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Reteaching Masters

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olt,

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NAME _________________________________________________ CLASS _______________ DATE ______________

Algebra 2 Reteaching 1.1 1

◆Skill A Identifying a linear relationship

Recall In a linear relationship, for each constant difference in the x-values, there is also a constant difference in the y-values.

◆ Example 1You can use the formula F � 1.8C � 32 to change temperature readings fromdegrees Celsius to degrees Fahrenheit.

Is this an example of a linear relationship?

◆ SolutionYes; notice that for each difference of 10 in Celsius values, there is acorresponding difference of 18 in Fahrenheit values.

◆ Example 2Make a table of x- and y-values for y � �3x � 5 with x-values of 0, 1, 2, 3, and 4.Is this a linear relationship?

◆ Solution

For each increase of 1 in x-values, there is a decrease of 3 in y-values. This is alinear relationship.

◆ Example 3Is y � x2 � 2 an example of a linear relationship?

◆ Solution

No; for each change of 2 in x-values, corresponding changes in y are not constant.

State whether each equation is linear.

1. y � 2x 2. y � 2x� 5 3. y � x

4. 5. y � 3 � x 6.

7. y � 1.5x 8. y � x2 � 1 9.

10. y � �x 11. y � 4 12. y � x2 � 2x

y �x � 1

2

y �2x

y �12

x � 1

Reteaching

1.1 Tables and Graphs of Linear Equations

°C 0 10 20 30 40

°F 32 50 68 86 104

x 0 1 2 3 4

y 5 2 �1 �4 �7

x �2 0 2 4 6

y 6 2 6 18 38

Graph each pair of linear equations on the same set of axes.

13. a. y � x � 2 14. a. y � 4 � x 15. a.

b. y � � 2x b. y � 3x � 2 b. y � x

For each table, write a linear equation that represents therelationship between x and y.

16. 17.

18. 19.

x

y

Ox

y

Ox

y

O

y �12

x � 1

Copyright ©

by Holt, R

inehart and Winston. A

ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________

2 Reteaching 1.1 Algebra 2

◆Skill B Graphing a linear equation

Recall If x and y are linearly related, this relationship can be expressed as a linear equationin the form y � mx � b.

◆ Example Graph the linear equation y � 2x �3.

◆ SolutionSince it takes two points to determine a line, picktwo x-values and use the equation to calculatethe corresponding y-values. Graph the orderedpairs (x, y) and draw a line. Use a third point tocheck that all three points are on the same line.

x

y

O–4

–4

–2–2

2

2 4

4

y =

2x –

3

x 0 1 2

y �3 �1 1

x 0 1 2 3

y 0 2 4 6

x �3 �2 �1 0

y �4 �3 �2 �1

x 0 2 4 6

y 0 1 2 3

x 1 2 3 4

y 2 5 8 11

Reteaching — Chapter 1

Lesson 1.1

1. linear relationship





6. not a linear relationship







13.

14.

15.

16.

17.

18.

19.

Lesson 1.2

1. slope: 1; y-intercept: 2


3. slope: �1; y-intercept: 4

x

y

O

x + y = 4

x

y

O

y =

2x

x

y

Oy =

x + 2

y � 3x � 1

y �12

x

y � x � 1

y � 2x

x

y

O

y = x

y = x + 1 12

x

y

O

y =

3x –

2

y = 4 – x

x

y

O

y = x

+ 2

y = –2x

ANSWERSC

opyr

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©by

Hol

t, R

ineh

art a

nd W

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on. A

ll rig

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.

Algebra 2 Answers 187

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olt,

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Graphing a linear equation using the slope and y-intercept

Recall The slope-intercept form of a line is y � mx � b, where m is the slope and b is the y-intercept.

◆ Example Graph the line with the equation 2x � y � �1.

◆ SolutionSolve for y. Write the equation in the form y � 2x � 1.When x = 0, then y = 1; the y-intercept is 1.Graph the point (0, 1).

The slope of 2 indicates that .

From the point (0, 1) “rise” 2 units and “run” tothe right 1 unit to locate the point (1, 3).

riserun

�21

Find the slope, m, and y-intercept, b, for each line. Then graph.

1. y � x � 2 2. y � 2x 3. x � y � 4

m: b: m: b: m: b:

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

m: b: m: b: m: b:

x

y

Ox

y

Ox

y

O

x

y

Ox

y

Ox

y

O

Reteaching

1.2 Slopes and Intercepts

x

y

O

2x –

y =

–1

(0, 1)

(1, 3)

Write an equation in slope-intercept form for each line.

7. 8. 9.

x

y

O

(1, 1)

(0, –2)

x

y

O

(0, 3)

(2, 0)x

y

O

(3, 2)

(0, 0)

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Writing the equation of a graphed line in the slope-intercept form

Recall The slope of the line containing (x1, y1) and (x2, y2) is given by .

◆ Example Write an equation for the line shown at right.

◆ SolutionThe line crosses the y-axis at (0, –1). Therefore,the y-intercept is –1.

The slope is .

The equation is .y �32

x � 1

2 � (�1)2 � 0

�32

y2 � y1

x2 � x1

(2, 2)

(0, –1)

x

y

O

◆Skill C Finding the x- and y-intercepts of the graph of a linear equation

Recall A line crosses the x-axis when y = 0; a line crosses the y-axis when x = 0.

◆ Example Find the x- and y-intercepts for the line given by 2x � y � 6.

◆ SolutionLet x � 0. Let y � 0.

2(0) � y � 6 2x � 0 � 6y � 6 x � 3

The x-intercept is 6 and the y-intercept is 3.

Find the x- and y-intercepts.

10. x � 2y � 8 11. �x � y � 5

12. 2x � 3y � 12 13. 3x � 2y � 0

14. 15. y � �2 y �12

x � 4


Lesson 1.1













13.

14.

15.

16.

17.

18.

19.

Lesson 1.2



3. slope: �1; y-intercept: 4

x

y

O

x + y = 4

x

y

O

y =

2x

x

y

Oy =

x + 2

y � 3x � 1

y �12

x

y � x � 1

y � 2x

x

y

O

y = x

y = x + 1 12

x

y

O

y =

3x –

2

y = 4 – x

x

y

O

y = x

+ 2

y = –2x

ANSWERSC

opyr

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©by

Hol

t, R

ineh

art a

nd W

inst

on. A

ll rig

hts

rese

rved

.


4. slope: ; y-intercept: 3

5. slope: 1; y-intercept: �1

6. slope: �2; y-intercept: �1

7.

8.

9.

10. x-intercept: 8; y-intercept: �4

11. x-intercept: �5; y-intercept: 5

12. x-intercept: 6; y-intercept: 4



15. x-intercept: none; y-intercept: �2

Lesson 1.3

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

Lesson 1.4

1.

2.

3.

4.

5.

6.

7. $104 8. $70 9. $10.60

10. $6.30 11. 12.

13. 14. 15.

16. 17. 80 18. $16.52

19. 32.5 minutes

y � 9

x � 17n � 12t � 813

a � 4x � 8

k � �15; y � �15x

k � �14

; y � �14

x

k �43

; y �43

x

k � 1.5; y � 1.5x

k �12

; y �12

x

k � 8; y � 8x

y � �52

x

y �25

x

y �12

x � 3

y � �2x � 13

y � �x � 4

y � x

y �13

x � 1

y � �3x � 1

y �23

x � 3; m �23

; �3

y � �2; m � 0; �2

y �47

x � 4; m �47

; �4

y � x; m � 1; 0

y � �12

x � 5; m � �12

; 5

y � �12

x; m � �12

; 0

y � 3x � 2

y � �32

x � 3

y �23

x

x

y

O 2x + y = –1

x

y

O

x – y

= 1

x

y

O

x + 2y = 6

�12

ANSWERSC

opyright ©by H

olt, Rinehart and W

inston. All rights reserved.

188 Answers Algebra 2

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olt,

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. All

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Writing a specified linear equation in slope-intercept form

Recall The point-slope form of the equation of a line with slope m that contains(x1, y1) is y � y1 � m(x � x1).

◆ Example Write an equation in slope-intercept form for the line containing the points(1, �1) and (2, 3).

◆ SolutionFind the slope of the line; .

Use the form y � y1 � m(x � x1). Replace m with 4, x1 with 2, and y1 with 3.

y � 3 � 4(x � 2)y � 3 � 4x � 8

y � 4x � 5 This is the slope-intercept form.

The slope is 4 and the y-intercept is �5.

Notice that if you had used the point (1,�1) rather than (2, 3), then you wouldhave the following:

y � (�1) � 4(x � 1)y � 4x � 5

m �3 � (�1)

2 � 1�

41

� 4

Write an equation for the line containing each pair of points.Then state the slope and the y-intercept of each line.

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

y � y � y �

slope: slope: slope:

y-intercept: y-intercept: y-intercept:

4. (7, 0) and (0, �4) 5. (�3, �2) and (5, �2) 6. (6, 1) and (0, �3)

y � y � y �

slope: slope: slope:

y-intercept: y-intercept: y-intercept:

Reteaching

1.3 Linear Equations in Two Variables

Write an equation in slope-intercept form for the line thatcontains the given point and is parallel to the given line. Thenwrite the equation for the line that contains the same point andis perpendicular to the given line.

(0, �1); y � �3x � 4

7. parallel to: 8. perpendicular to:

(2, 2); y � x � 5


(4, 5); 2x � y � 6


(0, 0); 2x �5y � 10


Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Writing equations for parallel or perpendicular lines

Recall Parallel lines have the same slope; perpendicular lines have slopes that are negativereciprocals of each other.

◆ Example 1Write an equation in slope-intercept form for the line that contains the point (2, 1) and is parallel to the line whose equation is y � 2x � 5.

◆ SolutionThe slope of the line with equation y � 2x � 5 is 2. Any line parallel to this linewill have a slope of 2. Use the point-slope form.

y � y1 � m(x – x1)y � 1 � 2(x – 2)y � 1 � 2x – 4

y � 2x � 3You can use a graphics calculator to check that these two lines are parallel.

◆ Example 2Write an equation in slope-intercept form for the line that contains the point (2, 1) and is perpendicular to the line whose equation is y � 2x � 5.

◆ SolutionSince the slope of the line with equation y � 2x � 5 is 2, the slope of the

perpendicular line will be the negative reciprocal of 2, which is .

y � �12

x � 2

y � 1 � �12

x � 1

y � 1 � �12

(x � 2)

�12




7.

8.

9.







Lesson 1.3

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

Lesson 1.4

1.

2.

3.

4.

5.

6.

7. $104 8. $70 9. $10.60

10. $6.30 11. 12.

13. 14. 15.

16. 17. 80 18. $16.52

19. 32.5 minutes

y � 9

x � 17n � 12t � 813

a � 4x � 8

k � �15; y � �15x

k � �14

; y � �14

x

k �43

; y �43

x

k � 1.5; y � 1.5x

k �12

; y �12

x

k � 8; y � 8x

y � �52

x

y �25

x

y �12

x � 3

y � �2x � 13

y � �x � 4

y � x

y �13

x � 1

y � �3x � 1

y �23

x � 3; m �23

; �3

y � �2; m � 0; �2

y �47

x � 4; m �47

; �4

y � x; m � 1; 0

y � �12

x � 5; m � �12

; 5

y � �12

x; m � �12

; 0

y � 3x � 2

y � �32

x � 3

y �23

x

x

y

O 2x + y = –1

x

y

O

x – y

= 1

x

y

O

x + 2y = 6

�12

ANSWERSC

opyright ©by H

olt, Rinehart and W



Cop

yrig

ht ©

by H

olt,

Rin

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Win

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. All

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serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Writing a direct variation equation

Recall An equation in the form y � kx is called a direct variation equation where k is theconstant of variation.

◆ Example The total cost of tickets to a local concert varies directly as the number of ticketsyou buy. If 4 tickets cost $72, find the constant of variation, k, and write anequation to show this direct variation. Then find the cost of 11 tickets.

◆ SolutionLet x represent the number of tickets and y represent the total cost. Then forsome value of k, y = kx.

72 � k(4)

k � 18Each ticket costs $18.An equation for the direct variation is y � 18x.To find the cost of 11 tickets, substitute 11 for x. Then find y.y � 18(11) � 198Thus, 11 tickets will cost $198.

724

� k

In Exercises 1–6, y varies directly as x. Find the constant ofvariation, and write an equation of direct variation that relatesthe two variables.

1. y � 40 when x � 5 2. y � 6 when x � 12 3. y � 15 when x � 10

k � k � k �

equation: equation: equation:

4. y � 1 when 5. y � 2 when x � �8 6. y � 3 when x � �0.2

k � k � k �

equation: equation: equation:

Use a direct variation equation to solve each problem.

7. If 5 tickets cost $65, find the cost of 8 tickets.

8. If 4 cassette tapes on sale cost $28, find the cost of 10 tapes.

9. If 6 colas cost $3.18, find the cost of 20 colas.

10. If 5 pens cost $1.75, find the cost of 18 pens.

x �34

Reteaching

1.4 Direct Variation and Proportion

Solve each proportion for the indicated variable. Check your answers.

11. 12. 13.

14. 15. 16.

Use a proportion to solve each problem.

17. If exactly 2 out of 3 students voted for candidate A, how many votes out

of the 120 cast were for candidate A?

18. If 8 gallons of gasoline cost $9.44, what is the cost for 14 gallons?

19. If Monica jogs at a constant speed and covers 2 miles in 13 minutes, how

long will it take her to jog 5 miles?

20. In the smaller of 2 similar right triangles, the hypotenuse measures 10centimeters and the shorter leg measures 3 centimeters. In the largertriangle the shorter leg measures 8 centimeters. How long is the

hypotenuse of the larger triangle?

y3

� y � 6x � 710

�x � 8

25n

24�

n � 330

16

�t

5010a

�2510

x12

�23

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Writing and solving proportions

Recall If , then ad � bc. (Cross Product Property)

◆ Example 1If 3 T-shirts cost $28.50, write and solve a proportion to find the cost of 5 T-shirts.

◆ Solution

3x � 5(28.50) Cross Product Property

3x � 142.50

x � 47.50Five T-shirts will cost $47.50.

◆ Example 2

Solve .

◆ Solution3(3a � 5) � 4a

9a � 15 � 4a5a � 15

a � 3

3a � 54

�a3

x �142.50

3

number of T- shirtstotal cost

→→

328.50

�5x

ab

�cd




7.

8.

9.







Lesson 1.3

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

Lesson 1.4

1.

2.

3.

4.

5.

6.

7. $104 8. $70 9. $10.60

10. $6.30 11. 12.

13. 14. 15.

16. 17. 80 18. $16.52

19. 32.5 minutes

y � 9

x � 17n � 12t � 813

a � 4x � 8

k � �15; y � �15x

k � �14

; y � �14

x

k �43

; y �43

x

k � 1.5; y � 1.5x

k �12

; y �12

x

k � 8; y � 8x

y � �52

x

y �25

x

y �12

x � 3

y � �2x � 13

y � �x � 4

y � x

y �13

x � 1

y � �3x � 1

y �23

x � 3; m �23

; �3

y � �2; m � 0; �2

y �47

x � 4; m �47

; �4

y � x; m � 1; 0

y � �12

x � 5; m � �12

; 5

y � �12

x; m � �12

; 0

y � 3x � 2

y � �32

x � 3

y �23

x

x

y

O 2x + y = –1

x

y

O

x – y

= 1

x

y

O

x + 2y = 6

�12

ANSWERSC

opyright ©by H

olt, Rinehart and W



20. centimeters

Lesson 1.5

1.

; moderate positive correlation

2.

; strong negative correlation

3.

4.

5.

6.

Lesson 1.6

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

Lesson 1.7

1.

2.

3.

4.

5.

6.

7.

8.

9.

0 2 4 6– 6 –4 –2

a �32

0 2 4 6– 6 –4 –2

x � �3

0 2 4 6– 6 –4 –2

y � 3

0 2 4 6– 6 –4 –2

c � �1

0 2 4 6– 6 –4 –2

m � �3

0 2 4 6– 6 –4 –2

t � 4

0 2 4 6– 6 –4 –2

y � 4

0 2 4 6– 6 –4 –2

a � �4

0 2 4 6– 6 –4 –2

x � �1

c �1

a � b

h �3Vπr2

m �Ec2

x �y

a � b � c

h �S

2πr� r

P �I

1 � rt

b � x(c � d) � a

r �C2π

a �p � 2b

2

h �2Ab

y � 0x � �8a � 3

x � �7.5c � �7r � 2

y � 2x � 9t � 3

a � 96y � 12x � 3

y � 0.1x � 5; y � 25; y � 70

y � �0.13x � 10.17; y � 4.97; y � �2.83

y � 0.77x � 13.9; y � 47; y � 21.6

y � 16x � 21.3; y � 117.3; y � 341.3

r � �0.995y � �0.73x � 8.87

x

y

O

2

4

6

8

10

2 4 6 8 10

r � 0.65y � 7.6x � 21.9

x

y

O

20

40

60

80

100

2 4 6 8

2623

ANSWERSC

opyr

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©by

Hol

t, R

ineh

art a

nd W

inst

on. A

ll rig

hts

rese

rved

.


y � y �

correlation correlation

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yrig

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olt,

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. All

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serv

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Using a scatter plot, least-squares line, and correlation coefficient to analyze data

Recall The least-squares line for a positive correlation will have a positive slope.

◆ Example Draw a scatter plot for the following data. Find an equation for the least-squaresline. Draw this line and describe the correlation.

◆ SolutionEnter the data in your calculator to find aleast-squares line of y ≈ 6.93x � 1.39 and acorrelation coefficient of r ≈ 0.994.

The graph and this value of r indicate a strongpositive correlation.

Create a scatter plot for the data in each table below. Describethe correlation. Then find an equation for the least-squares line.Draw this line on the scatter plot.

1. 2.

x

y

x

y

Reteaching

1.5 Scatter Plots and Least-Square Lines

X 2 3 4 5 6 7 8 9

Y 14 24 30 33 43 52 58 62

x

y

20

40

60

80

2 4 6 8 10

x 3 4 5 6 7 8

y 56 32 74 50 92 78

x 1 3 5 7 9 11

y 8 7 5 4 2 1

Use a graphics calculator to find the equation of the least-squares line for each set of data. Then find each value of y.Round answers to the nearest hundredth.

3. 4.

y � y �

If x � 6, then y � If x � 43, then y �


5. 6.

y � y �



Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Using a least-squares line to predict or estimate values of a variable

Recall Once you have found an equation for the least-squares line, you can use substitution to estimate or make a prediction of the value of the second variable.

◆ Example Randy used the following table to record miles he had driven and the amount of gas used.

Estimate the amount of gas needed for a trip of 275 miles and for a trip of 800 miles.

◆ SolutionUse a graphics calculator to find the equation of the least-squares line for this data.

y ≈ 0.037x � 0.2

Evaluate y ≈ 0.037x � 0.2 with x � 275. Then y ≈ 10.4.The 275-mile trip will require approximately 10.4 gallons of gas.

Evaluate y ≈ 0.037x � 0.2 with x � 800. Then y � 29.8.The 800-mile trip will require approximately 29.8 gallons of gas.

miles driven (x) 208 315 244 280 168 195 302

gallons of gas (y) 7.7 12.1 9.0 10.9 6.3 8.1 11.3

x 1 3 5 7 9 11

y 40 65 100 131 178 190

x 30 40 50 60 70 80

y 38 43 53 60 67 76

x 18 22 29 35 46 58

y 8 7 6.5 5 4.5 2.3

x 50 100 200 250 400 500

y 0 7 13 20 33 47

20. centimeters

Lesson 1.5

1.


2.


3.

4.

5.

6.

Lesson 1.6

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

Lesson 1.7

1.

2.

3.

4.

5.

6.

7.

8.

9.

0 2 4 6– 6 –4 –2

a �32

0 2 4 6– 6 –4 –2

x � �3

0 2 4 6– 6 –4 –2

y � 3

0 2 4 6– 6 –4 –2

c � �1

0 2 4 6– 6 –4 –2

m � �3

0 2 4 6– 6 –4 –2

t � 4

0 2 4 6– 6 –4 –2

y � 4

0 2 4 6– 6 –4 –2

a � �4

0 2 4 6– 6 –4 –2

x � �1

c �1

a � b

h �3Vπr2

m �Ec2

x �y

a � b � c

h �S

2πr� r

P �I

1 � rt

b � x(c � d) � a

r �C2π

a �p � 2b

2

h �2Ab

y � 0x � �8a � 3

x � �7.5c � �7r � 2

y � 2x � 9t � 3

a � 96y � 12x � 3

y � 0.1x � 5; y � 25; y � 70

y � �0.13x � 10.17; y � 4.97; y � �2.83

y � 0.77x � 13.9; y � 47; y � 21.6

y � 16x � 21.3; y � 117.3; y � 341.3

r � �0.995y � �0.73x � 8.87

x

y

O

2

4

6

8

10

2 4 6 8 10

r � 0.65y � 7.6x � 21.9

x

y

O

20

40

60

80

100

2 4 6 8

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by H

olt,

Rin

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Win

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serv

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Solving linear equations in one variable

Recall An equation is solved by using inverse operations.

◆ Example 1The total cost for a set of 4 CDs, including shipping and handling charges, is$49.50. If the shipping and handling charges are $3.50, what is the cost of each CD?

◆ Solution4n � 3.5 � 49.5 where n is the cost of 1 CD

4n � 3.5 � 3.5 � 49.5 � 3.5 Subtract 3.5 from each side of the equation.4n � 46

Divide each side by 4.n � 11.5

Each CD costs $11.50.To check your answer, show that 4(11.5) + 3.5 = 49.5.

◆ Example 2Solve 3x � 5 � 7x � 12.

◆ Solution3x � 5 � 7x � 12

3x � 5 � 3x � 7x � 12 � 3x Subtract 3x from each side of the equation.�5 � 4x � 12 Combine like terms.

�5 � 12 � 4x � 12 � 12�17 � 4x

Thus, x � , or .�414

�174

�174

� x

�174

�4x4

4n4

�464

Solve each equation.

1. 3x � 4 � 5 2.

3. 4. 1.5t � 3 � 7.5

5. 2(x � 3) � x � 3 6. 3(y � 1) � �(y � 5)

7. 2r � 1 � r � 5 � 3r 8.

9. 0.2(x � 5) � x � 5 10. 2a � (1 � a) � 11 � a

11. 12. 3(5y � 1) � 5(3y� 2) � 725

(x � 2) � x � 4

13

(c � 2) � c � 4

15 �16

a � �1

23

y � 8 � 0

Reteaching

1.6 Introduction to Solving Equations

Solve each literal equation for the indicated variable.

13. 14. p � 2a � 2b for a

15. C � 2π r for r 16.

17. I � P � Prt for P 18. S � 2πr (r � h) for h

19. y � ax � bx � cx for x 20. E � mc2 for m

21. 22. a � b �1c

for cV �13

πr2h for h

x �a � bc � d

for b

A �12

bh for h

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Solving literal equations for a specified variable

Recall Many formulas are literal equations that contain two or more variables.

◆ Example 1The formula for changing Celsius to Fahrenheit temperature is . To find a formula for changing Fahrenheit to Celsius, solvethis equation for C.

◆ Solution

Subtract 32 from each side of the equation.

Multiply each side of the equation by .

Thus, the equation for changing Fahrenheit temperature to Celsius is

◆ Example 2Solve y � a � abc for a.

◆ Solutiony � a � abcy � a(1 � bc) Distributive Property

Divide each side of the equation by 1 � bc.

y1 � bc

� a

y1 � bc

�a(1 � bc)

1 � bc

C �59

(F � 32).

59

(F � 32) � C

95

59

(F � 32) �59

�95

C

F � 32 �95

C

F � 32 �95

C � 32 � 32

F �95

C � 32

F �95

C � 32

20. centimeters

Lesson 1.5

1.


2.


3.

4.

5.

6.

Lesson 1.6

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

Lesson 1.7

1.

2.

3.

4.

5.

6.

7.

8.

9.

0 2 4 6– 6 –4 –2

a �32

0 2 4 6– 6 –4 –2

x � �3

0 2 4 6– 6 –4 –2

y � 3

0 2 4 6– 6 –4 –2

c � �1

0 2 4 6– 6 –4 –2

m � �3

0 2 4 6– 6 –4 –2

t � 4

0 2 4 6– 6 –4 –2

y � 4

0 2 4 6– 6 –4 –2

a � �4

0 2 4 6– 6 –4 –2

x � �1

c �1

a � b

h �3Vπr2

m �Ec2

x �y

a � b � c

h �S

2πr� r

P �I

1 � rt

b � x(c � d) � a

r �C2π

a �p � 2b

2

h �2Ab

y � 0x � �8a � 3

x � �7.5c � �7r � 2

y � 2x � 9t � 3

a � 96y � 12x � 3

y � 0.1x � 5; y � 25; y � 70

y � �0.13x � 10.17; y � 4.97; y � �2.83

y � 0.77x � 13.9; y � 47; y � 21.6

y � 16x � 21.3; y � 117.3; y � 341.3

r � �0.995y � �0.73x � 8.87

x

y

O

2

4

6

8

10

2 4 6 8 10

r � 0.65y � 7.6x � 21.9

x

y

O

20

40

60

80

100

2 4 6 8

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by H

olt,

Rin

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Win

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. All

right

s re

serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Solving linear inequalities in one variable and graphing solutions on a number line

Recall When you multiply or divide each side of an inequality by a positive number, theinequality sign remains the same.

◆ Example 1Solve 5x � 7 � 2x � 11. Graph the solution on a number line.

◆ Solution5x � 7 � 2x � 11

5x � 7 � 2x � 2x � 11 � 2x3x � 7 � 11

3x � 7 � 7 � 11 � 73x � 18

x � 6Notice the open circle at 6 because the number 6 is not in the solution set.

Recall When you multiply or divide each side of an inequality by a negative number, reverse the inequality symbol.

◆ Example 2Solve �4a � 10 � 20. Graph the solution on a number line.

◆ Solution�4a � 10 � 20

�4a � 10 � 10 � 20 � 10�4a � 30

a � �7.5Notice the filled circle at �7.5 because the number �7.5 is in the solution set.

�4a�4

�30�4

3xx

�183

Solve each inequality and graph the solution on the number line.

1. 3x � 1 � �4 2. 3a � 6a � 12 3. y � 7y � 24

4. 5. 6. 2 � c � 4 � c

7. �12 � �4y 8. 2x � 5 � �1 9. 3(a � 2) � 2 � a � 5

0 2 4 6– 6 –4 –20 2 4 6– 6 –4 –20 2 4 6– 6 –4 –2

0 2 4 6– 6 –4 –20 2 4 6– 6 –4 –20 2 4 6– 6 –4 –2

�m2

�32

�34

t � �3

0 2 4 6– 6 –4 –20 2 4 6– 6 –4 –20 2 4 6– 6 –4 –2

Reteaching

1.7 Introduction to Solving Inequalities

0 4 8–8 –4

0 4 8–8 –4

Graph the solution of each compound inequality on a number line.

10. and 11. and

12. and 13. and

14. or 15. or

16. or 17. or

0 2 4 6 8 10–10 –8 –6 –4 –20 2 4 6 8 10–10 –8 –6 –4 –2

a � 4a � 2x � 3x � �2

0 2 4 6 8 10–10 –8 –6 –4 –20 2 4 6 8 10–10 –8 –6 –4 –2

5y � 3 � 123y � 4 � 73z � 5 � 102z � 1 � 5

0 2 4 6 8 10–10 –8 –6 –4 –20 2 4 6 8 10–10 –8 –6 –4 –2

x � 5x � �4y � 5y � �4

0 2 4 6 8 10–10 –8 –6 –4 –20 2 4 6 8 10–10 –8 –6 –4 –2

3x � 2 � 74x � 3 � �53x � 2 � 83x � 2 � �1

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Solving and graphing compound linear inequalities in one variable

Recall An inequality involving and is true only if both parts of the inequality are true.

◆ Example 1Solve and . Graph the solution on a number line.

◆ Solution

and

◆ Example 2Graph the solution for and .

◆ Solution

Both inequalities are true only if .

Recall An inequality involving or is true if at least one part of the inequality is true.

◆ Example 3Solve and graph the solution for or .

◆ Solution

or

0 2 4 6– 6 –4 –2

x � 2x � 35x � 102x � 6

5x � 4 � 62x � 7 � 13

5x � 4 � 62x � 7 � 13

x � 2

0 2 4 6– 6 –4 –2

x � 2x � �1

0 2 4 6– 6 –4 –2

x � 3x � �15x � 152x � �2

5x � 9 � 62x � 3 � 1

5x � 9 � 62x � 3 � 1

20. centimeters

Lesson 1.5

1.


2.


3.

4.

5.

6.

Lesson 1.6

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

Lesson 1.7

1.

2.

3.

4.

5.

6.

7.

8.

9.

0 2 4 6– 6 –4 –2

a �32

0 2 4 6– 6 –4 –2

x � �3

0 2 4 6– 6 –4 –2

y � 3

0 2 4 6– 6 –4 –2

c � �1

0 2 4 6– 6 –4 –2

m � �3

0 2 4 6– 6 –4 –2

t � 4

0 2 4 6– 6 –4 –2

y � 4

0 2 4 6– 6 –4 –2

a � �4

0 2 4 6– 6 –4 –2

x � �1

c �1

a � b

h �3Vπr2

m �Ec2

x �y

a � b � c

h �S

2πr� r

P �I

1 � rt

b � x(c � d) � a

r �C2π

a �p � 2b

2

h �2Ab

y � 0x � �8a � 3

x � �7.5c � �7r � 2

y � 2x � 9t � 3

a � 96y � 12x � 3

y � 0.1x � 5; y � 25; y � 70

y � �0.13x � 10.17; y � 4.97; y � �2.83

y � 0.77x � 13.9; y � 47; y � 21.6

y � 16x � 21.3; y � 117.3; y � 341.3

r � �0.995y � �0.73x � 8.87

x

y

O

2

4

6

8

10

2 4 6 8 10

r � 0.65y � 7.6x � 21.9

x

y

O

20

40

60

80

100

2 4 6 8

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10.

11.

12. no solution

13.

14.

15. or

16.

17.

Lesson 1.8

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12. no solution

13.

14.

15.

16.

17.

0 2 4 6 8 10–10 –8 –6 –4 –2

b � �4 or b � �1

0 2 4 6 8 10–10 –8 –6 –4 –2

a � �4 and a � �2

0 2 4 6 8 10–10 –8 –6 –4 –2

a � �4 or a � �2

0 2 4 6 8 10–10 –8 –6 –4 –2

x � �3 and x � 3

0 2 4 6 8 10–10 –8 –6 –4 –2

x � �3 or x � 3

0 2 4 6– 6 –4 –2

0 2 4 6– 6 –4 –2

m � �3 or m � 5

0 2 4 6– 6 –4 –2

x � 0 or x � 6

0 2 4 6– 6 –4 –2

t � �6 or t � �2

0 2 4 6– 6 –4 –2

x � �2 or x �23

0 2 4 6– 6 –4 –2

a � 2.5

0 2 4 6– 6 –4 –2

t � �4 or t � �1

0 2 4 6– 6 –4 –2

y � �5 or y � �1

0 2 4 6– 6 –4 –2

x � 4 or x � 6

0 2 4 6– 6 –4 –2

b � 0 or b � 6

0 2 4 6– 6 –4 –2

a � �3 or a � �1

0 2 4 6– 6 –4 –2

x � 3 or x � �3

0 2 4 6 8 10–10 –8 –6 –4 –2

a � 2 or a � 4

0 2 4 6 8 10–10 –8 –6 –4 –2

x � 3

0 2 4 6 8 10–10 –8 –6 –4 –2

y � 3y � 1

0 2 4 6 8 10–10 –8 –6 –4 –2

z � 3 or z � 5

0 2 4 6 8 10–10 –8 –6 –4 –2

x � �4 and x � 5

0 2 4 6 8 10–10 –8 –6 –4 –2

0 2 4 6 8 10–10 –8 –6 –4 –2

x � �2 and x � �3

0 2 4 6 8 10–10 –8 –6 –4 –2

x � �1 and x � 2

ANSWERSC

opyright ©by H

olt, Rinehart and W



Cop

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olt,

Rin

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Win

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serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Solving absolute-value equations and graphing solutions on a number line

Recall The absolute value of x is the distance between x and 0 on the number line.

◆ Example 1Solve . Graph the solution on a number line.

◆ Solutionor


◆ Solutionor

or x � �1x � 42x � �22x � 8

0 2 4 6– 6 –4 –22x � 3 � �52x � 3 � 5

�2x � 3� � 5

0 2 4 6– 6 –4 –2x � �4x � 4

�x� � 4

Solve each equation. Graph the solution on a number line.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

0 2 4 6– 6 –4 –20 2 4 6– 6 –4 –2

�6 � 2r� � �2�1 � m� � 3 � 1

0 2 4 6– 6 –4 –20 2 4 6– 6 –4 –2

�x � 3� � 2 � 5�14 t � 1� �12

0 2 4 6– 6 –4 –20 2 4 6– 6 –4 –2

�3x � 2� � 4�2a � 5� � 0

0 2 4 6– 6 –4 –20 2 4 6– 6 –4 –2

�2t � 5� � 3�3 � y� � 2

0 2 4 6– 6 –4 –20 2 4 6– 6 –4 –2

�5 � x� � 1�b � 3� � 3

0 2 4 6– 6 –4 –20 2 4 6– 6 –4 –2

�a � 2� � 1�x� � 3

Reteaching

1.8 Solving Absolute-Value Equations and Inequalities

Solve each inequality. Graph the solution on the number line.

13. 14.

15. 16.

17. 18.

19. 20.

0 2 4 6 8 10–10 –8 –6 –4 –20 2 4 6 8 10–10 –8 –6 –4 –2

�6 � 3y� � 9�4c � 2� � 10

0 2 4 6 8 10–10 –8 –6 –4 –20 2 4 6 8 10–10 –8 –6 –4 –2

�4 � 2x� � 8�2b � 5� � 3

0 2 4 6 8 10–10 –8 –6 –4 –20 2 4 6 8 10–10 –8 –6 –4 –2

�a � 3� � 1�a � 3� � 1

0 2 4 6 8 10–10 –8 –6 –4 –20 2 4 6 8 10–10 –8 –6 –4 –2

�x� � 3�x� � 3

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Solving absolute-value inequalities and graphing solutions on a number line

Recall The solution of , where a is nonnegative, is all real numbers less than a andgreater than �a.


◆ Solutionand

and

Recall The solution of , where a is nonnegative, is all real numbers less than �a orgreater than a.


◆ Solution

or

or

0 2 4 6– 6 –4 –2

a � 4a � �1�2a � �8�2a � 2

3 � 2a � �53 � 2a � 5�3 � 2a� � 5

�3 � 2a� � 4 � 1

�3 � 2a� � 4 � 1

�x� � a

0 2 4 6– 6 –4 –2

x � �223

x � 4

3x � �83x � 123x � 2 � �103x � 2 � 10

�3x � 2� � 10

�x� � a

10.

11.

12. no solution

13.

14.

15. or

16.

17.

Lesson 1.8

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12. no solution

13.

14.

15.

16.

17.

0 2 4 6 8 10–10 –8 –6 –4 –2

b � �4 or b � �1

0 2 4 6 8 10–10 –8 –6 –4 –2

a � �4 and a � �2

0 2 4 6 8 10–10 –8 –6 –4 –2

a � �4 or a � �2

0 2 4 6 8 10–10 –8 –6 –4 –2

x � �3 and x � 3

0 2 4 6 8 10–10 –8 –6 –4 –2

x � �3 or x � 3

0 2 4 6– 6 –4 –2

0 2 4 6– 6 –4 –2

m � �3 or m � 5

0 2 4 6– 6 –4 –2

x � 0 or x � 6

0 2 4 6– 6 –4 –2

t � �6 or t � �2

0 2 4 6– 6 –4 –2

x � �2 or x �23

0 2 4 6– 6 –4 –2

a � 2.5

0 2 4 6– 6 –4 –2

t � �4 or t � �1

0 2 4 6– 6 –4 –2

y � �5 or y � �1

0 2 4 6– 6 –4 –2

x � 4 or x � 6

0 2 4 6– 6 –4 –2

b � 0 or b � 6

0 2 4 6– 6 –4 –2

a � �3 or a � �1

0 2 4 6– 6 –4 –2

x � 3 or x � �3

0 2 4 6 8 10–10 –8 –6 –4 –2

a � 2 or a � 4

0 2 4 6 8 10–10 –8 –6 –4 –2

x � 3

0 2 4 6 8 10–10 –8 –6 –4 –2

y � 3y � 1

0 2 4 6 8 10–10 –8 –6 –4 –2

z � 3 or z � 5

0 2 4 6 8 10–10 –8 –6 –4 –2

x � �4 and x � 5

0 2 4 6 8 10–10 –8 –6 –4 –2

0 2 4 6 8 10–10 –8 –6 –4 –2

x � �2 and x � �3

0 2 4 6 8 10–10 –8 –6 –4 –2

x � �1 and x � 2

ANSWERSC

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olt, Rinehart and W



18.

