 9 inches.


Your Notes



3. A right triangle has one leg that is 2 centimeters shorter than the other leg. The length of the hypotenuse is 10 centimeters. Find the unknown lengths.


Tell whether the triangle with the given side lengths is a right triangle.

a. 10, 11, 15 b. 3, 4, 5

102 1 112 0 152 32 1 42 0 52

1 0 1 0

The triangle a The triangle aright triangle. right triangle.

Example 3 Determine right triangles

Homework

Checkpoint Tell whether the triangle with the given side lengths is a right triangle.

4. 9, 40, 41 5. 10, 15, 18

6. A triangular mirror has side lengths of 1.2 meters, 1.6 meters, and 2 meters. Is the mirror a right triangle? Explain.

CONVERSE OF THE PYTHAGOREAN THEOREM

If a triangle has side lengths a, b, and c such that , then the triangle is a triangle.


Your Notes


3. A right triangle has one leg that is 2 centimeters shorter than the other leg. The length of the hypotenuse is 10 centimeters. Find the unknown lengths.

6 cm, 8 cm


Tell whether the triangle with the given side lengths is a right triangle.

a. 10, 11, 15 b. 3, 4, 5

102 1 112 0 152 32 1 42 0 52

100 1 121 0 225 9 1 16 0 25

221 Þ 225 25 5 25

The triangle is not a The triangle is aright triangle. right triangle.

Example 3 Determine right triangles

Homework

Checkpoint Tell whether the triangle with the given side lengths is a right triangle.

4. 9, 40, 41 5. 10, 15, 18

The triangle is a The triangle is not aright triangle. right triangle.

6. A triangular mirror has side lengths of 1.2 meters, 1.6 meters, and 2 meters. Is the mirror a right triangle? Explain.

Yes, The sum of the squares of the lengths of the legs equals the square of the length of the hypotenuse.

CONVERSE OF THE PYTHAGOREAN THEOREM

If a triangle has side lengths a, b, and c such that a2 1 b2 5 c2 , then the triangle is a right triangle.


Your Notes


11.5 Apply the Distance andMidpoint FormulasGoal p Use the distance and midpoint formulas.

VOCABULARY

Distance formula

Midpoint

Midpoint formula

Find the distance between (4, 23) and (27, 2).

Let (x1, y1) 5 (4, 23) and (x2, y2) 5 (27, 2).

d 5 Ï}}}

(x2 2 x1)2 1 (y2 2 y1)2 Distance formula

5 Ï}}}}

( 2 )2 1 ( 2 )2 Substitute.

5 Ï}}

( )2 1 ( )2 5 Simplify.

The distance between the points is units.

Example 1 Find the distance between two points

THE DISTANCE FORMULA

x

y

(x1, y1)d

�x2 2 x1�

�y2 2 y1�

(x2, y2)The distance between any two points (x1, y1) and (x2, y2) is

.

Your Notes



11.5 Apply the Distance andMidpoint FormulasGoal p Use the distance and midpoint formulas.

VOCABULARY

Distance formula The distance between any two points (x1, y1) and (x2, y2) is

d 5 Ï}}}

(x2 2 x1)2 1 (y2 2 y1)2 .

Midpoint The midpoint of a line segment is the point on the segment that is equidistant from the endpoints.

Midpoint formula The midpoint M of the line segment with endpoints A(x1, y1) and B(x2, y2) is

M 1 x1 1 x2 } 2 ,

y1 1 y2 } 2 2 .

Find the distance between (4, 23) and (27, 2).

Let (x1, y1) 5 (4, 23) and (x2, y2) 5 (27, 2).

d 5 Ï}}}


5 Ï}}}}

( 27 2 4 )2 1 ( 2 2 (23) )2 Substitute.

5 Ï}}

( 211 )2 1 ( 5 )2 5 Ï}

146 Simplify.

The distance between the points is Ï}

146 units.

Example 1 Find the distance between two points

THE DISTANCE FORMULA

x

y

(x1, y1)d

�x2 2 x1�

�y2 2 y1�

(x2, y2)The distance between any two points (x1, y1) and (x2, y2) is

d 5 Ï}}}

(x2 2 x1)2 1 (y2 2 y1)2 .

Your Notes




1. Find the distance 2. The distance betweenbetween (2, 23) (21, 2) and (3, b) is

and (5, 1). Ï}

41 units. Find the value of b.


The distance between (5, a) and (9, 6) is 4 Ï}

2 units. Find the value of a.

SolutionUse the distance formula with d 5 4 Ï

}

2 . Let (x1, y1) 5 (5, a) and (x2, y2) 5 (9, 6).

d 5 Ï}}}


5 Ï}}}

( 2 )2 1 ( 2 )2 Substitute.

5 Ï}}}

Multiply.

5 Ï}}

Simplify.

5 Square each side.


0 5 Factor.


a 5 or a 5 Solve for a.

The value of a is or .

Example 2 Find a missing coordinateYour Notes



1. Find the distance 2. The distance betweenbetween (2, 23) (21, 2) and (3, b) is

and (5, 1). Ï}

41 units. Find the 5 value of b.

23 or 7


The distance between (5, a) and (9, 6) is 4 Ï}

2 units. Find the value of a.

SolutionUse the distance formula with d 5 4 Ï

}

2 . Let (x1, y1) 5 (5, a) and (x2, y2) 5 (9, 6).

d 5 Ï}}}


4 Ï}

2 5 Ï}}}

( 9 2 5 )2 1 ( 6 2 a )2 Substitute.

4 Ï}

2 5 Ï}}}

16 1 a2 2 12a 1 36 Multiply.

4 Ï}

2 5 Ï}}

a2 2 12a 1 52 Simplify.

32 5 a2 2 12a 1 52 Square each side.

0 5 a2 2 12a 1 20 Write in standard form.

0 5 (a 2 10)(a 2 2) Factor.

a 2 10 5 0 or a 2 2 5 0 Zero-product property

a 5 10 or a 5 2 Solve for a.

The value of a is 10 or 2 .

Example 2 Find a missing coordinateYour Notes


Homework

Find the midpoint of the line segment with endpoints (23, 7) and (21, 11).

SolutionLet (x1, y1) 5 (23, 7) and (x2, y2) 5 (21, 11).

1 x1 1 x2 } 2 , y1 1 y2 } 2 2 5 1 1

, 1

2 5 ( , )

The midpoint is ( , ).

Example 3 Find the midpoint between two points

Checkpoint Find the midpoint of the line segment with the given endpoints.

3. (1, 22), (5, 24) 4. (5, 12), (13, 8)

THE MIDPOINT FORMULA

x

yB(x2, y2)

M ,

x2x1

y1

y2

x1 1 x2

2

y1 1 y2

2

y1 1 y2

2 ))A(x1, y1)

x1 1 x2

2

The midpoint M of the line segment with endpoints A(x1, y1) and B(x2, y2) is

M 1 , 2 .


Your Notes


Homework

Find the midpoint of the line segment with endpoints (23, 7) and (21, 11).

SolutionLet (x1, y1) 5 (23, 7) and (x2, y2) 5 (21, 11).

1 x1 1 x2 } 2 , y1 1 y2 } 2 2 5 1 23

2

1 21,

7

2

1 11 2

5 ( 22 , 9 )

The midpoint is ( 22 , 9 ).

Example 3 Find the midpoint between two points

Checkpoint Find the midpoint of the line segment with the given endpoints.

3. (1, 22), (5, 24) 4. (5, 12), (13, 8)

(3, 23) (9, 10)

THE MIDPOINT FORMULA

x

yB(x2, y2)

M ,

x2x1

y1

y2

x1 1 x2

2

y1 1 y2

2

y1 1 y2

2 ))A(x1, y1)

x1 1 x2

2

The midpoint M of the line segment with endpoints A(x1, y1) and B(x2, y2) is

M 1 x1 1 x2 }

2 ,

y1 1 y2 } 2 2 .