19.

20.


Lesson 2.1

1–10.

11. True; Associative Property of Multiplication

12. True; Inverse Property of Addition

13. False 14. False

15. True; Commutative Property ofMultiplication

16. True; Identity for Multiplication

17. True; Distributive Property

18. True; Commutative Property of Addition

19. 38 20. 2 21. 9 22. 2 23. 10

24. 1 25. 65 26. 0 27. 0

Lesson 2.2

1. 25 2. 32 3. 6 4. 1 5. 36 6. 81

7. 8. 9. 10. 11.

12. 13. 14. 4 15. 64 16. 9

17. 7 18. 19. 25 20. 20 21. 210

22. 2�1 23. 2�3 24. 25.

26. �2�4 27. not possible 28. 22 29. a9

30. 25x5 31. 3x3y3 32. �6a4b5 33. c8d9

34. 72x12y9 35. 36. 37.

38.

Lesson 2.3

1. no; domain: {1, 2}; range: {�2, �1, 1, 2}

2. yes; domain: {0, 1, 2, 3}; range: {0, 1, 2, 3}

3. yes; domain: {1, 2, 3, 4}; range: {�4, �3, �2, –1}

4. no; domain: {4, 9}; range: {�3, �2, 2, 3}

5. no; domain: x � 0; range: all real numbers

6. yes; domain: {�5, �3, �2, 0, 1, 3, 4}; range: {�4, �1, 0, 1, 3, 4}

7. 1 8. 5 9. 10. 11. �4

12. 13. 14. 15. 0 16. 9

17. f (t) � 60t

18. f (h) � 10 + 0.6h

19. f (e) � 6e2

20. f (r) � r3

Lesson 2.4

1. 3x2 � 2x � 1

2. 2x2 � 5x � 2

3. x2 � 3x � 1

4. 2x3 � 9x2 � 5x

5. 2x3 � 9x2 � 5x

4π3

1813

413

158

312

423

108x21y13

25z4

x16

y10y

4x2�z15

y10

2132

52

127

125

�125

94

32

49

18

12

Real Numbers

Rational

Integers

25

Whole

Natural

Irrational23

05

311

6.8 135

–1.6

–8 –√25

√2 3π

0 2 4 6 8 10–10 –8 –6 –4 –2

y � �5 or y � 1

0 2 4 6 8 10–10 –8 –6 –4 –2

c � �2 and c � 3

0 2 4 6 8 10–10 –8 –6 –4 –2

x � �2 and x � 6

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◆Skill A Classifying real numbers

Recall You can classify a real number as belonging to the natural numbers, wholenumbers, integers, rational numbers, or irrational numbers. A real number canbelong to more than one set of numbers.

◆ ExampleClassify �4 in as many ways as possible.

◆ Solution4 is not a natural number because natural numbers are positive whole

numbers.4 is not a whole number because whole numbers are either positive or 0.4 is an integer because integers are all the whole numbers and their opposites.4 is a rational number because it can be written as the terminating decimal �4.0.4 is a real number.

The number 4 is an integer, a rational number, and a real number.�

�

�

�

�

�

Use the diagram to classify each number in as many ways as possible by writing it in the smallest

rectangle in which it belongs. For example, is

placed in the rectangle labeled rational.

1. �8 2. 25 3. 6.8

4. 5. 6.

7. 8. 9.

10. 3π

05

��25311

�1.6513

�2

23

Reteaching

2.1 Operations With Numbers

Real Numbers

Rational

Integers

Whole

Natural

Irrational23

◆Skill B Identifying properties of real numbers

Recall The real numbers are characterized by the Commutative and Associative Propertiesof Addition and Multiplication and by the Distributive Property.

◆ ExampleTell if the statement is true or false. Justify your response.

a. b. c.

◆ Solution

a. True Commutative Property of Multiplicationb. False Subtraction is not associative.c. True Distributive Property

�4(a � b) � �4a � 4bx � (y � z) � (x � y) � z3ab2 � 3b2a

Tell whether each statement is true or false. State the propertythat is illustrated in each true statement. All variables representreal numbers.

11. 12.

13. 14.

15. 16.

17. 18. 5(3 � y) � 5(y � 3)3(x � wy) � 3x � 3wy

1 � ax � axabd � adb

5 � x � x � 57x( 17x) � 0

5x � (�5x) � 0(16a)b � 16(ab)

Copyright ©

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ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


Simplify each expression. Use a calculator to check.

19. 20.

21. 22.

23. 24.

25. 26.

27. 6 � {6 � [6 � (6 � 2)] � 2}

(7 � 3)2

7 � 3� 448 � 6 2 � 5 � 4

3 � 15

32 � 12 � 5(3�2)

8 � 25 � 2

18 6 � 3

(10 � 3) � (2 � 3)5 � 32 � 7

◆Skill C Simplifying numerical expressions by using the order of operations

Recall The order of operations can be remembered by using the following sentence.

Please Excuse My Dear Aunt Sally

Parentheses, Exponents, Multiplication and Division, Addition and Subtraction

◆ Example

Simplify .

◆ Solution

Work inside parentheses first.

The fraction bar is a grouping symbol.

Perform exponentiation.

Check: Enter into a calculator. Note the use of parenthesesaround . The display will show 9.6 � (3 � 1)

2�3/(6 � (3 � 1)) � 5

� 9� 4 � 5

�82

� 5

�23

2� 5

�23

6 � 4� 523

6 � (3 � 1)� 5

23

6 � (3 � 1)� 5

18.

19.

20.


Lesson 2.1

1–10.



13. False 14. False





19. 38 20. 2 21. 9 22. 2 23. 10

24. 1 25. 65 26. 0 27. 0

Lesson 2.2

1. 25 2. 32 3. 6 4. 1 5. 36 6. 81

7. 8. 9. 10. 11.

12. 13. 14. 4 15. 64 16. 9

17. 7 18. 19. 25 20. 20 21. 210

22. 2�1 23. 2�3 24. 25.

26. �2�4 27. not possible 28. 22 29. a9

30. 25x5 31. 3x3y3 32. �6a4b5 33. c8d9

34. 72x12y9 35. 36. 37.

38.

Lesson 2.3

1. no; domain: {1, 2}; range: {�2, �1, 1, 2}

2. yes; domain: {0, 1, 2, 3}; range: {0, 1, 2, 3}

3. yes; domain: {1, 2, 3, 4}; range: {�4, �3, �2, –1}

4. no; domain: {4, 9}; range: {�3, �2, 2, 3}


6. yes; domain: {�5, �3, �2, 0, 1, 3, 4}; range: {�4, �1, 0, 1, 3, 4}

7. 1 8. 5 9. 10. 11. �4

12. 13. 14. 15. 0 16. 9

17. f (t) � 60t

18. f (h) � 10 + 0.6h

19. f (e) � 6e2

20. f (r) � r3

Lesson 2.4

1. 3x2 � 2x � 1

2. 2x2 � 5x � 2

3. x2 � 3x � 1

4. 2x3 � 9x2 � 5x

5. 2x3 � 9x2 � 5x

4π3

1813

413

158

312

423

108x21y13

25z4

x16

y10y

4x2�z15

y10

2132

52

127

125

�125

94

32

49

18

12

Real Numbers

Rational

Integers

25

Whole

Natural

Irrational23

05

311

6.8 135

–1.6

–8 –√25

√2 3π

0 2 4 6 8 10–10 –8 –6 –4 –2

y � �5 or y � 1

0 2 4 6 8 10–10 –8 –6 –4 –2

c � �2 and c � 3

0 2 4 6 8 10–10 –8 –6 –4 –2

x � �2 and x � 6

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◆Skill A Evaluating numerical expressions with exponents

Recall You can associate a negative exponent with a reciprocal.

◆ Example 1Without using a calculator, evaluate each expression.

a. 43 b. 40 c. 4�1 d. 4�3 e. f.

◆ Solutiona.b. For any nonzero number a, a0 � 1.

c. A negative exponent indicates a reciprocal.

Notice that the result is not negative.

d.

e.

f.

◆ Example 2Express each number as a power of 3.

a. 81 b. 1 c. 93 d. e. f.

◆ Solutiona. b. c.

d. e. f. �27 � �33 � 332�3 � 3

121

27�

133 � 3�3

93 � (32)3 � 361 � 3081 � 34

�27�3127

amn � ( n�a)m4

32 � (�4)3 � 23 � 8

a1n �

n�a412 � �4 � 2

4�3 �143 �

164

4�1 �141 �

14

40 � 143 � 4 � 4 � 4 � 64

4324

14

Evaluate each expression.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

13. 14. 15.

16. 17. 18.

Express each number as a power of 2, if possible.

19. 32 20. 1 21. 45 22. 23.

24. 25. 26. 27. 0 28. 823�

116

3�2�32

18

12

9�

3249

3627

23

163216

14(�

15)2

(15)2(2

3)�2(23)�1

(23)2

2�32�1

30 � 33 � 3(2 � 3)2(6x)0

6x02552

Reteaching

2.2 Properties of Exponents

Simplify each expression, assuming that no variable equals zero.Write your answer using positive exponents only.

29. 30.

31. 32.

33. 34.

35. 36.

37. 38. ( 3xy�3)3(5x�10yz2

2x�1y3 )�2(x�5y�1)�2(x2y�4)3

(4x2y�3

y�2 )�1(�z3

y2)5

(2x2y3)3(3x3y0)2(c2d)3(cd 3)2

(a2b)(�3ab3)(2ab)(3x2y)(xy2)

x3(5x)2a2a4a3

Copyright ©

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ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Simplifying algebraic expressions involving exponents

Recall When you simplify an algebraic expression involving exponents, use Properties ofExponents.

◆ Example 1

Simplify . Write your answer using positive exponents only.

◆ Solution

Power of a Quotient:

Power of a Power:

Quotient of Powers:

◆ Example 2

Simplify . Write your answer using positive exponents only.

◆ Solution

Power of a Product: (ab)n � anbn

Power of Powers: (am)n � amn

Definition of negative exponent: a�n �1an�

5x4

y2

� 5x4y�2

� 5x(y�2x3)5x(yx�

32)�2

5x(yx�

32)�2

�x32

16y16

am

an � am�n�x20�(�12)

16y16

(a m)n � a mn�x20

16x�12y16

(ab)n

�an

bn�(�x5)4

(2x�3y4)4( �x5

2x�3y4)4

( �x5

2x�3y4)4

18.

19.

20.


Lesson 2.1

1–10.



13. False 14. False





19. 38 20. 2 21. 9 22. 2 23. 10

24. 1 25. 65 26. 0 27. 0

Lesson 2.2

1. 25 2. 32 3. 6 4. 1 5. 36 6. 81

7. 8. 9. 10. 11.

12. 13. 14. 4 15. 64 16. 9

17. 7 18. 19. 25 20. 20 21. 210

22. 2�1 23. 2�3 24. 25.

26. �2�4 27. not possible 28. 22 29. a9

30. 25x5 31. 3x3y3 32. �6a4b5 33. c8d9

34. 72x12y9 35. 36. 37.

38.

Lesson 2.3

1. no; domain: {1, 2}; range: {�2, �1, 1, 2}

2. yes; domain: {0, 1, 2, 3}; range: {0, 1, 2, 3}

3. yes; domain: {1, 2, 3, 4}; range: {�4, �3, �2, –1}

4. no; domain: {4, 9}; range: {�3, �2, 2, 3}


6. yes; domain: {�5, �3, �2, 0, 1, 3, 4}; range: {�4, �1, 0, 1, 3, 4}

7. 1 8. 5 9. 10. 11. �4

12. 13. 14. 15. 0 16. 9

17. f (t) � 60t

18. f (h) � 10 + 0.6h

19. f (e) � 6e2

20. f (r) � r3

Lesson 2.4

1. 3x2 � 2x � 1

2. 2x2 � 5x � 2

3. x2 � 3x � 1

4. 2x3 � 9x2 � 5x

5. 2x3 � 9x2 � 5x

4π3

1813

413

158

312

423

108x21y13

25z4

x16

y10y

4x2�z15

y10

2132

52

127

125

�125

94

32

49

18

12

Real Numbers

Rational

Integers

25

Whole

Natural

Irrational23

05

311

6.8 135

–1.6

–8 –√25

√2 3π

0 2 4 6 8 10–10 –8 –6 –4 –2

y � �5 or y � 1

0 2 4 6 8 10–10 –8 –6 –4 –2

c � �2 and c � 3

0 2 4 6 8 10–10 –8 –6 –4 –2

x � �2 and x � 6

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◆Skill A Identifying that a given relation is a function

Recall A function is a relation in which each value in the domain is paired with exactlyone value in the range.

◆ Example 1Does the table at right represent a function?

◆ SolutionYes; for each value of x there is only one value of y. Notice that the two x-values of �2 and 2 have the same y-value, 4. This is allowed in the definition of function.

◆ Example 2Does the solid-line graph shown represent a function? State the domain and range.

◆ SolutionYes; notice that any vertical line will intersect the graph in no more than 1 point.

domain: range: y � �4x � �2

State whether each relation represents a function and give thedomain and range.

1. 2.

domain: range: domain: range:

3. 4.


5. 6.


x

y

O

2

4

2 4–4

–4

–2–2

x

y

O

2

4

2 4–4

–4

–2–2

{(0, 0), (1, 1), (2, 2), (3, 3)}{(2, �2), (1, �1), (1, 1), (2, 2)}

Reteaching

2.3 Introduction to Functions

x

y

2

4

2 4–4

–4

–2–2

x 1 2 3 4

y �4 �3 �2 �1

x �4 �2 0 2

y 16 4 0 4

x 4 4 9 9

y 2 �2 3 �3

Let and . Evaluate each function.

7. 8.

9. 10.

11. 12.

13. 14.

15. 16.

Write each situation in function notation.

17. Driving at 60 miles per hour, the distance you travel depends on the number of hours spent driving.

18. The charge for electric service is $10.00 plus $0.60 for each kilowatt-hour of electricity that you use each month.

19. The surface area of a cube is 6 times the square of thelength of one edge.

20. The volume of a sphere is times the radius cubed.4π3

g(�6) � f(�6)f(0) � g(0)

g(4) � f(5)f(1) � g(0)

g(12)g(�2)

g(1)f(12)

f(0)f(6)

g(x) �12 x2 � 3xf(x) � 5 �

2x3

Copyright ©

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ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Writing and evaluating functions

Recall The value of depends on the value of x.

◆ Example 1Sarah uses an internet server which charges $12.50 per month plus $0.60 foreach hour over 20 hours that she uses it during the month. Write this relation infunction notation. How much will she be charged for using the service for 38hours in April?

◆ SolutionLet h � number of hours over 20. Thus, the function is as follows.

where h � 18

The charge for April will be $35.30.

◆ Example 2If , find .

◆ Solutionmeans replace x with the value and evaluate .

Thus, .� 10g(�5)� 10� 25 � 15� (�5)2 � 3(�5)g(�5)

g(x)�5g(�5)

g(�5)g(x) � x2 � 3x

f(18) � 23.30f(18) � 12.50 � 0.60(18)

f(h) � 12.50 � 0.60h

f(x) � x2 � 5

18.

19.

20.


Lesson 2.1

1–10.



13. False 14. False





19. 38 20. 2 21. 9 22. 2 23. 10

24. 1 25. 65 26. 0 27. 0

Lesson 2.2

1. 25 2. 32 3. 6 4. 1 5. 36 6. 81

7. 8. 9. 10. 11.

12. 13. 14. 4 15. 64 16. 9

17. 7 18. 19. 25 20. 20 21. 210

22. 2�1 23. 2�3 24. 25.

26. �2�4 27. not possible 28. 22 29. a9

30. 25x5 31. 3x3y3 32. �6a4b5 33. c8d9

34. 72x12y9 35. 36. 37.

38.

Lesson 2.3

1. no; domain: {1, 2}; range: {�2, �1, 1, 2}

2. yes; domain: {0, 1, 2, 3}; range: {0, 1, 2, 3}

3. yes; domain: {1, 2, 3, 4}; range: {�4, �3, �2, –1}

4. no; domain: {4, 9}; range: {�3, �2, 2, 3}


6. yes; domain: {�5, �3, �2, 0, 1, 3, 4}; range: {�4, �1, 0, 1, 3, 4}

7. 1 8. 5 9. 10. 11. �4

12. 13. 14. 15. 0 16. 9

17. f (t) � 60t

18. f (h) � 10 + 0.6h

19. f (e) � 6e2

20. f (r) � r3

Lesson 2.4

1. 3x2 � 2x � 1

2. 2x2 � 5x � 2

3. x2 � 3x � 1

4. 2x3 � 9x2 � 5x

5. 2x3 � 9x2 � 5x

4π3

1813

413

158

312

423

108x21y13

25z4

x16

y10y

4x2�z15

y10

2132

52

127

125

�125

94

32

49

18

12

Real Numbers

Rational

Integers

25

Whole

Natural

Irrational23

05

311

6.8 135

–1.6

–8 –√25

√2 3π

0 2 4 6 8 10–10 –8 –6 –4 –2

y � �5 or y � 1

0 2 4 6 8 10–10 –8 –6 –4 –2

c � �2 and c � 3

0 2 4 6 8 10–10 –8 –6 –4 –2

x � �2 and x � 6

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◆Skill A Using the four basic operations on functions to write new functions

Recall To write the sum, difference, product, or quotient of two functions, f and g, writethe sum, difference, product, or quotient of the expressions that define f and g.Then simplify.

◆ ExampleLet and . Write an expression for each function.

a. b. c. d.

◆ Solutiona.

Combine like terms.

b.

Combine like terms.

c.

Distributive Property

d. , where

, where x 15

�x2 � 3x � 2

5x � 1

g(x) 0�f(x)g(x)( f

g)(x)

� 5x3 � 14x2 � 7x � 2� 5x3 � 15x2 � 10x � x2 � 3x � 2� (x2 � 3x � 2)(5x) � (x2 � 3x � 2)(�1)� (x2 � 3x � 2)(5x � 1)� f(x) � g(x)(fg)(x)

� x2 � 2x � 3� x2 � 3x � 2 � 5x � 1� (x2 � 3x � 2) � (5x � 1)� f(x) � g(x)(f � g)(x)

� x2 � 8x � 1� (x2 � 3x � 2) � (5x � 1)� f(x) � g(x)(f � g)(x)

( fg)(x)(fg)(x)(f � g)(x)(f � g)(x)

5x � 1g(x) �x2 � 3x � 2f(x) �

Let , and . Find each newfunction, and state any domain restrictions.

1. 2.

3. 4.

5. 6.

7. 8. (hg)(x)( f

g)(x)

(f � h)(x)(hg)(x)

(gh)(x)(h � g)(x)

(f � h)(x)(f � g)(x)

h(x) � x2 � 5xf(x) � 3x2 � 2, g(x) � 2x � 1

Reteaching

2.4 Operations With Functions

Let , and . Find each compositefunction.

9. (x) 10. (x)

11. (x) 12. (x)

13. (x) 14. (x)

15. 16.

17. 18. (g � (g � g))(5)(f � (g � h))(1)

(f � f )(�3)(g � h)(4)

(h � h)(g � g)

(h � g)(h � f )

(g � f )(f � g)

h(x) � 5 � xf(x) � x2 � 1, g(x) � 3x

Copyright ©

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ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Finding the composite of two functions

Recall To write an expression for the composite function (x), replace each x in theexpression for f with the expression defining g. Then simplify the result.

◆ ExampleLet and . Find (2) and (2). Then writeexpressions for (x) and (x).

◆ Solution(2):

Thus, (2) .

(2):

Thus, (2) .

To write expressions for (x) and (x), use the variable x instead of aparticular number.

(x) (x)

� 50x2 � 3� 10x2 � 15

� 2(5x)2 � 3� 5(2x2 � 3)

� g(5x)� f(2x2 � 3)

� g(f(x))(g � f )� f(g(x))(f � g)

(g � f )(f � g)

� 197(g � f )

g(f(2)) � g(10) � 2(10)2 � 3 � 197f(2) � 5(2) � 10(g � f )

� 25(f � g)

f(g(2)) � f(5) � 5(5) � 25g(2) � 2(2)2 � 3 � 5(f � g)

(g � f )(f � g)(g � f )(f � g)g(x) � 2x2 � 3f(x) � 5x

(f � g)

18.

19.

20.


Lesson 2.1

1–10.



13. False 14. False





19. 38 20. 2 21. 9 22. 2 23. 10

24. 1 25. 65 26. 0 27. 0

Lesson 2.2

1. 25 2. 32 3. 6 4. 1 5. 36 6. 81

7. 8. 9. 10. 11.

12. 13. 14. 4 15. 64 16. 9

17. 7 18. 19. 25 20. 20 21. 210

22. 2�1 23. 2�3 24. 25.

26. �2�4 27. not possible 28. 22 29. a9

30. 25x5 31. 3x3y3 32. �6a4b5 33. c8d9

34. 72x12y9 35. 36. 37.

38.

Lesson 2.3

1. no; domain: {1, 2}; range: {�2, �1, 1, 2}

2. yes; domain: {0, 1, 2, 3}; range: {0, 1, 2, 3}

3. yes; domain: {1, 2, 3, 4}; range: {�4, �3, �2, –1}

4. no; domain: {4, 9}; range: {�3, �2, 2, 3}


6. yes; domain: {�5, �3, �2, 0, 1, 3, 4}; range: {�4, �1, 0, 1, 3, 4}

7. 1 8. 5 9. 10. 11. �4

12. 13. 14. 15. 0 16. 9

17. f (t) � 60t

18. f (h) � 10 + 0.6h

19. f (e) � 6e2

20. f (r) � r3

Lesson 2.4

1. 3x2 � 2x � 1

2. 2x2 � 5x � 2

3. x2 � 3x � 1

4. 2x3 � 9x2 � 5x

5. 2x3 � 9x2 � 5x

4π3

1813

413

158

312

423

108x21y13

25z4

x16

y10y

4x2�z15

y10

2132

52

127

125

�125

94

32

49

18

12

Real Numbers

Rational

Integers

25

Whole

Natural

Irrational23

05

311

6.8 135

–1.6

–8 –√25

√2 3π

0 2 4 6 8 10–10 –8 –6 –4 –2

y � �5 or y � 1

0 2 4 6 8 10–10 –8 –6 –4 –2

c � �2 and c � 3

0 2 4 6 8 10–10 –8 –6 –4 –2

x � �2 and x � 6

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6. 4x2 � 5x � 2

7.

8.

9. 9x2 � 1

10. 3x2 � 3

11. 6 � x2

12. 5 � 3x

13. 9x 14. x 15. 3 16. 63 17. 143

18. 135

Lesson 2.5

1. f �1(x) � {(�1, �5), (3, 0), (5, 3), (6, 5)}; 5; 5

2. f �1(x) � 2x � 6; 5; 5

3. f �1(x) � x � ; 5; 5

4. f �1(x) � x � ; 5; 5

5. {(�2, �4), (�1, �1), (0, 2), (1, 5), (2, 8)}

The inverse is a function.

6. {(�2, 3), (�1, 1), (0, �1), (1, 1), (2, 3)}

The inverse is not a function.

Lesson 2.6

1. 4.2 2. 4 3. 5 4. 1.8 5. �2

6. �1 7. 9 8. �5 9. �5 10. 6

11. 0 12. 11.5

13. f (m) � 0.12[m]; $1.92

14. f (h) � 50 � 35h; $190

15.

16.

17.

18.

x

y

O

1

1

x

y

O

1

1

x

y

O

1

1

x

y

O

1

1

x

y

O

2

2

f–1(x)

f(x)

x

y

O

1

1

f–1(x)f(x)

12

52

47

17

x2 � 5x2x � 1

, x 12

3x2 � 22x � 1

, x 12

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◆Skill A Finding inverses of functions

Recall The inverse of a relation is found by interchanging x and y and then solving for y.

◆ Example 1Find the inverse of the function given by {(2, 7), (5, 13), (7, 19), (9, 25)}. Tellwhether the inverse relation is a function.

◆ SolutionInterchange x and y. {(7, 2), (13, 5), (19, 7), (25, 9)}The inverse is a function because there is only one y-value for each x-value.

◆ Example 2Find the inverse, , of . The find and .

◆ SolutionReplace with y.Interchange x and y.Solve for y.

� x� x

� x �52

�52

� x � 5 � 5

�12

(2x � 5) �52

� 2(12

x �52) � 5

� f �1(2x � 5)f �1(f(x))� f (12

x �52)f(f �1(x))

y �12

x �52

x � 52

� y

x � 5 � 2yx � 2y � 5y � 2x � 5f(x)

f �1(f(x))f(f �1(x))f(x) � 2x � 5f �1

Find the inverse, , of each function. Then find and.

1. 2.

3. 4.

f �1(f(5)) �f(f �1(5)) �f �1(f(5)) �f(f �1(5)) �

f �1(x) �f �1(x) �

f(x) �2x � 1

5f(x) � 7x � 4

f �1(f(5)) �f(f �1(5)) �f �1(f(5)) �f(f �1(5)) �

f �1(x) �f �1(x) �

f(x) �12

x � 3f(x) � {(�5, �1), (0, 3), (3, 5), (5, 6)}

f �1(f(5))f(f �1(5))f �1(x)

Reteaching

2.5 Inverses of Functions

Use x-values of �2, �1, 0, 1, and 2 to find ordered pairs and grapheach function. On the same coordinate grid, sketch the inverserelation and determine whether it is a function.

5. 6. f(x) � x2 � 1f(x) � 3x � 2

Copyright ©

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ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Using the horizontal-line test and graphing the inverse of a function

Recall The inverse of a function is also a function if and only if every horizontal line inter-sects the graph of the given function in no more than one point.

◆ Examplea. Graph the function given by . Then use the horizontal-line test

to find out if its inverse relation is also a function.b. Sketch the line and the inverse relation on the same set of axes.

◆ Solutiona. Use several values of x to find ordered pairs.

{(�2, 5), (�1, 2), (0, 1), (1, 2), (2, 5)}

Graph these points and connect them with a smooth curve.

The horizontal line drawn shows that this function does not pass the horizontal-line test.Therefore, the inverse relation is not a function.

b. Use {(5, �2), (2, �1), (1, 0), (2, 1), (5, 2)} to graph the inverse relation. Notice that the inverse is the reflection of the graph of

across the line y � x.

Also observe that the inverse relation will not pass the vertical-line test. This confirms that the inverse relation is not a function.

f(x) � x2 � 1

y � x

f(x) � x2 � 1

x

y

O

2

4

2 4–4 –2–2

x

y

O

y = x

2

4

2 4–4 –2–2

x

y

Ox

y

O

6. 4x2 � 5x � 2

7.

8.

9. 9x2 � 1

10. 3x2 � 3

11. 6 � x2

12. 5 � 3x

13. 9x 14. x 15. 3 16. 63 17. 143

18. 135

Lesson 2.5

1. f �1(x) � {(�1, �5), (3, 0), (5, 3), (6, 5)}; 5; 5

2. f �1(x) � 2x � 6; 5; 5

3. f �1(x) � x � ; 5; 5

4. f �1(x) � x � ; 5; 5

5. {(�2, �4), (�1, �1), (0, 2), (1, 5), (2, 8)}


6. {(�2, 3), (�1, 1), (0, �1), (1, 1), (2, 3)}


Lesson 2.6

1. 4.2 2. 4 3. 5 4. 1.8 5. �2

6. �1 7. 9 8. �5 9. �5 10. 6

11. 0 12. 11.5

13. f (m) � 0.12[m]; $1.92

14. f (h) � 50 � 35h; $190

15.

16.

17.

18.

x

y

O

1

1

x

y

O

1

1

x

y

O

1

1

x

y

O

1

1

x

y

O

2

2

f–1(x)

f(x)

x

y

O

1

1

f–1(x)f(x)

12

52

47

17

x2 � 5x2x � 1

, x 12

3x2 � 22x � 1

, x 12

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◆Skill A Evaluating and applying rounding-up, rounding-down, and absolute valuefunctions

Recall The rounding-up function rounds a decimal value to the next highest integer.

◆ Example 1A long distance telephone company advertises that weekend calls cost $0.10 perminute. Each fraction of a minute is rounded up to the next whole minute.Write this as a rounding-up function. Then find the cost of a 23.5-minute call.

◆ Solutionwhere m � number of minutes

24 is the next highest integer after 23.5

The call will cost $2.40.

Recall The rounding-down function rounds a decimal value to the next lowest integer.

◆ Example 2Evaluate .

◆ Solution�4 is the next integer to the left of �3.7

Recall Absolute value means distance from 0.

◆ Example 3Evaluate .

◆ Solution�12.3 is at a distance 12.3 units from 0� 12.3��12.3�

��12.3�

��2.3� � 2.3

� �4[�3.7]

[�3.7]

[�2.3] � �3

� 2.40� 0.1(24)

f(23.5) � 0.123.5f(m) � 0.1m

�2.3 � �2

Evaluate.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

Write a function for each problem. Solve the problem.

13. Another phone company charges $0.12 per minute, but does not charge for the next minute unless you use the full minute. What is the charge for a 16.5 minute call?

14. A local plumber charges $50 for a house call plus $35 per hour or any fraction of an hour. What is the charge for a 3.75 hour house call?

��8.5� � [3.7]�3 � [�3]2.7 � �3.4

[�2.3] � [�1.8]�2� � ��7��2� � ��7�

�1.8[�1.8]��1.8�

4.2[4.2]�4.2�

Reteaching

2.6 Special Functions

Graph each function.

15. 16.

17. 18.

x

y

Ox

y

O

f(x) � �[x] if �2 � x � 1x if 1 � x � 4

f(x) � � �x� if x � 12 � �x � 2� if x � 1

x

y

Ox

y

O

f(x) � �12

x if �4 � x � 2

2x � 3 if x � 2f(x) � �x � 3, if x � 0

�2x � 5 if x � 0

Copyright ©

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ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Graphing piecewise, step, and absolute-value functions

Recall A piecewise function in x is a function defined by different expressions in xon different intervals for x.

◆ ExampleGraph this piecewise function.

◆ Solution

f(x) � � �x�,[x],2x � 5,

if �5 � x � �2if �2 � x � 2if 2 � x � 5

2

4

4–4

–4

–2x

y

O

x �2 �1.5 �1 �0.5 0 1

y � [x] �2 �2 �1 �1 0 1

x �5 �4 �3 �2.5

y � x 5 4 3 2.5��

x 2 2.5 3 4 5

y � 2x � 5 2 0 1 3 5

6. 4x2 � 5x � 2

7.

8.

9. 9x2 � 1

10. 3x2 � 3

11. 6 � x2

12. 5 � 3x

13. 9x 14. x 15. 3 16. 63 17. 143

18. 135

Lesson 2.5

1. f �1(x) � {(�1, �5), (3, 0), (5, 3), (6, 5)}; 5; 5

2. f �1(x) � 2x � 6; 5; 5

3. f �1(x) � x � ; 5; 5

4. f �1(x) � x � ; 5; 5

5. {(�2, �4), (�1, �1), (0, 2), (1, 5), (2, 8)}


6. {(�2, 3), (�1, 1), (0, �1), (1, 1), (2, 3)}


Lesson 2.6

1. 4.2 2. 4 3. 5 4. 1.8 5. �2

6. �1 7. 9 8. �5 9. �5 10. 6

11. 0 12. 11.5

13. f (m) � 0.12[m]; $1.92

14. f (h) � 50 � 35h; $190

15.

16.

17.

18.

x

y

O

1

1

x

y

O

1

1

x

y

O

1

1

x

y

O

1

1

x

y

O

2

2

f–1(x)

f(x)

x

y

O

1

1

f–1(x)f(x)

12

52

47

17

x2 � 5x2x � 1

, x 12

3x2 � 22x � 1

, x 12

ANSWERSC

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olt, Rinehart and W



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◆Skill A Identifying and describing transformations of a function

Recall The functions , , and are called parent functions. If youhave forgotten what their graphs look like, use a calculator to refresh your memory.

◆ Example 1Graph , , and on the same set of axes. Describe the transformations of f that give g and h.

◆ SolutionFunction g: translation 3 units downFunction h: translation 2 units to the right

◆ Example 2Graph , , and , where , on the same set of axes. Describe the transformations of f that give g and h.

◆ SolutionFunction g: reflection across the x-axisFunction h: reflection across the y-axis

◆ Example 3

Graph and

on the same set of axes. Describe the transformations of f that give g and h.

◆ SolutionFunction g: vertical stretch by a factor of 2Function h: horizontal stretch by a factor of 3 followed by reflection across the x-axis

h(x) � �13

x2f(x) � x2, g(x) � (2x)2,

x � 0h(x) � ��xg(x) � ��xf(x) � �x

h(x) � �x � 2�g(x) � �x� � 3f(x) � �x�

f(x) � �xf(x) � �x�f(x) � x2

Identify the parent function of f and describe thetransformations needed to graph f.

1.

2.

3.

4.

5.

6. �2�x � 4, x � �4f(x) �

�2x�f(x) �

2��x, x � 0f(x) �

(x � 2)2 � 3f(x) �

�x � 3�f(x) �

�x2 � 5f(x) �

Reteaching

2.7 A Preview of Transformations

O

f(x)h(x)

g(x)

2

4

2 4–4 –2–2

x

y

O

f(x)h(x)

g(x)

2

2 4–4 –2–2

x

y

f(x)

h(x)

g(x)

42

2

–4

–4

–2–2

x

y

O

Use the graph of f to help you write an equation for the graph of g.

7. 8. 9.

10. 11. 12.

x

y

O

f(x)

g(x)

2

4

2 4–4

–4

–2–2

x

y

O

f(x)g(x) 2

4

2 4–4

–4

–2–2

x

y

O

f(x)

g(x)

2

4

2 4–4

–4

–2–2

g(x) �g(x) �g(x) �

�xf(x) ��xf(x) ��x�f(x) �

x

y

O

f(x)g(x)

2

4

2 4–4

–4

–2–2

x

y

O

f(x)

g(x)

2

4

2 4–4

–4

–2–2

x

y

O

f(x)

g(x)

2

4

2 4–4

–4

–2–2

g(x) �g(x) �g(x) �

�x�f(x) �x2f(x) �x2f(x) �

Copyright ©

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ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Writing an equation for a function whose graph is given

Recall The transformations below show how to transform the graph of a function horizontally.x → x → kx where h and k are numbers

These transformations show how to transform the graph of a function f vertically.f(x) → f(x) → where h and k are numbers

◆ ExampleGiven the graph of f, write an equation for the graph of g.

◆ Solutiontranslation 3 units to the right: translation 2 units up: reflection across the x-axis:

The transformations above give .g(x) � �(x � 3)2 � 2

�f(x) � �x2f(x) � 2 � x2 � 2

f(x � 3) � (x � 3)2

k(f(x))f(x) � h

x � h

x

y

O

f(x)

g(x)

2

4

2 4–4

–4

–2–2

Lesson 2.7

1. f (x) � x2; reflection across the x-axis andvertical translation 5 units up

2. f (x) � |x|; horizontal translation 3 units tothe right

3. f (x) � x2; horizontal translation 2 units tothe left and vertical translation 3 units up

4. f (x) � ; reflection across the y-axis andvertical stretch by a factor of 2

5. f (x) � |x|; horizontal compression by a

factor of

6. f (x) � ; reflection across the x-axis,vertical stretch by a factor of 2, and ahorizontal translation 4 units to the left

7. g(x) � (x � 3)2 � 2

8. g(x) � �x2 � 2

9. g(x) � |x �3| � 1

10. g(x) � |2x| – 3, or g(x) = 2|x| – 3

11. g(x) �

12. g(x) � �1 �


Lesson 3.1

1.

intersecting, consistent, independent(2, �4)

2.

same line, consistent, dependentall (x, y) such that x � 2y � 2

3.

parallel lines, inconsistent, independentno solution

4. (1, 2) 5. (3, 1) 6. (5, 7)

7. (−1, 1) 8. (3, −4) 9. (2, 5)

10. (1, 1) 11. (5, 1) 12. (6, 2)

Lesson 3.2

1. (1, 4) 2. (3, 1) 3. (5, 1) 4. (�4, 3)

5. (�2, �3) 6. (�2, 1) 7. (1, �1)

8. (�1, �4)

9. inconsistent

10. dependent

11. consistent, (�2, 0)

12. consistent, (−3, 4)

13. inconsistent

14. dependent

15. consistent, (2, �2)

16. dependent

x

y

O 2

2

x

y

O 2

2

x

y

O

(2, –4)

1

1

�x � 3

�x � 2

�x

12

�x

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◆Skill A Solving a system of linear equations by graphing

Recall A system of equations may be consistent or inconsistent. If it is consistent, theequations may be dependent or independent.

◆ Example Graph each system. Classify the graphs as intersecting lines, parallel lines, or the same line. Then classify the system as consistent or inconsistent. If it isconsistent, classify it as dependent or independent and find the solution.

a. b. c.

◆ SolutionThe first equation in each system is already solved for y. Solve the second equation in each system for y and graph.