Your Notes



Radical expression



Radical equation

Hypotenuse

Radical function



Extraneous solution





Radical expression

Ï}

x 2 2


y 5 Ï}

x 2 4 2 2


2x Ï}

3

Radical equation

x 5 Ï}

2x 1 15

Hypotenuse

In the triangle, side c is the hypotenuse.

a

b

c

Radical function

y 5 Ï}

x 2 2


y 5 Ï}

x


Ï}

2 } Ï

}

5 p Ï

}

5 } Ï

}

5 5 Ï

}

10 } 5

Extraneous solution

23 is an extraneous solution of x 5 Ï

}

2x 1 15


In the triangle, sides a and b are the legs.

a

b

c




Pythagorean theorem

Midpoint

Distance formula

Midpoint formula




Pythagorean theorem

a2 1 b2 5 c2

Midpoint

The midpoint of the line segment with endpoints (23, 7) and (21, 11) is (22, 9).

Distance formula

d 5 Ï}}}

(x2 2 x1)2 1 (y2 2 y1)2

Midpoint formula

1 x1 1 x2 } 2 ,

y1 1 y2 } 2 2



Model Inverse VariationGoal p Write and graph inverse variation equations.

VOCABULARY

Inverse variation

Constant of variation

Hyperbola

Branches of a hyperbola

Asymptotes of a hyperbola

Tell whether the equation represents direct variation, inverse variation, or neither.

a. xy 5 22 Write original equation.

y 5 Divide each side by .

Because xy 5 22 be written in the form y 5 a } x , xy 5 22 represents . The constant of variation is .

b. y }

4 5 x Write original equation.

y 5 Multiply each side by .

Because y } 4 5 x be written in the form y 5 ax,

y } 4 5 x represents .

Example 1 Identify direct and inverse variation


12.1

Your Notes


Model Inverse VariationGoal p Write and graph inverse variation equations.

VOCABULARY

Inverse variation The variables x and y show inversevariation if y 5 a }

x and a Þ 0.

Constant of variation The nonzero number a in the equation y 5 a }

x

Hyperbola The graph of the inverse variation equation y 5 a }

x (a Þ 0) is a hyperbola.

Branches of a hyperbola The two symmetrical parts of a hyperbola

Asymptotes of a hyperbola A line that the hyperbola approaches but does not intersect

Tell whether the equation represents direct variation, inverse variation, or neither.

a. xy 5 22 Write original equation.

y 5 22 } x Divide each side by x .

Because xy 5 22 can be written in the form y 5 a } x , xy 5 22 represents inverse variation . The constant of variation is 22 .

b. y }

4 5 x Write original equation.

y 5 4x Multiply each side by 4 .

Because y } 4 5 x can be written in the form y 5 ax,

y } 4 5 x represents direct variation .

Example 1 Identify direct and inverse variation


12.1

Your Notes


Checkpoint Tell whether the equation represents direct variation, inverse variation, or neither.

1. y }

25 5 x 2. y 5 3x 2 1 3. xy 5 8

Graph y 5 22 } x . Then find the domain and range for the function.

Step 1 Make a table by choosing several integer values of x and finding the values of y. Then plot the points. To see how the function behaves for values of x very close to 0 and very far from 0, make a second table for such values and plot the points.

x y

24

22

21

0

1

2

4

x y

210

25

20.5

20.2

0.2

0.5

5

10

x

y

3

9

3 92323

29

29

Step 2 Connect the points in Quadrant II by drawing a smooth curve through them. Repeat for points in Quadrant IV.

Both the domain and range for the function are all real numbers except 0.

Example 2 Graph an inverse variation equation


Note that y is undefined when x 5 0. There is no point (0, y) on the graph. Also, there is no value of x for which y 5 0, so there is no point (x, 0) on the graph. Neither the domain nor the range of an inverse variation function includes 0.

Your Notes


Checkpoint Tell whether the equation represents direct variation, inverse variation, or neither.

1. y }

25 5 x 2. y 5 3x 2 1 3. xy 5 8

direct neither inversevariation variation

Graph y 5 22 } x . Then find the domain and range for the function.

Step 1 Make a table by choosing several integer values of x and finding the values of y. Then plot the points. To see how the function behaves for values of x very close to 0 and very far from 0, make a second table for such values and plot the points.

x y

24 0.5

22 1

21 2

0 undef.

1 22

2 21

4 20.5

x y

210 0.2

25 0.4

20.5 4

20.2 10

0.2 210

0.5 24

5 20.4

10 20.2

x

y

3

9

3 92323

29

29

Step 2 Connect the points in Quadrant II by drawing a smooth curve through them. Repeat for points in Quadrant IV.

Both the domain and range for the function are all real numbers except 0.

Example 2 Graph an inverse variation equation


Note that y is undefined when x 5 0. There is no point (0, y) on the graph. Also, there is no value of x for which y 5 0, so there is no point (x, 0) on the graph. Neither the domain nor the range of an inverse variation function includes 0.

Your Notes


GRAPHS OF DIRECT VARIATION AND INVERSE VARIATION EQUATIONS

Direct Variation

x

y

x

y

y 5 ax, a > 0 y 5 ax, a < 0

Inverse Variation

x

y

x

y

y 5 a } x , a > 0 y 5 a } x , a < 0

The variables x and y vary inversely, and y 5 24 when x 5 6. Write an inverse variation equation that relates x and y. Find the value of y when x 5 3.

Solution

Because y varies with x, the equation has the form y 5 a } x . Use the fact that x 5 6 and y 5 24 to find

the value of a.

y 5 a } x Write inverse variation equation.

5 a Substitute for x and for y.

5 a Multiply each side by .

An equation that relates x and y is y 5 .

When x 5 3, y 5 5 .

Example 3 Use an inverse variation equation


Your Notes


GRAPHS OF DIRECT VARIATION AND INVERSE VARIATION EQUATIONS

Direct Variation

x

y

x

y

y 5 ax, a > 0 y 5 ax, a < 0

Inverse Variation

x

y

x

y

y 5 a } x , a > 0 y 5 a } x , a < 0

The variables x and y vary inversely, and y 5 24 when x 5 6. Write an inverse variation equation that relates x and y. Find the value of y when x 5 3.

Solution

Because y varies inversely with x, the equation has the form y 5 a } x . Use the fact that x 5 6 and y 5 24 to find

the value of a.

y 5 a } x Write inverse variation equation.

24 5 a

6 Substitute 6 for x and 24 for y.

224 5 a Multiply each side by 6 .

An equation that relates x and y is y 5 224 } x .

When x 5 3, y 5 3

224 5 28 .

Example 3 Use an inverse variation equation


Your Notes


Tell whether the ordered pairs (25, 1.2), (22, 3), (1.5, 24), (8, 20.75), (10, 20.6) represent inverse variation. If so, write the inverse variation equation.

SolutionFind the products xy for all pairs (x, y):

25(1.2) 5 , 22(3) 5 , 1.5(24) 5 ,

8(20.75) 5 , 10(20.6) 5

The products are equal to the same number, . So, .

The inverse variation equation is xy 5 , or y 5 .

Example 4 Write an inverse variation equation

4. Graph y 5 3 } x . Then find the

domain and range.

x

y

3

9

3 9

29

2329 23

5. The variables x and y vary inversely, and y 5 5 when x 5 23. Write an inverse variation equation that relates x and y. Then find the value of y when x 5 9.

6. Tell whether the ordered pairs (220, 23), (212, 25), (10, 6), (15, 4), (40, 1.5) represent inverse variation. If so, write the inverse variation equation.


Homework


Your Notes


Tell whether the ordered pairs (25, 1.2), (22, 3), (1.5, 24), (8, 20.75), (10, 20.6) represent inverse variation. If so, write the inverse variation equation.

SolutionFind the products xy for all pairs (x, y):

25(1.2) 5 26 , 22(3) 5 26 , 1.5(24) 5 26 ,

8(20.75) 5 26 , 10(20.6) 5 26

The products are equal to the same number, 26 . So, y varies inversely with x .

The inverse variation equation is xy 5 26 , or y 5 26 } x .

Example 4 Write an inverse variation equation

4. Graph y 5 3 } x . Then find the

domain and range.

x

y

3

9

3 9

29

2329 23

Both the domain and range are all real numbers except zero.

5. The variables x and y vary inversely, and y 5 5 when x 5 23. Write an inverse variation equation that relates x and y. Then find the value of y when x 5 9.

y 5 215 } x ; y 5 2 5 } 3

6. Tell whether the ordered pairs (220, 23), (212, 25), (10, 6), (15, 4), (40, 1.5) represent inverse variation. If so, write the inverse variation equation.

y varies inversely with x;

y 5 60 } x


Homework


Your Notes


12.2 Graph Rational FunctionsGoal p Graph rational functions.

VOCABULARY

Rational function

PARENT RATIONAL FUNCTION

The function y 5 1 } x is the

x

y

y 5 1x for any rational

function whose numerator has degree 0 or 1 and whose denominator has degree 1. The function and its graph has the following characteristics:

• The domain and range are all real numbers.

• The horizontal asymptote is the -axis. The vertical asymptote is the -axis.

The graph of y 5 21 } 2x is a vertical

x

y

y 52 12x y 5 1x with a reflection in the

of the graph of y 5 1 } x .

Example 1 Compare graph of y 5 a } x with graph of y 5 1 } x


Your Notes


12.2 Graph Rational FunctionsGoal p Graph rational functions.

VOCABULARY

Rational function A rational function has a rule given by a fraction whose numerator and denominator are polynomials and whose denominator is not 0.

PARENT RATIONAL FUNCTION

The function y 5 1 } x is the

x

y

y 5 1x parent function for any rational

function whose numerator has degree 0 or 1 and whose denominator has degree 1. The function and its graph has the following characteristics:

• The domain and range are all nonzero real numbers.

• The horizontal asymptote is the x -axis. The vertical asymptote is the y -axis.

The graph of y 5 21 } 2x is a vertical

x

y

y 52 12x y 5 1x shrink with a reflection in the

x-axis of the graph of y 5 1 } x .

Example 1 Compare graph of y 5 a } x with graph of y 5 1 } x


Your Notes


1. Identify the domain and range

x

y

3

1

1

3

y 5 14x

y 5 1x

of y 5 1 } 4x . Compare the graph

with the graph of y 5 1 } x .


Graph y 5 1 } x 2 2 and identify its domain and range.

Compare the graph with the graph of y 5 1 } x .

SolutionGraph the function using a table x y

22

21

20.5

0

0.5

1

2

of values. The domain is all real numbers except . The range is all real numbers except .

The graph of y 5 1 } x 2 2 is a

translation (of units

) of the graph of y 5 1 } x .

y

x

3

1

1 32121

23

25

23

Example 2 Graph y 5 1 } x 1 k


Your Notes


1. Identify the domain and range

x

y

3

1

1

3

y 5 14x

y 5 1x

of y 5 1 } 4x . Compare the graph

with the graph of y 5 1 } x .

The domain is all real numbers except 0. The range is all real numbersexcept 0. The graph is a

vertical shrink of the graph of y 5 1 } x .





22 22.5

21 23

20.5 24

0 undef.

0.5 0

1 21

2 21.5

of values. The domain is all real numbers except 0 . The range is all real numbers except 22 .


vertical translation (of 2 units

down ) of the graph of y 5 1 } x .

y

x

3

1

1 32121

23

25

23

y 5 1x

y 5 22 1x

Example 2 Graph y 5 1 } x 1 k


Your Notes


2. Graph y 5 1 } x 1 2 and identify

x

y

1

3

5

1 323

23

2121

its domain and range. Compare the graph with the graph of

y 5 1 } x .





25

24

23.5

23

22.5

22

21

of values. The domain is all real numbers except . The range is all real numbers except .


translation (of units ) of the graph

of y 5 1 } x .

x

y

1

3

5

12121

23

2325

Example 3 Graph y 5 1 } x 2 h


Your Notes



x

y

1

3

5

1 323

23

2121

y 5 1x

y 5 12 1x


y 5 1 } x .

The domain is all real numbers except 0. The range is all real numbers except 2. The graph is a vertical translation (of 2 units up) of the graph

of y 5 1 } x .





25 20.5

24 21

23.5 22

23 undef.

22.5 2

22 1

21 0.5

of values. The domain is all real numbers except 23 . The range is all real numbers except 0 .


horizontal translation (of 3 units to the left ) of the graph

of y 5 1 } x .

x

y

1

3

5

12121

23

2325

y 5 1

x � 3y 5 1

x

Example 3 Graph y 5 1 } x 2 h


Your Notes



x

y

12121

23

23

1

3

3


y 5 1 } x .


GRAPH OF y 5 1 } x 2 h

1 k

x

y

y 5 k

x 5 h

The function y 5 a } x 2 h 1 k is

a that has the following characteristics:

• If⏐a⏐> 1, the graph is a vertical

of the graph of y 5 1 } x .

If 0 <⏐a⏐< 1, the graph is a vertical of the

graph of y 5 1 } x . If⏐a⏐< 0, the graph is a reflection in

the of the graph of y 5 1 } x .

• The horizontal asymptote is y 5 . The vertical asymptote is x 5 .

The domain of the function is all real numbers except x 5 . The range is all real numbers except y 5 .


Your Notes



x

y

12121

23

23

1

3

3

y 5 1x

y 5 1x � 1 its domain and range. Compare the graph with the graph of

y 5 1 } x .

The domain is all real numbers except 1. The range is all real numbers except 0. The graph is a horizontal translation (of 1 unit to the right) of

the graph of y 5 1 } x .


GRAPH OF y 5 1 } x 2 h

1 k

x

y

y 5 k

x 5 h

The function y 5 a } x 2 h 1 k is

a hyperbola that has the following characteristics:

• If⏐a⏐> 1, the graph is a vertical

stretch of the graph of y 5 1 } x .

If 0 <⏐a⏐< 1, the graph is a vertical shrink of the

graph of y 5 1 } x . If⏐a⏐< 0, the graph is a reflection in

the x-axis of the graph of y 5 1 } x .

• The horizontal asymptote is y 5 k . The vertical asymptote is x 5 h .

The domain of the function is all real numbers except x 5 h . The range is all real numbers except y 5 k .


Your Notes


Graph y 5 2 } x 2 3

1 4.

SolutionStep 1 Identify the asymptotes of the graph. The vertical

asymptote is x 5 . The horizontal asymptote is y 5 .

Step 2 Plot several points on each side of the asymptote.

Step 3 Graph two branches that pass through the plotted points and approach the .

x

y

9

15

3

92329 3 15

Example 4 Graph y 5 a } x 2 h 1 k

4. Graph y 5 3 } x 1 2 2 1.

x23

29

29 9

y

3

9

323


Homework


Your Notes


Graph y 5 2 } x 2 3

1 4.

SolutionStep 1 Identify the asymptotes of the graph. The vertical

asymptote is x 5 3 . The horizontal asymptote is y 5 4 .

Step 2 Plot several points on each side of the vertical asymptote.

Step 3 Graph two branches that pass through the plotted points and approach the asymptotes .

x

y

9

15

3

92329

2 x 2 3

y 5 1 4

3 15

Example 4 Graph y 5 a } x 2 h 1 k

4. Graph y 5 3 } x 1 2 2 1.

x23

29

29 9

3 x 1 2

y

3

9

y 5 2 1

323


Homework


Your Notes


12.3 Divide PolynomialsGoal p Divide polynomials.

1. Divide (12x3 1 9x2 2 3x) by 3x.


Divide 10x3 2 25x2 1 15x by 5x.

SolutionMethod 1: Write the division as a fraction.