→

→

→

a. intersecting lines; consistent; independent; (2, 2)b. parallel lines; inconsistent; no solutionc. same line; consistent; dependent; all ordered pairs

(x, y) such that y �12

x � 1

y �12

x � 1x � 2y � �2

y �12

x � 2y � 2 �12

x

y � �x � 4x � y � 4

�y �12

x � 1

x � 2y � �2�y �12

x � 1

y � 2 �12

x�y �12

x � 1

x � y � 4

Graph and classify each system. Then find the solution from the graph.

1. 2. 3.

x

y

Ox

y

Ox

y

O

�y � 2x � 12x � y � �3��3x � 6y � �6

x � 2y � 2�x � y � �2x � y � 6

Reteaching

3.1 Solving Systems by Graphing or Substitution

x

y

O

x + y = 4

y = x – 21

2y = x + 11

2

Use substitution to solve each system of equations. Check your solution.

4. 5. 6.

7. 8. 9.

10. 11. 12. �13

x �12

y � 1

�4y � 10 � �3x�5x � 9y � 16x � 2 � 3y�2x � 5y � �3

3y � 4x � �1

�5x � 2y � 0x � 3y � 17�4x � y � 16

3x � 2y � 1�3x � 4y � 1x � 3y � �4

�y � 2x � 3x � 2y � 9�x � y � 4

3x � 2y � 7�y � x � 1x � y � 3

Copyright ©

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ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Solving a system by substitution

Recall Your objective is to obtain a single equation in one variable which is easily solved.

◆ Example 1Use substitution to solve this system.

◆ SolutionIn the second equation, substitute for x.

Solve for x.

The solution is (5, 8). Check to see that the ordered pair (5, 8) satisfies each ofthe original equations.

◆ Example 2Use substitution to solve this system.

◆ SolutionIt is easier to solve the second equation for y, and then substitute the result intothe first equation. → Now, use to find y.

Thus, the solution is (3, �2).

x � 311x � 33

y � �23x � 8x � 16 � 17y � �2(3) � 43x � 4(�2x � 4) � 17y � �2x � 43x � 4y � 17

x � 3y � �2x � 42x � y � 4

�3x � 4y � 172x � y � 4

x � 5x � 8 � 3x � y � 3

y � 83y � 24

2y � 6 � y � 182(y � 3) � y � 18

y � 3

�x � y � 32x � y � 18

Lesson 2.7






factor of


7. g(x) � (x � 3)2 � 2

8. g(x) � �x2 � 2

9. g(x) � |x �3| � 1

10. g(x) � |2x| – 3, or g(x) = 2|x| – 3

11. g(x) �

12. g(x) � �1 �


Lesson 3.1

1.


2.


3.


4. (1, 2) 5. (3, 1) 6. (5, 7)

7. (−1, 1) 8. (3, −4) 9. (2, 5)

10. (1, 1) 11. (5, 1) 12. (6, 2)

Lesson 3.2

1. (1, 4) 2. (3, 1) 3. (5, 1) 4. (�4, 3)

5. (�2, �3) 6. (�2, 1) 7. (1, �1)

8. (�1, �4)

9. inconsistent

10. dependent



13. inconsistent

14. dependent


16. dependent

x

y

O 2

2

x

y

O 2

2

x

y

O

(2, –4)

1

1

�x � 3

�x � 2

�x

12

�x

ANSWERSC

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inst

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Solving a consistent and independent system of equations by elimination

Recall Two lines with different slopes represent a consistent and independent system of equations. Since these lines intersect in one point, there is one solution to the system.

◆ Example

Solve by using the elimination method.

◆ SolutionTo eliminate the y terms multiply each side of the first equation by 3.

Then multiply each side of the second equation by 2.

The system that results is shown below.

Addition Property of Equality

To find y, replace x with 3 in the first equation.

Check the ordered pair (3, 2) in the second original equation.

12 � 6 � 64(3) � 3(2) � 6

4x � 3y � 6

y � 2

2y � 43(3) � 2y � 13

3x � 2y � 13

x � 3

17x � 51

8x � 6y � 129x � 6y � 39

4x � 3y � 6 → 2(4x � 3y) � 2(6) → 8x � 6y � 12

3x � 2y � 13 → 3(3x � 2y) � 3(13) → 9x � 6y � 39

�3x � 2y � 134x � 3y � 6

Use elimination to solve each system of equations.Check your solution.

1. 2.

3. 4.

5. 6.

7. 8. �8x � 5y � �28�3x � 2y � �5�3x � 4y � 7

�6x � y � �7

�3x � 2y � �44x � 3y � �5�2x � 5y � 11

�x � 3y � �7

�x � 4 � 8�2x � 5y � 23�7x � 3y � 32

2x � y � 11

�x � 3y � 6x � y � 2�x � y � �3

x � y � 5

Reteaching

3.2 Solving Systems by Elimination

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Text

Recall Text

◆ ExampleSBHCT

Classify each system as consistent or inconsistent, independentor dependent. If the system is consistent, find the solution.

9. 10.

11. 12.

13. 14.

15. 16. �7x � 14y � 21x � 2y � 3�3x � 5y � 16

�x � 4y � �10

�2x � 7y � �16x � 21y � �3�4x � 3y � 15

12x � 9y � 36

�x � y � 1x � 5y � �23�2x � 7y � �4

�x � 8y � 2

�2x � y � 1810x � 5y � 90�x � 7y � 13

�x � 7y � �7

◆Skill B Classifying a system as dependent or inconsistent

Recall If the same line represents two different equations, the system is dependent.If parallel lines represent two different equations, the system is inconsistent.

◆ Example 1Classify the system as consistent or inconsistent, independent

or dependent.

◆ Solution

Since is a false statement, this system represents a pair of parallel lines. There is no solution because this system is inconsistent.

◆ Example 2Classify the system as consistent or inconsistent, independent

or dependent.

◆ Solution

Since is a true statement, these are both the same line. This system is aconsistent and dependent system.

Any solution to the first equation will also be a solution to the second equation.

Any solution to the second equation will also be a solution to the first equation.

0 � 00 � 0

�2x � 2y � 6�2x � 2y � �6�x � y � 3

�2x � 2y � �6→ �2(x � y) � 2(3)

�2x � 2y � �6→

�x � y � 3�2x � 2y � �6

0 � 70 � 7

�2x � y � 3�2x � y � 4�2x � y � 3

2x � y � �4→ �2x � y � 3

(�1)(2x � y) � (�1)(�4)→

�2x � y � 32x � y � �4

Lesson 2.7






factor of


7. g(x) � (x � 3)2 � 2

8. g(x) � �x2 � 2

9. g(x) � |x �3| � 1

10. g(x) � |2x| – 3, or g(x) = 2|x| – 3

11. g(x) �

12. g(x) � �1 �


Lesson 3.1

1.


2.


3.


4. (1, 2) 5. (3, 1) 6. (5, 7)

7. (−1, 1) 8. (3, −4) 9. (2, 5)

10. (1, 1) 11. (5, 1) 12. (6, 2)

Lesson 3.2

1. (1, 4) 2. (3, 1) 3. (5, 1) 4. (�4, 3)

5. (�2, �3) 6. (�2, 1) 7. (1, �1)

8. (�1, �4)

9. inconsistent

10. dependent



13. inconsistent

14. dependent


16. dependent

x

y

O 2

2

x

y

O 2

2

x

y

O

(2, –4)

1

1

�x � 3

�x � 2

�x

12

�x

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◆Skill A Text

Recall Text

◆ ExampleSBHCT

Reteaching

3.3 Linear Inequalities in Two Variables

◆Skill A Writing an inequality in two variables for a given graph

Recall A broken boundary line in the graph of an inequality indicates that points on theline are not part of the solution set.

◆ ExampleThe school drama club is selling tickets fortheir next production. Adult tickets cost $5and student tickets are $3. They must sell atleast $600 worth of tickets to cover expenses.The shaded part of the graph contains all theordered pairs of numbers of adult and studenttickets, (a, s), that will earn them at least $600.Determine which of the following orderedpairs are in the shaded region. Then write aninequality in two variables for the shadedportion of the graph.

(50, 100), (100, 50), (90, 90)

◆ SolutionThe point (50, 100) is not in the shaded region.Notice that ($5)(50) � ($3)(100) � $550 and $550 � $600.

The point (100, 50) is in the shaded region.($5)(100) � ($3)(50) � $600

The point (90, 90) is in the shaded region.($5)(90) � ($3)(90) � $600

The shaded region contains all the points where .5a � 3s � 600

Determine whether the point (2, 3) is in the shaded region ofeach graph. Then write an inequality in two variables for thegraph shown.

1. 2. 3.

(2, 3): (2, 3): (2, 3):

inequality: inequality: inequality:

x

y

2

4

2 4–4

–4

–2–2

x

y

2

4

2 4–4

–4

–2–2

a

s

O 50 100 150 200 250

50

100

150

200 (0, 200)

(120, 0)

x

y

2

4

2 4–4

–4

–2–2

Graph each linear inequality.

4. 5. 6.

x

y

Ox

y

O

2x � y � 1�2x � 3y � �6y � �x � 1

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Graphing linear inequalities in 2 variables

Recall When you multiply or divide each member of an inequality by a negative number,you must reverse the inequality sign.

◆ Example 1Graph .

◆ Solution

The “�” indicates that you shade above the solid boundary line .


◆ Solution

Shade the half plane where the x-coordinate of each point is less than �3.

Notice that the y-coordinate of each point can be any real number.

The broken line indicates that points on the line are not in the solution set.

x � �32x � �6

2x � �6

y � 2x � 3

y � 2x � 3� y � �2x � 3

2x � y � 3

2x � y � 3

x

y

2

4

2 4–4

–4

–2–2

x

y

2

4

2 4–4

–4

–2–2

x

y

O

Lesson 3.3

1. yes;

2. no;

3. no;

4.

5.

6.

Lesson 3.4

1.

2.

3.

4.

5.

6.

7. �y �14

x � 4

y � �2x � 1

�y �23

x � 2

y �23

x � 3

�x � �1y � 1y � �2x � 4

�y � �x � 3y � 2x � 2

x

y

O 1

1

x

y

O 1

1

x

y

O 2

2

x

y

O 1

1

x

y

O 1

1

x

y

O 1

1

y �12

x � 2

y � �12

x � 5

y � 2x

ANSWERSC

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olt, Rinehart and W



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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Graphing a system of linear inequalities in two variables

Recall If x � 0 and y � 0 in a system of inequalities, then all solutions are in the firstquadrant.

◆ Example

Graph the following system of inequalities.

◆ SolutionGraph the line and the broken line

.

Since and , shading will be in the first quadrant.

Shade all points in the first quadrant that

are above the line and below the

broken line .

Check the point (1, 1), which is in the shaded region.

The pair (1, 1) satisfies all four of the inequalities.

y � �x � 41 � �1 � 4

y �12

x

1 �12

(1)

y � 01 � 0

x � 01 � 0

y � x � 4

y �12

x

y � 0x � 0

y � x � 4

y �12

x

�x � 0y � 0

y �12

y � �x � 4

Graph the solution to each system of linear inequalities.

1. 2. 3.

x

y

Ox

y

O

�2x � 3y � 63x � 2y � 6�x � 0

y � 0x � y � 4�x � 3

y �12

x � 1

Reteaching

3.4 Systems of Linear Inequalities

x

y

2

4

2 4–4

–4

–2

–2

x

y

O

Write the system of inequalities whose solution is graphed.Assume each vertex has integer coordinates.

4. 5.

6. 7.

x

y

2

4

2 4–4 –2–2

x

y

2

4

2 4–4 –2–2

x

y

2

4

2 4–4 –2–2

x

y

2

4

2 4–4 –2–2

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Writing a system of linear inequalities in two variables for a given region in the plane

Recall The linear equation has slope m and y-intercept b.

◆ Example Write a system of inequalities to represent the shaded region at right.

◆ SolutionThe broken line has slope 1 and y-intercept 0. An equation for the line is y � x.

The solid line has slope �3 and y-intercept 4. An equation for the line is .

The shaded region can be described as:to the right of x � 0above the broken line y � x

and below the line

Thus, is a system representing the shaded region.�x � 0y � xy � �3x � 4

⇒ y � �3x � 4y � �3x � 4⇒ y � x⇒ x � 0

y � �3x � 4

y � mx � b

x

y

2

4

2 4–2

–2

Lesson 3.3

1. yes;

2. no;

3. no;

4.

5.

6.

Lesson 3.4

1.

2.

3.

4.

5.

6.

7. �y �14

x � 4

y � �2x � 1

�y �23

x � 2

y �23

x � 3

�x � �1y � 1y � �2x � 4

�y � �x � 3y � 2x � 2

x

y

O 1

1

x

y

O 1

1

x

y

O 2

2

x

y

O 1

1

x

y

O 1

1

x

y

O 1

1

y �12

x � 2

y � �12

x � 5

y � 2x

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◆Skill A Finding vertices of a polygon

Recall The vertices of a polygon occur at the intersection of the segments which form thesides of the polygon.

◆ Example

Graph . Then find the vertices of the shaded polygon.

◆ SolutionIn this example, if you draw your boundary lines carefully, you can identify the intersection points by observation.

For this graph, these points appear to be (1, 1), (6, 1), and (3, 3).

To check you can set pairs of expressions for yequal to each other.

For example, since and are

boundary lines, set the 2 expressions for y equal to each other and solve for x.

Since y � x, then y � 3 and the point of intersection is (3, 3).

53

x � 5 → 35

�53

x � 5 �35

→ x � 3

x �23

x � 5

x � �23

x � 5

y � �23

x � 5y � x

�y � 1y � x

y � �23

x � 5

Graph the feasible region and identify the vertices for each setof constraints.

1. 2. 3.

x

y

Ox

y

O

�y � 4x � 4y � �x � 1

y �23

x � 4�x � �1

y � 4y � x�x � 0

y � 0y � �x � 5

Reteaching

3.5 Linear Programming

x

y

O

(1, 1)

(3, 3)

(6, 1)

x

y

O

Find the maximum and minimum values, if they exist, of eachobjective function for each feasible region.

4. 5.

6. 7.

x

y

O

(4, 1)(0, 1)

(2, 0)

(3, 5)(1, 4)

x

y

O

(4, 0)(–2, 0)

(0, –3)

(0, 5)

P � x � yP � 3x � 4y

x

y

O

(–2, 4)

(–2, –2)

(3, 0)

(3, 3)

x

y

O

(–1, 1)

(4, –1)

(4, 3)

P � x � 2yP � x � y

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Finding the maximum and minimum values of an objective function

Recall The maximum and minimum values of an objective function will each occur at avertex of the feasible region (polygon).

◆ Example Find the maximum and minimum values of

for the feasible region shown at right.

◆ Solutionvertex P � 3x � 2y (objective function)(1, 2)(2, 4)(6, 5)(7, 2)

The maximum value of is 17 and occurs at (7, 2).

The minimum value of is �2 and occurs at (2, 4).P � 3x � 2y

P � 3x � 2y

P � 3(7) � 2(2) � 17P � 3(6) � 2(5) � 8P � 3(2) � 2(4) � �2P � 3(1) � 2(2) � �1

P � 3x � 2y

x

y

O 2 4 6 8 10

2

4

6

8

10

(7, 2)(1, 2)

(2, 4)(6, 5)

Lesson 3.5

1.

(0, 0), (5, 0), (0, 5)

2.

(�1, �1), (�1, 4), (4, 4)

3.

(�1, 0), (�3, 2), (0, 4)

4. maximum: 7; minimum: 0

5. maximum: 3; minimum: �10



Lesson 3.6

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12. �x(t) � t2 � ty(t) � t

�x(t) � 2ty(t) � t2

�x(t) � t � 6y(t) � 2t � 5

y � �2x � 4

y �23

(x � 5)

y � �12

x � 7

y � �12

x �12

y � x � 8

y �12

x � 1

x

y

1

1

x

y

1

1

x

y

1

1

x

y

2

2

(–3, 2) (0, 4)

(–1, 0)

x

y

(–1, 4)

(4, 4)

(–1, –1)

x

y

3

3 (5, 0)

(0, 5)

(0, 0)

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Graphing a pair of parametric equations

Recall In a pair of parametric equations, such as and , the values ofboth x and y are dependent upon the value of the parameter t.

◆ Example(You will need a graphics calculator.)Your cat chases a bird that it sees on the lawn. The bird takes flight and itshorizontal distance increases by 10 feet per second; . At the same timeits vertical distance can be modeled by . Sketch the path of the bird for the first 5 seconds as it escapes from your cat.

◆ SolutionUse your calculator to complete the table. Plot the points and connect them with a smooth curve.

Check your graph by using your graphics calculator in parametric mode.

y(t) � 3�tx(t) � 10t

y(t) � 3�tx(t) � 10t

Graph each pair of parametric equations for .

1. 2. 3.

x

y

x

y

�x(t) �12

t2

y(t) � �t � 6�x(t) � t

y(t) � �12

t � 7�x(t) � 2ty(t) � t � 1

0 � t � 5

Reteaching

3.6 Parametric Equations

x

y

O 10 20 30 40 50

2

4

6

8

10t

0 0 0

1 10 3

2 20 4.24

3 30 5.20

4 40 6

5 50 6.71

y(t) � 3�tx(t) � 10t

x

y

Write each pair of parametric equations as a single equation in x and y.

4. 5. 6.

7. 8. 9. �x(t) �12

t � 1

y(t) � �t � 6�x(t) � 3t � 5y(t) � 2t�x(t) � t

y(t) � �12

t � 7

�x(t) � 2t � 1y(t) � �t�x(t) � t � 3

y(t) � t � 5�x(t) � 2ty(t) � t � 1

Copyright ©

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ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Writing a pair of parametric equations as a single equation

◆ ExampleWrite the pair of parametric equations as a single equation

involving only x and y.

◆ SolutionSolve each equation for t.

Set the two resulting expressions for t equal to each other and solve for y.

y � 2x � 132x � 12 � y � 1

x � 6 �y � 1

2

t �y � 1

2t � x � 6

y � 2t � 1x � t � 6

�x(t) � t � 6y(t) � 2t � 1

◆Skill C Using parametric equations to find the inverse of a function

Recall To find the inverse of a function, switch the x and y variables.

◆ Example 1Find the inverse for

◆ Solution

Use your graphics calculator to check that these two functions are reflectionsacross the line y � x. To graph y � x in parametric mode use and .y(t) � tx(t) � t

�x(t) � 2t � 1y(t) � t

�x(t) � ty(t) � 2t � 1

Find the inverse of each pair of parametric equations. Use agraphics calculator to check.

10. 11. 12. �x(t) � ty(t) � t2 � t�x(t) � t2

y(t) � 2t�x(t) � 2t � 5y(t) � t � 6

Lesson 3.5

1.

(0, 0), (5, 0), (0, 5)

2.

(�1, �1), (�1, 4), (4, 4)

3.

(�1, 0), (�3, 2), (0, 4)

4. maximum: 7; minimum: 0




Lesson 3.6

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12. �x(t) � t2 � ty(t) � t

�x(t) � 2ty(t) � t2

�x(t) � t � 6y(t) � 2t � 5

y � �2x � 4

y �23

(x � 5)

y � �12

x � 7

y � �12

x �12

y � x � 8

y �12

x � 1

x

y

1

1

x

y

1

1

x

y

1

1

x

y

2

2

(–3, 2) (0, 4)

(–1, 0)

x

y

(–1, 4)

(4, 4)

(–1, –1)

x

y

3

3 (5, 0)

(0, 5)

(0, 0)

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◆Skill A Writing a matrix and reading data from a matrix

Recall The rows of a matrix run across the page and the columns run down the page.

◆ ExampleThe following table represents Kelli’s grades for the first semester.

Use matrix M to represent this set of data. Find the elements m32 and m23, anddescribe the data these two elements represent.

◆ Solution

M � �9295828884

8796799182

8894859387

�Use the data for woodwind instruments to complete theexercises that follow.

1. Represent these data in a matrix W.

Reteaching

4.1 Using Matrices to Represent Data

First 6 weeks Second 6 weeks Third 6 weeks

Biology 92 87 88

World history 95 96 94

English 82 79 85

Algebra 2 88 91 93

Spanish 84 82 87

m32 is the element in the third row and second column.m32 � 79; Kelli’s English grade for the second 6 weeks was 79.

m23 is the element in the second row and third column.m23 � 94; Kelli’s history grade for the third 6 weeks was 94.

Concert band Pep band Orchestra

Flutes 8 4 3

Oboes 4 0 2

Clarinets 20 6 2

Saxophones 4 6 0

Bassoons 3 0 1

Use matrices D, E, and F to perform the indicated operations.

8. D � F 9. D � F 10. F � D

11. 2D 12. �E 13. E � F

F � � 20�18

420�E � � 6

12�9

04

�8�D � �1222

�1037�

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Finding a sum or difference of two matrices and finding a scalar product

Recall To add or subtract two matrices, they must have the same dimensions.

◆ Example

Find a. A � B b. B � A c. �2A

◆ Solutiona. Add the elements in the same position in each matrix. For example, 6 + 3 = 9.

b. Subtract the elements in the same position in each matrix. For example, 3 � 6 � �3.

c. Multiply each element of matrix A by the scalar −2.

�2A � �2�60

37

�25� � ��12

0�6

�144

�10�

B � A � � 3�3

�711

6�4� � �6

037

�25� � ��3

�3�10

48

�9�

A � B � �60

37

�25� � � 3

�3�711

6�4� � � 9

�3�418

41�

B � � 3�3

�711

6�4�A � �6

037

�25�

Give the entry at the indicated address.

2. w21 3. w33 4. w53

Describe the data in the given location.

5. w42

6. w23

7. w32


Lesson 4.1

1.

2. 4 3. 2 4. 1

5. 6 saxophones in the pep band

6. 2 oboes in the orchestra

7. 6 clarinets in the pep band

8.

9.

10.

11.

12.

13. not possible

Lesson 4.2

1.

2.

3. not possible

4.

5.

6.

7.

reflection across the x-axis

8.

reflection in the origin

9.

reflection across the line y � x

x

y

O

C’

A’

B’

x

y

OC’

B’

A’

x

y

OA’

B’

C’

��155

�1951�

�27

�30�45�18

�21203318

�121521

6

37�35�58�32

��30

9�32

�9�9�3

5716�

��237

0�29�

��32�10

91�

� �6�12

90

�48�

�2444

�2074�

� 8�40

52�37�

��840

�5237�

�324

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W � �84

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opyright ©by H

olt, Rinehart and W



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by H

olt,

Rin

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Win

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. All

right

s re

serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Multiplying matrices

Recall When multiplying matrices, you multiply elements in a row of the first matrix by corresponding elements in a column of the second matrix. The sum of these products gives one element in the product matrix.

◆ Example

Find the matrix products AB and BA, if they exist.

◆ Solution

Multiply each element in the first row of A by the corresponding element in thefirst column of B. Then add these products.

1st row � 1st column: (2)(1) � (�1)(�4) � 6

Follow the same procedure with the first row of Aand the second column of B, and then the third column of B.

1st row � 2nd column: (2)(4) � (�1)(0) � 81st row � 3rd column: (2)(�2) � (�1)(�3) � �1

Then follow the same procedure with the second rowof A and each column of B.

2nd row � 1st column: (5)(1) � (3)(�4) � �72nd row � 2nd column: (5)(4) � (3)(0) � 202nd row � 3rd column: (5)(�2) � (3)(�3) � �19

The product BA does not exist because the 3 elements in each row of B will not“match” with the 2 elements in each column of A.

� 6�7

820

�1�19�

AB � �25

�13�� 1

�440

�2�3�

B � � 1�4

40

�2�3�A � �2

5�1

3�

Find each product, if it exists.

1. CD 2. DC 3. CE

4. CF 5. EF 6. FE

F � � 3�6

�54

03

9�7�E � �

10

�1�2

�4572�D � ��2

865�C � �4

1�3�1�

Reteaching

4.2 Matrix Multiplication

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Using matrices to perform transformations in the plane

Recall A matrix can be used to represent points in the plane. The first row contains the x-coordinates and the second row contains the y-coordinates of a figure.

◆ Example a. Sketch the triangle with vertices given by the matrix .

b. Multiply this matrix of vertices by the matrix to graph and describe a transformation of the original triangle.

◆ Solution

a. is graphed as A(�1, 1),

B(3, 5) and C(3, 0).

b.

The result has vertices (1, 1), (�3, 5),and (�3, 0).Triangle A'B'C' is the reflection of triangle ABC across the y-axis.

��10

01��1

135

30� � �1

1�3

5�3

0�

��11

35

30�

��10

01��1

135

30�

x

y

O

A

B

C

C’

B’

A’

x

y

O

Use the triangle given in the example above. Graph and describe the transformation that results from multiplying by each of thefollowing matrices.

7. 8. 9.

x

y

Ox

y

Ox

y

O

�01

10��1

00

�1��10

0�1�


Lesson 4.1

1.

2. 4 3. 2 4. 1

5. 6 saxophones in the pep band

6. 2 oboes in the orchestra

7. 6 clarinets in the pep band

8.

9.

10.

11.

12.

13. not possible

Lesson 4.2

1.

2.

3. not possible

4.

5.

6.

7.

reflection across the x-axis

8.

reflection in the origin

9.

reflection across the line y � x

x

y

O

C’

A’

B’

x

y

OC’

B’

A’

x

y

OA’

B’

C’

��155

�1951�

�27

�30�45�18

�21203318

�121521

6

37�35�58�32

��30

9�32

�9�9�3

5716�

��237

0�29�

��32�10

91�

� �6�12

90

�48�

�2444

�2074�

� 8�40

52�37�

��840

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ANSWERSC

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by H

olt,

Rin

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t and

Win

ston

. All

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serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Finding the inverse of a matrix with a graphics calculator

Recall Inverse functions “undo” each other;

◆ Example 1Use a graphics calculator to find the inverse of each matrix, if it exists.

a. b.

◆ SolutionEnter each matrix and use the key to find the inverses A�1 and B�1.

a. b. B�1 does not exist

◆ Example 2Determine if matrices C and D are inverses of each other.

◆ Solution

and

Since each product is the identity matrix, C and D are inverses of each other.

DC � �10

01�CD � �1

001�

D � � 0.4�0.6

�0.30.7�C � �7

634�

A�1 � � 4�3

�54�

x–1

B � �104

52�A � �4

354�

f ( f �1(x)) � x

Determine whether each pair of matrices are inverses of each other.

1. 2.

AB � CD �

BA � DC �

Use a graphics calculator to find the inverse of each matrix, if it exists. If no inverse exists, write no inverse.

3. 4. 5. G � �121

1�1

2

112�F � �1

001�E � �2

45

11�

D � ��52

2�3�C � �3

225�B � ��6

55

�4�A � �45

56�

Reteaching

4.3 The Inverse of a Matrix

Use a determinant to find the inverse of each matrix, if it exists. Then use your graphics calculator to check your answer.

6. 7.

8. 9.

10. 11. A � �52

84�B � �7

3115�

F � ��45

9�11�E � �3

648�

D � �56

45�C � �2

45

11�

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆ Skill B Using a determinant to find the inverse of a matrix

Recall The determinant of matrix is written as and its value is .

If and , then .

◆ Example Find the inverse of the matrix by using its determinant.

◆ Solution

Check:

� �10

01��4

523�� 1.5

�2.5�1

2��45

23�� 1.5

�2.5�1

2� � �10

01�

� � 1.5�2.5

�12�

�12� 3

�5�2

4�

A�1 �1

(4)(3) � (2)(5)� 3�5

� 24�

A � �45

23�

A�1 �1

ad � bc� d�c

�ba�ad � bc 0�a

cbd�A �

ad � bc�ac bd��a

cbd�

Lesson 4.3

1. inverses 2. not inverses

3.

4.

5.

6.

7.

8. The inverse does not exist.

9.

10.

11.

Lesson 4.4

1.

2.

3.

4.

5.

6.

7. (1, 1) 8. (4, �5) 9.

10. (3, 3, �4)

Lesson 4.5

1.

2.

3. (2, 1)

4.

5. (2, 4)

6. dependent system

7. (1.1, 0.3) 8. (0, 2, �3)


Lesson 5.1

1.

2.

3.

4.

5.

6.

7. ; a � 1, b � �5, c � �24

8. ; a � 1, b � �16, c � 63

9. ; a � 4, b � 0, c � �49

10. ; a � 12, b � 1, c � �1

11. ; a � 49, b � 42, c � 9

12. ; a � 9, b � 0, c � �121

13. a. opens upb. minimumc. x-intercepts: �1 and 5d. (2, �9)

f(x) � 9x2 � 121

f(x) � 49x2 � 42x � 9

f(x) � 12x2 � x � 1

f(x) � 4x2 � 49

f(x) � x2 � 16x � 63

f(x) � x2 � 5x � 24

16x2 � 8x � 1

4x2 � 12x � 9

6x2 � 13x � 5

x2 � 64

2x2 � 13x � 6

x2 � 4x � 21

(�38

, 1112)

�10

01

4�1��

�10

01

3�2��

(52

, �32)

�210

102

0�4

1� � �abc � � � 17

�315�

� 3�2

1

4�1�6

15

�1� � �xyz� � � 11

3�8�

�11

1�1� � �x

y� � �91�

�121

�61� � �r

t� � �07�

��39

1�8� � �a

b� � �74�

�51

3�5� � �x

y� � � 8�2�

� 1�0.5

�21.25�

� 2.5�1.5

�5.53.5�

�115

94�

� 5�6

�45�

�5.5�2

�2.51�

� 21.5

�2.5

0�0.5

0.5

�1�0.5

1.5��10

01�

�5.5�2

�2.51�

ANSWERSC

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t, R

ineh

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.


Cop

yrig

ht ©

by H

olt,

Rin

ehar

t and

Win

ston

. All

right

s re

serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Writing a matrix equation to represent a system of equations

Recall The inverse of matrix A is written as A�1.

◆ Example Randy saved gas by only using his car to travel to and from work and the grocerystore. One week he made 5 round trips to work and 2 round trips to the store. Henoticed on the odometer that he had traveled 61.2 miles. The next week 4 roundtrips to work and 3 round trips to the store gave a total mileage of 55.4. Write amatrix equation that represents this information.

◆ SolutionA system of equations for this problem is

where w represents the number of miles in a trip to and from work, and s represents the number of miles in a trip to and from the store.

Let ,

Then the matrix equation AX � B is and represents thesame system of equations.

A � �54

23��w

s � � �61.255.4�

A � �54

23�, X � �w

s � and B � �61.255.4�

�5w � 2s � 61.24w � 3s � 55.4

Write the matrix equation that represents each system.

1. 2.

3. 4.

5. 6. �2a � b � 7a � 4c � �32b � c � 15�3x � 4y � z � 11

�2x � y � 5z � 3x � 6y � z � �8

�x � y � 9x � y � 1�12r � 6t � 0

r � t � 7

��3a � b � 79a � 8b � 4�5x � 3y � 8

x � 5y � �2

Reteaching

4.4 Solving Systems With Matrix Equations

Solve each system by writing and solving a matrix equation.Write the matrix equation that represents each system, andsolve the system, if possible, by using a matrix equation.

7. 8.

9. 10. �x � 4y � 3z � �215x � 2y � z � 253x � y � 2z � �2�1

5c �

12

d �54

35

c �12

d �34

�4a � 5b � �93a � 2b � 22�2x � 5y � 7

4x � 3y � 1

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Solving systems of equations by using matrices

Recall The identity matrix, denoted by I, is .

Associative Property of MultiplicationIdentity Property of Multiplication

◆ Example Use the example given in Skill A of this lesson to find the miles in a one-way tripto work and a one-way trip to the store.

◆ SolutionWrite a matrix equation.

Enter matrices A and B into a graphics calculator. Then find . Your result

should be .

Since these numbers represent round trip distances, the distance to work is

5.2 miles and the distance to the store is 2.3 miles.12

(4.6) �12

(10.4) �

�10.44.6�

A�1B

�54

23��w

s � � �61.255.4�

X � A�1BIX � A�1B

(A�1A)X � A�1BA�1(AX) � A�1B

AX � B�10

01�2 � 2

Lesson 4.3


3.

4.

5.

6.

7.


9.

10.

11.

Lesson 4.4

1.

2.

3.

4.

5.

6.

7. (1, 1) 8. (4, �5) 9.

10. (3, 3, �4)

Lesson 4.5

1.

2.

3. (2, 1)

4.

5. (2, 4)

6. dependent system

7. (1.1, 0.3) 8. (0, 2, �3)


Lesson 5.1

1.

2.

3.

4.

5.

6.

7. ; a � 1, b � �5, c � �24

8. ; a � 1, b � �16, c � 63

9. ; a � 4, b � 0, c � �49

10. ; a � 12, b � 1, c � �1

11. ; a � 49, b � 42, c � 9

12. ; a � 9, b � 0, c � �121


f(x) � 9x2 � 121

f(x) � 49x2 � 42x � 9

f(x) � 12x2 � x � 1

f(x) � 4x2 � 49

f(x) � x2 � 16x � 63

f(x) � x2 � 5x � 24

16x2 � 8x � 1

4x2 � 12x � 9

6x2 � 13x � 5

x2 � 64

2x2 � 13x � 6

x2 � 4x � 21

(�38

, 1112)

�10

01

4�1��

�10

01

3�2��

(52

, �32)

�210

102

0�4

1� � �abc � � � 17

�315�

� 3�2

1

4�1�6

15

�1� � �xyz� � � 11

3�8�

�11

1�1� � �x

y� � �91�

�121

�61� � �r

t� � �07�

��39

1�8� � �a

b� � �74�

�51

3�5� � �x

y� � � 8�2�

� 1�0.5

�21.25�

� 2.5�1.5

�5.53.5�

�115

94�

� 5�6

�45�

�5.5�2

�2.51�

� 21.5

�2.5

0�0.5

0.5

�1�0.5

1.5��10

01�

�5.5�2

�2.51�

ANSWERSC

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©by

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t, R

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.


Cop

yrig

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by H

olt,

Rin

ehar

t and

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. All

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serv

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Finding the reduced row-echelon form of a matrix

Recall Multiplying by an identity matrix, such as , has the same effect as multiplying a real number by 1.

◆ ExampleUse row operations to change to reduced row-echelon form.

◆ SolutionThe first element in the 1st row needs to be a 1, so multiply row 1 by .

The first element in the 2nd row needs to be 0, so subtract 3 times row 1 from row 2.

The second element in the 2nd row needs to be 1, so multiply row 2 by .

The second element in the 1st row needs to be 0, so add times row 2 to row 1.

The matrix at right above is in reduced row-echelon form.

�1

0

0

1

32

�1��R1 �32

R2 → R1�1

0

�321

3

�1��

32

�1

0

�321

3

�1��217

R2 → R2�1

0

�32

172

3

�172��

217

�1

0

�32

172

3

�172��R2 � 3R1 → R2�1

3

�324

312��

�1

3

�324

312��1

2R1 → R1�23 �3

4

612��

12

�2

3

�3

4

612��

�10

01�

Find the reduced row-echelon form of each matrix.

1. 2.

Hint: (exchange rows)

R1 � 2R2 → R1

110

R2 → R2

R2 � 3R1 → R2

R1 ↔ R2

�13

3�2

114��3

14

�217��

Reteaching

4.5 Using Matrix Row Operations

Use an augmented matrix to solve each system of equations. If the system is inconsistent or dependent, so state.

3. 4. 5.

6. 7. 8. �2x � 3y � 3z � �153x � 2y � 5z � 195x � 4y � 2z � �2

�0.2x � 0.5y � 0.070.8x � 0.3y � 0.79�3x � 6y � �9

�2x � 4y � 6

�3x � y � 2x � 2y � 10��2x � 3y � �2

�4x � 6y � 7�x � 4y � �2�2x � y � �3 C

opyright ©by H

olt, Rinehart and W


NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Using the row-reduction method to solve a system of equations(You will need a graphics calculator.)

Recall A dependent system has an infinite number of solutions. An inconsistent systemhas no solution.

◆ ExampleSolve each system or determine why it has no unique solution.

a. b. c.

◆ SolutionWrite an augmented matrix for each system. Then use a graphics calculator tofind the reduced row-echelon form.

a. → The solution is x = 5 and y = 2.

b. →

The last row indicates that 0y � 0, which is true for any y. This system isdependent.

c. →

The last row indicates that 0y � 1, which is false. This system is inconsistent.

�10

20

01�� 1

�22

�46

�6��

�10

�0.50

20�� 2

�6�1

34

�12��

�10

01

52��1

4�3�6

�18��

�x � 2y � 6�2x � 4y � �6�2x � y � 4

�6x � 3y � �12�x � 3y � �14x � 6y � 8

Lesson 4.3


3.

4.

5.

6.

7.


9.

10.

11.

Lesson 4.4

1.

2.

3.

4.

5.

6.

7. (1, 1) 8. (4, �5) 9.

10. (3, 3, �4)

Lesson 4.5

1.

2.

3. (2, 1)

4.

5. (2, 4)

6. dependent system

7. (1.1, 0.3) 8. (0, 2, �3)


Lesson 5.1

1.

2.

3.

4.

5.

6.