(10x3 2 25x2 1 15x) 4 5x

5 Write as a fraction.

5 2 1 Divide each term by .

5 Simplify.

Method 2: Use long division.

Think:

10x3 4 5x 5 ?

Think:

225x2 4 5x 5 ?

Think:

15x 4 5x 5 ?

2 1

5x qww 10x3 2 25x2 1 15x

(10x3 2 25x2 1 15x) 4 5x 5

Example 1 Divide a polynomial by a monomial

To check your answer, multiply the quotient by the divisor.


Your Notes


12.3 Divide PolynomialsGoal p Divide polynomials.

1. Divide (12x3 1 9x2 2 3x) by 3x.

4x2 1 3x 2 1


Divide 10x3 2 25x2 1 15x by 5x.

SolutionMethod 1: Write the division as a fraction.

(10x3 2 25x2 1 15x) 4 5x

5 5x

10x3 2 25x2 1 15x Write as a fraction.

5 5x

10x3 2

5x

25x2 1

5x

15x Divide each term by 5x .

5 2x2 2 5x 1 3 Simplify.

Method 2: Use long division.

Think:

10x3 4 5x 5 ?

Think:

225x2 4 5x 5 ?

Think:

15x 4 5x 5 ?

2x2 2 5x 1 3

5x qww 10x3 2 25x2 1 15x

(10x3 2 25x2 1 15x) 4 5x 5 2x2 2 5x 1 3

Example 1 Divide a polynomial by a monomial

To check your answer, multiply the quotient by the divisor.


Your Notes


Divide 4x2 2 4x 2 3 by 2x 1 1.

SolutionStep 1 Divide the first term of 4x2 2 4x 2 3 by the first

term of 2x 1 1.

2x 1 1 qww 4x2 2 4x 2 3 Think: 4x2 4 2x 5 ?

Multiply and .

Subtract.

Step 2 Bring down . Then divide the first term of by the first term of 2x 1 1.

2x 1 1 qww 4x2 2 4x 2 3

Think: 26x 4 2x 5 ?

Multiply and .

Subtract.

(4x2 2 4x 2 3) 4 (2x 1 1) 5

Example 2 Divide a polynomial by a binomial

Divide 2x2 1 9x 2 6 by 2x 1 3.

Solution

2x 1 3 qww 2x2 1 9x 2 6

Multiply and .

Subtract . Bring down .

Multiply and .

Subtract .

(2x2 1 9x 2 6) 4 (2x 1 3) 5



Your Notes


Divide 4x2 2 4x 2 3 by 2x 1 1.

SolutionStep 1 Divide the first term of 4x2 2 4x 2 3 by the first

term of 2x 1 1.

2x

2x 1 1 qww 4x2 2 4x 2 3 Think: 4x2 4 2x 5 ?

4x2 1 2x Multiply 2x and 2x 1 1 .

26x Subtract.

Step 2 Bring down 23 . Then divide the first term of 26x 2 3 by the first term of 2x 1 1.

2x 2 3

2x 1 1 qww 4x2 2 4x 2 3

4x2 1 2x

26x 2 3 Think: 26x 4 2x 5 ?

26x 2 3 Multiply 23 and 2x 1 1 .

0 Subtract.

(4x2 2 4x 2 3) 4 (2x 1 1) 5 2x 2 3


Divide 2x2 1 9x 2 6 by 2x 1 3.

Solution

x 1 3

2x 1 3 qww 2x2 1 9x 2 6

2x2 1 3x Multiply x and 2x 1 3 .

6x 2 6 Subtract 2x2 1 3x . Bring down 26 .


215 Subtract 6x 1 9 .

(2x2 1 9x 2 6) 4 (2x 1 3) 5 x 1 3 1 215 } 2x 1 3



Your Notes


2. (3x2 2 x 2 14) 4 (3x 2 7)

3. (6x2 2 13x 1 11) 4 (3x 2 5)

Checkpoint Divide.

Divide 2x 1 2 1 3x2 by 1 1 x.

x 1 1 qww 3x2 1 2x 1 2 Rewrite polynomials.

Multiply and .


Multiply and .

Subtract.

(2x 1 2 1 3x2) 4 (1 1 x) 5

Example 4 Rewrite polynomials

Divide 224 1 6x2 by 26 1 3x.

3x 2 6 qww 6x2 1 0x 2 24 Rewrite polynomials. Insert missing term.

Multiply and .


Multiply and .

Subtract.

(224 1 6x2) 4 (26 1 3x) 5

Example 5 Insert missing terms


Your Notes


2. (3x2 2 x 2 14) 4 (3x 2 7)

x 1 2

3. (6x2 2 13x 1 11) 4 (3x 2 5)

2x 2 1 1 6 } 3x 2 5

Checkpoint Divide.

Divide 2x 1 2 1 3x2 by 1 1 x.

3x 2 1

x 1 1 qww 3x2 1 2x 1 2 Rewrite polynomials.

3x2 1 3x Multiply 3x and x 1 1 .


2x 2 1 Multiply 21 and x 1 1 .

3 Subtract.

(2x 1 2 1 3x2) 4 (1 1 x) 5 3x 2 1 1 3 } x 1 1

Example 4 Rewrite polynomials

Divide 224 1 6x2 by 26 1 3x.

2x 1 4

3x 2 6 qww 6x2 1 0x 2 24 Rewrite polynomials. Insert missing term.

6x2 2 12x Multiply 2x and 3x 2 6 .



0 Subtract.

(224 1 6x2) 4 (26 1 3x) 5 2x 1 4

Example 5 Insert missing terms


Your Notes


4. (6 2 2x 1 x2) 4 (2 1 x)

5. (211 1 3x2) 4 (23 1 x)

Checkpoint Divide.

6. Graph y 5 5x 1 13

} x 1 3 .

x

y

9

15

3 9

3

29215 23


Homework


Graph y 5 4x 2 3 } x 2 1

.

SolutionStep 1 Rewrite the rational Step 2 Graph the function. function in the form

x

y

3 51

1

3

5

7

2123

y 5 a } x 2 h 1 k.

x 2 1 qw 4x 2 3

So, y 5 .

Example 6 Rewrite and graph a rational function

Your Notes


4. (6 2 2x 1 x2) 4 (2 1 x)

x 2 4 1 14 } x 1 2

5. (211 1 3x2) 4 (23 1 x)

3x 1 9 1 16 } x 2 3

Checkpoint Divide.

6. Graph y 5 5x 1 13

} x 1 3 .

x

y

9

15

3 9

3

29215 23

22x 1 3

y 5 1 5


Homework


Graph y 5 4x 2 3 } x 2 1

.

SolutionStep 1 Rewrite the rational Step 2 Graph the function. function in the form

x

y

3 51

1

3

5

7

2123

1x 2 1

y 5 1 4

y 5 a } x 2 h 1 k.

4

x 2 1 qw 4x 2 3

4x 2 4

1

So, y 5 1 } x 2 1

1 4 .

Example 6 Rewrite and graph a rational function

Your Notes


Goal p Use synthetic division to divide polynomials.

VOCABULARY

Synthetic Division

Use Synthetic Division

Divide 2x3 1 4x2 2 9x 1 3 by x 2 1 using synthetic division.

Step 1 Write the value of k and the coefficients of the dividend in order of descending exponents.

1 2 4 29

Step 2 Bring down the leading coefficient and multiply it by the k-value. Write the product under the second coefficient. Add.

1 2 4 29

2

2

Step 3 Multiply the previous sum by the k-value, and write the product under the next coefficient. Add. Repeat for all other coefficients

1 2 4 29

2

2

Step 4 Identify the quotient and the remainder. The coefficients of the quotient are 2, , and . The remainder is .

(2x3 1 4x2 2 9x 1 3) 4 (x 2 1) = 2x2 1 x 2

Example 1 Use synthetic division

To check your answer, multiply the quotient by the divisor, then add the remainder to the product.



Your Notes


Goal p Use synthetic division to divide polynomials.

VOCABULARY

Synthetic Division A method for dividing a polynomial by a binomial of the form x – k where k is a constant.

Use Synthetic Division

Divide 2x3 1 4x2 2 9x 1 3 by x 2 1 using synthetic division.

Step 1 Write the value of k and the coefficients of the dividend in order of descending exponents.

1 2 4 29 3

Step 2 Bring down the leading coefficient and multiply it by the k-value. Write the product under the second coefficient. Add.

1 2 4 29 3

2

2 6

Step 3 Multiply the previous sum by the k-value, and write the product under the next coefficient. Add. Repeat for all other coefficients

1 2 4 29 3

2 6 23

2 6 23 0

Step 4 Identify the quotient and the remainder. The coefficients of the quotient are 2, 6, and –3. The remainder is 0.

(2x3 1 4x2 2 9x 1 3) 4 (x 2 1) = 2x2 1 6 x 2 3


To check your answer, multiply the quotient by the divisor, then add the remainder to the product.



Your Notes


Divide x3 2 x2 1 7 by x 1 2 using synthetic division.

Solution

1 0

The coefficients of the quotient are , and . The

remainder is .

(x3 2 x2 2 7) 4 (x 1 2) 5


Remember to use a 0 as the coefficient of any missing terms.

Checkpoint Divide using synthetic division. 1. (x3 2 6x2 1 5x 1 3) 4 (x 2 4)

2. (x3 1 8x 1 9) 4 (x 1 1)

3. (3x3 1 16x2 1 2x 2 2) 4 (x 1 5)


Homework

Your Notes


Divide x3 2 x2 1 7 by x 1 2 using synthetic division.

Solution

22 1 21 0 7

22 6 212

1 23 6 25

The coefficients of the quotient are 1 , 23 and 6 . The

remainder is 25.

(x3 2 x2 2 7) 4 (x 1 2) 5 x2 2 3x 1 6 2 5 } x 1 2


Remember to use a 0 as the coefficient of any missing terms.

Checkpoint Divide using synthetic division. 1. (x3 2 6x2 1 5x 1 3) 4 (x 2 4)

x2 2 2x 2 3 2 9 } x 2 4

2. (x3 1 8x 1 9) 4 (x 1 1)

x2 2 x 1 9

3. (3x3 1 16x2 1 2x 2 2) 4 (x 1 5)

3x2 1 x 2 3 1 13 } x 1 5


Homework

Your Notes


12.4 Simplify Rational ExpressionsGoal p Simplify rational expressions.

VOCABULARY

Rational expression

Excluded value

Simplest form of a rational expression

Find the excluded values, if any, of the expression.

a. x } 4x 2 8 b. 3x } x2 2 16

Solution

a. The expression x } 4x 2 8 is undefined when

5 0, or x 5 . The excluded value is .

b. The expression 3x } x2 2 16

is undefined when

5 0, or ( )( ) 5 0. The solutions of the equation are and . The excluded values are and .

Example 1 Find excluded values

Checkpoint Find the excluded values, if any, of the expression.

1. x 1 6 } 14x 2. 9x 1 1 } x2 2 x 2 20


Your Notes


12.4 Simplify Rational ExpressionsGoal p Simplify rational expressions.

VOCABULARY

Rational expression An expression that can be written as a ratio of two polynomials

Excluded value A number that makes a rational expression undefined

Simplest form of a rational expression A rational expression is in simplest form if the numerator and denominator have no factors in common other than 1.

Find the excluded values, if any, of the expression.

a. x } 4x 2 8 b. 3x } x2 2 16

Solution

a. The expression x } 4x 2 8 is undefined when

4x 2 8 5 0, or x 5 2 . The excluded value is 2 .

b. The expression 3x } x2 2 16

is undefined when

x2 2 16 5 0, or ( x 1 4 )( x 2 4 ) 5 0. The solutions of the equation are 24 and 4 . The excluded values are 24 and 4 .

Example 1 Find excluded values

Checkpoint Find the excluded values, if any, of the expression.

1. x 1 6 } 14x 2. 9x 1 1 } x2 2 x 2 20

0 24, 5


Your Notes



Simplify the rational expression, if possible. State the excluded values.

a. 18x } 6x2 5 Divide out common factors.

5 Simplify.

The excluded value is .

b. 12x2 2 6x } 24x 5 Factor numerator and denominator.

5 Divide out common factors.

5 Simplify.

The excluded value is .

Example 2 Simplify expressions by dividing out monomials

Checkpoint Simplify the rational expression, if possible. State the excluded values.

3. 7 } 5x 1 3 4. 5x } 5x2 2 25

5. 6x3 } 2x 1 4

SIMPLIFYING RATIONAL EXPRESSIONS

Let a, b, and c be polynomials where b Þ 0 and c Þ 0.

Algebra Example

ac } bc 5 5 3x 2 9 } 4x 2 12 5 5

Your Notes



Simplify the rational expression, if possible. State the excluded values.

a. 18x } 6x2 5

6 p x p x

6 p 3 p x Divide out common factors.

5 3 } x Simplify.

The excluded value is 0 .

b. 12x2 2 6x } 24x 5 6 p 4 p x

6x(2x 2 1) Factor numerator and

denominator.

5 6 p 4 p x

6x(2x 2 1) Divide out common factors.

5 2x 2 1 } 4 Simplify.

The excluded value is 0 .

Example 2 Simplify expressions by dividing out monomials

Checkpoint Simplify the rational expression, if possible. State the excluded values.

3. 7 } 5x 1 3 4. 5x } 5x2 2 25

5. 6x3 } 2x 1 4

2 3 } 5 x } x 2 2 5

; 6 Ï}

5 3x 3 }

x 1 2 ; 22

SIMPLIFYING RATIONAL EXPRESSIONS

Let a, b, and c be polynomials where b Þ 0 and c Þ 0.

Algebra Example

ac } bc 5 b p c

a p c 5 a }

b 3x 2 9 } 4x 2 12 5

4(x 2 3)

3(x 2 3) 5 3 }

4

Your Notes


Homework

Simplify x2 1 x 2 12 }}

x2 2 5x 1 6 . State the excluded values.

x2 1 x 2 12

}} x2 2 5x 1 6

5 Factor and divide out common factor.

5 Simplify.

The excluded values are and .

Example 3 Simplify an expression by dividing out binomials

Checkpoint Simplify the rational expression. State the excluded values.

6. x2 1 7x 1 6

}} x2 1 3x 2 18

7. 4 2 x2

}} x2 1 5x 2 14

Simplify 10 1 3x 2 x2 }}

x2 2 25 . State the excluded values.

10 1 3x 2 x2

}} x2 2 25

5 Factor numerator and denominator.

5 Rewrite as .

5 Divide out common factor.

5 5 Simplify.

The excluded values are and .

Example 4 Recognize opposites


Your Notes


Homework

Simplify x2 1 x 2 12 }}

x2 2 5x 1 6 . State the excluded values.

x2 1 x 2 12

}} x2 2 5x 1 6

5 (x 2 3)(x 2 2)

(x 2 3)(x 1 4) Factor and divide

out common factor.

5 (x 1 4) }

(x 2 2) Simplify.

The excluded values are 3 and 2 .

Example 3 Simplify an expression by dividing out binomials

Checkpoint Simplify the rational expression. State the excluded values.

6. x2 1 7x 1 6

}} x2 1 3x 2 18

7. 4 2 x2

}} x2 1 5x 2 14

x 1 1

} x 2 3 ; 26 and 3 2 x 1 2 } x 1 7 ; 2 and 27

Simplify 10 1 3x 2 x2 }}

x2 2 25 . State the excluded values.

10 1 3x 2 x2

}} x2 2 25

5 (x 2 5)(x 1 5)

(5 2 x)(2 1 x) Factor numerator

and denominator.

5 (x 2 5)(x 1 5)

2(x 2 5)(2 1 x) Rewrite 5 2 x

as 2(x 2 5) .