7. ; a � 1, b � �5, c � �24

8. ; a � 1, b � �16, c � 63

9. ; a � 4, b � 0, c � �49

10. ; a � 12, b � 1, c � �1

11. ; a � 49, b � 42, c � 9

12. ; a � 9, b � 0, c � �121


f(x) � 9x2 � 121

f(x) � 49x2 � 42x � 9

f(x) � 12x2 � x � 1

f(x) � 4x2 � 49

f(x) � x2 � 16x � 63

f(x) � x2 � 5x � 24

16x2 � 8x � 1

4x2 � 12x � 9

6x2 � 13x � 5

x2 � 64

2x2 � 13x � 6

x2 � 4x � 21

(�38

, 1112)

�10

01

4�1��

�10

01

3�2��

(52

, �32)

�210

102

0�4

1� � �abc � � � 17

�315�

� 3�2

1

4�1�6

15

�1� � �xyz� � � 11

3�8�

�11

1�1� � �x

y� � �91�

�121

�61� � �r

t� � �07�

��39

1�8� � �a

b� � �74�

�51

3�5� � �x

y� � � 8�2�

� 1�0.5

�21.25�

� 2.5�1.5

�5.53.5�

�115

94�

� 5�6

�45�

�5.5�2

�2.51�

� 21.5

�2.5

0�0.5

0.5

�1�0.5

1.5��10

01�

�5.5�2

�2.51�

ANSWERSC

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©by

Hol

t, R

ineh

art a

nd W

inst

on. A

ll rig

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rese

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.


Cop

yrig

ht ©

by H

olt,

Rin

ehar

t and

Win

ston

. All

right

s re

serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Writing a quadratic function in the form

One method for multiplying two binomials is the “FOIL” method.

1. Multiply First terms.2. Multiply Outside terms.3. Multiply Inside terms.4. Multiply Last terms.

◆ Example 1Use the “FOIL” method to find the product .

◆ Solution(2x � 3)(x � 5) F: product of first terms(2x � 3)(x � 5) O: product of outside terms(2x � 3)(x � 5) I: product of inside terms(2x � 3)(x � 5) L: product of last terms

Thus, .

Recall The Distributive Property states that .

◆ Example 2Write in the form . Identify a, b, and c.

◆ Solution

Distributive PropertyDistributive Property Combine like terms.

Thus, and , , and .c � �15b � �7a � 2f(x) � 2x2 � 7x � 15� 2x2 � 7x � 15� 2x2 � 3x � 10x � 15� (2x � 3)x � (2x � 3)(�5)

f(x) � (2x � 3)(x � 5)

f(x) � ax2 � bx � cf(x) � (2x � 3)(x � 5)

a(b � c) � ab � ac

(2x � 3)(x � 5) � 2x2 � 10x � 3x � 15 � 2x2 � 7x � 15(3)(�5) � �15

(3)(x) � 3x(2x)(�5) � �10x

(2x)(x) � 2x2

(2x � 3)(x � 5)

f(x) � ax2 � bx � c

Use the “FOIL” method to find each product.

1. 2. 3.

4. 5. 6.

Write each function in the form . Identify a, b, and c.

7. 8. 9.

10. 11. 12. (3x � 11)(3x � 11)f(x) �(7x � 3)2f(x) �(3x � 1)(4x � 1)f(x) �

(2x � 7)(2x � 7)f(x) �(x � 9)(x � 7)f(x) �f(x) � (x � 8)(x � 3)

f(x) � ax2 � bx � c

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

(x � 8)(x � 8)(2x � 1)(x � 6)(x � 7)(x � 3)

Reteaching

5.1 Introduction to Quadratic Functions

For each parabola, tell a. whether it opens up or down; b. whether it has a maximum or minimum value; c. where itcrosses the x-axis; and d. the coordinates of the vertex.

13. 14. 15.

a. a. a.

b. b. b.

c. c. c.

d. d. d.

16. 17. 18.

a. a. a.

b. b. b.

c. c. c.

d. d. d.

3x2 � 6x � 24f(x) ��x2 � 2x � 8f(x) �x2 � 4x � 5f(x) �

(x � 1)(x � 2)f(x) ��(x � 2)(x � 2)f(x) �(x � 1)(x � 5)f(x) �

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Graphing a quadratic function and finding its maximum or minimum value

Recall In the form , if , the parabola opens up; if , theparabola opens down.

◆ ExampleDetermine for the graph of a. whether it opens up or down.b. whether it has a maximum or minimum value.c. where it crosses the x-axis.d. the coordinates of the vertex.

◆ Solutiona.

Since and , the parabola opens up.b. A parabola that opens up has a minimum value.c. If , then .

If , then .When 0, the graph crosses the x-axis. So this graph crosses the x-axis when and when .

d. The x-coordinate of the vertex is halfway between 2 and 4, so the axis of symmetry is the line .Substitute 1 for x in .

The coordinates of the vertex are (1, �9).Check with a graphics calculator.

� �9f(1) � (1 � 2)(1 � 4)

(x � 2)(x � 4)f(x) �x � 1

�

x � �2x � 4

f(x) �(x � 2)(x � 4) � (�2 � 2)(�2 � 4) � 0x � �2

(x � 2)(x � 4) � (6)(0) � 0x � 4

1 � 0a � 1f(x) � (x � 2)(x � 4) � x2 � 2x � 8

(x � 2)(x � 4)f(x) �

a � 0a � 0ax2 � bx � cf(x) �

x

y

O

2

6

10

–10

2 6–6 10–10

Lesson 4.3


3.

4.

5.

6.

7.


9.

10.

11.

Lesson 4.4

1.

2.

3.

4.

5.

6.

7. (1, 1) 8. (4, �5) 9.

10. (3, 3, �4)

Lesson 4.5

1.

2.

3. (2, 1)

4.

5. (2, 4)

6. dependent system

7. (1.1, 0.3) 8. (0, 2, �3)


Lesson 5.1

1.

2.

3.

4.

5.

6.

7. ; a � 1, b � �5, c � �24

8. ; a � 1, b � �16, c � 63

9. ; a � 4, b � 0, c � �49

10. ; a � 12, b � 1, c � �1

11. ; a � 49, b � 42, c � 9

12. ; a � 9, b � 0, c � �121


f(x) � 9x2 � 121

f(x) � 49x2 � 42x � 9

f(x) � 12x2 � x � 1

f(x) � 4x2 � 49

f(x) � x2 � 16x � 63

f(x) � x2 � 5x � 24

16x2 � 8x � 1

4x2 � 12x � 9

6x2 � 13x � 5

x2 � 64

2x2 � 13x � 6

x2 � 4x � 21

(�38

, 1112)

�10

01

4�1��

�10

01

3�2��

(52

, �32)

�210

102

0�4

1� � �abc � � � 17

�315�

� 3�2

1

4�1�6

15

�1� � �xyz� � � 11

3�8�

�11

1�1� � �x

y� � �91�

�121

�61� � �r

t� � �07�

��39

1�8� � �a

b� � �74�

�51

3�5� � �x

y� � � 8�2�

� 1�0.5

�21.25�

� 2.5�1.5

�5.53.5�

�115

94�

� 5�6

�45�

�5.5�2

�2.51�

� 21.5

�2.5

0�0.5

0.5

�1�0.5

1.5��10

01�

�5.5�2

�2.51�

ANSWERSC

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©by

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t, R

ineh

art a

nd W

inst

on. A

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.


14. a. opens downb. maximumc. x-intercepts: �2 and 2d. (0, 4)

15. a. opens upb. minimumc. x-intercepts: 1 and 2d. (1.5, �0.25)

16. a. opens upb. minimumc. x-intercepts: 1 and �5d. (�2, �9)

17. a. opens downb. maximumc. x-intercepts: �4 and 2d. (�1, 9)


Lesson 5.2

1.

2.

3.

4. 2 and �10

5. and ; 12.75 and �2.75

6. 3 and �5

7.

8.

9.

10.

11.

12.

Lesson 5.3

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13. 3 and 4

14. 0 and 2

15. 3 and 5

16. �5 and 0

17.

18.

19. 0 and

20. �7 and 7

21. �5 and 0

22. �6 and �2

23.

24. �4 and 4

25. 1 and 2

26. �3 and 3

27. �3 and 17

x � �5

�43

a � 1

x � �32

(d � 15)(d � 5)

(3x � 1)(3x � 1)

(5r � 13)(r � 2)

5y(y � 16)

(x � 8)(x � 5)

(2z � 7)(z � 4)

5(c � 4)(c � 4)

3(a � 2)(a � 1)

(x � 1)2

(a � 7)2

2(c � 7)(c � 2)

(x � 5)(x � 4)

x � 14.14

x � 21.66

x � 13.08

x � 12

x � 20.59

x � 13

5 � �605 � �60

�10 and ��10; 3.16 and �3.16

�13 and ��13; 3.61 and �3.61

�60 and ��60; 7.75 and �7.75

ANSWERSC

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olt, Rinehart and W



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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Solving a quadratic equation by taking square roots

Recall If , then and 4 or 4. The solutions are �4 and 4.

◆ Example 1Solve . Give exact solutions and approximate solutions to thenearest hundredth.

◆ Solution

Add 12 to each side.Divide each side by 5.Take the square root of each side.Use a calculator.

You can check on a graphics calculator by graphing and 188 onthe same screen. The x-coordinates of the intersections of these two graphs willbe the two solutions for the original equation.

◆ Example 2

Solve . Give exact solutions and approximate solutions to the

nearest hundredth.

◆ Solution

Multiply each side by 2.

Take the square root of each side.

or exact solutions

or approximate solutionsx � �0.16x � 6.16

3 � �10x � 3 � �10

x � 3 �10

x � 3 � �10

�(x � 3)2 � 10

(x � 3)2 � 10

12

(x � 3)2 � 5

12

(x � 3)2 � 5

y �y � 5x2 � 12

� 6.32x � �40

x2 � 405x2 � 200

5x2 � 12 � 188

5x2 � 12 � 188

x � �x �x � �16x2 � 16

Solve each equation. Give both exact solutions and approximateirrational solutions to the nearest hundredth.

1. 2. 3.

4. 5. 6. 5(x � 1)2 � 8012

(x � 5)2 � 3013

(x � 4)2 � 12

3x2 � 8 � 225x2 � 65x2 � 60

Reteaching

5.2 Introduction to Solving Quadratic Equations

Find the unknown length in each right triangle. Round answersto the nearest hundredth.

7. 8.

9. 10.

11. 12.

10 10

x

37

30

x

145

x

16

20x

1018

x

5

12

x

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Using the Pythagorean Theorem to find the lengths of the sides of a right triangle

Recall In a right triangle with legs of lengths a and b and hypotenuse of length c,.

◆ Example 1Approximate the length of diameter to the nearest hundredth.

◆ SolutionPythagorean Theorem

Choose the positive root for a length.Use a calculator.

◆ Example 2Find the unknown length in the triangle shown. Round your answer to the nearest hundredth.

◆ SolutionLet FD � 8 and DE � 20.

Use a calculator.FE � 18.33(FE)2 � �202 � 82(FE)2 � 202 � 82

(FE)2 � 82 � 202(FE)2 � (FD)2 � (DE)2

AB � 11.18

AB � �125125 � (AB)2

100 � 25 � (AB)2102 � 52 � (AB)2

(AC)2 � (BC)2 � (AB)2

AB

a2 � b2 � c2

0

510

A

C

B

820

Fd

D

E






Lesson 5.2

1.

2.

3.

4. 2 and �10

5. and ; 12.75 and �2.75

6. 3 and �5

7.

8.

9.

10.

11.

12.

Lesson 5.3

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13. 3 and 4

14. 0 and 2

15. 3 and 5

16. �5 and 0

17.

18.

19. 0 and

20. �7 and 7

21. �5 and 0

22. �6 and �2

23.

24. �4 and 4

25. 1 and 2

26. �3 and 3

27. �3 and 17

x � �5

�43

a � 1

x � �32

(d � 15)(d � 5)

(3x � 1)(3x � 1)

(5r � 13)(r � 2)

5y(y � 16)

(x � 8)(x � 5)

(2z � 7)(z � 4)

5(c � 4)(c � 4)

3(a � 2)(a � 1)

(x � 1)2

(a � 7)2

2(c � 7)(c � 2)

(x � 5)(x � 4)

x � 14.14

x � 21.66

x � 13.08

x � 12

x � 20.59

x � 13

5 � �605 � �60

�10 and ��10; 3.16 and �3.16

�13 and ��13; 3.61 and �3.61

�60 and ��60; 7.75 and �7.75

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◆Skill A Factoring a quadratic expression

Recall To factor a quadratic expression, start by looking for the following patterns.

(1) common monomial factor(2) difference of two squares(3) perfect square trinomial

◆ ExampleFactor each expression.a. b. c.

◆ Solutiona. common monomial factor

difference of 2 squaresb. common monomial factor

perfect square trinomialc. None of the basic 3 patterns occurs.

Find the factors of 24. → 1 24, 2 12, 3 8, 4 6Find a pair such that one factor is negative, so that the product is �24,and the sum of the factors is 5.

and

So, .

Check by using “FOIL.”

x2 � 5x � 24 � (x � 3)(x � 8)

�3 � 8 � 5�3 � 8 � �24

��

� 2(x � 4)22x2 � 16x � 32 � 2(x2 � 8x � 16)

� 3(x � 3)(x � 3)� 3(x2 � 9)3x2 � 27

x2 � 5x � 242x2 � 16x � 323x2 � 27

x2 � 10x � 25 � (x � 5)24a2 � 49 � (2a � 7)(2a � 7)4a2 � 20a � 4a(a � 5)

Factor each quadratic trinomial.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12. d2 � 10d � 759x2 � 1

5r 2 � 23r � 265y2 � 80y

x2 � 3x � 402z2 � z � 28

5c2 � 803a2 � 3a � 6

x2 � 2x � 1a2 � 14a � 49

2c2 � 10c � 28x2 � x � 20

Reteaching

5.3 Factoring Quadratic Expressions

Solve each quadratic equation by factoring and applying theZero-Product Property.

13. 14. 15.

16. 17. 18.

19. 20. 21.

Use factoring and the Zero-Product Property to find the zeros ofeach function.

22. 23. 24.

25. 26. 27. f(x) � x2 � 14x � 51f(x) � �5x2 � 45f(x) � 7x2 � 21x � 14

f(x) � 3x2 � 48f(x) � x2 � 10x � 25f(x) � x2 � 8x � 12

�8m2 � 40mx2 � 49 � 012b2 � �16b

a2 � 2a � 1 � 04x2 � 9 � �12xx2 � 5x � 0

x2 � 8x � 15 � 03x2 � 6x � 0x2 � 7x � 12 � 0

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Using factoring to solve a quadratic equation

Recall If , then either or . Zero Product Property

◆ Example 1Solve .

◆ Solution

common monomial factorreverse of “FOIL”

Since ,either or . Zero Product Property

or The solutions are �2 and 5.

◆ Example 2Find the zeros of .

◆ Solutionset equal to 0common monomial factor

or Zero Product Property

or

Thus, the zeros are �3.5 and 0.

x � �72

x � 0

2x � 7 � 0x � 0x(2x � 7) � 02x2 � 7x � 0

f(x) � 2x2 � 7x

x � �2x � 5x � 2 � 0x � 5 � 0

2 02(x � 5)(x � 2) � 0

2(x2 � 3x � 10) � 02x2 � 6x � 20 � 0

2x2 � 6x � 20 � 0

x � 5 � 0x � 3 � 0(x � 3)(x � 5) � 0






Lesson 5.2

1.

2.

3.

4. 2 and �10

5. and ; 12.75 and �2.75

6. 3 and �5

7.

8.

9.

10.

11.

12.

Lesson 5.3

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13. 3 and 4

14. 0 and 2

15. 3 and 5

16. �5 and 0

17.

18.

19. 0 and

20. �7 and 7

21. �5 and 0

22. �6 and �2

23.

24. �4 and 4

25. 1 and 2

26. �3 and 3

27. �3 and 17

x � �5

�43

a � 1

x � �32

(d � 15)(d � 5)

(3x � 1)(3x � 1)

(5r � 13)(r � 2)

5y(y � 16)

(x � 8)(x � 5)

(2z � 7)(z � 4)

5(c � 4)(c � 4)

3(a � 2)(a � 1)

(x � 1)2

(a � 7)2

2(c � 7)(c � 2)

(x � 5)(x � 4)

x � 14.14

x � 21.66

x � 13.08

x � 12

x � 20.59

x � 13

5 � �605 � �60

�10 and ��10; 3.16 and �3.16

�13 and ��13; 3.61 and �3.61

�60 and ��60; 7.75 and �7.75

ANSWERSC

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olt, Rinehart and W



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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Solving a quadratic equation by completing the square

Recall A perfect square trinomial can be factored into 2 identical factors.

◆ Example 1Add a term to to make a perfect square trinomial.

◆ SolutionUse the coefficient of the x term, �12.

Take of this coefficient and square it: .

The perfect square trinomial is .Notice that

◆ Example 2Solve by completing the square. Give exact solutions andsolutions rounded to the nearest hundredth.

◆ Solution

Add to both sides.

Factor.

oror Use a calculator.x � �0.73x � 2.73

x � 1 � �3x � 1 � �3x � 1 � �3

(x � 1)2 � 3

�12

(�2)�2x2 � 2x � (�1)2 � 2 � (�1)2

x2 � 2x � 2x2 � 2x � 2 � 0

x2 � 2x � 2 � 0

x2 � 12x � 36 � (x � 6)2x2 � 12x � 36

�12

(�12)�2� (�6)2 � 361

2

x2 � 12x

Add a term to complete the square.

1. 2. 3.

Solve by completing the square. Give exact solutions andapproximate solutions to the nearest hundredth.

4. 5.

6. 7.

8. 9. x2 � 10x � 25 � 36x2 � 12x � 9 � 0

x2 � 10x � �13x2 � 2x � 15

x2 � 6x � 6 � 0x2 � 12x � 4 � 0

x2 � 5x �x2 � 16x �x2 � 6x �

Reteaching

5.4 Completing the Square

Write each quadratic function in vertex form. Give thecoordinates of the vertex and the axis of symmetry.

10. 11. 12.

13. 14. 15. y � x2 � 3x � 4y � x2 � 6x � 9y � x2 � 2x � 3

y � x2 � 4xy � �x2 � 5y � 2x2

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Writing a quadratic function in the vertex form

Recall The equation can be written in the form , where(h, k) is the vertex of the parabola.

◆ ExampleGiven ,a. write the equation in vertex form.b. identify the transformations of the graph of .c. give the coordinates of the vertex.d. write the equation of the axis of symmetry.

◆ Solutiona.

Complete the square for by adding 4. To keep the equation in balance you must also add �4.

Notice that the standard vertex form contains �h. Therefore, in this problem , and .

b. This is a translation of by 2 units to the right and 3 units down.

c. The vertex of the parabola is at (h, k) � (2, �3).d. The axis of symmetry is the vertical line containing the vertex, .x � 2

x2f(x) �

k � �3h � 2a � 1

(x � 2)2 � 3g(x) �

(x2 � 4x � 4) � 1 � 4g(x) �

x2 � 4x

x2 � 4x � 1g(x) �

x2f(x) �

x2 � 4x � 1g(x) �

y � a(x � h)2 � ky � ax2 � bx � c

x

y

O

2

2–6–10

–6

–10

6 10

Lesson 5.4

1. 9 2. 64 3.

4. ;

5. ; 1.27 and 4.73

6. �5 and 3

7. ; 1.54 and 8.46

8. ; 0.80 and 11.20

9. �11 and 1

10. vertex form: vertex: (0, 0)axis of symmetry:


12. vertex form: vertex: (�2, �4)axis of symmetry:

13. vertex form: vertex: (1, �4)axis of symmetry:

14. vertex form: vertex: (�3, 0)axis of symmetry:

15. vertex form:

vertex:

axis of symmetry:

Lesson 5.5

1. 2.24 and �6.24

2. 1 and �3.5

3. 1.85 and �4.85

4. 0.68 and �0.88

5. 2.04 and �1.75

6. 0.75 and �0.5

7. 1.25 and �1

8. 0.59 and 5.08

9. x-intercepts: �3 and 5axis of symmetry: vertex: (1, �16)

10. x-intercepts: �5 and �1axis of symmetry: vertex: (�3, �4)

11. x-intercepts: 0 and 3axis of symmetry: vertex: (1.5, 2.25)

12. x-intercepts: 0.5 and �2.5axis of symmetry: vertex: (�1, �9)

13. x-intercepts: �2 and 3axis of symmetry: vertex: (0.5, 6.25)

14. x-intercept: 1.5axis of symmetry: vertex: (1.5, 0)

Lesson 5.6

1. 24; 2 real solutions

2. �4; 2 nonreal solutions


4. 0; 1 real solution


x � 1.5

x � 0.5

x � �1

x � 1.5

x � �3

x � 1

x � �32

(�32

, � 254 )

�254

y � (x �32)2

x � �3

y � (x � 3)2 � 0

x � 1

y � (x � 1)2 � 4

x � �2

y � (x � 2)2 � 4

x � 0

y � �(x � 0)2 � 5

x � 0

y � 2(x � 0)2 � 0

6 � �27 and 6 � �27

5 � �12 and 5 � �12

3 � �3 and 3 � �3

�11.66 and �0.34�6 � �32 and �6 � �32

254

ANSWERSC

opyr

ight

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t, R

ineh

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nd W

inst

on. A

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Cop

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olt,

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Using the quadratic formula to solve a quadratic equation

Recall If , and , then .

◆ Example 1Use the quadratic formula to find the roots of .

◆ Solution

Writing the formula will help you to remember it.

and

or

or or

Therefore, the solutions are �3.5 and 4.

◆ Example 2Solve . Give exact solutions and solutions rounded to the nearest hundredth.

◆ SolutionWrite in the form ax 2 � bx � c.

or or x � �0.27x � 7.27x �14 � �228

4x �

14 � �2284

a � 2, b � �14, c � �4x ��(�14) �(�14)2 � 4(2)(�4)

2(2)

x ��b �b2 � 4ac

2a

2x2 � 14x � 4 � 0

2x2 � 14x � 4

x � �32

x � 4x �5 � 11

4x �

5 � 114

x �5 � �121

4x �

5 � �1214

c � 12a � 2, b � �5,x ��(�5) �(�5)2 � 4(2)(�12)

2(2)

x ��b �b2 � 4ac

2a

2x2 � 5x � 12 � 0

x ��b �b2 � 4ac

2aa 0ax2 � bx � c � 0

Use the quadratic formula to solve each equation. Round youranswer to the nearest hundredth if the solutions are irrational.

1. 2.

3. 4.

5. 6.

7. 8. 17x � 3x2 � 94x2 � 5 � x � 0

8x2 � 2x � 3 � 07x2 � 2x � 25

5x2 � 3 � �x3x2 � 9x � 27

2x2 � 5x � 7 � 0x2 � 4x � 14 � 0

Reteaching

5.5 The Quadratic Formula

Find the x-intercepts, the equation for the axis of symmetry, andthe coordinates of the vertex for each parabola.

9. 10. 11.

12. 13. 14. y � 4x2 � 12x � 9y � �x2 � x � 6y � 4x2 � 8x � 5

y � �x2 � 3xy � x2 � 6x � 5y � x2 � 2x � 15

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Finding the vertex and the equation of the axis of symmetry for a parabola given inthe form

Recall The sum of the two roots of is , and the axis of symmetry

crosses the x-axis at the average of these two roots. Therefore, the axis of symmetry

is .

◆ ExampleFor the equation , finda. the x-intercepts.b. an equation of the axis of symmetry.c. the coordinates of the vertex.

◆ Solutiona. The parabola crosses the x-axis when .

Set equal to 0 and use the quadratic formula to solve.

→ → or

The x-intercepts are �2 and 8.

b. The axis of symmetry is halfway between these two points.

Notice that this is . So the axis of symmetry is .

c. Substitute 3 for x in .

The coordinates of the vertex are (3, �25).

Check by using a graphics calculator to graph the function.

32 � 6(3) � 16 � �25y � x2 � 6x � 16

x � 3�b

2a�

�(�6)2(1)

� 3

�2 � 82

� 3

x � �2x � 8x �6 �100

2x �

�(�6) �(�6)2 � 4(1)(�16)2(1)

a � 1, b � �6, c � �16� 0x2 � 6x � 16x2 � 6x � 16

y � 0

y � x2 � 6x � 16

x � �b

2a

�ba

ax2 � bx � c � 0

y � ax2 � bx � c

Lesson 5.4

1. 9 2. 64 3.

4. ;

5. ; 1.27 and 4.73

6. �5 and 3

7. ; 1.54 and 8.46

8. ; 0.80 and 11.20

9. �11 and 1






15. vertex form:

vertex:

axis of symmetry:

Lesson 5.5

1. 2.24 and �6.24

2. 1 and �3.5

3. 1.85 and �4.85

4. 0.68 and �0.88

5. 2.04 and �1.75

6. 0.75 and �0.5

7. 1.25 and �1

8. 0.59 and 5.08







Lesson 5.6






x � 1.5

x � 0.5

x � �1

x � 1.5

x � �3

x � 1

x � �32

(�32

, � 254 )

�254

y � (x �32)2

x � �3

y � (x � 3)2 � 0

x � 1

y � (x � 1)2 � 4

x � �2

y � (x � 2)2 � 4

x � 0

y � �(x � 0)2 � 5

x � 0

y � 2(x � 0)2 � 0

6 � �27 and 6 � �27

5 � �12 and 5 � �12

3 � �3 and 3 � �3

�11.66 and �0.34�6 � �32 and �6 � �32

254

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olt,

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. All

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s re

serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Using the discriminant to find the number and nature of the roots of a quadratic equation

Recall The discriminant of the equation is .

◆ Example 1Find the discriminant for each equation. Then tell the number and nature of the solutions.a. b. c.

◆ Solution

a.Since , this equation has 2 real solutions.

b.Since the discriminant equals 0, this equation has 1 real solution.

c.Since , this equation has 2 nonreal solutions.

Recall and , where

◆ Example 2Use the quadratic formula to solve .

◆ Solution

→

or where ��31 � i�31x �34

� i�314

x �34

� i�314

x �3 ��31

4x �

�(�3) �(�3)2 � 4(2)(5)2(2)

2x2 � 3x � 5 � 0

r � 0��r � i�r��1 � i

�36 � 0b2 � 4ac � 122 � 4(3)(15) � �36

b2 � 4ac � 122 � 4(3)(12) � 0

48 � 0b2 � 4ac � 122 � 4(3)(8) � 48

3x2 � 12x � 15 � 03x2 � 12x � 12 � 03x2 � 12x � 8 � 0

b2 � 4acax2 � bx � c

Find the discriminant, and determine the number of realsolutions.

1. 2. 3.

4. 5. 6.

Use the quadratic formula to solve each equation.

7. 8. 9. 6x2 � 7x � �3x2 � 12x � 32 � 0x2 � 10x � 34 � 0

x2 � 4x � 5 � 03x2 � 4x � 2 � 0x2 � 8x � 16 � 0

x2 � 5x � 5 � 0x2 � 6x � 10 � 0x2 � 6x � 3 � 0

Reteaching

5.6 Quadratic Equations and Complex Numbers

Perform the indicated operation.

10. 11.

12. 13.

14. 15.

16. 17.

Write the conjugate and find the absolute value of each complex number.

18. 19. 20. �5 � 5i�2 � i8 � 5i

(3 � 4i) (1 � i)(5 � i) (2 � i)

(5 � 2i)(5 � 2i)(3 � i)(�4 � 2i)

(�7 � 2i) � (4 � 5i)(2 � i) � (3 � 2i)

(�8 � 5i) � (�8 � 5i)(�3 � i) � (6 � 3i)

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Performing operations with complex numbers

Recall Any power of i can be simplified to i, �1, �i, or 1.

i2 � �1, i3 � �i, i4 � 1, i5 � i, i6 � �1, i7 � �1, i8 � 1, . . .◆ Example 1

Find each sum, difference, product, and quotient.a. b.c. d.

◆ Solutiona. b.

c. d.

Recall The absolute value of , denoted , is given by .

◆ Example 2Graph and its conjugate on the complex plane. Then find the absolute value of .

◆ SolutionThe graphs of and are shown at right.

The length of the segment from the origin to the point representing is .�133 � 2i

�3 � 2i� � �32 � 22 � �13

3 � 2i3 � 2i

2 � 3i3 � 2i

�a2 � b2�a � bi�a � bi

� 11 � 10i� (8 � 3) � (�2 � 12)i� �2i � 12i � 3(�1)� 8 � 2i � 12i � 3i2

(2 � 3i) (4 � i)(2 � 3i)(4 � i)

� �2 � 4i� 6 � 2i� (2 � 4) � (3i � (�i))� (2 � 4) � (3i � i)

(2 � 3i) � (4 � i)(2 � 3i) � (4 � i)

(2 � 3i) (4 � i)(2 � 3i)(4 � i)(2 � 3i) � (4 � i)(2 � 3i) � (4 � i)

imaginary

real

3 – 2i

3 + 2i2

4

2 4–2–2

��10 � 11i

�15�

23

�1115

i

�2 � 8i � 3i � 12i2

�15

�2 � 3i1 � 4i

�1 � 4i1 � 4i

Lesson 5.4

1. 9 2. 64 3.

4. ;

5. ; 1.27 and 4.73

6. �5 and 3

7. ; 1.54 and 8.46

8. ; 0.80 and 11.20

9. �11 and 1






15. vertex form:

vertex:

axis of symmetry:

Lesson 5.5

1. 2.24 and �6.24

2. 1 and �3.5

3. 1.85 and �4.85

4. 0.68 and �0.88

5. 2.04 and �1.75

6. 0.75 and �0.5

7. 1.25 and �1

8. 0.59 and 5.08







Lesson 5.6






x � 1.5

x � 0.5

x � �1

x � 1.5

x � �3

x � 1

x � �32

(�32

, � 254 )

�254

y � (x �32)2

x � �3

y � (x � 3)2 � 0

x � 1

y � (x � 1)2 � 4

x � �2

y � (x � 2)2 � 4

x � 0

y � �(x � 0)2 � 5

x � 0

y � 2(x � 0)2 � 0

6 � �27 and 6 � �27

5 � �12 and 5 � �12

3 � �3 and 3 � �3

�11.66 and �0.34�6 � �32 and �6 � �32

254

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7. 5 � 3i and 5 � 3i

8. 4 and 8

9. and

10.

11. �16

12.

13.

14.

15. 29

16.

17.

18.

19.

20.

Lesson 5.7

1.

2.

3.

4.

5.

6.

7.

Lesson 5.8

1.

2. or

3. or

4.

5.

6. or

0 2 4 6 8 10–2–4–6–8–10x

x � 3x � �3

0 2 4 6 8 10–2–4–6–8–10x

�7 � x � �1

0 2 4 6 8 10–2–4–6–8–10x

�2 � x � 4

0 2 4 6 8 10–2–4–6–8–10x

x � 3x � �5

0 2 4 6 8 10–2–4–6–8–10x

x � 4x � �2

0 2 4 6 8 10–2–4–6–8–10x

�3 � x � 1

–102

x

y

f(x) � �1.31x2 � 6.55x � 4.95

O

10

2

y

x

f(x) � 0.52x2 � 3.10x � 2.69

O

1

1

y

x

f(x) � 3.5x2 � 18.3x � 23.5

f(x) � �x2 � x � 1

f(x) � �2x2 � 3

f(x) � 5x2 � 7x

f(x) � x2 � 5x � 8

�5 � 5i; �50

�2 � i; �5

8 � 5i; �89

�12

�72

i

95

�75

i

�14 � 2i

�11 � 7i

�1 � 3i

3 � 2i

�712

� i�2312

�712

� i�2312

ANSWERSC

opyright ©by H

olt, Rinehart and W



Cop

yrig

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by H

olt,

Rin

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. All

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serv

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Finding a quadratic function that contains 3 given noncollinear points in the plane(You will need a graphics calculator.)

Recall If a system of linear equations has coefficient matrix A, variable matrix X, and constant matrix B, then and .

◆ ExampleFind a quadratic function whose graph contains the points (�1, 14), (1, 0), and (2, �5).

◆ SolutionEach of the points must be a solution for .

→

So . Use a graphics calculator to solve for matrix X.

�

Thus, , and .

The function is .

Use a graphics calculator to check that the three given points are on this parabola.

f(x) � 2x2 � 7x � 5

c � 5a � 2, b � �7X � � 2�7

5�

� 140

�1��16

�0.513

�0.5

0.5

1

130

�13�� 14

0�1�X � �1

14

�112

111�

�1

X � A�1B

� � 140

�1��abc��1

14

�112

111�AX � B

f(x) � ax2 � bx � c

X � A�1BAX � B

Solve a system of equations to find a quadratic function that fitseach set of data points exactly.

1. (�2, 22), (0, 8), and (3, 2) 2. (�1, 12), (1, �2), and (2, 6)

3. (0, 3), (1, 1), and (2, �5) 4. (�4, �21), (�1, �3), and (3, �7)

Reteaching

5.7 Curve Fitting With Quadratic Models

point

14(1, 0) , 0(2, �1) , �1 4a � 2b � c � �1a(2)2 � b(2) � c � �1f(x) �x � 2

a � b � c � 0a(1)2 � b(1) � c � 0f(x) �x � 1a � b � c � 14a(�1)2 � b(�1) � c � 14f(x) �x � �1,(�1, 14)

ax2 � bx � c � f(x)

Use a graphics calculator to find the best quadratic model torepresent each set of data. If necessary, round coefficients to thenearest hundredth. Then graph the data and the regressionequation.

5. (1, 8) , (2, 3) , (3, �2), 6. (�5, 28) , (�2, �1) , (1, 3), 7. (�3, �38) , (0, 0) , (3, �2),and (4, 7) and (3, �11) and (6, �11)

x

y

Ox

y

Ox

y

O

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Finding the quadratic model that best represents a set of data(You will need a graphics calculator.)

Recall A quadratic model will be in the form .

◆ ExampleLynda dives off the high diving board. You and Carlo use a stopwatch andobserve markings on the side of the board to calculate Lynda’s height above thewater at various times. As Carlo clocks times of 0.5 seconds, 1 second, 1.5seconds, and 2 seconds, you estimate Lynda’s altitude as 24 meters, 25 meters, 22 meters, and 17 meters respectively. Find the quadratic regression equation that best fits these data.

◆ SolutionUse a graphics calculator and enter the data points (0.5, 24), (1, 25), (1.5, 22), and (2, 17).Then use the quadratic regression feature to find a quadratic model.

Graph the data points and the regression equation. Notice that the data do not exactly fit the curve, but your estimates of height were actually quite accurate.

f(x) � �6x2 � 10.2x � 20.5

f(x) � ax2 � bx � c

15


7. 5 � 3i and 5 � 3i

8. 4 and 8

9. and

10.

11. �16

12.

13.

14.

15. 29

16.

17.

18.

19.

20.

Lesson 5.7

1.

2.

3.

4.

5.

6.

7.

Lesson 5.8

1.

2. or

3. or

4.

5.

6. or

0 2 4 6 8 10–2–4–6–8–10x

x � 3x � �3

0 2 4 6 8 10–2–4–6–8–10x

�7 � x � �1

0 2 4 6 8 10–2–4–6–8–10x

�2 � x � 4

0 2 4 6 8 10–2–4–6–8–10x

x � 3x � �5

0 2 4 6 8 10–2–4–6–8–10x

x � 4x � �2

0 2 4 6 8 10–2–4–6–8–10x

�3 � x � 1

–102

x

y

f(x) � �1.31x2 � 6.55x � 4.95

O

10

2

y

x

f(x) � 0.52x2 � 3.10x � 2.69

O

1

1

y

x

f(x) � 3.5x2 � 18.3x � 23.5

f(x) � �x2 � x � 1

f(x) � �2x2 � 3

f(x) � 5x2 � 7x

f(x) � x2 � 5x � 8

�5 � 5i; �50

�2 � i; �5

8 � 5i; �89

�12

�72

i

95

�75

i

�14 � 2i

�11 � 7i

�1 � 3i

3 � 2i

�712

� i�2312

�712

� i�2312

ANSWERSC

opyright ©by H

olt, Rinehart and W



Cop

yrig

ht ©

by H

olt,

Rin

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Win

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. All

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serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Graphing a quadratic inequality in one variable

Recall The product of two factors is positive only if both factors have the same sign.

◆ Example 1Graph the solution to on a number line.

◆ Solution

Since this product is 0 if or , mark these points on the number line.The product is positive if both factors are positive or both factors are negative. Sofind the values of x that cause both factors to have the same sign.

only when .

only when .

only whenor .

Therefore the product is greater than or equal to 0, that is only if

or . The graph is shown below.

◆ Example 2Graph the solution to on a number line.

◆ SolutionUse the chart in Example 1 to see that the product is negative, that is

, only if .