5 (x 2 5)(x 1 5)

2(x 2 5)(2 1 x) Divide out

common factor.

5 (x 1 5)

2(2 1 x) 5 2 x 1 2

} x 1 5 Simplify.

The excluded values are 5 and 25 .

Example 4 Recognize opposites


Your Notes


12.5 Multiply and Divide RationalExpressionsGoal p Multiply and divide rational expressions.

MULTIPLYING AND DIVIDING RATIONAL EXPRESSIONS

Let a, b, c, and d be polynomials.

Algebra

a } b p c } d 5 where b Þ 0 and d Þ 0

a } b 4 c } d 5 a } b p 5 where b Þ 0, c Þ 0, and d Þ 0

Examples

2x } x 1 1 p x } 5 5 3 } x2 4 x } 5 5 3 }

x2 p 5

Find the product 3x4 }

4x3 p 2x2

} 5x3

.

Solution

3x4 }

4x3 p 2x2

} 5x3 5 Multiply numerators and

denominators.

5 Product of powers property

5 Factor and divide out common factors.

5 Simplify.

Example 1 Multiply rational expressions involving monomials


Your Notes


12.5 Multiply and Divide RationalExpressionsGoal p Multiply and divide rational expressions.

MULTIPLYING AND DIVIDING RATIONAL EXPRESSIONS

Let a, b, c, and d be polynomials.

Algebra

a } b p c } d 5 bd

ac where b Þ 0 and d Þ 0

a } b 4 c } d 5 a } b p c

d 5

bc

ad where b Þ 0, c Þ 0, and d Þ 0

Examples

2x } x 1 1 p x } 5 5 5(x 1 1)

2x2 3 }

x2 4 x } 5 5 3 } x2 p

x

5 5

x3

15

Find the product 3x4 }

4x3 p 2x2

} 5x3

.

Solution

3x4 }

4x3 p 2x2

} 5x3 5

(4x3)(5x3)

(3x4)(2x2) Multiply numerators and

denominators.

5 20x6

6x6 Product of powers property

5 10 p 2 p x6

3 p 2 p x6 Factor and divide out

common factors.

5 3 } 10

Simplify.

Example 1 Multiply rational expressions involving monomials


Your Notes


Find the product x }} 5x2 2 6x 2 8

p 2x2 2 4x } 7x2

.

Solution

x }} 5x2 2 6x 2 8

p 2x2 2 4x } 7x2

5 Multiply numerators and denominators.


5 Simplify.

Example 2 Multiply rational expressions involving polynomials

Find the product 4x }} x2 2 x 2 12

p (x 2 4).

Solution

4x } x2 2 x 2 12

p (x 2 4)

5 4x } x2 2 x 2 12

p Rewrite polynomial as a fraction.



5 Simplify.

Example 3 Multiply a rational expression by a polynomial


Your Notes


Find the product x }} 5x2 2 6x 2 8

p 2x2 2 4x } 7x2

.

Solution

x }} 5x2 2 6x 2 8

p 2x2 2 4x } 7x2

5 (5x2 2 6x 2 8)(7x2)

x (2x2 2 4x ) Multiply numerators and

denominators.

5 7x2(5x 1 4)(x 2 2)

2x2(x 2 2) Factor and divide out common

factors.

5 2 } 7(5x 1 4)

Simplify.

Example 2 Multiply rational expressions involving polynomials

Find the product 4x }} x2 2 x 2 12

p (x 2 4).

Solution

4x } x2 2 x 2 12

p (x 2 4)

5 4x } x2 2 x 2 12

p 1

x 2 4 Rewrite polynomial as

a fraction.

5 x2 2 x 2 12

4x(x 2 4) Multiply numerators and

denominators.

5 (x 2 4)(x 1 3)

4x(x 2 4) Factor and divide out common

factor.

5 4x } x 1 3

Simplify.

Example 3 Multiply a rational expression by a polynomial


Your Notes


1. 2x4 }

5x2 p 6x } 3x3

2. x2 2 5x 1 4 } 3x2 2 12x

p 2x2 1 2 } x2 1 6x 2 7

3. 2x }} x2 1 5x 2 24

p (x 1 8)


Find the quotient x2 1 5x 2 24 }} x2 1 9x 1 8

4 x2 2 9 }

6x 2 18 .

Solution

x2 1 5x 2 24 }} x2 1 9x 1 8

4 x2 2 9 } 6x 2 18

5 x2 1 5x 2 24 }} x2 1 9x 1 8

p Multiply by multiplicative inverse.



5 Simplify.

Example 4 Divide rational expressions involving polynomials


Your Notes


1. 2x4 }

5x2 p 6x } 3x3

4 } 5

2. x2 2 5x 1 4 } 3x2 2 12x

p 2x2 1 2 } x2 1 6x 2 7

2(x2 1 1) } 3x(x 1 7)

3. 2x }} x2 1 5x 2 24

p (x 1 8)

2x } x 2 3


Find the quotient x2 1 5x 2 24 }} x2 1 9x 1 8

4 x2 2 9 }

6x 2 18 .

Solution

x2 1 5x 2 24 }} x2 1 9x 1 8

4 x2 2 9 } 6x 2 18

5 x2 1 5x 2 24 }} x2 1 9x 1 8

p x2 2 9

6x 2 18 Multiply by

multiplicative inverse.

5 (x2 1 9x 1 8)(x2 2 9)

(x2 1 5x 2 24)(6x 2 18) Multiply numerators

and denominators.

5 (x 1 8)(x 1 1)(x 1 3)(x 2 3)

6(x 2 3)(x 2 3)(x 1 8) Factor and divide out

common factors.

5 6(x 2 3) }} (x 1 1)(x 1 3)

Simplify.

Example 4 Divide rational expressions involving polynomials


Your Notes


Homework

4. x2 1 2x 2 15 }} x2 1 4x 2 5

4 x2 2 4 } 7x 2 14

5. x2 1 8x 1 7 }}

x2 2 1 4 (x 1 7)

Checkpoint Find the quotient.

Find the quotient x2 2 25 } x 2 3

4 (x 2 5).

Solution

x2 2 25 } x 2 3

4 (x 2 5)

5 x2 2 25 } x 2 3

4 Rewrite polynomial as fraction.

5 x2 2 25 } x 2 3

p Multiply by multiplicative inverse.



5 Simplify.

Example 5 Divide a rational expression by a polynomial


Your Notes


Homework

4. x2 1 2x 2 15 }} x2 1 4x 2 5

4 x2 2 4 } 7x 2 14

7(x 2 3) }} (x 1 2)(x 2 1)

5. x2 1 8x 1 7 }}

x2 2 1 4 (x 1 7)

1 } x 2 1

Checkpoint Find the quotient.

Find the quotient x2 2 25 } x 2 3

4 (x 2 5).

Solution

x2 2 25 } x 2 3

4 (x 2 5)

5 x2 2 25 } x 2 3

4 1

x 2 5 Rewrite polynomial as fraction.

5 x2 2 25 } x 2 3

p x 2 5

1 Multiply by multiplicative inverse.

5 (x 2 3)(x 2 5)

x2 2 25 Multiply numerators and

denominators.

5 (x 2 3)(x 2 5)

(x 1 5)(x 2 5) Factor and divide out common

factors.

5 x 1 5 } x 2 3

Simplify.

Example 5 Divide a rational expression by a polynomial


Your Notes


Simplify Complex Fractions

Goal p Simplify complex fractions.

VOCABULARY

Complex fraction

Simplify 2x 1 4 }

2x2 }

3 .

Solution

2x 1 4 }

2x2

} 3

5 1 2x 1 4 2 4 } Write fraction as quotient.

5 1 2x 1 4 2 • } Multiply by multiplicative

inverse.

5 }

Multiply numerators and denominators.

5 } Simplify.

Example 1 Simplify a complex fraction

The widest fraction bar separates the numerator of a complex fraction from the denominator.