0 2 4 6 8 10–2–4–6–8–10x

�6 � x � 2(x � 6)(x � 2) � 0

x2 � 4x � 12 � 0

0 2 4 6 8 10–2–4–6–8–10x

x � 2x � �6(x � 6)(x � 2) � 0

x � 2x � 6(x � 6)(x � 2) � 0

x � 2x � 2 � 0

x � �6x � 6 � 0

x � 2x � �6(x � 6)(x � 2) � 0x2 � 4x � 12 � 0

x2 � 4x � 12 � 0

Graph the solution to each inequality on a number line.

1. 2.

3. 4.

5. 6.

0 2 4 6 8 10–2–4–6–8–10x

0 2 4 6 8 10–2–4–6–8–10x

x2 � 9 � 0x2 � 8x � 7 � 0

0 2 4 6 8 10–2–4–6–8–10x

0 2 4 6 8 10–2–4–6–8–10x

x2 � 6x � 8 � 0x2 � 2x � 15 � 0

0 2 4 6 8 10–2–4–6–8–10x

0 2 4 6 8 10–2–4–6–8–10x

(x � 4)(x � 2) � 0(x � 3)(x � 1) � 0

Reteaching

5.8 Solving Quadratic Inequalities

0 2 4 6 8 10–2–4–6–8–10

(x + 6) – + +

(x – 2) – – +

(x + 6)(x – 2) + – +

x

Graph each inequality.

7. 8. 9.

x

y

Ox

y

Ox

y

O

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

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Graphing a quadratic inequality in two variables

Recall If a quadratic function is written in the form , the vertex is atthe point (h, k).


◆ SolutionStart with the equation .

Complete the square.vertex form

The vertex is at (�1, �3).

For shade above the curve.The shaded region is where the y-coordinate of each point is greater than for the x-coordinate of that point.


◆ SolutionDraw the parabola from Example 1 with a broken curve and shade below the curve.Test the point (0, 0).Since , this point is not in the shaded region.

0 � 02 � 2 � 0 � 2

y � x2 � 2x � 2

x2 � 2x � 2

y � x2 � 2x � 2

y � (x � 1)2 � 3

y � (x2 � 2x � 1) � 2 � 1y � x2 � 2x � 2

y � x2 � 2x � 2

a(x � h)2 � kf(x) �

x

y

O

2

6

10

2 106–10

–10

–6

–6

x

y

O

2

2

6

10

106–10

–10

–6

–6


7. 5 � 3i and 5 � 3i

8. 4 and 8

9. and

10.

11. �16

12.

13.

14.

15. 29

16.

17.

18.

19.

20.

Lesson 5.7

1.

2.

3.

4.

5.

6.

7.

Lesson 5.8

1.

2. or

3. or

4.

5.

6. or

0 2 4 6 8 10–2–4–6–8–10x

x � 3x � �3

0 2 4 6 8 10–2–4–6–8–10x

�7 � x � �1

0 2 4 6 8 10–2–4–6–8–10x

�2 � x � 4

0 2 4 6 8 10–2–4–6–8–10x

x � 3x � �5

0 2 4 6 8 10–2–4–6–8–10x

x � 4x � �2

0 2 4 6 8 10–2–4–6–8–10x

�3 � x � 1

–102

x

y

f(x) � �1.31x2 � 6.55x � 4.95

O

10

2

y

x

f(x) � 0.52x2 � 3.10x � 2.69

O

1

1

y

x

f(x) � 3.5x2 � 18.3x � 23.5

f(x) � �x2 � x � 1

f(x) � �2x2 � 3

f(x) � 5x2 � 7x

f(x) � x2 � 5x � 8

�5 � 5i; �50

�2 � i; �5

8 � 5i; �89

�12

�72

i

95

�75

i

�14 � 2i

�11 � 7i

�1 � 3i

3 � 2i

�712

� i�2312

�712

� i�2312

ANSWERSC

opyright ©by H

olt, Rinehart and W



7.

8.

9.


Lesson 6.1

1. 1.12 2. 0.75 3. 0.925

4. 1.082 5. 1.01 6. 0.995

7. approximately 7350


9. approximately 13,900




13. about 6

14. about 4

15. about 6

Lesson 6.2

1. linear

2. quadratic

3. exponential

4. linear

5. exponential

6. quadratic

7. exponential decay

8. exponential growth

xO

1

1

y

xO

1

1

y

x

y

O

2

2

x

y

O

2

2

x

y

O

2

2

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Cop

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olt,

Rin

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Win

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. All

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serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Finding the multiplier for growth or decay

Recall A multiplier greater than 1 models growth. A multiplier between 0 and 1 modelsdecay.

◆ ExampleFind the multiplier for each situation.a. 5% growth b. 8% decay

◆ Solutiona. Add the growth rate to 100%.

100% � 5% � 105% or 1.05The multiplier is 1.05.

b. Subtract the rate of decay from 100%.100% � 8% � 92% or 0.92

The multiplier is 0.92.

Find the multiplier for each situation.

1. 12% growth 2. 25% decay 3. 7.5% decay

4. 8.2% growth 5. 1% growth 6. 0.5% decay

Reteaching

6.1 Exponential Growth and Decay

◆Skill B Writing and evaluating an exponential expression that models growth or decay(You will need a calculator.)

Recall Any growth or decay rate related to a natural event assumes that the rate remainsconstant, and a prediction based on this rate will give approximate results.

◆ Example The population of a small town of 10,000 people is growing at the rate of about5.2% per year. Predict the approximate population 10 years from now.

◆ SolutionThe multiplier is 100% � 5.2% � 105.2% or 1.052.

10,000(1.052)10 ≈ 16,602.The predicted population is about 16,600.

Use a growth or decay model to solve each problem.A new school district is experiencing an annual growth rate of 9.5%. Theschool population is now 5600 students. What is the approximate predictedpopulation

7. 3 years from now? 8. 5 years from now? 9. 10 years from now?

Use a calculator and table to solve each problem.

13. After 2 hours, only 75% of a new medication remains in your body. If you take an 80-milligram tablet, and this rate of decay is constant, in approximately how many hours will less than 15 milligrams remain in your system?

14. You invest $5000 in an account that earns interest at an effective rate of 8.4% per year. In how many years will you have over $6800 in the account?

15. If you invest $50,000 in a high interest account that earns interest at an effective rate of 13.8% per year, how many years will it take to double your money?

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill C Using a table to find a specific value for an exponential function(You will need a calculator.)

◆ Example Your doctor prescribes a medication for your allergies. After each 1 hour interval,only 90% of the medication present 1 hour ago remains in your system. If youtake a 100-milligram tablet, in approximately how many hours will only 50% ofthe medication remain in your system?

◆ SolutionThe multiplier is 100% � 10% � 90%, or 0.9.Make a table for 100(0.9)x, where x is a positive integer. Use a calculator.

50% of the medication will be left in your system between 6 and 7 hours after theinitial dose.

x 1 2 3 4 5 6 7

100(0.9)x 90 81 72.9 65.61 59.05 53.14 47.83

The rate in the number of reported cases of robbery is dropping at about 7%per year in a given region of the country. The number of cases reported thisyear was approximately 156,000. If the number continues to drop at this rate,what is the approximate predicted number of cases

10. 1 year from now? 11. 3 years from now? 12. 5 years from now?

7.

8.

9.


Lesson 6.1

1. 1.12 2. 0.75 3. 0.925

4. 1.082 5. 1.01 6. 0.995







13. about 6

14. about 4

15. about 6

Lesson 6.2

1. linear

2. quadratic

3. exponential

4. linear

5. exponential

6. quadratic



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olt,

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Identifying a function as linear, quadratic, or exponential

Recall An exponential function contains a variable in an exponent.

◆ Example Classify each function as linear, quadratic, or exponential.a. f(x) � 1.5x2 � 2.5 b. g(x) � 7x � 5 c. h(x) � 5(1.3)x

◆ Solutiona. f(x) is a quadratic function because the variable x is a base and its highest

exponent is 2.b. g(x) is a linear function because the variable x is a base and its highest

exponent is 1.c. h(x) is an exponential function because the variable x is used as an exponent.

Identify each function as linear, quadratic, or exponential.

1. f(x) � 0.7x � 12 2. f(x) � 25 � 1.8 x2 3. f(x) � 7(0.5)x

4. f(x) � 3x 5. f(x) � 70.5x 6. f(x) � (8 � x)2

Reteaching

6.2 Exponential Functions

◆Skill B Classifying an exponential function as exponential growth or exponential decay

Recall The function f(x) � bx represents exponential growth if b > 1 and exponential decay if 0 < b < 1.

◆ Example Identify each function as exponential growth or decay and then graph each function.a. b.

◆ Solutiona. represents exponential growthbecause the base 3 is greater than 1. Thegraph is shown at right.

b. represents exponential decay

because the base is between 0 and 1.

The graph is shown at right.

13

g(x) � (13)x

f(x) � 3x

g(x) � (13)x

f(x) � 3x

xO

3

6

9

g(x) f(x)

3–3

y

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill C Calculating compound interest

Recall The formula for compound interest is , where P is the principal, r is

the annual interest rate, n is the number of times interest is compounded each year,and t is the time in years.

◆ Example Find the amount of a $500 investment after 8 years at 6% interest compoundeda. annually. b. quarterly. c. monthly.

◆ Solution

a. Annually means once a year (n � 1).

≈ 796.92 or $796.92

b. Quarterly means 4 times a year (n � 4).

≈ 805.16 or $805.16

c. (n � 12)

≈ 807.07 or $807.07

A(8) � 500(1 �0.0612 )12�8

A(8) � 500(1 �0.06

4 )4�8

A(8) � 500(1 �0.06

1 )1�8

A(t) � P(1 �rn)nt

Identify each function as exponential growth or exponential decay.Then graph each function.

7. 8. 9.

x

y

Ox

y

Ox

y

O

f(x) � 3(12)x

f(x) � (32)x

f(x) � (23)x

Find the final value of each investment.

10. $1000 at 4.5% compounded annually for 5 years

11. $800 at 6.2% compounded monthly for 10 years

12. $2300 at 8% compounded daily for 7 years

13. $10,000 at 7.7% compounded monthly for 3 years

7.

8.

9.


Lesson 6.1

1. 1.12 2. 0.75 3. 0.925

4. 1.082 5. 1.01 6. 0.995







13. about 6

14. about 4

15. about 6

Lesson 6.2

1. linear

2. quadratic

3. exponential

4. linear

5. exponential

6. quadratic



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10. $1246.18 11. $1484.77

12. $4026.30 13. $12,589.30

Lesson 6.3

1.

2.

3.

4.

5.

6.

7.

8.

9. 10. 11.

12. 13. n � 2 14. n � 4

15. n � 1 16. n � 81 17. n � 32

18. n � 1 19. n � 5 20. n � 10

21. n � 22. n � 0 23. n � 3

24. n � 25. n � 7 26. n � �4

27. n � 5 28. n � 3

Lesson 6.4

1. log2 40

2. log5 10

3. log8 160

4. log3 3

5. log3 256

6.

7. 1.1761 8. �0.2219 9. 1.398

10. 1.4313 11. 1.8751 12. 2.1303

13. 3 14. 13 15. 5 16. 8

17. –8 and 3 18. –4 and 4 19. 4 and 7

Lesson 6.5

1.

2.

3.

4. 5. 6.

7. x � 5 8. x 1.14 9. x 2.90

10. 11. 12.

13. 3 14. 15.

16. 1.83 17. 4.73 18. �2.1

Lesson 6.6

1. 4 2. 15 3. 9 4. 9 5. 5 6. 15

7.

8.

9.

10.

11. x � 6.15

x � 12.18

x � 4.08

x � 3.69

x � 3.53

��

� �4.58� 2.33

� 8.64� 2.86� 1.61

��

x � �1x � 7.64x � 2.32

x � 4

x � 1.43

x � 2.25

log6( x5

y3)

164

12

50 � 1

81 � 82�3 �18

26 � 64

104 � 10,000

72 � 49

log 164

18

�12

log5 125

� �2

log16 4 �12

log5 625 � 4

log3 27 � 3

log8 64 � 2

xO

1

1

y

ANSWERSC

opyright ©by H

olt, Rinehart and W



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yrig

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olt,

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Writing equations in equivalent exponential or logarithmic form

Recall If b is positive and b ≠ 1, then x � logb y if and only if bx � y.

◆ Example 1Write each equation in exponential form.

a. log5 25 � 2 b. log10 1000 � 3 c.

◆ Solutiona. log5 25 � 2 indicates that you must raise the base 5 to the power 2 to get 25.

52 � 25b. log10 1000 � 3 is equivalent to 103 � 1000.

c. is equivalent to . Recall that

◆ Example 2Write each equation in logarithmic form.

a. 25 � 32 b. c.

◆ Solution

a. 25 � 32 is equivalent to log2 32 � 5

b.

c. 813 � ⇒ log8 2 �

13

4�2 �1

16 ⇒

log4

116

� �2

813 � 24�2 �

116

3�2 �132 �

19

3�2 �19

log3 19

� �2

log3 19

� �2

Write each equation in logarithmic form.

1. 82 � 64 2. 33 � 27 3. 54 � 625

4. 5. 6.

Write each equation in exponential form.

7. log7 49 � 2 8. log10 10,000 � 4 9. log2 64 � 6

10. 11. log8 8 � 1 12. log5 1 � 0log2 18

� �3

( 164)

12 �

18

5�2 �1

2516

12 � 4

Reteaching

6.3 Logarithmic Functions

Find the value of n in each equation.

13. n � log10 100 14. n � log2 16

15. n � log8 8 16. log9 n � 2

17. log2 n � 5 18. log4 n � 0

19. logn 125 � 3 20. logn 1,000,000 � 6

21. 22. n � log7 1

23. logn 243 � 5 24.

25. 26.

27. log5 n � 1 28. n � log15

125

n � log12

16logn 149

� �2

log14

n � 3

logn 18

� 3

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Solving equations using the definitions of exponential and logarithmic functions

Recall The One-to-One Property of Exponential Functions states that if bx � by, then x � y.

◆ Example Find the value of n in each equation.a. n � log5 125 b. 4 � log2 n c. logn 100,000 � 5

◆ Solutiona. n � log5 125 5n � 125

5n � 53

n � 3

b. 4 � log2 n 24 � n16 � n

c. logn 100,000 � 5 n5 � 100,000n5 � 105

n � 10

→

→

→


10. $1246.18 11. $1484.77

12. $4026.30 13. $12,589.30

Lesson 6.3

1.

2.

3.

4.

5.

6.

7.

8.

9. 10. 11.

12. 13. n � 2 14. n � 4

15. n � 1 16. n � 81 17. n � 32

18. n � 1 19. n � 5 20. n � 10

21. n � 22. n � 0 23. n � 3

24. n � 25. n � 7 26. n � �4

27. n � 5 28. n � 3

Lesson 6.4

1. log2 40

2. log5 10

3. log8 160

4. log3 3

5. log3 256

6.

7. 1.1761 8. �0.2219 9. 1.398

10. 1.4313 11. 1.8751 12. 2.1303

13. 3 14. 13 15. 5 16. 8

17. –8 and 3 18. –4 and 4 19. 4 and 7

Lesson 6.5

1.

2.

3.

4. 5. 6.

7. x � 5 8. x 1.14 9. x 2.90

10. 11. 12.

13. 3 14. 15.

16. 1.83 17. 4.73 18. �2.1

Lesson 6.6

1. 4 2. 15 3. 9 4. 9 5. 5 6. 15

7.

8.

9.

10.

11. x � 6.15

x � 12.18

x � 4.08

x � 3.69

x � 3.53

��

� �4.58� 2.33

� 8.64� 2.86� 1.61

��

x � �1x � 7.64x � 2.32

x � 4

x � 1.43

x � 2.25

log6( x5

y3)

164

12

50 � 1

81 � 82�3 �18

26 � 64

104 � 10,000

72 � 49

log 164

18

�12

log5 125

� �2

log16 4 �12

log5 625 � 4

log3 27 � 3

log8 64 � 2

xO

1

1

y

ANSWERSC

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olt, Rinehart and W



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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Using the properties of logarithms to write equivalent logarithmic expressions

Recall Three properties of logarithms are:

logb(mn) � logb m � logb n logb mp � p logb m

◆ Example 1Write each expression as a single logarithm.a. log3 9 � log3 3 b. log5 625 � log5 125 c. 3 log2 r � 5 log2 t

◆ Solutiona. log3 9 � log3 3 � log3(9 · 3) Product Property of Logarithms

� log3 27b. Quotient Property of Logarithms

� log5 5c. 3 log2 r � 5 log2 t � log2 r3 � log2 t5 Power Property of Logarithms

� log2 r3t5 Product Property of Logarithms

◆ Example 2Given log10 2 ≈ 0.3010 and log10 3 ≈ 0.4771, find each logarithm.a. log10 6 b. log10 15 c. log10 18

◆ Solutiona. log10 6 � log10(2 · 3) b. c. log10 18 � log(2 · 32)

� log10 2 � log10 3 � log10 3 � log10 2 � log10 2 � 2 log10 3

≈ 0.3010 � 0.4771 ≈ 0.4771 � 0.3010 ≈ 0.3010 � 2(0.4771)

≈ 0.7781 ≈ 0.1761 ≈ 1.2552

log10 1.5 � log10(32)

log5 625 � log5 125 � log5(625125)

logb (mn ) � logb m � logb n

Write each expression as a single logarithm. Then simplify if possible.

1. log2 5 � log2 8 2. log5 20 � log5 2 3. log8 16 � log8 10

4. log3 15 � log3 5 5. 2 log3 2 � 3 log3 4 6. 5 log6 x � 3 log6 y

Use log10 3 ≈ 0.4771 and log10 5 ≈ 0.6990 to evaluate eachlogarithm without using a calculator.

7. log10 15 ≈ 8. log10 0.6 ≈ 9. log10 25 ≈

10. log10 27 ≈ 11. log10 75 ≈ 12. log10 135 ≈

Reteaching

6.4 Properties of Logarithmic Functions

Using the inverse property of exponents and logarithms,simplify each expression.

13. log6 63 14. 15. 16. log10 108 12log12 55log5 13

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Using the inverse property of exponents and logarithms

Recall Since f(x) � bx and g(x) �logb x are inverse functions,

.

◆ ExampleSimplify each expression.a. log5 54 b.

◆ Solutiona. log5 54 � 4 b.

Notice in this example that Notice in this example thatlog5 54 � log5 625

� 4 � 93log3 9 � 32

3log3 9 � 9

3log3 9

logb bx � x and blogb x � x for x � 0

◆Skill C Solving an equation involving logarithms.

Recall The One-to-One Property of Logarithmic Functions states that if logb x � logb y, then x � y.

◆ Example Solve log5(x

2 � 10) � log5 3x for x, and check your answers.

◆ Solutionlog5(x

2 � 10) � log5 3xx2 � 10 � 3x One-to-One Property of Logarithms

x2 � 3x � 10 � 0(x � 5)(x � 2) � 0

x � 5 or x � �2Check:log5(5

2 �10) log5(3 · 5) log5((�2)2 � 10) log5(3 · (�2))log5 15 � log5 15 log5(�6) log5(�6)

true undefined

Since is undefined and is true, the solution is x � 5.x � 5x � �2

Solve each equation for x.

17. log3(x2 � 5x) � log3 24 18. log8 x2 � log8 16 19. log2(x

2 � 28) � log2 11x


10. $1246.18 11. $1484.77

12. $4026.30 13. $12,589.30

Lesson 6.3

1.

2.

3.

4.

5.

6.

7.

8.

9. 10. 11.

12. 13. n � 2 14. n � 4

15. n � 1 16. n � 81 17. n � 32

18. n � 1 19. n � 5 20. n � 10

21. n � 22. n � 0 23. n � 3

24. n � 25. n � 7 26. n � �4

27. n � 5 28. n � 3

Lesson 6.4

1. log2 40

2. log5 10

3. log8 160

4. log3 3

5. log3 256

6.

7. 1.1761 8. �0.2219 9. 1.398

10. 1.4313 11. 1.8751 12. 2.1303

13. 3 14. 13 15. 5 16. 8

17. –8 and 3 18. –4 and 4 19. 4 and 7

Lesson 6.5

1.

2.

3.

4. 5. 6.

7. x � 5 8. x 1.14 9. x 2.90

10. 11. 12.

13. 3 14. 15.

16. 1.83 17. 4.73 18. �2.1

Lesson 6.6

1. 4 2. 15 3. 9 4. 9 5. 5 6. 15

7.

8.

9.

10.

11. x � 6.15

x � 12.18

x � 4.08

x � 3.69

x � 3.53

��

� �4.58� 2.33

� 8.64� 2.86� 1.61

��

x � �1x � 7.64x � 2.32

x � 4

x � 1.43

x � 2.25

log6( x5

y3)

164

12

50 � 1

81 � 82�3 �18

26 � 64

104 � 10,000

72 � 49

log 164

18

�12

log5 125

� �2

log16 4 �12

log5 625 � 4

log3 27 � 3

log8 64 � 2

xO

1

1

y

ANSWERSC

opyright ©by H

olt, Rinehart and W



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yrig

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olt,

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Using logarithms to solve exponential equations

Recall The common logarithm, log10 x, is usually written as log x.

◆ Example Solve each equation.a. 3x � 81 b. 5x � 75 c. 7x � 1 � 150

◆ Solutiona. 3x � 81Since 81 is a power of 3, use powers of 3.

3x � 34

x � 4 One-to-One Property of Exponential Functionsb. 5x � 75Since 75 is not a power of 5, use logarithms to solve this equation.log 5x � log 75

x log 5 � log 75 Power Property of Logarithms

x ≈ 2.68Check: 52.68 ≈ 75c. 7x � 1 � 150

log 7x � 1 � log 150(x � 1)log 7 � log 150

x ≈ 1.57

x �log 150log 7

� 1

x � 1 �log 150

log 7

x �log 75log 5

Solve each equation. Round your answers to the nearest hundredth.

1. 7 x � 80 2. 5x � 10 3. 6 x � 1296

4. 4x � 1 � 100 5. 2x � 3 � 25 6. 3x � 4 � 27

7. 62x � 7 � 216 8. 53x � 1 � 49 9. 10x � 5 � 125

Reteaching

6.5 Applications of Common Logarithms

Evaluate each logarithmic expression to the nearest hundredth.

10. log6 18 11. log5 100 12. log2 400

13. log8 512 14. log10 215 15.

16. log13 110 17. log2.5 76 18. log 116

329

log12

24

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Finding logarithms with bases other than 10(You will need a calculator.)

Recall The equation x � logb y is equivalent to bx � y.

◆ Example 1Find log3 40 using your calculator.

◆ SolutionSince calculators do not work in base 3, you can change this problem to a base 10logarithm problem.

log 3x � log 40 base 10 logarithmsx log 3 � log 40

x ≈ 3.36 Use a calculator.

Recall The change-of-base formula is , where a can be any permissible

logarithmic base.

◆ Example 2Find log5 68 using your calculator.

◆ SolutionTo find log5 68, use logarithms with base 10. That is, use a � 10. Then you canuse a calculator’s built-in base 10 logarithms.

x ≈ 2.62

x � log5 68 ⇒ x �log 68log 5

logb x �loga xloga b

x �log 40log 3

x � log3 40 → 3x � 40


10. $1246.18 11. $1484.77

12. $4026.30 13. $12,589.30

Lesson 6.3

1.

2.

3.

4.

5.

6.

7.

8.

9. 10. 11.

12. 13. n � 2 14. n � 4

15. n � 1 16. n � 81 17. n � 32

18. n � 1 19. n � 5 20. n � 10

21. n � 22. n � 0 23. n � 3

24. n � 25. n � 7 26. n � �4

27. n � 5 28. n � 3

Lesson 6.4

1. log2 40

2. log5 10

3. log8 160

4. log3 3

5. log3 256

6.

7. 1.1761 8. �0.2219 9. 1.398

10. 1.4313 11. 1.8751 12. 2.1303

13. 3 14. 13 15. 5 16. 8

17. –8 and 3 18. –4 and 4 19. 4 and 7

Lesson 6.5

1.

2.

3.

4. 5. 6.

7. x � 5 8. x 1.14 9. x 2.90

10. 11. 12.

13. 3 14. 15.

16. 1.83 17. 4.73 18. �2.1

Lesson 6.6

1. 4 2. 15 3. 9 4. 9 5. 5 6. 15

7.

8.

9.

10.

11. x � 6.15

x � 12.18

x � 4.08

x � 3.69

x � 3.53

��

� �4.58� 2.33

� 8.64� 2.86� 1.61

��

x � �1x � 7.64x � 2.32

x � 4

x � 1.43

x � 2.25

log6( x5

y3)

164

12

50 � 1

81 � 82�3 �18

26 � 64

104 � 10,000

72 � 49

log 164

18

�12

log5 125

� �2

log16 4 �12

log5 625 � 4

log3 27 � 3

log8 64 � 2

xO

1

1

y

ANSWERSC

opyright ©by H

olt, Rinehart and W



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olt,

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Using the inverse functions f(x) � ex and g(x) � ln x to solve equations

Recall ln e is the base e logarithm of x. Therefore, ln e � 1, just like log 10 � 1 (base 10).

◆ Example 1Simplify each expression.a. eln 4 b. ln e2

◆ Solutiona. Since y � ex and y � ln x are inverse functions, eln x � x. So, eln 4 � 4.b. Because of inverse functions, ln ex � x. So ln e2 � 2.

◆ Example 2Solve for x.a. 2e2x � 1 � 60 b. ln x � 3.2

◆ Solutiona. 2e2x � 1 � 60

e2x � 1 � 30ln e2x � 1 � ln 30

2x � 1 � ln 30

x ≈ 1.20

b. ln x � 3.2eln x � e3.2

x ≈ e3.2

x ≈ 24.53

x �ln 30 � 1

2

Simplify each expression.

1. e ln 4 2. e ln 15 3. e2 ln 3

4. ln e9 5. ln e5 6. 5 ln e3

Solve each equation for x by using the natural logarithmicfunction.

7. ex � 34 8. 3ex � 120 9. ex � 8 � 51

10. ln x � 2.5 11. ln(3x � 2) � 2.8 12. ln ex � 5

Reteaching

6.6 The Natural Base, e

Find the amount, A, by using the formula A = Pert forcontinuously compounded interest.

13. $1000 at 4% for 10 years 14. $800 at 7% for 3 years 15. $2500 at 5.2% for 6 years

16. $10,000 at 12% for 10 years 17. $8000 at 8.9% for 2 years 18. $50,000 at 15% for 7 years

19. How long will it take to double your money if you deposit $1200 at an

annual rate of 6.9% compounded continuously?

20. How long will it take to triple your money if you deposit $2000 at an

annual rate of 8.5% compounded continuously?

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Finding interest compounded continuously

Recall If the principal, P, is invested at an annual interest rate of r, compounded continuously, the amount, A, in the investment after t years is A � Pert.

◆ Example 1Find the amount in an account if $1500 is invested at an annual rate of 5.8% andinterest is compounded continuously for 7 years.

◆ SolutionA � Pert

A � 1500e(0.058)(7) 5.8% � 0.058A � $2251.20 Use a calculator.

◆ Example 2How long will it take to double your money if you deposit $500 at an annual rateof 7.2% compounded continuously?

◆ SolutionA � Pert

1000 � 500e(0.072)t $500 doubles to $10002 � e(0.072)t Divide each side by 500.

ln 2 � ln e(0.072)t Take the natural logarithm of each side.ln 2 � 0.072t inverse functions: ln ex � x

t ≈ 9.63It will take about 9 years and 7.5 months to double your money.

t �ln 2

0.072


10. $1246.18 11. $1484.77

12. $4026.30 13. $12,589.30

Lesson 6.3

1.

2.

3.

4.

5.

6.

7.

8.

9. 10. 11.

12. 13. n � 2 14. n � 4

15. n � 1 16. n � 81 17. n � 32

18. n � 1 19. n � 5 20. n � 10

21. n � 22. n � 0 23. n � 3

24. n � 25. n � 7 26. n � �4

27. n � 5 28. n � 3

Lesson 6.4

1. log2 40

2. log5 10

3. log8 160

4. log3 3

5. log3 256

6.

7. 1.1761 8. �0.2219 9. 1.398

10. 1.4313 11. 1.8751 12. 2.1303

13. 3 14. 13 15. 5 16. 8

17. –8 and 3 18. –4 and 4 19. 4 and 7

Lesson 6.5

1.

2.

3.

4. 5. 6.

7. x � 5 8. x 1.14 9. x 2.90

10. 11. 12.

13. 3 14. 15.

16. 1.83 17. 4.73 18. �2.1

Lesson 6.6

1. 4 2. 15 3. 9 4. 9 5. 5 6. 15

7.

8.

9.

10.

11. x � 6.15

x � 12.18

x � 4.08

x � 3.69

x � 3.53

��

� �4.58� 2.33

� 8.64� 2.86� 1.61

��

x � �1x � 7.64x � 2.32

x � 4

x � 1.43

x � 2.25

log6( x5

y3)

164

12

50 � 1

81 � 82�3 �18

26 � 64

104 � 10,000

72 � 49

log 164

18

�12

log5 125

� �2

log16 4 �12

log5 625 � 4

log3 27 � 3

log8 64 � 2

xO

1

1

y

ANSWERSC

opyright ©by H

olt, Rinehart and W



12.

13. $1491.82 14. $986.94

15. $3415.39 16. $33,201.17

17. $9558.60 18. $142,882.56

19. approximately 10 years


Lesson 6.7

1. ; approximately 39,000

2. ; approximately $9480

3.

4.

5.


Lesson 7.1

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11. 16 12. �20 13. 151 14. 0

15.

quartic function; 3 turning points atapproximately �1, 0, and 2

x

y

O

2

2

x2 � 5x � 3

x2 � 6x � 16

5a4 � a3 � 6a2 � a � 4

5x3 � 2x � 8

11x5 � 4x3 � 2x � 6

�3w4 � 3w2 � 4w

7y3 � 9y2 � 2y � 10

2a2 � 23a � 14

10x3 � 3x2 � 3x � 14

5x3 � 7x2 � 3x � 1

xO

10

1

y

x � 0.71

O

1

1

y

x

x � 4

O

1

1

y

x

x � 2

P(t) � 5000e0.128t

P(t) � 1000e0.229t

x � 5

ANSWERSC

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©by

Hol

t, R

ineh

art a

nd W

inst

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olt,

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ehar

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. All

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serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Solving problems using an exponential growth function

Recall The continuous exponential growth function is P(t) � P0ekt, where population

depends on time and P0 is the initial population.

◆ Example 1A population of bacteria grows exponentially. An initial population of 5000bacteria grows to 18,000 in 3 hours. Write an exponential function for thispopulation of bacteria in terms of time.

◆ SolutionP(t) � P0e

kt where P0 � 5000Since the population is 18,000 after 3 hours, t � 3 and P(3) � 18,000.18,000 � 5000ek(3)

Divide each side by 5000 and simplify.

k ≈ 0.427Therefore, P(t) ≈ 5000e0.427t.

◆ Example 2If the bacteria continue to grow at this rate, predict the approximate populationof bacteria in above example 24 hours after the initial time.

◆ SolutionP(t) ≈ 5000e0.427t

P(24) ≈ 5000e0.427(24) t � 24≈ 141,130,167 or approximately 140,000,000.

k �ln( 18

5 )3

ln(185 ) � 3k

ln(185 ) � ln e3k

185

� e3k

Use the exponential growth function, P(t) = P0ekt, to complete

the following.

1. A population of bacteria grows exponentially. An initial population of 1000 bacteria grows to 2500 in 4 hours. Write the function for this population of bacteria as a function of time. Then estimate the population at the end of 16 hours.

2. A special savings account grew exponentially due to interest from$5000 to $5682 in one year. If the account continues to grow at this rate, how much will be in the account 5 years later?

Reteaching

6.7 Solving Equations and Modeling

Solve each equation algebraically and graphically.

3. log2 x � log2(x � 2) � 3 4. log4 x � log4(x � 3) � 1 5. 5ex � 2 � 75

x

y

Ox

y

Ox

y

O

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Solving logarithmic and exponential equations algebraically and graphically

Recall To solve exponential equations, use logarithms; to solve logarithmic equations,use exponents.

◆ Example 1Solve each equation algebraically.a. log2 x � log2(x � 6) � 4 b. 2ex � 4 � 48

◆ Solutiona. log2 x � log2(x � 6) � 4

log2[x(x � 6)] � 4 Product Property of Logarithmsx(x � 6) � 16

x2 � 6x � 16 � 0(x � 8)(x � 2) � 0

x � 8 or x � �2Check: log2 8 � log2(8 � 6) � 4

log2 23 � log2 2 � 4 log2(�2) � log2(�2 � 6) � 4log2(�2) � log2(�2) � 4 log2(�2) � log2(�8) � 4 undefined

3 � 1 � 4 trueSince the domain of a logarithmic function excludes negative numbers, the only solution is 8.b. 2ex � 4 � 48

ex � 4 � 24ln ex � 4 � ln 24

x � 4 � ln 24 ln ex � xx � 4 � ln 24x ≈ 7.18

Recall To graph y � log2 x, enter in your calculator.

◆ Example 2Solve log2 x � log2(x � 6) � 4 graphically.

◆ SolutionGraph y1 � log2 x � log2(x � 6) and y2 � 4.The coordinates of the point of intersection are(8, 4). So, the solution is x � 8.

y �log xlog 2

logb x � y ⇒ x � by

2

2

y

x

12.

13. $1491.82 14. $986.94

15. $3415.39 16. $33,201.17

17. $9558.60 18. $142,882.56



Lesson 6.7



3.

4.

5.


Lesson 7.1

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11. 16 12. �20 13. 151 14. 0

15.


x

y

O

2

2

x2 � 5x � 3

x2 � 6x � 16

5a4 � a3 � 6a2 � a � 4

5x3 � 2x � 8

11x5 � 4x3 � 2x � 6

�3w4 � 3w2 � 4w

7y3 � 9y2 � 2y � 10

2a2 � 23a � 14

10x3 � 3x2 � 3x � 14

5x3 � 7x2 � 3x � 1

xO

10

1

y

x � 0.71

O

1

1

y

x

x � 4

O

1

1

y

x

x � 2

P(t) � 5000e0.128t

P(t) � 1000e0.229t

x � 5

ANSWERSC

opyr

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©by

Hol

t, R

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art a

nd W

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Cop

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olt,

Rin

ehar

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Win

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. All

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serv

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Simplifying and classifying polynomials

Recall To add or subtract polynomials, combine like terms.

◆ ExampleSimplify each expression, write the result in standard form, and then classify thepolynomial by degree and number of terms.

a. b.

◆ Solutiona.

Combine like terms.Write in standard form.

Since the greatest exponent of x is 4, and the polynomial has 3 terms, this is aquartic, or fourth degree, trinomial.

b.Distribute the �1.

Since the greatest exponent of x is 2, and the polynomial has 2 terms, this is aquadratic binomial.

� 2x2 � 8x� (x3 � x3) � 2x2 � 8x � (11 � 11)� x3 � 8x � 11 � x3 � 2x2 � 11

(x3 � 8x � 11) � (x3 � 2x2 � 11)

� �x4 � 7x � 3� (�2x4 � x4) � (5x2 � 5x2) � 7x � 3

(�2x4 � 5x2 � 7x) � (3 � 5x2 � x4)

(x3 � 8x � 11) � (x3 � 2x2 � 11)(�2x4 � 5x2 � 7x) � (3 � 5x2 � x4)

Write each sum or difference as a polynomial in standard form.Classify the polynomial by degree and number of terms.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10. (2x2 � 5x) � (x2 � 3)(3x2 � 2x � 10) � (2x2 � 4x � 6)

(5a3 � 6a2 � 3a � 1) � (5a4 � 6a3 � 2a � 5)(�2x3 � 4) � (�7x3 � 2x � 4)

(9x5 � x3 � 2x) � (�2x5 � 5x3 � 6)(�3w4 � 6w2 � 2w � 1) � (1 � 2w � 3w2)

(7y3 � 2y2 � 6) � (7y2 � 2y � 4)(7a2 � 13a � 20) � (5a2 � 10a � 6)

(8x3 � 3x � 6) � (3x2 � 8 � 2x3)(5x3 � 4x2 � 5x) � (3x2 � 2x � 1)

Reteaching

7.1 An Introduction to Polynomials

Evaluate each polynomial expression for .

11. 12.

13. 14. �2x3 � 5x2 � 4x � 3�3x3 � 7x2 � 3x � 2

x3 � x2 � x � 1x2 � 2x � 1

x � �3

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Evaluating polynomial expressions

Recall You must use the correct order of operations to evaluate expressions.

◆ ExampleEvaluate for .

◆ Solution

� 60� �2(�8) � 5(4) � 7(�2) � 10

�2x3 � 5x2 � 7x � 10 � �2(�2)3 � 5(�2)2 � 7(�2) � 10

x � �2�2x3 � 5x2 � 7x � 10

◆Skill C Describing the general shape of a graph(You will need a graphics calculator.)

Recall A turning point in the graph of a function occurs when there is a change in the vertical direction.

◆ ExampleUse a calculator to graph

and then describe the general shape of the graph.