SIMPLIFYING A COMPLEX FRACTION

Let a, b, c, and d be polynomials where b ≠ 0, c ≠ 0,

and d ≠ 0.

Algebra a } b

} c } d = } b 4 } d =


Focus On

Operations


Your Notes


Simplify Complex Fractions

Goal p Simplify complex fractions.

VOCABULARY

Complex fraction A fraction that contains a fraction in its numerator, denominator, or both

Simplify 2x 1 4 }

2x2 }

3 .

Solution

2x 1 4 }

2x2

} 3

5 1 2x 1 4 2 4 2x2

}

3 Write fraction as quotient.

5 1 2x 1 4 2 • 3

}

2x2 Multiply by multiplicative

inverse.

5 6x 1 12

}

2x2

Multiply numerators and denominators.

5 3x 1 6

}

x2 Simplify.


The widest fraction bar separates the numerator of a complex fraction from the denominator.

SIMPLIFYING A COMPLEX FRACTION

Let a, b, c, and d be polynomials where b ≠ 0, c ≠ 0,

and d ≠ 0.

Algebra a } b

} c } d =

a } b 4

c } d = a }

b •

d } c


Focus On

Operations


Your Notes


Simplify 5x2 1 5x

} 3x

}

x2 2 3x 2 4 }}

x 2 4

.

Solution

5x2 1 5x }

4 }}

Write fraction as quotient.

5 5x2 1 5x }

• }} Multiply by .

5 }} Multiply numerators and denominators.

5 }} Factor and divide out common factors.

5 } Simplify.


Checkpoint Simplify the complex fraction.

1. 28

} 7x }

16x2

3. x

2 1 3x 2 10 }}

x2 2 9 }}

x2 1 10x 1 25

}} 4x 1 12

2. x2 - 36 }

2x2 - 12x } 3x2

326 12.5 Focus On Operations` • Algebra 1 Notetaking Guide Copyright © Holt McDougal. All rights reserved.

Homework

Your Notes


Simplify 5x2 1 5x

} 3x

}

x2 2 3x 2 4 }}

x 2 4

.

Solution

5x2 1 5x }

3x

4 x2 2 3x 2 4

}}

x 2 4

Write fraction as quotient.

5 5x2 1 5x }

3x

• x 2 4 }}

x2 2 3x 2 4 Multiply by multiplicative

inverse.

5 1 5x2 1 5x 2 1 x2 4 2

}}

3x 1 x2 2 3x 2 4 2 Multiply numerators and

denominators.

5 5x 1 x 1 1 2 1 x 2 4 2

}}

3x 1 x 1 1 2 1 x 2 4 2 Factor and divide out common

factors.

5 5 }

3 Simplify.


Checkpoint Simplify the complex fraction.

1. 28

} 7x }

16x2

2 1 } 14x 3

3. x

2 1 3x 2 10 }}

x2 2 9 }}

x2 1 10x 1 25

}} 4x 1 12

2. x2 - 36 }

2x2 - 12x } 3x2

3x(x 1 6) }

2

4(x 2 2) }}

(x 2 3)(x 1 5)

326 12.5 Focus On Operations` • Algebra 1 Notetaking Guide Copyright © Holt McDougal. All rights reserved.

Homework

Your Notes


12.6 Add and Subtract RationalExpressionsGoal p Add and subtract rational expressions.

VOCABULARY

Least common denominator of rational expressions (LCD)


ADDING AND SUBTRACTING RATIONAL EXPRESSIONS WITH THE SAME DENOMINATOR

Let a, b, and c be polynomials where c Þ 0.

Algebra

a } c 1 b } c 5 a } c 2 b } c 5

a. 3 } 8x 1 4 } 8x 5 8x

Add numerators.

5 Simplify.

b. 2x 1 9 } x 1 1 2 7 } x 1 1 5 x 1 1

Subtract numerators.

5 x 1 1

Simplify.


5 Simplify.

Example 1 Add and subtract with the same denominator

Your Notes


12.6 Add and Subtract RationalExpressionsGoal p Add and subtract rational expressions.

VOCABULARY

Least common denominator of rational expressions (LCD) The least common denominator of two or more rational expressions is the product of the factors of the rational expressions with each common factor used only once.


ADDING AND SUBTRACTING RATIONAL EXPRESSIONS WITH THE SAME DENOMINATOR

Let a, b, and c be polynomials where c Þ 0.

Algebra

a } c 1 b } c 5 c

a 1 b a } c 2 b } c 5

c

a 2 b

a. 3 } 8x 1 4 } 8x 5 8x

3 1 4 Add numerators.

5 7 } 8x

Simplify.

b. 2x 1 9 } x 1 1 2 7 } x 1 1 5 x 1 1

(2x 1 9) 2 7 Subtract numerators.

5 x 1 1

2x 1 2 Simplify.

5 x 1 1

2(x 1 1) Factor and divide out

common factor.

5 2 Simplify.

Example 1 Add and subtract with the same denominator

Your Notes


1. x 1 8 } 4x 1 3 } 4x 2. 6x 2 5 } x 2 2x 2 5 } x


Find the LCD of the rational expressions.

a. 1 } 3x3 , 5 }

4x4 b. 7 } x2 2 4

, x 1 3 } x2 1 x 2 2

Solution

a. Find the of 3x3 and 4x4.

3x3 5

4x4 5

LCM 5 5

The LCD of 1 } 3x3 and 5 }

4x4 is .

b. Find the of x2 2 4 and x2 1 x 2 2.

x2 2 4 5

x2 1 x 2 2 5

LCM 5

The LCD of 7 } x2 2 4

and x 1 3 } x2 1 x 2 2

is

.

Example 2 Find the LCD of rational expressions

3. 5 } 36x , x 1 2 } 4x3 4. 7x } x 2 8 , x 2 1 } x 1 3

Checkpoint Find the LCD of the rational expressions.


Your Notes


1. x 1 8 } 4x 1 3 } 4x 2. 6x 2 5 } x 2 2x 2 5 } x

x 1 11 } 4x 4


Find the LCD of the rational expressions.

a. 1 } 3x3 , 5 }

4x4 b. 7 } x2 2 4

, x 1 3 } x2 1 x 2 2

Solution

a. Find the least common multiple of 3x3 and 4x4.

3x3 5 3 p x p x p x

4x4 5 4 p x p x p x p x

LCM 5 x p x p x p 3 p 4 p x 5 12x4

The LCD of 1 } 3x3 and 5 }

4x4 is 12x4 .

b. Find the least common multiple of x2 2 4 and x2 1 x 2 2.

x2 2 4 5 (x 2 2) p (x 1 2)

x2 1 x 2 2 5 (x 2 1) p (x 1 2)

LCM 5 (x 1 2) p (x 2 2) p (x 2 1)

The LCD of 7 } x2 2 4

and x 1 3 } x2 1 x 2 2

is

(x 1 2)(x 2 2)(x 2 1) .

Example 2 Find the LCD of rational expressions

3. 5 } 36x , x 1 2 } 4x3 4. 7x } x 2 8 , x 2 1 } x 1 3

36x3 (x 2 8)(x 1 3)

Checkpoint Find the LCD of the rational expressions.


Your Notes



Find the sum 1 } 3x3

1 5 } 4x4 .

Solution

1 } 3x3 1 5 }

4x4

5 3x3 p

1 p 1

4x4 p

5 p Rewrite fractions using LCD,

.

5 1 Simplify numerators and denominators.

5 Add fractions.

Example 3 Add expressions with different denominators

Find the difference x 1 1 }} x2 1 5x 1 6

2 x 2 4 } x2 2 9

.

Solution

x 1 1 }} x2 1 5x 1 6

2 x 2 4 } x2 2 9

5 x 1 1 ( (( (

5 (x 1 1) 2 (x 2 4)

5

5

5

Example 4 Subtract expressions with different denominators

( (

( (

( (

( (

2 x 2 4 ( (( (

Your Notes



Find the sum 1 } 3x3

1 5 } 4x4 .