◆ SolutionThis is a graph of a cubic function. The graph has 2 turning points at approximately x � 0 and x � 2.

f(x) � �x3 � 3x2 � 4–10 10

10

–10

Graph each function. Describe its general shape.

15. 16. 17.

x

y

Ox

y

Ox

y

O

f(x) � x3 � 4x2 � x � 1f(x) � 2x2 � 7x � 4f(x) � �x4 � 2x3 � 4x2 � 8

12.

13. $1491.82 14. $986.94

15. $3415.39 16. $33,201.17

17. $9558.60 18. $142,882.56



Lesson 6.7



3.

4.

5.


Lesson 7.1

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11. 16 12. �20 13. 151 14. 0

15.


x

y

O

2

2

x2 � 5x � 3

x2 � 6x � 16

5a4 � a3 � 6a2 � a � 4

5x3 � 2x � 8

11x5 � 4x3 � 2x � 6

�3w4 � 3w2 � 4w

7y3 � 9y2 � 2y � 10

2a2 � 23a � 14

10x3 � 3x2 � 3x � 14

5x3 � 7x2 � 3x � 1

xO

10

1

y

x � 0.71

O

1

1

y

x

x � 4

O

1

1

y

x

x � 2

P(t) � 5000e0.128t

P(t) � 1000e0.229t

x � 5

ANSWERSC

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16.

quadratic function; 1 turning point atapproximately 2

17.

cubic function; 2 turning points atapproximately 0 and 2.5

Lesson 7.2

1.

maximum at (�2.7, 2.5); minimum at (0, �7); x � �2.7, x > 0;�2.7 � x � 0

2.

maximum at (0, 5); minimum at (�1.2, 2.8) and (1.2, 2.8);�1.2 � x � 0 and x � 1.2; x � �1.2 and 0 � x � 1.2

3.

maximum at (1.8, 8.1); no minimum; x � 1.8; x � 1.8

4. rising on the left and the right,2 turning points

5. falling on the left and the right,2 turning points

6. falling on the left, rising on the right,1 turning point

7. rising on the left, falling on the right,1 turning point




x

y

O

2

2

x

y

O

2

2

x

y

O

2

2

x

y

O

2

2

x

y

O

2

2

ANSWERSC

opyright ©by H

olt, Rinehart and W



Cop

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by H

olt,

Rin

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Win

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Describing the graph of a polynomial function(You will need a graphics calculator.)

Recall Graphs are always read from left to right.A function is increasing from left to right if the y-values are increasing.A function is decreasing from left to right if the y-values are decreasing.

◆ Example Given :a. graph the function,b. give the approximate coordinates of each local minimum and maximum,c. find the intervals over which the function is increasing, andd. find the intervals over which the function is decreasing.

◆ Solutiona. The graph is shown at right.b. Local minimum: approximately

(�0.8, �6.2)For all values of x close to �0.8, the y-coordinate is greater than �6.2.Local maximum: approximately (2.1, 6.1)For all values of x close to 2.1, the y-coordinate is less than 6.1.

c. increasing when d. decreasing when and also when x � 2.1x � �0.8

�0.8 � x � 2.1

P(x) � �x3 � 2x2 � 5x � 4

Graph each function. Find any local maxima or minima to the nearest tenth, and the intervals over which the function is increasing and decreasing.

1. 2. 3.

x

y

Ox

y

Ox

y

O

P(x) � �3x2 � 11x � 2P(x) � x4 � 3x2 � 5P(x) � x3 � 4x2 � 7

Reteaching

7.2 Polynomial Functions and Their Graphs

–10 10

10

–10

Describe the end behavior and number of turning points for thegraph of each function.

4. 5. 6.

7. 8. 9.

Without using a calculator, answer each question. Then checkusing a graphics calculator.

f(x) � �x4 � 5x � 1f(x) � x5 � 5x3 � 4xf(x) � �x2 � x � 5

f(x) � x2 � 5x � 1f(x) � �x3 � 2x2 � 4f(x) � x3 � 7x � 2

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Describing the end behavior and the number of turning points of the graph of apolynomial function (You will need a graphics calculator.)

Recall The degree of a polynomial is determined by the greatest power of the variable.

◆ Example Describe the end behavior and state the number of turning points for each graph shown below.

◆ Solutionf(x): As you read across the graph from

left to right, this function is rising on the left and on the right. There are 2 turning points in between.

Cubic functions have at most 2 turning points.

g(x): As you read across the graph from left to right, this function is rising on the left but falling on the right with one turning point.

Quadratic functions have one turning point.

–10 10

10

–10

f(x)

g(x)

10. Exercise 4 is an odd-degree function(degree 3) and the coefficient of x3 ispositive (�1). Using this as a pattern,predict the end behavior and greatestpossible number of turning points of thegraph of the function

.

12. Using the even-degree function in Exercise6 as a model, predict the end behavior andgreatest possible number of turning pointsof the graph of the function

.

11. Exercise 5 is an odd-degree function andthe coefficient of x3 is negative. Using thisas a pattern, predict the end behavior andthe greatest possible number of turningpoints of the graph of the function

.

13. Using the even-degree function in Exercise7 as a model, predict the end behavior andgreatest possible number of turning pointsof the graph of the function

.f(x) � �x4 � 3x

f(x) � �x5 � 2x2 � 1

f(x) � x4 � 4x3 � 2x2 � 12x � 1

f(x) � x5 � 15x4 � 85x3 � 225x2 � 274x � 120

16.

quadratic function; 1 turning point atapproximately 2

17.

cubic function; 2 turning points atapproximately 0 and 2.5

Lesson 7.2

1.

maximum at (�2.7, 2.5); minimum at (0, �7); x � �2.7, x > 0;�2.7 � x � 0

2.

maximum at (0, 5); minimum at (�1.2, 2.8) and (1.2, 2.8);�1.2 � x � 0 and x � 1.2; x � �1.2 and 0 � x � 1.2

3.

maximum at (1.8, 8.1); no minimum; x � 1.8; x � 1.8


5. falling on the left and the right,2 turning points

6. falling on the left, rising on the right,1 turning point





x

y

O

2

2

x

y

O

2

2

x

y

O

2

2

x

y

O

2

2

x

y

O

2

2

ANSWERSC

opyright ©by H

olt, Rinehart and W



11. falling on the left and the right,4 possible turning points, but only 2 forthis function

12. falling on the left, rising on the right,3 turning points

13. rising on the left, falling on the right,3 possible turning points, but only 1 forthis function

Lesson 7.3

1.

2.

3.

4.

5.

6.

7. yes 8. yes 9. no

10.

11.

12. 121

13. 12

Lesson 7.4

1.

2. 1, 2, and �3

3.

4. 5 occurring 3 times

5. 8 and �2 occurring twice

6. 0, 4, and �4

7. 2, �2, 3, and �3

8.

9. �1.2, 1.8, and 3.4

10. �0.6 and 4.6

Lesson 7.5

1. �3, 1 � 2i, and 1 � 2i

2. �2, 3, �1 � i, and �1 � i

3.

4. 1, �1, 2, and �3

5.

6. �0.53, 0.65, and 2.88

7.

8.


Lesson 8.1

1. direct variation

2. inverse variation

3. direct variation

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14. F varies jointly as m1 and m2 and inverselyas r squared.

y � 0.504

y � 60

y � 5

y � 245

y �12

y � 0.004

y � 0.9375

y � 14

y � 0.25

y � 25

P(x) � x4 � 16

P(x) � �5x2 � 10x � 15

x � 1.32

2, 1 � �3, and 1 � �3

3, �3, �52

, and ��52

�6, 4, and 32

�3 and 12

occurring twice

x � 7

x2 � 5x � 5

(2x � 3)(x � 1)(x � 1)

2x(2x � 7)(2x � 7)

x3(x � 1)(x2 � x � 1)

2x3 � 11x2 � 2x � 15

x4 � 6x3 � 10x2 � 2x � 5

3x3 � 5x2 � 17x � 15

ANSWERSC

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olt,

Rin

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t and

Win

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. All

right

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serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Multiplying and factoring polynomials

Recall and

◆ Example 1Multiply . Express the answer in standard form.

◆ Solution

◆ Example 2Factor each polynomial. a. b.

◆ Solutiona. common monomial factor

b. groups in pairscommon monomial factors(x � 5) is common to bothterms.

◆ Example 3Determine if is a factor of .

◆ SolutionLet . Then is a factor only if .

Therefore, is a factor of .2x3 � x2 � 2x � 3x � 1

P(�1) � 0P(�1) � 2(�1)3 � (�1)2 � 2(�1) � 3

P(�1) � 0x � 1P(x) � 2x3 � x2 � 2x � 3

2x3 � x2 � 2x � 3x � 1

� (x2 � 3)(x � 5)� x2(x � 5) � (�3)(x � 5)

x3 � 5x2 � 3x � 15 � (x3 � 5x2) � (�3x � 15)

x3 � 8 � x3 � 23� x2(x � 2)(x2 � 2x � 4)x5 � 8x2 � x2(x3 � 8)

x3 � 5x2 � 3x � 15x5 � 8x2

� 2x3 � 3x2 � x � 10x2 � 15x � 5 � 2x3 � 13x2 � 16x � 5(x � 5)(2x2 � 3x � 1) � x(2x2 � 3x � 1) � (�5)(2x2 � 3x � 1)

(x � 5)(2x2 � 3x � 1)

a3 � b3 � (a � b)(a2 � ab � b2)a3 � b3 � (a � b)(a2 � ab � b2)

Write each product as a polynomial in standard form.

1. 2. 3.

Factor each polynomial.

4. 5. 6.

Use substitution to determine whether (x � 2) is a factor of thefollowing polynomials.

7. 8. 9. x3 � 10x2 � 16x � 2x4 � 2x3 � 2x � 42x3 � 7x2 � 11x � 10

2x3 � 3x2 � 2x � 38x3 � 98xx6 � x3

(2x � 3)(x � 1)(x � 5)(x2 � 2x � 1)(x2 � 8x � 5)(x � 3)(3x2 � 4x � 5)

Reteaching

7.3 Products and Factors of Polynomials

Divide by long division and by synthetic division.

10. 11.

For each function below, use synthetic division and substitution to find the indicated value.

12. 13. P(x) � x3 � 8x � 5; P(�1)P(x) � x4 � x3 � x2 � x � 1; P(�3)

(x2 � x � 42) (x � 6)(x3 � 2x2 � 10x � 15) (x � 3)

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Dividing polynomials

Recall To divide a polynomial by , you can divide synthetically by �4.

◆ Example 1Divide by , using a. long division; b. synthetic division.

◆ Solution

a.

0

b. 1 2 0 32

↓ �4 8 �32

1 �2 8 0

↓quotient:

◆ Example 2Given , show that when is divided by

, the remainder is equal to P(5).

◆ Solution5 4 �13 �32

20 354 7 3

Thus, by using synthetic division and substitution, you can see that theremainder is equal to P(5).

� 3P(5) � 4(5)2 � 13(5) � 32

P(x) � 4x2 � 13x � 32

x � 54x2 � 13x � 32P(x) � 4x2 � 13x � 32

x2 � 2x � 8

�4

�(8x � 32)

8x � 32

�(� 2x2 � 8x)

� 2x2 � 0x � 32

�(x3 � 4x2)

x � 4�x3 � 2x2 � 0x � 32x2 � 2x � 8

x � 4x3 � 2x2 � 32

x � 4

Hold place with 0x.x3 x � 2(x � 4)(x 2) � x 3 � 4x 2

Subtract.�2x 2 x � �2x(x � 4)(�2x ) � �2x 2 � 8xSubtract.8x x � 8(x � 4)(8) � 8x � 32Subtract.

Bring down 1.Multiply: (�4)(1) � �4Add: 2 � (�4) � �2Multiply: (�4)(�2) � 8Add: 0 � 8 � 8Multiply: (�4)(8) � �32Add: 32 � (�32) � 0




Lesson 7.3

1.

2.

3.

4.

5.

6.

7. yes 8. yes 9. no

10.

11.

12. 121

13. 12

Lesson 7.4

1.

2. 1, 2, and �3

3.



6. 0, 4, and �4

7. 2, �2, 3, and �3

8.

9. �1.2, 1.8, and 3.4

10. �0.6 and 4.6

Lesson 7.5

1. �3, 1 � 2i, and 1 � 2i

2. �2, 3, �1 � i, and �1 � i

3.

4. 1, �1, 2, and �3

5.

6. �0.53, 0.65, and 2.88

7.

8.


Lesson 8.1

1. direct variation


3. direct variation

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.


y � 0.504

y � 60

y � 5

y � 245

y �12

y � 0.004

y � 0.9375

y � 14

y � 0.25

y � 25

P(x) � x4 � 16

P(x) � �5x2 � 10x � 15

x � 1.32

2, 1 � �3, and 1 � �3

3, �3, �52

, and ��52

�6, 4, and 32

�3 and 12

occurring twice

x � 7

x2 � 5x � 5

(2x � 3)(x � 1)(x � 1)

2x(2x � 7)(2x � 7)

x3(x � 1)(x2 � x � 1)

2x3 � 11x2 � 2x � 15

x4 � 6x3 � 10x2 � 2x � 5

3x3 � 5x2 � 17x � 15

ANSWERSC

opyr

ight

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t, R

ineh

art a

nd W

inst

on. A

ll rig

hts

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rved

.


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yrig

ht ©

by H

olt,

Rin

ehar

t and

Win

ston

. All

right

s re

serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Using graphs, synthetic division, and factoring to find rational roots

Recall If 3 is a zero of , then is a solution (root) ofand is a factor of .

◆ Example Find the roots of .

◆ SolutionGraph the function and check for zeros of the function. (These occur where the graph crosses the x-axis.) The graph indicates that 2 may be a root of the given equation.

Use synthetic division to check if 2 is a root.

2 1 0 �3 �22 4 2

1 2 1 0 ← The zero remainderindicates that 2 is a root.

This means that is a factor of .

Thus, . (You read the coefficients of thesecond factor, , from the first three numbers, 1, 2, and 1, found in thelast line of the synthetic division above.)

Factor

Solve .

Zero-Product Property

The roots are 2 and �1, with �1 occurring twice.

x � �1x � �1x � 2x � 2 � 0 or x � 1 � 0 or x � 1 � 0(x � 2)(x � 1)2 � 0(x � 2)(x � 1)2 � 0

x2 � 2x � 1: x2 � 2x � 1 � (x � 1)2

x2 � 2x � 1x3 � 3x � 2 � (x � 2)(x2 � 2x � 1)

x3 � 3x � 2x � 2

f(x) � x3 � 3x � 2

x3 � 3x � 2 � 0

x2 � x � 12x � 3x2 � x � 12 � 0x � 3P(x) � x2 � x � 12

Use a graph, synthetic division, and factoring to find all of theroots of each equation.

1. 2.

3. 4.

5. 6. x3 � 16x � 0x3 � 4x2 � 28x � 32 � 0

x3 � 15x2 � 75x � 125 � 02x3 � x2 � 54x � 72 � 0

x3 � 7x � 6 � 04x3 � 8x2 � 11x � 3 � 0

Reteaching

7.4 Solving Polynomial Equations

–10 10

10

–10

Use variable substitution to find all of the roots of each equation.

7. 8. 2x4 � 23x2 � 45 � 0x4 � 13x2 � 36 � 0

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Solving equations by variable substitution

Recall by using the Distributive Property or “FOIL.”

◆ Example Solve by using variable substitution.

◆ SolutionLet ; so .

Now replace u with and solve for x using the Zero-Product Property.

or

The roots of are 2, �2, , and .��5�5x4 � 9x2 � 20 � 0

x � �5x � 2x2 � 5x2 � 4

x2

(u � 4)(u � 5) � 0x4 � 9x2 � 20 � 0 → u2 � 9u � 20 � 0

u2 � (x2)2 � x4u � x2

x4 � 9x2 � 20 � 0

(x2 � 1)(x2 � 2) � x4 � 3x2 � 2

◆Skill C Using the Location Principle to approximate real zeros of a function

Recall Real zeros of a function occur where its graph crosses the x-axis.

◆ Example Find the approximate zeros of .

◆ SolutionThe graph indicates a zero between 0 and 1. Make a table to approximate that zero.

Since f(x) changes sign when x is between 0.6 and 0.7, it appears that one zero is approximately 0.7.

The graph indicates a zero between 4 and 5. The next table shows a sign changein f(x) when x is between 4.3 and 4.4.

The second real zero is approximately 4.3.

f(x) � x2 � 5x � 3

–10 10

10

–10

x 0.4 0.5 0.6 0.7 0.8

f(x) 1.16 0.75 0.36 �0.01 �0.36

x 4.2 4.3 4.4 4.5

f(x) �0.36 �0.01 0.36 0.75

Use a graph and the Location Principle to find the real zeros of each function. Give approximate values to the nearest hundredth if necessary.

9. 10. f(x) � �x2 � 4x � 3f(x) � x3 � 4x2 � 7




Lesson 7.3

1.

2.

3.

4.

5.

6.

7. yes 8. yes 9. no

10.

11.

12. 121

13. 12

Lesson 7.4

1.

2. 1, 2, and �3

3.



6. 0, 4, and �4

7. 2, �2, 3, and �3

8.

9. �1.2, 1.8, and 3.4

10. �0.6 and 4.6

Lesson 7.5

1. �3, 1 � 2i, and 1 � 2i

2. �2, 3, �1 � i, and �1 � i

3.

4. 1, �1, 2, and �3

5.

6. �0.53, 0.65, and 2.88

7.

8.


Lesson 8.1

1. direct variation


3. direct variation

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.


y � 0.504

y � 60

y � 5

y � 245

y �12

y � 0.004

y � 0.9375

y � 14

y � 0.25

y � 25

P(x) � x4 � 16

P(x) � �5x2 � 10x � 15

x � 1.32

2, 1 � �3, and 1 � �3

3, �3, �52

, and ��52

�6, 4, and 32

�3 and 12

occurring twice

x � 7

x2 � 5x � 5

(2x � 3)(x � 1)(x � 1)

2x(2x � 7)(2x � 7)

x3(x � 1)(x2 � x � 1)

2x3 � 11x2 � 2x � 15

x4 � 6x3 � 10x2 � 2x � 5

3x3 � 5x2 � 17x � 15

ANSWERSC

opyr

ight

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Hol

t, R

ineh

art a

nd W

inst

on. A

ll rig

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rese

rved

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Cop

yrig

ht ©

by H

olt,

Rin

ehar

t and

Win

ston

. All

right

s re

serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Finding all the zeros of a polynomial function

Recall A corollary to the Fundamental Theorem of Algebra states that an nth-degreepolynomial function will have exactly n complex zeros.

◆ Example Find all the zeros of .

◆ SolutionThe only possible rational roots are those found by using a factor of −9 (constantterm) as the numerator of a fraction and a factor of 2 (coefficient of x4) as thedenominator.

possibilities:

The graph shows that one zero appears tobe �1.

�1 2 �1 3 �3 �9�2 3 �6 9

2 �3 6 �9 0

The zero remainder indicates that �1 is a zero.The graph shows that there is another zero between 2 and 3. The only possible rational root from the list above between

2 and 3 is , or 1.5.

To see if 1.5 is a root, apply synthetic division to found fromthe last line of the synthetic division above.

1.5 2 �3 6 �93 0 9

2 0 6 0 ← zero remainder

So 1.5 is also a zero.

This leaves a last row (without the remainder) which represents .

Use the quadratic formula where a � 1, b � 0, and c � 6.

The four zeros of are −1, 1.5, , and .�i�3i�3f(x) � 2x4 � x3 � 3x2 � 3x � 9

x ��0 �02 � 4(2)(6)

2(2)�

��484

� 4i�3

4� i�3

2x2 � 6

2x3 � 3x2 � 6x � 9

32

11

, 12

, 31

, 32

, 91

, 92

f(x) � 2x4 � x3 � 3x2 � 3x � 9

Find all of the rational roots of each polynomial equation.

1. 2.

3. 4. f(x) � x4 � x3 � 7x2 � x � 6f(x) � x3 � 4x2 � 2x � 4

f(x) � x4 � x3 � 6x2 � 14x � 12f(x) � x3 � x2 � x � 15

Reteaching

7.5 Zeros of Polynomial Functions

–10 10

10

–10

Find all zeros of each polynomial function. Round to the nearesthundredth if necessary.

5. 6. �x3 � 3x2 � 2 � 3x3 � 4x2 � 7 � x2 � 8x � 11

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Graphing to solve equations where both members are polynomials(You will need a graphics calculator.)

Recall The coordinates of the point of intersection of two graphs satisfy both equations.

◆ Example Find all the real values of x for which .

◆ SolutionGraph each member as a polynomial function.

Find all points of intersection.(�2.30, �4.91) and (1.30, 5.91)

The solutions are approximately �2.30 and 1.30.

y2 � �x2 � 2x � 5y1 � x2 � 4x � 1

x2 � 4x � 1 � �x2 � 2x � 5

◆Skill C Writing a polynomial function when given its zeros

Recall If a � bi is a zero of a polynomial function, then a � bi is also a zero.

◆ Example Write the simplest polynomial function for which 3 and �2 � i are zeros and

.

◆ SolutionSince �2 � i must also be a zero, the factors are , , and

.

Allow for a stretch factor of a and write what follows.

Since , substitute 0 for x and 45 for .

Thus, , or .P(x) � �3x3 � 3x2 � 21x � 45P(x) � �3(x � 3)(x2 � 4x � 5)

�3 � a45 � �15a45 � a(0 � 3)(02 � 4 � 0 � 5)

P(x)P(0) � 45

� a(x � 3)(x2 � 4x � 5)� a(x � 3[x2 � x(�2 � i) � (�2 � i)x � (�2 � i)(�2 � i)]

P(x) � a(x � 3)[x � (�2 � i)][x � (�2 � i)]

[x � (�2 � i)][x � (�2 � i)]x � 3

P(0) � 45

–10 10

10

–10

Write a polynomial function in standard form by using the given information.

7. zeros �1 and 3; 8. zeros 2, �2, 2i, and �2i; P(0) � �16P(0) � 15




Lesson 7.3

1.

2.

3.

4.

5.

6.

7. yes 8. yes 9. no

10.

11.

12. 121

13. 12

Lesson 7.4

1.

2. 1, 2, and �3

3.



6. 0, 4, and �4

7. 2, �2, 3, and �3

8.

9. �1.2, 1.8, and 3.4

10. �0.6 and 4.6

Lesson 7.5

1. �3, 1 � 2i, and 1 � 2i

2. �2, 3, �1 � i, and �1 � i

3.

4. 1, �1, 2, and �3

5.

6. �0.53, 0.65, and 2.88

7.

8.


Lesson 8.1

1. direct variation


3. direct variation

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.


y � 0.504

y � 60

y � 5

y � 245

y �12

y � 0.004

y � 0.9375

y � 14

y � 0.25

y � 25

P(x) � x4 � 16

P(x) � �5x2 � 10x � 15

x � 1.32

2, 1 � �3, and 1 � �3

3, �3, �52

, and ��52

�6, 4, and 32

�3 and 12

occurring twice

x � 7

x2 � 5x � 5

(2x � 3)(x � 1)(x � 1)

2x(2x � 7)(2x � 7)

x3(x � 1)(x2 � x � 1)

2x3 � 11x2 � 2x � 15

x4 � 6x3 � 10x2 � 2x � 5

3x3 � 5x2 � 17x � 15

ANSWERSC

opyr

ight

©by

Hol

t, R

ineh

art a

nd W

inst

on. A

ll rig

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rese

rved

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Cop

yrig

ht ©

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olt,

Rin

ehar

t and

Win

ston

. All

right

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serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Identifying and using direct and inverse variations

Recall Direct variation can be modeled by ; inverse variation can be modeled by

.

◆ Example 1Determine which type of variation is demonstrated by each of the following.a. the distance you travel when riding at a constant speedb. the time it takes to get to your destination if you accelerate

◆ Solutiona. Your distance increases as your time increases. This is an example of direct

variation.b. The time it takes to get to your destination decreases as your speed increases.

This is an example of inverse variation.

◆ Example 2If y varies inversely as x, and 15 when 4, find the value of y when 10.

◆ Solution

or

, so .

When 6.�x � 10, y �6010

y �60x

k � 4 � 15 � 60

k � xyy �kx

x �x �y �

y �kx

y � kx

Determine whether each of the following is an example of directvariation or inverse variation.

1. the number of words you can type in 30 minutes as typing speed increases

2. the number of tickets you can buy for $100 as the price per ticket increases

3. the number of credit-hours you have as you take more courses

For each problem, assume that y varies inversely as x.

4. If 10 when 10, 5. If 1 when 12, 6. If y � 7 when x � 14,find y when 4. find y when 48. find y when 7.

7. If 2.5 when 1.5, 8. If 100 when 0.4 9. If when ,find y when 4. find y when 10,000.

find y when .x �13

x �x �x �

14

y �23

x �y �x �y �

x �x �x �x �y �x �y �

Reteaching

8.1 Inverse, Joint, and Combined Variation

Solve each variation problem.

10. If y varies directly as x2, and 20 11. If y varies inversely as , and 10 when 2, find y when 7. when 25, find y when 100.

12. If y varies jointly as x and z, and 49 13. If y varies directly as x and inversely as ,when 2.8 and 3.5, find y when and 4 when 12 and 3, find y

1.6 and 7.5. when 7 and 5.

14. Newton’s Universal Law of Gravity is given as . If G is a

constant, explain how F varies in relation to the variables m1, m2, and r.

F � Gm1m2

r2

z �x �z �x �z �x �y �z �x �

z3y �

x �x �x �x �y ��xy �

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Using joint and combined variations

Recall is an example of joint variation; is an example of combinedvariation.

◆ ExampleGiven that when and , find y.a. y varies directly as and 5.b. y varies inversely as and 2.c. y varies jointly as w and z, 20, and 1.

d. y varies directly as s and inversely as r, and .

◆ Solutiona.

When

c.

When and ,

y � 5y � 0.4(20)(1)

z � 1w � 20k � 0.4

16 � 40k16 � k(5)(8)

y � kwz

y � 250t � 5, y � 2 � 53

k � 216 � 8k16 � k � 23

y � kt3

r � 223

s � 212

z �w �x �x2

t �t3z � 8r � 2, s � 6, t � 2, w � 5, x � 0.5y � 16

y �kxz

y � kxz

b.

When and .

d.

If and , .� 5(16

3 )(52)

83

y �r � 223

s � 212

k �163

16 � 3k

16 �k(6)

2

y �ksr

y � 1x � 2, y �422

k � 4

16 �k

0.25

16 �k

(0.5)2

y �kx2




Lesson 7.3

1.

2.

3.

4.

5.

6.

7. yes 8. yes 9. no

10.

11.

12. 121

13. 12

Lesson 7.4

1.

2. 1, 2, and �3

3.



6. 0, 4, and �4

7. 2, �2, 3, and �3

8.

9. �1.2, 1.8, and 3.4

10. �0.6 and 4.6

Lesson 7.5

1. �3, 1 � 2i, and 1 � 2i

2. �2, 3, �1 � i, and �1 � i

3.

4. 1, �1, 2, and �3

5.

6. �0.53, 0.65, and 2.88

7.

8.


Lesson 8.1

1. direct variation


3. direct variation

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.


y � 0.504

y � 60

y � 5

y � 245

y �12

y � 0.004

y � 0.9375

y � 14

y � 0.25

y � 25

P(x) � x4 � 16

P(x) � �5x2 � 10x � 15

x � 1.32

2, 1 � �3, and 1 � �3

3, �3, �52

, and ��52

�6, 4, and 32

�3 and 12

occurring twice

x � 7

x2 � 5x � 5

(2x � 3)(x � 1)(x � 1)

2x(2x � 7)(2x � 7)

x3(x � 1)(x2 � x � 1)

2x3 � 11x2 � 2x � 15

x4 � 6x3 � 10x2 � 2x � 5

3x3 � 5x2 � 17x � 15

ANSWERSC

opyr

ight

©by

Hol

t, R

ineh

art a

nd W

inst

on. A

ll rig

hts

rese

rved

.


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yrig

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olt,

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ehar

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. All

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serv

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Finding the domain of a rational function

Recall Division by zero is not allowed; it is undefined.

◆ Example

Find the domain of .

◆ SolutionYou must exclude from the domain any values of x which cause the denominatorto have a value of 0.Set equal to 0.

orThe domain is all real numbers except �5 and 2.

x � 2x � �5(x � 5)(x � 2) � 0

x2 � 3x � 10

g(x) �x2 � 8x

x2 � 3x � 10

Find the domain of each rational function.

1. 2. 3. x2 � 5x2 � 4x � 21

5xx2 � 7x

x � 3x2 � 16

Reteaching

8.2 Rational Functions and Their Graphs

◆Skill B Identifying vertical asymptotes and holes in the graph of a rational function

Recall If is a factor in both the numerator and denominator, there will be a hole inthe graph at . If is a factor of the denominator but not a factor of thenumerator, there will be a vertical asymptote of .

◆ Example

For the rational function

a. identify the x-coordinates of any holes in the graph.b. write the equations of any vertical asymptotes.

◆ Solution

a.

Since is a factor of both the numerator and denominator, there will be a hole at .

b. Since is a factor of the denominator but not the numerator, and has a value of 0 when 3, there will be a vertical asymptote at

3.x �

x �

x � 3x � �2

x � 2

2x2 � 3x � 2x2 � x � 6

�(2x � 1)(x � 2)(x � 2)(x � 3)

f(x) �2x2 � 3x � 2x2 � x � 6

x � b(x � b)x � b

(x � b)

–10 10

10

–10

Identify all holes and asymptotes in the graph of each rationalfunction.

4. 5.

6. 7. x2 � 6x � 7x � 1

f(x) �x2 � 4x � 3x2 � x � 6

f(x) �

x � 5(x � 1)(x � 4)

f(x) �(x � 3)(x � 2)(x � 3)(x � 2)

f(x) �

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill C Writing an equation for the horizontal asymptote of a graph

Recall The degree of a polynomial is the greatest degree of its terms.

◆ ExampleWrite the equation of the horizontal asymptote for each of the followingfunctions.

a. b. c.

◆ Solutiona. Since the degree of the numerator and denominator are the same, divide the

coefficient of the term with the greatest degree in the numerator by thecoefficient of the like term in the denominator.

(coefficients of the terms)

Thus, 2 is a horizontal asymptote. The graph is shown on the preceding page.

b. Since the degree of the numerator is less than the degree of the denominator,the horizontal asymptote is 0. (The graph is shown at left below.)

c. Since the degree of the numerator is greater than the degree of thedenominator, there is no horizontal asymptote. (The graph is shown at right below.)

y �

y �

x221

� 2

h(x) ��x2

x � 2g(x) �

3xx2 � 5

f(x) �2x2 � 3x � 2

x2 � x � 6

Identify any horizontal asymptotes for the following functions.Use a graphics calculator to check your answer.

8. 9. 10. x � 5(x � 1)(x � 4)

f(x) �x2 � 5x � 6

x � 2f(x) �

5x2 � 82x2 � 3x

f(x) �

–10 10

10

–10

–10 10

10

–10

Lesson 8.2

1. all real numbers,



4. hole at x � �2; vertical asymptote: x � �3

5. no holes; vertical asymptotes: x � �4 and x � 1

6. hole at x � 3; vertical asymptote: x � �2

7. hole at x � 1; no vertical asymptotes

8.

9. no horizontal asymptote

10.

Lesson 8.3

1.

2.

3.

4. 4a4

5.

6.

7.

8. 1

9.

10.

11.

12.

Lesson 8.4

1. 2 2. x � 4

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

Lesson 8.5

1.

2.

3.

4.

5.

6.

7.

8.

9.

10. �2.4 � x � 0 or x � 0.4

�3 � x � 0 or x � 1

x � �4 or x � �2

�2.5 � x � 0

x � 5

x � �5

x � �12

, 6

x � 5

x � �72

x � �9

�(x � 1)x(x � 1)

2(x � 2)(x � 4)

5(a � 4)a(a � 2)(a � 2)

11x � 21(x � 6)(2x � 3)

�3y2(y � 2)

4142x

x(x � 3)(x � 7)(x � 7)

(x � 2)(x � 1)(x � 5)

5(x � 1)(x � 1)(x � 7)

5a(a � 2)(a � 2)

2y(y � 2)

3x � 3

3x � 7x � 5

x � 52

12xx � 3

4(a � 3)a(a � 5)

x(x � 1)3(x � 1)

45x2

y � 34

x(x � 5)2(x � 4)

715x2

y � 5y � 4

3x � 3

y � 0

y �52

x 7, �3

x 0, �7

x 4, �4

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olt, Rinehart and W



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◆Skill A Simplifying rational expressions

Recall To simplify rational expressions, start by factoring the numerator and denominator.

◆ Example 1

Simplify.

◆ Solution

Factor numerator and denominator.

Divide numerator and denominator by x � 5.

◆ Example 2

Simplify .

◆ Solution

◆ Example 3

Simplify .

◆ Solution

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

�x � 3x � 5

x2 � 5x � 6x2 � x � 20

�x2 � 3x � 4x2 � x � 2

x2 � 5x � 6x2 � x � 20

�x2 � 3x � 4x2 � x � 2

�y

21

�1y4

21y3

�5 � 2 � 3 � y4

3 � 3 � 2 � 5 � 7 � y2 � y5

9y2 �2y4

10�

37y

59y2 �

2y4

10�

37y

�x � 5x � 2

�(x � 5)(x � 5)(x � 5)(x � 2)

x2 � 25x2 � 3x � 10

x2 � 25x2 � 3x � 10

Simplify each rational expression.

1. 2.

3. 4.

5. 6.y2 � 6y � 9

2y2 � 18�

3y2 � 276y � 18

x2 � 25x2 � 16

�x2 � 4x2x � 10

a3

�5�

a5

3� �

4a

�15a3

4x2

5�

712x4

y2 � 2y � 15y2 � y � 12

3x � 6x2 � 5x � 6

Reteaching

8.3 Multiplying and Dividing Rational Expressions

Simplify each expression.

7. 8.

9. 10.

11. 12.

x2 � 2x � 152x � 8

x2 � x � 12x2 � 16

3x2 � 9xx � 2x2 � 94x � 8

a2 � 5a � 6a3

a2 � 3a � 104a2

x2

x2 � 2x � 1

3xx2 � 1

3y � 122y � 4

6y � 244y � 8

3x3

5

6x5

8

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Dividing rational expressions

Recall To divide, multiply by the reciprocal of the divisor.

◆ Example 1

Simplify.

◆ Solution

◆ Example 2

Simplify the complex fraction.

◆ Solution

�1

x � 1

�(x � 1)(x � 1)(x � 2)

(x � 1)(x � 2)(x � 1)(x � 1)

�x2 � 1

x2 � 3x � 2�

x � 2x2 � 2x � 1

x2 � 1x2 � 3x � 2x2 � 2x � 1

x � 2

x2 � 1x2 � 3x � 2x2 � 2x � 1

x � 2

�(x � 3)(x � 1)

x

�(x � 3)(x � 3)(x � 1)(x � 1)

x(x � 1)(x � 3)

�x2 � 9x2 � x

�x2 � 1x � 3

x � 3x2 � 1

x2 � 9x2 � x

x � 3x2 � 1

x2 � 9x2 � x

Lesson 8.2








8.


10.

Lesson 8.3

1.

2.

3.

4. 4a4

5.

6.

7.

8. 1

9.

10.

11.

12.

Lesson 8.4

1. 2 2. x � 4

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

Lesson 8.5

1.

2.

3.

4.

5.

6.

7.

8.

9.

10. �2.4 � x � 0 or x � 0.4

�3 � x � 0 or x � 1

x � �4 or x � �2

�2.5 � x � 0

x � 5

x � �5

x � �12

, 6

x � 5

x � �72

x � �9

�(x � 1)x(x � 1)

2(x � 2)(x � 4)

5(a � 4)a(a � 2)(a � 2)

11x � 21(x � 6)(2x � 3)

�3y2(y � 2)

4142x

x(x � 3)(x � 7)(x � 7)

(x � 2)(x � 1)(x � 5)

5(x � 1)(x � 1)(x � 7)

5a(a � 2)(a � 2)

2y(y � 2)

3x � 3

3x � 7x � 5

x � 52

12xx � 3

4(a � 3)a(a � 5)

x(x � 1)3(x � 1)

45x2

y � 34

x(x � 5)2(x � 4)

715x2

y � 5y � 4

3x � 3

y � 0

y �52

x 7, �3

x 0, �7

x 4, �4

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◆Skill A Adding and subtracting rational expressions with like denominators

Recall You never add or subtract the denominators of fractions.

◆ Example

Simplify a. b. c.

◆ Solution

a. Add numerators.

b. Subtract numerators.

Factor.

Divide out common factors.

c.

�5x � 6x � 2

�5x

x � 2�

�6x � 2

5xx � 2

�6

2 � x�

5xx � 2

�(�1)6

(�1)(2 � x)

�1

x � 1

�x � 1

(x � 1)(x � 1)

�x � 1x2 � 1

x � 3x2 � 1

�4

x2 � 1

�3x � 4x � 5

xx � 5

�2x � 4x � 5

5xx � 2

�6

2 � xx � 3x2 � 1

�4

x2 � 1x

x � 5�

2x � 4x � 5

Simplify.