Solution

1 } 3x3 1 5 }

4x4

5 4x3x3 p

4x1 p 1

34x4 p

35 p Rewrite fractions using LCD,

12x4 .

5 12x4

4x 1

12x4

15 Simplify numerators and

denominators.

5 4x 1 15 } 12x4

Add fractions.

Example 3 Add expressions with different denominators

Find the difference x 1 1 }} x2 1 5x 1 6

2 x 2 4 } x2 2 9

.

Solution

x 1 1 }} x2 1 5x 1 6

2 x 2 4 } x2 2 9

5 x 1 1

x 1 2

x 1 3( (( (

5 (x 1 1)

(x 1 2)(x 1 3) 2 (x 2 4)

(x 2 3)(x 1 3)

5 (x 1 3)(x 1 2)(x 2 3)

(x 1 1)(x 2 3) 2 (x 2 4)(x 1 2)

5 (x 1 3)(x 1 2)(x 2 3)

x2 2 2x 2 3 2 (x2 2 2x 2 8)

5 (x 1 3)(x 1 2)(x 2 3)

5

Example 4 Subtract expressions with different denominators

x 2 3( (

x 2 3( (

x 1 2( (

x 1 2( (

2 x 2 4

x 2 3

x 1 3( (( (

Your Notes


Homework

5. 9 } x 2 1 2 15 } 3x 1 1

6. 12 } 5x 1 3x } x 2 4

7. x 2 1 }} x2 2 2x 2 24

1 4 } x2 2 5x 2 6

8. x 1 2 }} x2 1 2x 2 15

2 x 2 6 }} x2 1 4x 2 21



Your Notes


Homework

5. 9 } x 2 1 2 15 } 3x 1 1

12x 1 24 }} (x 2 1)(3x 1 1)

6. 12 } 5x 1 3x } x 2 4

15x2 1 12x 2 48 }} 5x(x 2 4)

7. x 2 1 }} x2 2 2x 2 24

1 4 } x2 2 5x 2 6

x2 1 4x 1 15 }} (x 2 6)(x 1 4)(x 1 1)

8. x 1 2 }} x2 1 2x 2 15

2 x 2 6 }} x2 1 4x 2 21

10x 1 44 }} (x 2 3)(x 1 5)(x 1 7)



Your Notes


12.7


Solve Rational EquationsGoal p Solve rational equations.

VOCABULARY

Rational equation

Example 1 Use the cross products property

Solve 5 } x 2 1

5 x } 4 . Check your solution.

Solution

5 } x 2 1

5 x } 4 Write original equation.

20 5 Cross products property

0 5 Subtract from each side.

0 5 ( )( ) Factor polynomial.



The solutions are and .

CHECK If x 5 : If x 5 :

5

2 10

4

5

2 1 0

4

5 5

Your Notes


12.7


Solve Rational EquationsGoal p Solve rational equations.

VOCABULARY

Rational equation An equation that contains one or more rational expressions

Example 1 Use the cross products property

Solve 5 } x 2 1

5 x } 4 . Check your solution.

Solution

5 } x 2 1

5 x } 4 Write original equation.

20 5 x2 2 x� Cross products property

0 5 x2 2 x 2 20 Subtract 20 from each side.

0 5 ( x 2 5 )( x 1 4 ) Factor polynomial.



The solutions are 5 and 24 .

CHECK If x 5 5 : If x 5 24 :

5

5

2 10

4

5

24

5

2 1 0

4

24

5 } 4 5 5 }

4 21 5 21

Your Notes


3. Solve 3 } x 2 3 2 1 } x 1 3 5 14 } x2 2 9

. Check your solution.


Solve x } x 1 6

2 1 } 2 5 4 }

x 1 6 .

Solution

x } x 1 6

2 1 } 2 5 4 } x 1 6

x } x 1 6 p 2 1 } 2 p 5 4 } x 1 6 p

x 1 6

2 2

5 x 1 6

5

5

x 5

The solution is .

Example 2 Multiply by the LCD

1. 22 } x 1 9 5 x } 7 2. 6 } x 2 4 5 3 } x

Checkpoint Solve the equation. Check your solution.


Your Notes


3. Solve 3 } x 2 3 2 1 } x 1 3 5 14 } x2 2 9

. Check your solution.

1


Solve x } x 1 6

2 1 } 2 5 4 }

x 1 6 .

Solution

x } x 1 6

2 1 } 2 5 4 } x 1 6

x } x 1 6 p 2(x 1 6) 2 1 } 2 p 2(x 1 6) 5 4 } x 1 6 p 2(x 1 6)

x 1 6

x p 2(x 1 6) 2

2

2(x 1 6) 5

x 1 64 p 2(x 1 6)

2x 2 x 2 6 5 8

x 2 6 5 8

x 5 14

The solution is 14 .

Example 2 Multiply by the LCD

1. 22 } x 1 9 5 x } 7 2. 6 } x 2 4 5 3 } x

x 5 22 or x 5 27 x 5 24

Checkpoint Solve the equation. Check your solution.


Your Notes



4. Solve 1 } x 1 6 1 2 5 x2 2 38 }} x2 1 2x 2 24

Checkpoint Complete the following exercise.Homework

Solve 3 } x 1 2

2 1 5 25 }} x2 2 3x2 10

.

Solution

Write each denominator in factored form. The LCD is .

3 } x 1 2 2 1 5 25 }} (x 1 2)(x 2 5)

x 1 23 p

2 1 p

5 (x 1 2)(x 2 5)

25 p

2

5

2 ( ) 5

5

5 0

( ) 5 0

5 0 or 5 0

5

x 5 or x 5

The solutions are and .

Example 3 Factor to find the LCDYour Notes



4. Solve 1 } x 1 6 1 2 5 x2 2 38 }} x2 1 2x 2 24

2, 27

Checkpoint Complete the following exercise.Homework

Solve 3 } x 1 2

2 1 5 25 }} x2 2 3x2 10

.

Solution

Write each denominator in factored form. The LCD is (x 1 2)(x 2 5) .

3 } x 1 2 2 1 5 25 }} (x 1 2)(x 2 5)

x 1 2

(x 1 2)(x 2 5)3 p 2 1 p (x 1 2)(x 2 5)

5 (x 1 2)(x 2 5)

(x 1 2)(x 2 5)25 p

(x 1 2)

3(x 1 2)(x 2 5) 2 (x 1 2)(x 2 5)

5 (x 1 2)(x 2 5)

25(x 1 2)(x 2 5)

3x 2 15 2 ( x2 2 3x 2 10 ) 5 25

2x2 1 6x 2 5 5 25

2x2 1 6x 5 0

x ( 2x 1 6 ) 5 0

x 5 0 or 2x 1 6 5 0

2x 5 26

x 5 0 or x 5 6

The solutions are 0 and 6 .

Example 3 Factor to find the LCDYour Notes


Inverse variation

Hyperbola


Synthetic division



Rational function

Rational expression




Inverse variation

y 5 4 } x

Hyperbola

x

y

Branches

Horizontal asymptote

Vertical asymptote


The two symmetrical parts of a hyperbola

Synthetic division

Divide x3 2 2x2 2 x 1 2 by x 22.

2 1 22 21 2

2 0 22 1 0 21 0


The number 4 in y 5 4 } x .


The asymptotes of the graph of y 5 4 }

x are the

x-axis and the y-axis.

Rational function

y 5 1 } x 2 2

Rational expression

x } 4x 2 8





Excluded value

Complex fraction

Rational equation


Least common denominator of rational expressions




Excluded value

The excluded value of

x } 4x 2 8

is 2.

Complex fraction

2 } 3x } x2

Rational equation

5 } x 2 1

5 x } 4


x 1 4 } x 2 2

Least common denominator of rational expressions

The LCD of 1 } 3x3

and

5 } 4x4

is 12x4.



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