1. 2. 3. 4. 2x � 11x2 � 9

�2 � x9 � x2

3xx � 5

�7

5 � xx2

x � 4�

8x � 16x � 4

2xx � 3

�6

x � 3

Reteaching

8.4 Adding and Subtracting Rational Expressions

◆Skill B Finding the least common multiple of two polynomial expressions

Recall The least common multiple, LCM, of 6 and 8 is 24 because 24 is the smallest number divisible by both 6 and 8.

◆ ExampleFind the least common multiple of the following polynomial expressions.

◆ SolutionFactor each expression.

The LCM is .(x � 1)2(x � 1)(x � 3)� 4x � 3 � (x � 3)(x � 1)x2

� 1 � (x � 1)(x � 1)x2� 2x � 1 � (x � 1)(x � 1)x2

x2 � 4x � 3x2 � 1x2 � 2x � 1

Simplify.

10. 11.

12. 13.

14. 15. 1x

�2

x � 2�

2x2 � 3x � 2

5x2 � 3x � 4

�3

x2 � x � 2

10a2 � 2a

�5

a2 � 43

x � 6�

52x � 3

3y6y � 12

�2y

y � 227x

�1114x

�2

21x

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill C Adding and subtracting rational expressions with unlike denominators

Recall To add or subtract two fractions, they must have a common denominator.

◆ Example

Simplify.

◆ SolutionFactor the denominators: and .Then find the LCM of and . The LCM is .Multiply each fraction by 1 so that they will both have the same denominator.

where and

��3x � 29

4(x � 3)(x � 3)

�4x � 8 � 7x � 21

4(x � 3)(x � 3)

x � 3x � 3

� 144

� 1�(x � 2) � 4 � 7(x � 3)

4(x � 3)(x � 3)

x � 2x2 � 9

�7

4x � 12�

x � 2(x � 3)(x � 3)

�44

�7

4(x � 3)�

x � 3x � 3

4(x � 3)(x � 3)4(x � 3)(x � 3)(x � 3)4(x � 3)(x � 3)(x � 3)

x � 2x2 � 9

�7

4x � 12

Find the least common multiple for each set of polynomials.

5. and 6. and

7. and 8. and

9. , and x2 � 4x � 21x2 � 3x, x2 � 49

x2 � 6x � 5x2 � 3x � 105x2 � 5x2 � 8x � 7

5a2 � 20a2 � 2ay2 � 2y2y � 4

Lesson 8.2








8.


10.

Lesson 8.3

1.

2.

3.

4. 4a4

5.

6.

7.

8. 1

9.

10.

11.

12.

Lesson 8.4

1. 2 2. x � 4

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

Lesson 8.5

1.

2.

3.

4.

5.

6.

7.

8.

9.

10. �2.4 � x � 0 or x � 0.4

�3 � x � 0 or x � 1

x � �4 or x � �2

�2.5 � x � 0

x � 5

x � �5

x � �12

, 6

x � 5

x � �72

x � �9

�(x � 1)x(x � 1)

2(x � 2)(x � 4)

5(a � 4)a(a � 2)(a � 2)

11x � 21(x � 6)(2x � 3)

�3y2(y � 2)

4142x

x(x � 3)(x � 7)(x � 7)

(x � 2)(x � 1)(x � 5)

5(x � 1)(x � 1)(x � 7)

5a(a � 2)(a � 2)

2y(y � 2)

3x � 3

3x � 7x � 5

x � 52

12xx � 3

4(a � 3)a(a � 5)

x(x � 1)3(x � 1)

45x2

y � 34

x(x � 5)2(x � 4)

715x2

y � 5y � 4

3x � 3

y � 0

y �52

x 7, �3

x 0, �7

x 4, �4

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◆Skill A Solving rational equations

Recall You should find the domain before solving the equation.

◆ Example

Solve.

◆ Solution1. Factor so that you can determine the domain.

The domain is all real numbers except 3 and �2.2. The common denominator for all of the fractions is .3. Multiply every term by the common denominator.

4. Simplify this equation so that no fractions remain.

orBut is not in the domain, so the solution is .

Check by making a table of values for

and .

Identify where .

The corresponding x-value is a solution.

y1 � y2

y2 �2

x � 2y1 �

xx � 3

�7x � 6

x2 � x � 6

x � 4x � 3x � 4x � 3

(x � 3)(x � 4) � 0x2 � 7x � 12 � 0

x2 � 2x � 7x � 6 � 2x � 6x(x � 2) � (7x � 6) � 2(x � 3)

xx � 3

(x � 3)(x � 2) �7x � 6

(x � 3)(x � 2)(x � 3)(x � 2) �

2x � 2

(x � 3)(x � 2)

(x � 3)(x � 2)

x2 � x � 6 � (x � 3)(x � 2)

xx � 3

�7x � 6

x2 � x � 6�

2x � 2

Solve each equation. Check your solution.

1.

2.

3.

4.

5.

6. 4xx2 � 9

�x � 1

x2 � 6x � 9�

2x � 3

7x � 2

�3

x � 4�

2x2 � 6x � 8

12x � 2

�3x

� 2

52x � 2

�15

x2 � 1

2x � 4

�3

x � 4� 10

14x

�34

�7x

Reteaching

8.5 Solving Rational Equations and Inequalities

Use a graphics calculator to solve each inequality.

7. 8.

9. 10. 2x �1x

� x � 2xx � 1

�x

x � 3

3x2x � 5

� 33x � 5x

� 5

Copyright ©

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ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Solving rational inequalities (You will need a graphics calculator.)

Recall If , then .

◆ Example 1

Use a graphics calculator to solve .

◆ Solution

Graph and .

The graph of is above the graph of when .

The solution is .

◆ Example 2

Solve .

◆ Solution

If and ,

it is not easy to see where the graph of is above the graph of .

Instead, graph and determine where . (above the x-axis)

From the graph, the solution is or .0 � x � 3x � �4

y � 0y �x � 1x � 4

�x

x � 3

y2y1

y2 �x

x � 3y1 �

x � 1x � 4

x � 1x � 4

�x

x � 3

x � 6

x � 6y1y2

y2 � 4y1 �2x

x � 3

2xx � 3

� 4

a � b � 0a � b

–10 10

10

–10

–10 10

10

–10

Lesson 8.2








8.


10.

Lesson 8.3

1.

2.

3.

4. 4a4

5.

6.

7.

8. 1

9.

10.

11.

12.

Lesson 8.4

1. 2 2. x � 4

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

Lesson 8.5

1.

2.

3.

4.

5.

6.

7.

8.

9.

10. �2.4 � x � 0 or x � 0.4

�3 � x � 0 or x � 1

x � �4 or x � �2

�2.5 � x � 0

x � 5

x � �5

x � �12

, 6

x � 5

x � �72

x � �9

�(x � 1)x(x � 1)

2(x � 2)(x � 4)

5(a � 4)a(a � 2)(a � 2)

11x � 21(x � 6)(2x � 3)

�3y2(y � 2)

4142x

x(x � 3)(x � 7)(x � 7)

(x � 2)(x � 1)(x � 5)

5(x � 1)(x � 1)(x � 7)

5a(a � 2)(a � 2)

2y(y � 2)

3x � 3

3x � 7x � 5

x � 52

12xx � 3

4(a � 3)a(a � 5)

x(x � 1)3(x � 1)

45x2

y � 34

x(x � 5)2(x � 4)

715x2

y � 5y � 4

3x � 3

y � 0

y �52

x 7, �3

x 0, �7

x 4, �4

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olt, Rinehart and W



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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Evaluating cube-root expressions

Recall and

◆ ExampleEvaluate each expression.

a. b. c.

◆ Solution

a. b. c.

� �15� 3� 5(�3)� 23

�13

� 3213

3�272� 5 3��275( 3��1084 )� 42 � 7( 3�64)2 � 7

13

3�2725( 3��1084 )( 3�64)2 � 7

x23 �

3�x2 � ( 3�x)2x13 �

3�x

Evaluate each expression without using a calculator.

1. 2. 3.

4. 5. 6. (274)13�1

2 3��216��11

3( 3�813 )

(�8)13 � 2 3�1252 ( 3�1252)

Reteaching

8.6 Radical Expressions and Radical Functions

◆Skill B Using transformations to graph square root functions

Recall The domain of a square root function is the set of all x such that the radicand(expression under the radical sign) is greater than or equal to 0.

◆ ExampleFor the function a. state the domain andb. use transformations to sketch a graph.

◆ Solutiona. Since the radicand must be

nonnegative, . The domain is .

b. Write in the form , where

is the vertical stretch/compression factor,is the horizontal stretch/compression factor,

h gives the horizontal translation, and k gives the vertical translation.In this function, , and .

Start with and stretch vertically by a factor of 3; reflect across the x-axis(since ); there is not a horizontal stretch; translate 1 unit to the left and 5units up.

a � 0�x

�3�1(x � (�1)) � 5g(x) �

k � 5a � �3, b � 1, h � �1

�b��a�

a�b(x � h) � kg(x) �

�3�x � 1 � 5g(x) �x � �1

x � 1 � 0

�3�x � 1 � 5g(x) �

–10 10

10

–10

g(x)

f(x)

Find the inverse of each quadratic function. Then graph thefunction and its inverse in the same coordinate plane.

9. 10.

x

y

Ox

y

O

f(x) � x2 � 3xf(x) � x2 � 4

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill C Finding and graphing the inverse of a quadratic function

Recall To find the inverse of a function, interchange x and y.

◆ ExampleFind the inverse of . Then graph both the function and its inverse.

◆ Solution

Let , and . Use the quadratic formula to solve for y.

Graph , and .y3 �3 � �4x � 25

2y1 � x2 � 3x � 4, y2 �

3 � �4x � 252

�3 �25 � 4x

2

y ��(�3) �(�3)2 � 4(1)(�4 � x)

2(1)

c � �4 � xa � 1, b � �3y2 � 3y � 4 � x � 0x � y2 � 3y � 4

f(x) � x2 � 3x � 4

Find the domain and sketch a graph for each radical function.Check with a graphics calculator.

7. 8.

x

y

Ox

y

O

�2�x � 5 � 2f(x) �2�3(x � 1) � 3f(x) �

–10 10

10

–10

Lesson 8.6

1. 25 2. 25 3. 0

4. 1 5. 6. 81

7.

8.

9.

10.

Lesson 8.7

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11. 6a

12.

13. �2

14.

15.

16.

17.

18.

19.

Lesson 8.8

1.

2.

3.

4.

5.

6. no solution

x � 7

x � �5 or x � 5

x � �13

x � 2

x � 11

12 � 8�2

�217

3�22

6y � 4

21 � 21�2

2x � 7�x � 4

3a5b3�2

3xy�x

5x2y4�3

6x3�x

6x2

2r2s2 5�r 2

3x2 3�2

5�a��b�c2�7b

5m6�6

2�x3�y2�2y

4�x�y2�3x

–16 16

10

–10

y ��3 �4x � 9

2

–16 16

10

–10

y � �x � 4

–10 10

10

–10

x � �5

–10 10

10

–10

x � 1

�13

ANSWERSC

opyr

ight

©by

Hol

t, R

ineh

art a

nd W

inst

on. A

ll rig

hts

rese

rved

.


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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Simplifying radicals

Recall If n is an even integer, then ; for example, .

If n is an odd integer, then ; for example, .

◆ Example

Simplify .

◆ Solution

Make one monomial a perfect square.

Product Property of Radicals

c2 is always nonnegative.Thus, absolute value is not needed.

� 5�a�c2�2ab

� �25a2c4 � �2ab

� �25a2c4 � 2ab�50a3bc4

�50a3bc4

3�(�2)3 � �2� a n�an

4�(�2)4 � ��2� � 2� �a� n�an

◆Skill B Simplifying products and quotients of radical expressions

Recall Product Property of Radicals:

Quotient Property of Radicals: where

◆ ExampleSimplify each expression. Assume that the value of each variable is positive.

a. b.

◆ Solution

a. Product Property of Radicals

No absolute value is necessary, since values of variables are assumed to be positive.

b.

� 3�8x3y5 � 3�8x3y3 � y2 � 3�8x3y3 � 3�y2 � 2xy 3�y2

� 3�24x3y7

3y2(24x3y7)

3�3y2

13

� 6xyz�z

� �36x2y2z2 � �z

� �36x2y2z2 � z

� �2xz3 � �18xy2 � 2xz3(18xy2)12

(24x3y7)

3�3y2

13

(18xy2)12 � �2xz3

b 0 n�a

n�b

n�ab

�

n�ab � n�a � n�b

Simplify each radical expression by using the Properties of nthRoots.

1. 2. 3.

4. 5. 6. 5�32r12x10 3�54x6�175a2b3c4

�150m12�8x6y5�48x3y4

Reteaching

8.7 Simplifying Radical Expressions

Simplify each product or quotient. Assume that the value ofeach variable is positive.

7. 8. 9.

10. 11. 12.3�108a15b10

(2b)13

(216a9) 3�a6

13�54x4y6

�6xy4

(15x4y2)12 � �5y6�3x4 � �12x3�6x2 � �6x2

Find each sum, difference, or product. Give your answer insimplest radical form. Assume that the value of each variable is positive.

13. 14.

15. 16.

Write each expression with a rational denominator and insimplest form.

17. 18. 19.4

3 � 2�2

3�7

7�3

3

�2

(3�y � 1)(2�y � 4) � 10�y3�7(�7 � �14)

(2�x � 1)(�x � 4)(�11 � �13)(�11 � �13)

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill C Performing operations on radical expressions

Recall By using “FOIL,” .

◆ Example 1Simplify by performing the indicated operations: .

◆ Solution

◆ Example 2

Rationalize the denominator of each expression. a. b.

◆ Solution

a. Multiply by . b.

�6�3 � 12

�1� 4�2

6�3 � 12

(�3)2 � (2)2��8�2

2

6

�3 � 2�

6

�3 � 2�

�3 � 2

�3 � 2�2�2

8

�2�

8

�2�

�2

�2

6

�3 � 2

8

�2

� 6 � 11�2

� 10 � 4 � �2(�2 � 10 � 3)

� 10 � 2�2 � 10�2 � 4 � 3�2

� 10 � 2�2 � 5�8 � �16 � 3�2(2 � �8)(5 � �2) � 3�2

(2 � �8)(5 � �2) � 3�2

(a � �b)(c � �d) � ac � a�d � c�b � �bd

Lesson 8.6

1. 25 2. 25 3. 0

4. 1 5. 6. 81

7.

8.

9.

10.

Lesson 8.7

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11. 6a

12.

13. �2

14.

15.

16.

17.

18.

19.

Lesson 8.8

1.

2.

3.

4.

5.

6. no solution

x � 7

x � �5 or x � 5

x � �13

x � 2

x � 11

12 � 8�2

�217

3�22

6y � 4

21 � 21�2

2x � 7�x � 4

3a5b3�2

3xy�x

5x2y4�3

6x3�x

6x2

2r2s2 5�r 2

3x2 3�2

5�a��b�c2�7b

5m6�6

2�x3�y2�2y

4�x�y2�3x

–16 16

10

–10

y ��3 �4x � 9

2

–16 16

10

–10

y � �x � 4

–10 10

10

–10

x � �5

–10 10

10

–10

x � 1

�13

ANSWERSC

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ineh

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Solving radical equations and inequalities without graphs

Recall When you raise both sides of an equation or inequality to a power, you must checkfor possible extraneous (false) solutions.

◆ Example 1Solve . Check your solution(s).

◆ Solution

Isolate the radical.

Square each side.

Collect like terms and set equal to 0.Factor.

oror

Check. � �

The only solution is .

◆ Example 2Solve . Check with a table.

◆ Solution

Multiply by �1 and reverse the inequality sign.

Check. Let and use a table.When .x � �7, y � 3

y � �2 � xx � �7

�x � 72 � x � 9

�2 � x � 3

�2 � x � 3

x � 36 � 62 0

2 � 3�8 � 3 � 1 � 12 � 0�8 � 0 � 1 � 1

x � 3x � 0x � 3 � 04x � 0

4x(x � 3) � 04x2 � 12x � 0

8x � 1 � 4x2 � 4x � 1(�8x � 1)2 � (2x � 1)2

�8x � 1 � 2x � 1

�8x � 1 � 1 � 2x

�8x � 1 � 1 � 2x

Solve each radical equation by using algebra. Check yoursolution(s).

1. 2. 3.

Solve each radical inequality by using algebra. Check yoursolution(s).

4. 5. 6. �3x � 5 � 3 3�4x � 1 � 3�x2 � 9 � 4

�7 � 6x � 2 � 3xx � �x � 1 � 3�2x � 3 � 5

Reteaching

8.8 Solving Radical Equations and Inequalities

Solve each radical equation by using a graph. Round solutions tothe nearest tenth. Check your solution by any method.

7.

8.

9.

Solve each radical inequality with a graphics calculator. Roundyour solution to the nearest tenth, if necessary.

10.

11. �x � �2x � 1

�3x � 2 � 1 � �x � 5

�3x � 28 � x

�x � 4 � 2x � 7

�x � 3 � 2 � 4

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Solving radical equations and inequalities using graphs

Recall If there is more than one radical, you must start by raising both sides to the same power.

◆ Example 1Solve . Check your solution with a graph.

◆ Solution

Use the quadratic formula.

There are no real solutions.

The graphs show that and will never intersect.

◆ Example 2Solve by graphing.

◆ SolutionGraph and .To the nearest tenth, the x-coordinate of the point of intersection is 1.9.When , the graph of

is above the graph of

. So the solution is approximately .x � 1.9y2 � 3�x � 4

2x � 5y1 �x � 1.9

y2 � 3�x � 4y1 � 2x � 5

2x � 5 � 3�x � 4

y2 � �x � 2y1 � �2x � 1

x ��(�2) �(�2)2 � 4(1)(9)

2(1)� 1 2i�2

x2 � 2x � 9 � 04 � 2x � x2 � 6x � 9

(2�2x)2 � (�x � 3)2

2�2x � �x � 3

2x � 2�2x � 1 � x � 2

(�2x � 1)2 � (�x � 2)2

�2x � 1 � �x � 2

�2x � 1 � �x � 2

–10 10

10

–10

–10 10

10

–10

Lesson 8.6

1. 25 2. 25 3. 0

4. 1 5. 6. 81

7.

8.

9.

10.

Lesson 8.7

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11. 6a

12.

13. �2

14.

15.

16.

17.

18.

19.

Lesson 8.8

1.

2.

3.

4.

5.

6. no solution

x � 7

x � �5 or x � 5

x � �13

x � 2

x � 11

12 � 8�2

�217

3�22

6y � 4

21 � 21�2

2x � 7�x � 4

3a5b3�2

3xy�x

5x2y4�3

6x3�x

6x2

2r2s2 5�r 2

3x2 3�2

5�a��b�c2�7b

5m6�6

2�x3�y2�2y

4�x�y2�3x

–16 16

10

–10

y ��3 �4x � 9

2

–16 16

10

–10

y � �x � 4

–10 10

10

–10

x � �5

–10 10

10

–10

x � 1

�13

ANSWERSC

opyr

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©by

Hol

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ineh

art a

nd W

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7.

8.

9.

10.

11.


Lesson 9.1

1. ;

parabola

2. ;

circle

3. ;

ellipse

4.

5. 9m � (�2, �0.5)

6.m � (�1, 2)

7. a. (�2, 2)b. m � 10πc. 25π

8. a. (2.5, 0)b. 13πc. 42.25π

9. a. (4, 4)b.c. 32π

10. a. (1.5, �4)b. 15πc. 56.25π

Lesson 9.2

1.

V(–2, 0)

F(–2, –1)

directrix y = 1

x

y

–8 4

3

–9

8�2π � 1131π

�80 � 8.94

m � (6.5, 1.5)�2 � 1.41

–10 10

10

–10

y � �19

(36 � 4x2)

–10 10

10

–10

y � �25 � x2

–10 10

10

–10

y �18

(x2 � 12)

0 � x � 5.8

x � 5.2

x � 7

x � 5

x � 7

ANSWERSC

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◆Skill A Identifying a conic section from its graph

Recall When you use a graphics calculator to graph conic sections, be sure to use a squareviewing window.

◆ ExampleSolve for y, graph the resulting equation, and identify the conic section.

◆ Solution

Use a graphics calculator to graph

and .The graph of is a hyperbola.y2 � 4x2 � 16

y2 � ��4x2 � 16y1 � �4x2 � 16

y � �4x2 � 16

y2 � 4x2 � 16y2 � 4x2 � 16

y2 � 4x2 � 16

Solve each equation for y, use a calculator to help you sketch thegraph of the resulting equation, and identify the conic section.

1. 2. 3.

x

y

Ox

y

Ox

y

O

4x2 � 9y2 � 36x2 � y2 � 25x2 � 8y � 12

Reteaching

9.1 Introduction to Conic Sections

–15.16 15.16

10

–10

◆Skill B Finding the length and midpoint of a segment

Recall The distance formula comes directly from the Pythagorean Theorem.

◆ ExampleGiven the points P(�3, �1) and Q(5, 2), finda. the distance PQ. b. the midpoint of .

◆ Solutiona.

b. Find the average of the x-coordinates and the average of the y-coordinates.

(5 � (�3)2

, 2 � (�1)2 ) � (1, 0.5)

d � �73 � 8.54d2 � 82 � 32d2 � (5 � (�3))2 � (2 � (�1))2

PQx

d

Q

P

y

O

2

82

3

–2

–2

–4 4


Find the center, the circumference, and the area of each circledescribed below.

7. The endpoints of a diameter are (2, �1) and 8. The endpoints of a diameter are (0, �6) (�6, 5). and (5, 6).

a. a.

b. b.

c. c.

9. The endpoints of a diameter are (0, 0) and 10. The endpoints of a diameter are (�3, �10)(8, 8). and (6, 2).

a. a.

b. b.

c. c.

Find the distance between P and Q and the coordinates of themidpoint of . Give exact answers and approximate answers tothe nearest hundredth when appropriate.

4. P(7, 1) and Q(6, 2) 5. P(�2, �5) and Q(�2, 4) 6. P(�3, 6) and Q(1, �2)

PQ

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill C Using the distance and midpoint formulas to find the center, circumference, andarea of a circle

Recall For any circle, and .

◆ ExampleThe endpoints of a diameter of a circle are A(�5, 4) and B(3, �2).Find the a. center; b. circumference; c. area.

◆ Solution

a. Midpoint Formula

The center is at (�1, 1).

For parts b and c use the distance formula to find the length of the radius.

b. c.

A � 25πC � 10πA � π(5)2C � 2π(5)A � πr2C � 2πr

� �25 � 5

MA � �(�1 � (�5))2 � (1 � 4)2

M(�5 � 32

, 4 � 22 )

A � πr2C � 2πr

7.

8.

9.

10.

11.


Lesson 9.1

1. ;

parabola

2. ;

circle

3. ;

ellipse

4.

5. 9m � (�2, �0.5)

6.m � (�1, 2)

7. a. (�2, 2)b. m � 10πc. 25π

8. a. (2.5, 0)b. 13πc. 42.25π

9. a. (4, 4)b.c. 32π

10. a. (1.5, �4)b. 15πc. 56.25π

Lesson 9.2

1.

V(–2, 0)

F(–2, –1)

directrix y = 1

x

y

–8 4

3

–9

8�2π � 1131π

�80 � 8.94

m � (6.5, 1.5)�2 � 1.41

–10 10

10

–10

y � �19

(36 � 4x2)

–10 10

10

–10

y � �25 � x2

–10 10

10

–10

y �18

(x2 � 12)

0 � x � 5.8

x � 5.2

x � 7

x � 5

x � 7

ANSWERSC

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olt, Rinehart and W



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◆Skill A Graphing a parabola and labeling its parts

Recall If the vertex of a parabola is at (h, k) and the distance between the vertex and thefocus is p, then the following are true.

◆ ExampleGraph . Label the vertex, focus, and directrix.

◆ Solution

Complete the square for the y-terms.

Divide by 12.

In this form you can see that the vertex is at

(�3, 2) and, since , the value of p is 3.

The graph opens to the right.Graph the vertex at (�3, 2) and mark the point 3 units to the right, (0, 2), as the focus.The directrix is also at a distance p from the vertex, so its equation is .

Make a table of x- and y-coordinates to find several other points.

x � �6

14p

�1

12

x � 3 �1

12(y � 2)2

12x � 36 � (y � 2)212x � 32 � y2 � 4y

y2 � 12x � 4y � 32 � 0

y2 � 12x � 4y � 32 � 0

Graph each equation. Label the vertex, focus, and directrix. Check with a graphics calculator.

1. 2. 3.

x

y

Ox

y

Ox

y

O

x2 � 8x � y � 18 � 0x � �18

(y � 1)2 � 2y �14

(x � 2)2

Reteaching

9.2 Parabolas

p � 0 opens up opens to the right

p � 0 opens down opens to the left

x � h �14p

(y � k)2y � k �14p

(x � h)2

x

y

O

2

–2

4

directrixx = –6

FV

2 4–4

Write the standard equation for each parabola below.

4. 5.

6. 7.

2

–2

4

6

8

2–2–6–8

y

x

directrix

V

x

4

–2

–4

8

2 4–4 –2

y

directrix

V

2

–2

–4

4

6

4–2–4 6

y

V F

xx

y

2

–2

–6

–4

4

6

4–4–6 6V

F

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Writing an equation for a graphed parabola

Recall The focus and the directrix are each at a distance of p from the vertex.

◆ ExampleWrite the standard equation for the parabola graphed at right.

◆ SolutionThis parabola opens down, so its equation can be written in the form

, where .

Since the distance between the vertex and focus is 2, .Using the vertex (h, k) (3, 0), you can write

or .

Check by graphing on your calculator.

y � �18

(x � 3)2y � 0 �1

4(�2)(x � 3)2

�p � �2

p � 0y � k �1

4p(x � h)2

xO

V

F

2

–2

–4

–6

–8

2–2 6 8

y

7.

8.

9.

10.

11.


Lesson 9.1

1. ;

parabola

2. ;

circle

3. ;

ellipse

4.

5. 9m � (�2, �0.5)

6.m � (�1, 2)

7. a. (�2, 2)b. m � 10πc. 25π

8. a. (2.5, 0)b. 13πc. 42.25π

9. a. (4, 4)b.c. 32π

10. a. (1.5, �4)b. 15πc. 56.25π

Lesson 9.2

1.

V(–2, 0)

F(–2, –1)

directrix y = 1

x

y

–8 4

3

–9

8�2π � 1131π

�80 � 8.94

m � (6.5, 1.5)�2 � 1.41

–10 10

10

–10

y � �19

(36 � 4x2)

–10 10

10

–10

y � �25 � x2

–10 10

10

–10

y �18

(x2 � 12)

0 � x � 5.8

x � 5.2

x � 7

x � 5

x � 7

ANSWERSC

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olt, Rinehart and W



2.

3.

4.

5.

6.

7.

Lesson 9.3

1.

2.

3.

4. ; ;

5.

6.

7. ; ;

8.

9.

10.

Lesson 9.4

1.

2.

3.

4. C:(0, 2); F:(0, ), (0, );V:(0, 8), (0, �4); Co-V: (4, 2), (4, �2)

5. C:(0, 0); F:(4, 0), (�4, 0); V:(5, 0), (�5, 0); Co-V: (0, 3), (0, �3)

2 � 2�52 � 2�5

(x � 2)2

4�

(y � 3)2

1� 1

(x � 2)2

1�

(y � 2)2

9� 1

x2

16�

(y � 2)2

4� 1

x

y

6

–6

6–6

x

y

6

–6

6–6

x

y

6

–6

6–6

r � 2�3C(5, �2)(x � 5)2 � (y � 2)2 � 12

x2 � (y � 7)2 � 36; C(0, �7); r � 6

(x � 3)2 � (y � 1)2 � 4; C(3, �1); r � 2

r � 6C(�2, 3)(x � 2)2 � (y � 3)2 � 36

x2 � y2 � 16

(x � 1)2 � (y � 2)2 � 25

(x � 2)2 � y2 � 4

x � 2 � �14

(y � 3)2

y � 2 � �1

16x2

x � 1 �14

(y � 1)2

y � 1 �18

x2

x–2 10

V(4, 2)F(4, 2.25)

directrixy = 1.75

y

10

–2

–4 8

5

–7

x

y

V(2, –1)F(0, –1)

directrixx = 4

ANSWERSC

opyr

ight

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t, R

ineh

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olt,

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. All

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serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Writing an equation for a graphed circle

Recall The standard equation for a circle with center at C(h, k) and radius r is.

◆ ExampleWrite the standard equation for the circle shown.

◆ SolutionThe center is at (�2, �1), so and

.You can see that the radius is 3, so .The equation is .To check with a graphics calculator,

and

y2 � �1 � �9 � (x � 2)2

y1 � �1 � �9 � (x � 2)2

y � 1 � �9 � (x � 2)2

(y � 1)2 � 9 � (x � 2)2

(x � 2)2 � (y � 1)2 � 9r � 3

k � �1h � �2

(x � h)2 � (y � k)2 � r2

Write the standard equation for each circle graphed below.

1. 2. 3.

x

y

OC

2

–22–2

2

–2

–4

2–2 4

y

x

Cx

y

C2

4

–4

–22–2–4

Reteaching

9.3 Circles

x

y

O

6

–6

–2

2

4

–6 –4 642C

◆Skill B Completing the square to write the standard equation of a circle

Recall You must complete the squares for both x and y.

◆ ExampleWrite an equation in standard form for the circle .Identify the center and length of the radius.

◆ Solution

The center is at (3, �5) and the radius is 6.(x � 3)2 � (y � 5)2 � 36

x2 � 6x � 9 � y2 � 10y � 25 � 2 � 9 � 25x2 � 6x � _ � y2 � 10y � _ � 2

x2 � y2 � 6x � 10y � 2 � 0

Graph each equation.

8. 9. 10.

x

y

Ox

y

Ox

y

O

x2 � y2 � 2x � 2y � 2 � 0x2 � y2 � 25(x � 3)2 � y2 � 9

Write the standard equation for each circle. Then state thecoordinates of its center and give its radius.

4. 5.

C( , ); C( , );

6. 7.

C( , ); C( , ); r �r �

x2 � y2 � 10x � 4y � 17 � 0x2 � y2 � 14y � 13 � 0

r �r �

x2 � y2 � 6x � 2y � 6 � 0x2 � y2 � 4x � 6y � 23 � 0

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill C Graphing a circle from its equation

Recall To graph a circle, you need the coordinates of its center and the length of its radius.

◆ ExampleGraph .

◆ SolutionLocate the center at (0, 2) and draw a circle with radius of length 4.

x2 � (y � 2)2 � 16

x

y

2

4

2 4–4 –2 6

C

2.

3.

4.

5.

6.

7.

Lesson 9.3

1.

2.

3.

4. ; ;

5.

6.

7. ; ;

8.

9.

10.

Lesson 9.4

1.

2.

3.

4. C:(0, 2); F:(0, ), (0, );V:(0, 8), (0, �4); Co-V: (4, 2), (4, �2)

5. C:(0, 0); F:(4, 0), (�4, 0); V:(5, 0), (�5, 0); Co-V: (0, 3), (0, �3)

2 � 2�52 � 2�5

(x � 2)2

4�

(y � 3)2

1� 1

(x � 2)2

1�

(y � 2)2

9� 1

x2

16�

(y � 2)2

4� 1

x

y

6

–6

6–6

x

y

6

–6

6–6

x

y

6

–6

6–6

r � 2�3C(5, �2)(x � 5)2 � (y � 2)2 � 12

x2 � (y � 7)2 � 36; C(0, �7); r � 6

(x � 3)2 � (y � 1)2 � 4; C(3, �1); r � 2

r � 6C(�2, 3)(x � 2)2 � (y � 3)2 � 36

x2 � y2 � 16

(x � 1)2 � (y � 2)2 � 25

(x � 2)2 � y2 � 4

x � 2 � �14

(y � 3)2

y � 2 � �1

16x2

x � 1 �14

(y � 1)2

y � 1 �18

x2

x–2 10

V(4, 2)F(4, 2.25)

directrixy = 1.75

y

10

–2

–4 8

5

–7

x

y

V(2, –1)F(0, –1)

directrixx = 4

ANSWERSC

opyr

ight

©by

Hol

t, R

ineh

art a

nd W

inst

on. A

ll rig

hts

rese

rved

.


Cop

yrig

ht ©

by H

olt,

Rin

ehar

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. All

right

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serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Writing an equation for a graphed ellipse

Recall Ellipse with a horizontal major axis:

Ellipse with a vertical major axis:In both cases, the center is at (h, k).

◆ ExampleWrite the standard equation for the ellipse shown at right.

◆ SolutionThe center is at (�1, 2), so and .

The major axis is 6 units long, so .

The minor axis is 4 units long, so .

Since the major axis is vertical:

The equation is .(x � 1)2

4�

(y � 2)2

9� 1

(x � (�1))2

22 �(y � 2)2

32 � 1

b �42

� 2

a �62

� 3

k � 2h � �1

(x � h)2

b2 �(y � k)2

a2 � 1

(x � h)2

a2 �(y � k)2

b2 � 1

Write the standard equation for each ellipse below.

1. 2. 3.

x

y

C

2

–4

4

2 4–4 –2x

y

O

C2

4

–4

–24–2–4

x

y

O

C2

–2

–4

2 4–4 –2

Reteaching

9.4 Ellipses

} where a2 � b2

xO

–2

–4

4

6

8

2–4 4

y

C

◆Skill B Identifying the center, foci, vertices, and co-vertices of an ellipse

Recall The foci are on the major axis at a distance c from the center, where .

◆ ExampleFind the coordinates of the center, foci, vertices, and co-vertices for the ellipse

.x2 � 9y2 � 4x � 18y � 4 � 0

c � �a2 � b2

Find the coordinates of the center, foci, vertices, and co-verticesof each ellipse.

4. 5. 9x2 � 25y2 � 225x2

16�

(y � 2)2

36� 1

Sketch the graph of each ellipse.

6. 7.

x

y

O

25x2 � (y � 1)2 � 25

x

y

O

x2

25�

y2

9� 1

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆ Solution

Complete the squares.

Standard Form

(Recall that a2 � b2.)Therefore, the center is at (2, �1).Since is in the x-term, the major axis is horizontal.The foci are at a horizontal distance c from the center:

and .

The vertices are at a horizontal distance a from the center: and (5, �1).The co-vertices are at a vertical distance b from the center: (2, 0) and (2, �2).

(�1, �1)

(2 � �8, �1) � (�0.8, �1)(2 � �8, �1) � (4.8, �1)

a2

h � 2, k � �1, a � 3, b � 1, c � �32 � 12 � �8

(x � 2)2

9�

(y � 1)2

1� 1

(x � 2)2 � 9(y � 1)2 � 9x2 � 4x � 4 � 9(y2 � 2y � 1) � �4 � 4 � 9 � 1x2 � 4x � _ � 9(y2 � 2y � _) � �4

◆Skill C Graphing an ellipse from its equation

Recall To sketch a graph of any conic, write its equation in standard form.

◆ Example

Sketch a graph of .

◆ SolutionThe center is at (�2, 0). Since 9 is in the x-term, mark points at a distance horizontally from the center; (�5, 0) and (1, 0) are the vertices.Since 25 is in the y-term, mark points at a distance vertically from the center; (�2, 5) and (�2, �5) are the co-vertices.

�25 � 5

�9 � 3

(x � 2)2

9�

y2

25� 1

x

y

2

6

–2

2–2–4–6C

2.

3.

4.

5.

6.

7.

Lesson 9.3

1.

2.

3.

4. ; ;

5.

6.

7. ; ;

8.

9.

10.

Lesson 9.4

1.

2.

3.

4. C:(0, 2); F:(0, ), (0, );V:(0, 8), (0, �4); Co-V: (4, 2), (4, �2)

5. C:(0, 0); F:(4, 0), (�4, 0); V:(5, 0), (�5, 0); Co-V: (0, 3), (0, �3)

2 � 2�52 � 2�5

(x � 2)2

4�

(y � 3)2

1� 1

(x � 2)2

1�

(y � 2)2

9� 1

x2

16�

(y � 2)2

4� 1

x

y

6

–6

6–6

x

y

6

–6

6–6

x

y

6

–6

6–6

r � 2�3C(5, �2)(x � 5)2 � (y � 2)2 � 12

x2 � (y � 7)2 � 36; C(0, �7); r � 6

(x � 3)2 � (y � 1)2 � 4; C(3, �1); r � 2

r � 6C(�2, 3)(x � 2)2 � (y � 3)2 � 36

x2 � y2 � 16

(x � 1)2 � (y � 2)2 � 25

(x � 2)2 � y2 � 4

x � 2 � �14

(y � 3)2

y � 2 � �1

16x2

x � 1 �14

(y � 1)2

y � 1 �18

x2

x–2 10

V(4, 2)F(4, 2.25)

directrixy = 1.75

y

10

–2

–4 8

5

–7

x

y

V(2, –1)F(0, –1)

directrixx = 4

ANSWERSC

opyr

ight

©by

Hol

t, R

ineh

art a

nd W

inst

on. A

ll rig

hts

rese

rved

.


6.

7.

Lesson 9.5

1. ;

2.

3.

4.

5.

6.

Lesson 9.6

1. (�3, �4), (4, 3)

2. (�2, 12), (3, �3)

3. (3, 4), (3, �4), (�3, 4), (�3, �4)

4. (5, 4), (�5, 4), (5, �4), (�5, �4)

5. (2, 0), (�2, 0)

6. (3, �1), (2, �2)

7. (1.9, 1.5), (�1.9, 1.5), (1.9, �1.5), (�1.9, �1.5)

8. (3, 0), (�3, 0)

9. no solution

10. ellipse 11. hyperbola 12. parabola

x

y

11

–1

Ox

5–7

y

8

–4

x

y

O

6

–6

6–6

y �43

x � 1, y � �43

x � 1

(y � 1)2

16�

x2

9� 1

y �12

x �12

, y � �12

x �12

(x � 1)2

16�

y2

4� 1

y � x, y � �x

y2

4�

x2

4� 1

x

y

6

–6

6–6

x

y

6

–6

6–6

ANSWERSC

opyright ©by H

olt, Rinehart and W



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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Writing an equation for a graphed hyperbola

Recall Hyperbola with a horizontal transverse axis:

Hyperbola with a vertical transverse axis:

◆ ExampleFor the graph at right, write an equation fora. the hyperbola. b. each asymptote.

◆ Solutiona. The center is at (h, k) (0, 2).

Since the transverse axis is horizontal,

.

This implies that a is the horizontal distance from the center to the vertices, so 2.This leaves 3 as the vertical distance from the center to the edge of the “rectangle.” The equation of the hyperbola is

.

b. The asymptotes both have a y-intercept of

2 and slopes of and .

The equations of the asymptotes are and .y � �32

x � 2y �32

x � 2

�ba

� �32

ba

�32

x2

4�

(y � 2)2

9� 1

b �a �

(x � 0)2

a2 �(y � 2)2

b2 � 1

�

(y � k)2

a2 �(x � h)2

b2 � 1

(x � h)2

a2 �(y � k)2

b2 � 1

Write the standard equation for each hyperbola and give theequations for the asymptotes.

1. 2. 3.

x642–6 –4 –2

y

4

2

–6

–4

y

4

–4

–6

x6–4

x

y

6

4

–6

–4

64–6 –4

Reteaching

9.5 Hyperbolas

x64–6 –4

y

8

6

–4

–2

Graph each hyperbola.

4. 5. 6.

x

y

Ox

y

Ox

y

O

x2 � y2 � 6x � 10y � 17 � 0(x � 1)2

16�

(y � 2)2

9� 1

y2

25�

x2

4� 1

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Graphing a hyperbola from its equation

Recall Drawing a rectangle and asymptotes can help you draw a hyperbola, but they are really not part of the graph of the hyperbola.

◆ Example

Sketch a graph of .

◆ SolutionUse one of the standard forms to identify h, k, a, and b.Since the y-term is positive, the form is

where , and .

Mark the center (�2, 1) and sketch a rectangle by moving 4 units above and below the center and 3 units to the left and to the right of the center.

Draw the diagonals of the rectangle and extend them to form the asymptotes.

Since the y-term is positive, sketch the two branches above and below the center.

b � 3h � �2, k � 1, a � 4

(y � k)2

a2 �(x � h)2

b2 � 1

(y � 1)2

16�

(x � 2)2

9� 1

y

2

4

–2

x42–2–4–6–8

6.

7.

Lesson 9.5

1. ;

2.

3.

4.

5.

6.

Lesson 9.6

1. (�3, �4), (4, 3)

2. (�2, 12), (3, �3)

3. (3, 4), (3, �4), (�3, 4), (�3, �4)

4. (5, 4), (�5, 4), (5, �4), (�5, �4)

5. (2, 0), (�2, 0)

6. (3, �1), (2, �2)

7. (1.9, 1.5), (�1.9, 1.5), (1.9, �1.5), (�1.9, �1.5)

8. (3, 0), (�3, 0)

9. no solution


x

y

11

–1

Ox

5–7

y

8

–4

x

y

O

6

–6

6–6

y �43

x � 1, y � �43

x � 1

(y � 1)2

16�

x2

9� 1

y �12

x �12

, y � �12

x �12

(x � 1)2

16�

y2

4� 1

y � x, y � �x

y2

4�

x2

4� 1

x

y

6

–6

6–6

x

y

6

–6

6–6

ANSWERSC

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olt, Rinehart and W



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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Solving a nonlinear system by substitution or elimination

Recall You can often use either the substitution or elimination method to obtain an equation in one variable.

◆ Example

Solve a. by substitution. b. by elimination.

◆ Solutiona. Solve the linear equation for y.

Substitute into the other equation.

or

Substitute these values for x into and solve for y.

Possible solutions are (0, 4), (0, �4), and (4, 0).Check each in the linear equation.0 � 4 � 4 True 0 � (�4) � 4 False

4 � 0 � 4 TrueThe only solutions are (0, 4) and (4, 0).

y � 0y2 � 0y � 4

42 � y2 � 1602 � y2 � 16

x2 � y2 � 16

x � 0x � 42x � 0

2x(x � 4) � 02x2 � 8x � 0

x2 � x2 � 8x � 16 � 16x2 � (�x � 4)2 � 16

x2 � y2 � 16

y � �x � 4

�2x2 � y2 � 65x2 � 2y2 � �3�x2 � y2 � 16

x � y � 4

Use any method to solve each system. If there is no solution,write no solution.

1. 2. 3.

4. 5. 6. �y � x � 4y � x2 � 4x � 2�9x2 � 4y2 � 36

x2 � y2 � 4�x2 � y2 � 9x2 � 9y2 � 169

�2x2 � y2 � 2x2 � y2 � 25�3x � y � 6

x2 � 4x � y�x2 � y2 � 25y � x � 1

Reteaching

9.6 Solving Nonlinear Systems

b. Multiply both sides of the firstequation by 2; then add the twoequations.

or

Substitute each of these values into the original first equation.

The four possible solutions are(1, 2), (1, �2), (�1, 2), and (�1, �2).

Check all four possible solutions inthe second given equation.

All four of these ordered pairs aresolutions to the original system.

y � 2y � 2y2 � 4y2 � 4

2(�1)2 � y2 � 62(1)2 � y2 � 6

x � �1x � 1x2 � 1

9x2 � 9

5x2 � 2y2 � �34x2 � 2y2 � 12

Solve each system with a graphics calculator. If there is nosolution, write no solution.

7. 8. 9. �x2 � 3y2 � 9y � �x2 � 4�x2 � y2 � 9

x2 � y2 � 9�4x2 � 9y2 � 3625x2 � 4y2 � 100

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Solving a nonlinear system by graphing

Recall A coordinate graph will show only the real-number solutions for a system.

◆ Example

Solve by graphing.

◆ SolutionSolve the first equation for y.

Use a graphics calculator to graph , and .Use the trace or intersection feature of your calculator to find solutions to the nearest tenth.The solutions are (2.9, 2.7), (�2.9, 2.7), (1.5, �3.7), and (�1.5, �3.7).

y3 � x2 � 6y1 � �16 � x2, y2 � ��16 � x2

y � �16 � x2

y2 � 16 � x2

�x2 � y2 � 16y � x2 � 6

–10 10

10

–10

Classify the conic section defined by each equation.

10. 11. 12. y2 � 6y � x � 4 � 0x2 � y2 � 1x2 � 4y2 � 6x � 7 � 0

◆Skill C Classifying a conic section from its equation

Recall The equation represents a conic section which isparallel to either the x-axis or the y-axis.

◆ Example

If: then the conic section is:

◆ ExampleName the conic section that is the graph of .

◆ SolutionSince and ; therefore, this is an equation for a hyperbola.C � 1, AC � 0A � �9

y2 � 9x2 � 8y � 36x � 29 � 0

Ax2 � Cy2 � Dx � Ey � F � 0

(both A and C have the same sign) an ellipse (or circle)

and a circle

(in other words, either or ) a parabola

(A and C have opposite signs) a hyperbolaAC � 0

C � 0A � 0AC � 0

A 0A � C

AC � 0

6.

7.

Lesson 9.5

1. ;

2.

3.

4.

5.

6.

Lesson 9.6

1. (�3, �4), (4, 3)

2. (�2, 12), (3, �3)

3. (3, 4), (3, �4), (�3, 4), (�3, �4)

4. (5, 4), (�5, 4), (5, �4), (�5, �4)

5. (2, 0), (�2, 0)

6. (3, �1), (2, �2)

7. (1.9, 1.5), (�1.9, 1.5), (1.9, �1.5), (�1.9, �1.5)

8. (3, 0), (�3, 0)

9. no solution


x

y

11

–1

Ox

5–7

y

8

–4

x

y

O

6

–6

6–6

y �43

x � 1, y � �43

x � 1

(y � 1)2

16�

x2

9� 1

y �12

x �12

, y � �12

x �12

(x � 1)2

16�

y2

4� 1

y � x, y � �x

y2

4�

x2

4� 1

x

y

6

–6

6–6

x

y

6

–6

6–6

ANSWERSC

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olt, Rinehart and W



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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Applying the Fundamental Counting Principle

Recall If there are m ways to make one selection and n ways to make another selection,there are ways to make both selections.

◆ Example 1In your daily commute to school you find that there are 5 different routes to theexpressway and 3 different routes from the expressway to school. How manydifferent trips could you take to get to school?

◆ Solution5 routes to expressway3 routes from expressway to school

There are different routes to school.

◆ Example 2Tim’s security password on a computer at school must consist of 4 letters. How many different passwords can he choose if he doesn’t use any letter more than once?

◆ Solutionchoices for 1st letter: 26choices for 2nd letter: 25choices for 3rd letter: 24choices for 4th letter: 23

There are or 358,800 different possible codes.26 � 25 � 24 � 23

5 � 3 � 15

m � n

Use the Fundamental Counting Principle to answer the following.

1. A pizza outlet offers 6 kinds of meat toppings and 10 different vegetabletoppings. How many different pizzas with 1 meat and 1 vegetable could you order?

2. In a word game, you may replace 2 of the letters on your tray on yourturn. The letters you have are E, X, J, U, F, A, and Z. In how many ways could you select 1 consonant and 1 vowel to discard?

3. A local department store will put a 3-letter monogram on your selection of towels. How many different monograms are possible?

4. You must decide how to appoint 9 people to the 9 different positions on a baseball team. In how many different ways could you do this?

5. How many different signals can be shown by arranging 3 flags in a row if 7 different flags are available?

Reteaching

10.1 Introduction to Probability

Find the probability of each event.

6. If you flip a quarter, what is the probability it will land a. heads up? b. tails up?

7. If you roll a number cube, what is the probability of rolling

a. 6? b. a number less than 6? c. 7? d. a number less than 7?

8. A 3-digit number is chosen at random. Assuming the first digit is not 0,what is the probability that the number is

a. a multiple of 5? b. a multiple of 5 and less than 400?

9. A multiple-choice test has 50 questions. Each question has 5 choices. For each question, you make a choice at random.

a. What is the probability that you will get the first answer correct?

b. Theoretically, how many answers might you expect to get correct?

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Finding the theoretical or experimental probability of an event

Recall The probability of an event is always between 0 and 1 inclusive; .

◆ Example 1A drawer contains 6 brown, 8 blue, and 4 tan single socks. If you randomly pullout 2 socks, a. what is the probability that the first sock is blue?b. if the first sock is blue, what is the probability that the second one is blue?

◆ Solutiona. Since there are a total of 18 socks, the theoretical probability of getting a blue

one is , or approximately 44%.

b. Since there are only 17 socks left and 7 of these are blue,

, or approximately 41%.

◆ Example 2A testing center has randomly assigned you a 4-digit number. Assuming the firstdigit is not 0, what is the probability that this number is even and is greater thanor equal to 3000?

◆ SolutionStart by using the Fundamental Counting Principle. Consider each digitseparately.

1st digit could be any digit from 3 to 9 7 choices2nd digit could be any of 10 digits 10 choices3rd digit could be any of 10 digits 10 choices4th digit must be 0, 2, 4, 6, or 8 5 choices

, or approximately 39%number of even numbers � 3000number of 4- digit numbers

�7 � 10 � 10 � 5

9 � 10 � 10 � 10�

718

number of blue sockstotal number of socks

�717

number of blue sockstotal number of socks

�8

18�

49

0 � P(event) � 1


Lesson 10.1

1. 60 2. 12 3. 17,576

4. 362,880 5. 210

6. a. b.

7. a. b. c. 0 d. 1

8. a. b.

9. a. b. 10

Lesson 10.2

1. 40,320 2. 6,497,400

3. 60 4. 362,880

5. If this were a permutation of 26 letterstaken 3 at a time, you could not use anyletter more than once on a license plate.

6. 95,040 7. 34,650 8. 5040

9. 39,916,800 10. 24 11. 12 12. 3

Lesson 10.3

1. a. 210b. 35c. 6

2. a. 1140b. 6840

3. 15 4. 270,725 5. 60

6. a. � 5.5%b. � 0.26%c. � 0.18%d. � 0.0000037%

7. a. � 6.9%b. � 2.2%

8. � 21%

9. a. � 29%b. � 71%

Lesson 10.4

1. a.

b.

c.

d. inclusive events

2. a.

b.

c. inclusive eventd. 0

e.

3.

4. a. (1d)

b. (2c)

5. the number of boys that have brown hair

6.

7. a.

b.

Lesson 10.5

1. a.

b.

2. 56.25%

3. a.

b.

c.

d.

4. a. 0.144b. 0.616

136

� 2.8%

236

� 5.6%

136

� 2.8%

136

� 2.8%

48196

� 24%

48196

� 24%

55150

� 37%

95150

� 63%

1760

� 28%

20732,598,960

12072,598,960

56

67,0672,598,960

7922,598,960

65,7802,598,960

29,900270,725

2270,725

1430270,725

15

115

15

56

16

12

12

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Cop

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by H

olt,

Rin

ehar

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Win

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serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Solving problems involving linear permutations

Recall A permutation is an arrangement of objects in a specific order.

◆ Example 1In how many ways can you arrange 7 books on a shelf?

◆ SolutionAny one of 7 books can occupy the 1st position. For each of these, any one of the6 remaining books can be in the 2nd position. Therefore, for all 7 books there are

arrangements.

There are 5040 different ways to arrange the books.

◆ Example 2A CD-player is set to choose the tracks of a 15-track CD at random. In how manyways could 5 tracks be played?

◆ SolutionThere are 15 choices for the 1st track and 14 choices for the 2nd track. For all 5tracks there are choices.

Notice that , where

There are 360,360 ways to play 5 of the 15 tracks.

◆ Example 3Suppose a “word” can be composed of any set of letters in a particular order. Howmany different 5-letter “words” can you make from the letters RADAR using allof the letter in each “word?”

◆ Solution

But since the 2 R’s are indistinguishable from each other, and likewise for the 2A’s, you must “divide out” these arrangements.

You can make 30 different five-letter “words.”5!2!2!

�1202 � 2

� 30

5P5 � 5! � 120

� 360,360� 15 � 14 � 13 � 12 � 11

15!10!

�15 � 14 � 13 � 12 � 11 � 10 � 9 � 8 � 7 � 6 � 5 � 4 � 3 � 2 � 1

10 � 9 � 8 � 7 � 6 � 5 � 4 � 3 � 2 � 1

15P5 �15!

(15 � 5)!�

15!10!

15 � 14 � 13 � 12 � 11

7P7 � 7! � 5040

7 � 6 � 5 � 4 � 3 � 2 � 1

Find the number of permutations for each problem.

1. In how many ways could you arrange 8 different shirts on hangers in the closet?

2. From a deck of 52 cards, how many different ways could you draw 4 cards,

if the order in which you draw them will make a difference?

Reteaching

10.2 Permutations

3. Find the number of permutations of all the letters in the word PEPPER.

4. How many different batting lineups are possible for 9 players on a baseball team?

5. Many license plates start with 3 letters of the alphabet. Consider all thepossible 3 letters that could be used. Why is this not equal to 26P3? (Hint: Consider the 3 letters AAA.)

6. The Hawaiian alphabet has only 12 letters. How many permutations could be made using 5 different letters?

7. Find the number of permutations of all the letters in the word MISSISSIPPI.

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Solving problems involving circular permutations

Recall A circular arrangement, unlike a linear arrangement, has no beginning or endingpoint.

◆ ExampleHow many different arrangements are there for seating 6 people around acircular table?

◆ SolutionSince a circular table has no “end,” one person could be anywhere around thetable. Relative to this person, the rest of the people could be arranged in 5!different ways.

There are 120 circular arrangements of the 6 people.

(n � 1)! � (6 � 1)! � 5! � 120

Find the number of permutations for each problem.

8. In how many orders can 8 children be standing on the playground merry-go-round?

9. A stage dance starts with 12 people in a circle. In how many different ways could these 12 dancers be arranged?

10. Five CDs are placed on a rotary tray in a CD-changer. In how many different ways could these CDs be arranged?

11. If 2 of the CDs in Exercise 10 were identical, in how many ways could all of the CD’s be arranged on the tray?

12. Two rings in a child’s playpen each have 4 large plastic beads strung onthem. One appears to have the beads in a red-blue-green-yellow orderand the other in a yellow-green-blue-red order. But you realize that ifyou flip the second ring over, it has exactly the same arrangement ofthese 4 colors as the other ring. Taking this fact into account, in how many different orders could these 4 beads be placed on the ring?


Lesson 10.1

1. 60 2. 12 3. 17,576

4. 362,880 5. 210

6. a. b.

7. a. b. c. 0 d. 1

8. a. b.

9. a. b. 10

Lesson 10.2

1. 40,320 2. 6,497,400

3. 60 4. 362,880


6. 95,040 7. 34,650 8. 5040

9. 39,916,800 10. 24 11. 12 12. 3

Lesson 10.3

1. a. 210b. 35c. 6

2. a. 1140b. 6840

3. 15 4. 270,725 5. 60

6. a. � 5.5%b. � 0.26%c. � 0.18%d. � 0.0000037%

7. a. � 6.9%b. � 2.2%

8. � 21%

9. a. � 29%b. � 71%

Lesson 10.4

1. a.

b.

c.

d. inclusive events

2. a.

b.


e.

3.

4. a. (1d)

b. (2c)


6.

7. a.

b.

Lesson 10.5

1. a.

b.

2. 56.25%

3. a.

b.

c.

d.

4. a. 0.144b. 0.616

136

� 2.8%

236

� 5.6%

136

� 2.8%

136

� 2.8%

48196

� 24%

48196

� 24%

55150

� 37%

95150

� 63%

1760

� 28%

20732,598,960

12072,598,960

56

67,0672,598,960

7922,598,960

65,7802,598,960

29,900270,725

2270,725

1430270,725

15

115

15

56

16

12

12

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Cop

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by H

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serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Solving problems involving combinations and permutations

Recall For 3 distinct objects there are 3! permutations (arrangements) but only 1combination of all 3.

◆ ExampleYou have decided to read 5 novels out of a list of 12 for this semester’s English class.a. How many different sets of 5 novels could you choose?b. In how many different orders could you read the 5 novels you choose?

◆ Solutiona. There are 12P5 � 95,040 different arrangements of 5 books out of 12. But since

order is not being considered here, any set of 5 books could be chosen in 5! � 120 different orders. Therefore, to find the combinations of 5 books youmust “divide out” the 120 orders.

There are 792 combinations of 5 novels to read.

b. Once you have chosen 5 novels, there is only one combination of these 5, butthere are 5! � 120 different orders in which you could read them.

12P5

5!�

95,040120

� 792

Solve each combination problem.

1. You have 7 books, but only 3 will fit on the shelf.a. How many different arrangements of 3 books out of 7 could you make?

b. How many different ways could you select 3 of the 7 books, without regard to order?

c. How many ways could you arrange one selection of 3 books on the shelf?

2. A science class has 20 students.a. In how many ways could a group of 3 students be selected from the

class?

b. In how many ways could the whole class elect a reader, a recorder, and a grader for a class project?

3. Six people wish to play bridge. But only 4 of them can play at one time. Ifthey decide to play a game for each possible group of 4 players, how many games will they need to play?

4. You are dealt 4 cards for a card game. How many different 4-card hands are possible when starting a game with a 52-card deck?

5. The choices for a sandwich are 4 different meats and 5 different cheeses. How many different sandwiches of 2 meats and 2 cheeses could you make?

Reteaching

10.3 Combinations

Solve each combination problem.

6. From a deck of 52 cards, 4 cards are drawn. What is the probability that

a. all 4 cards are black? (A deck has 26 black cards.)

b. all 4 cards are spades? (A deck has 13 spades.)

c. all 4 cards are face cards? (A deck has 12 face cards.)

d. all 4 cards are aces? (A deck has 4 aces.)

7. Five apples, 7 oranges, and 4 peaches are mixed in a fruit bin. If 4 pieces offruit are picked out at random, what is the probability of picking

a. 2 oranges and 2 peaches? b. 4 oranges or 4 apples?

8. Out of 40 sketches submitted, 8 were picked at random. If you submitted5 sketches, what is the probability that exactly 2 of your sketches were picked?

9. From X, P, O, Z, E, A, D, F, M, and B, 3 letters are picked at random. What is the probability that

a. 3 consonants are picked? b. at least 1 vowel is picked?

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Using combinations and probability

Recall .

◆ ExampleFrom a small club of 8 girls and 6 boys, four students are to be chosen asrepresentatives to a city-wide convention. If the committee of 4 students ischosen at random with a hat drawing, what is the probability that it will havea. exactly 2 girls and 2 boys? b. at least 3 girls?

◆ Solutiona. There are 14C4 � 1001 combinations of 14 students taken 4 at a time.

There are 8C2 � 28 combinations of 2 girls and 6C2 � 15 combinations of 2 boys.

The probability of picking exactly 2 girls and 2 boys is approximately 42%

b. “At least 3 girls” means there could be 3 girls and 1 boy or 4 girls and no boys.Combinations of 3 girls and 1 boy � 8C3 � 6C1. (Note: “and” impliesmultiplication.)

Combinations of 4 girls and 0 boys � 8C4 � 6C0.

Number of combinations of 3 girls and 1 boy or 4 girls and 0 boys(8C3)(6C1) � (8C3)(6C0) (Note: “or” implies addition.)

Therefore, .�56 � 6 � 70 � 1

1001� 0.41, or 41%

(8C3)(6C1) � (8C3)(6C0)14C4

4201001

� 0.428C2 � 6C2

14C4�

(ways to choose 2 girls) � (ways to choose 2 boys)ways to choose 4 students

�

Probability �number of desired outcomes

total number of possible outcomes


Lesson 10.1

1. 60 2. 12 3. 17,576

4. 362,880 5. 210

6. a. b.

7. a. b. c. 0 d. 1

8. a. b.

9. a. b. 10

Lesson 10.2

1. 40,320 2. 6,497,400

3. 60 4. 362,880


6. 95,040 7. 34,650 8. 5040

9. 39,916,800 10. 24 11. 12 12. 3

Lesson 10.3

1. a. 210b. 35c. 6

2. a. 1140b. 6840

3. 15 4. 270,725 5. 60

6. a. � 5.5%b. � 0.26%c. � 0.18%d. � 0.0000037%

7. a. � 6.9%b. � 2.2%

8. � 21%

9. a. � 29%b. � 71%

Lesson 10.4

1. a.

b.

c.

d. inclusive events

2. a.

b.


e.

3.

4. a. (1d)

b. (2c)


6.

7. a.

b.

Lesson 10.5

1. a.

b.

2. 56.25%

3. a.

b.

c.

d.

4. a. 0.144b. 0.616

136

� 2.8%

236

� 5.6%

136

� 2.8%

136

� 2.8%

48196

� 24%

48196

� 24%

55150

� 37%

95150

� 63%

1760

� 28%

20732,598,960

12072,598,960

56

67,0672,598,960

7922,598,960

65,7802,598,960

29,900270,725

2270,725

1430270,725

15

115

15

56

16

12

12

ANSWERSC

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Finding the probability of mutually exclusive events

Recall Two mutually exclusive events, A and B, cannot both occur at the same time. Insuch a situation the probability of A or B happening is the sum of their individualprobabilities.

◆ ExampleA number cube is tossed. What is the probability of it landing witha. a 4 or 6 on top? b. an even number or 5 on top?c. an even number or a number less than 5 on top?

◆ Solution

a. . Therefore, P(4 or 6) � .

b. . Therefore, P(even or 5) � .

c. These 2 events are not mutually exclusive, because 2 and 4 are even numbersand less than 5. These are inclusive events. Save this problem until youcomplete Skill B for this lesson.

36

�16

�23

P(5) �16

P(even) �36

16

�16

�13

P(6) �16

P(4) �16

Solve the problems that involve mutually exclusive events; giveeach answer as a fraction. Label the others as “inclusive events”and wait to solve them until after you have studied Skill B.

Note: A deck of 52 cards has the following cards.face cards

black

red

1. If there are 270,725 possible 4-card hands, what is the probability thatyou will be dealt a 4-card hand of exactly

a. 4 spades or 4 clubs? b. 4 aces or 4 kings?

c. 4 red cards or 4 black cards? d. 4 face cards or 4 hearts?

2. If there are 2,598,960 possible 5-card hands, what is the probability thatyou will be dealt a 5-card hand of exactly

a. 5 red cards?

b. 5 face cards?

c. 5 red cards or 5 face cards?

d. 5 aces or 5 kings?

e. 5 black cards or 5 diamonds?

A A

22

33

44

55

66

77

88

99

1010

JJ

QQ

KK�hearts

diamonds

A A

22

33

44

55

66

77

88

99

1010

JJ

QQ

KK�spades

clubs

Reteaching

10.4 Using Addition With Probability

Solve each problem.

3. If a number cube is tossed, what is the probability that either an evennumber or a number less than 5 will be on top?

4. Two parts in Exercises 1 and 2 for Skill A of this lesson involved inclusive events. Identify these two Exercises and solve them.

a.

b.

5. A math class of 25 students has 12 boys and 14 students with brown hair. To find the probability of randomly selecting one student that is aboy or has brown hair, what other piece of information would you need to know?

6. A guide shows that out of 60 cable channels available on TV at 8:00 P.M., 8 are premium channels, 6 of which are showing movies. It also showsthat a total of 15 channels are showing movies at this time. If youchannel-surf to a random station, what is the probability that it will be a movie or a premium channel?

7. A surveyor found that in a neighborhood of 150 houses, 55 houses had atleast one dog, 70 houses had a least one cat, and of these houses, 30 had atleast one dog and one cat. What is the probability that a house picked atrandom has

a. a dog or a cat? b. neither a dog nor a cat?

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Finding the probability of inclusive events

Recall To find the probability of A or B, when A and B are inclusive events, you mustsubtract the “overlapping” probability. P(A or B) � P(A) � P(B) � P(A and B)

◆ ExampleScott has found some old books in the attic. Out of the 28 books, he notices that8 are novels and 6 are about history. However, 3 out of these 14 books areactually historical novels. If he picks one book to read at random from the 28,what is the probability that it is a novel or history?

◆ Solution

P(novel)

P(history)

P(novel and history)

P(novel or history)

, or approximately 39%

There is a 39% chance that Scott’s choice is a novel or history (or both).

�1128

�828

�628

�328

�3

28

�6

28

�828


Lesson 10.1

1. 60 2. 12 3. 17,576

4. 362,880 5. 210

6. a. b.

7. a. b. c. 0 d. 1

8. a. b.

9. a. b. 10

Lesson 10.2

1. 40,320 2. 6,497,400

3. 60 4. 362,880


6. 95,040 7. 34,650 8. 5040

9. 39,916,800 10. 24 11. 12 12. 3

Lesson 10.3

1. a. 210b. 35c. 6

2. a. 1140b. 6840

3. 15 4. 270,725 5. 60

6. a. � 5.5%b. � 0.26%c. � 0.18%d. � 0.0000037%

7. a. � 6.9%b. � 2.2%

8. � 21%

9. a. � 29%b. � 71%

Lesson 10.4

1. a.

b.

c.

d. inclusive events

2. a.

b.


e.

3.

4. a. (1d)

b. (2c)


6.

7. a.

b.

Lesson 10.5

1. a.

b.

2. 56.25%

3. a.

b.

c.

d.

4. a. 0.144b. 0.616

136

� 2.8%

236

� 5.6%

136

� 2.8%

136

� 2.8%

48196

� 24%

48196

� 24%

55150

� 37%

95150

� 63%

1760

� 28%

20732,598,960

12072,598,960

56

67,0672,598,960

7922,598,960

65,7802,598,960

29,900270,725

2270,725

1430270,725

15

115

15

56

16

12

12

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s re

serv

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Finding the probability of independent events using individual probabilities

Recall If A and B are independent events, P(A and B) � P(A) � P(B)

◆ Example 1A bag contains 5 blue marbles and 8 red marbles. If you draw 2 marbles from thebag, describe each of the following as dependent or independent events.a. drawing a blue marble, replacing it in the bag, and then drawing a blue marbleb. drawing a blue marble, and without replacing it, drawing another blue marble

◆ Solutiona. Since you replace the first marble drawn, the first draw has no effect on the

second draw. These are independent events.b. The first draw changes the number of marbles in the bag and therefore, the

probability of drawing a blue marble on the second draw. These aredependent events.

◆ Example 2What is the probability of tossing 4 coins such that they all land heads up?

◆ SolutionThe way one coin lands has no effect on the other coins. The 4 landings are

independent events. Since P(heads) for each coin,

. The probability of getting 4 heads is .116

12

�12

�12

�12

�116

�12

Find each probability.

1. A bag contains 8 peppermint and 6 spearmint candies.a. What is the probability of picking a spearmint, replacing it, and then

picking a peppermint?

b. What is the probability of picking a peppermint, replacing it, and then picking a spearmint?

2. The probability that you will get an A in French is 0.75. The probabilityfor getting an A in history is 0.75. What is the probability that you will get an A in both courses?

3. You roll a red number cube and a blue number cube. What is theprobability that you will get

a. a red 5 and a blue 6? b. a red 6 and a blue 5?

c. a total of 11 for both cubes? d. a total of 12 for both cubes?

4. A and B are independent events, P(A) � 0.40, and P(B) � 0.36.

a. Find P(A and B). b. Find P(A or B).

Reteaching

10.5 Independent Events

5. A and B are mutually exclusive events, P(A) � 0.40, and P(B) � 0.36.

a. Find P(A and B). b. Find P(A or B).

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Finding the probability of independent events using the complement

Recall The probability of an event that is certain to happen is 1, or 100%. However, for an uncertain event A, it must either happen or not happen. This means that P(A) � P(not A) � 1, or P(not A) � 1 � P(A).

◆ ExampleIf 4 coins are tossed, what is the probability that 0, 1, 2, or 3 coins will land heads up?

◆ SolutionThis includes all possibilities except that of having 4 heads. Since the probability

of all 4 landing heads up is (see Skill A, example 2), the probability of 0, 1, 2,

or 3 heads is .1 �116

�1516

116


6. Smith and Jones work in the same office. The probability that Smith is onthe phone is 0.5, while the probability that Jones is on the phone is 0.75.Assuming they are not talking to each other, what is the probability at anygiven time that

a. both Smith and Jones are on the phone?

b. Smith or Jones is on the phone?

c. neither is on the phone?

d. Smith is on the phone but Jones is not?

e. Jones is on the phone but Smith is not?

f. only one of them is on the phone?

7. The next 3 batters have batting averages of 0.275, 0.215, and 0.305respectively. What is the probability that

a. all 3 will get a hit? b. none of them gets a hit?

8. The weather report says that the probability of rain is 30% on Mondayand 50% on Tuesday. If these are independent events, what is theprobability that it will rain

a. both days? b. neither day?

c. Monday, but not Tuesday? d. Tuesday, but not Monday?


Lesson 10.1

1. 60 2. 12 3. 17,576

4. 362,880 5. 210

6. a. b.

7. a. b. c. 0 d. 1

8. a. b.

9. a. b. 10

Lesson 10.2

1. 40,320 2. 6,497,400

3. 60 4. 362,880


6. 95,040 7. 34,650 8. 5040

9. 39,916,800 10. 24 11. 12 12. 3

Lesson 10.3

1. a. 210b. 35c. 6

2. a. 1140b. 6840

3. 15 4. 270,725 5. 60

6. a. � 5.5%b. � 0.26%c. � 0.18%d. � 0.0000037%

7. a. � 6.9%b. � 2.2%

8. � 21%

9. a. � 29%b. � 71%

Lesson 10.4

1. a.

b.

c.

d. inclusive events

2. a.

b.


e.

3.

4. a. (1d)

b. (2c)


6.

7. a.

b.

Lesson 10.5

1. a.

b.

2. 56.25%

3. a.

b.

c.

d.

4. a. 0.144b. 0.616

136

� 2.8%

236

� 5.6%

136

� 2.8%

136

� 2.8%

48196

� 24%

48196

� 24%

55150

� 37%

95150

� 63%

1760

� 28%

20732,598,960

12072,598,960

56

67,0672,598,960

7922,598,960

65,7802,598,960

29,900270,725

2270,725

1430270,725

15

115

15

56

16

12

12

ANSWERSC

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5. a. 0b. 0.76

6. a. 0.375b. 0.875c. 0.125d. 0.125e. 0.375f. 0.5

7. a. � 1.8%b. � 39.6%

8. a. 15%b. 35%c. 15%d. 35%

Lesson 10.6

1. 0.15 2. 0.7

3. a. � 0.38b. 0.625c. 0.79

4. 0.1375

5. a. 0.625b. � 0.417

6. a. � 0.528b. � 0.544

7. a. 0.36

b.

8. a.

b. 10C2 represents the number of ways todraw 2 marbles and 6C2 represents thenumber of ways to draw 2 blue marbles.

Lesson 10.7

1.

2. Answers may vary.

3.




Lesson 11.1

1. �3, �2, �1, 0, 1, 2

2. 5, 10, 15, 20, 25, 30

3. 2, 5, 8, 11, 14, 17

4. 2, 5, 10, 17, 26, 37

5.

6. 4, 9, 16, 25, 36, 49

7. 1, �3, �7, �11

8. 4, 14, 24, 34

9. �2, �4, �8, �16

10. 5, �5, 5, �5

�12

, �112

, �212

, �312

, �412

, �512

0001–2418

2419–5774

5775–8588

8589–9769

9770–10,000

001–198

199–259

260–279

280–311

312–429

430–693

694–880

881–1000

13

13

ANSWERSC

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill Finding conditional probability

Recall The probability of two dependent events is called conditional probability.means the probability of event B happening if event A has already occurred.

◆ Example 1Suppose that 65% of the students in the junior class are taking chemistry, 40%are taking German, and 20% are taking both of these courses. What is theprobability that a student selected at randoma. is taking chemistry, if it is known this student is taking German?b. is taking German, if it is known this student is taking chemistry?

◆ SolutionLet C be the event “taking chemistry” and G be the event “taking German.”

a.

If a student is in German, there is a 50% chance this student is also in chemistry.

b.

If a student is in chemistry, there is a 31% chance this student is also in German.

Recall If events A and B are independent, P(A and B) � P(A) � P(B).But, if events A and B are dependent, P(A and B) � P(A) � P( ), which is just

another form of .

◆ Example 2A drawer contains 6 brown, 8 blue, and 4 tan single socks. If you randomly pullout 2 socks, what is the probability that both socks are blue?

◆ SolutionLet F be the event “first sock is blue” and S be the event “second sock is blue.”The probability that both socks are blue is P(F and S) � P(F) � P( ), where

. If the first sock is blue, there are only 7 blue socks left in a drawer of

17 socks, so .

P(F and S) � P(F) � P( )

� 18.3%

�8

18�

717

S�F

P(S�F) �7

17

P(F) �8

18

S�F

P(B�A) �P (A and B)

P(A)

B�A

�0.200.65

� 31%

P(G�C) �P (C and G)

P(C)

�0.200.40

� 50%

P(C�G) �P (C and G)

P(G)

P(B�A) �P (A and B)

P (A)

P(B�A)

Reteaching

10.6 Conditional Probability


1. The probability that the stock market will go up on Wednesday is 0.25. Ifit goes up on Wednesday, the probability it will go up on Thursday is 0.6.Find the probability the stock market will go up on both Wednesday and Thursday.

2. Recall that P(A or B) � P(A) � P(B) � P(A and B). In Exercise 1, what is the probability that the stock market will go up on Wednesday or Thursday?

3. A survey of a city’s high school students showed that 65% liked the sportsprogram, 40% liked the general activities program, and 25% liked both.What is the probability that a student selected at random from this highschool

a. likes the general activities program if it is known the student likes the sports program?

b. likes the sports program if it is known the student likes the general activities program?

c. likes the sports program or the general activities program?

4. Suppose that P(A) � 0.25 and P( ) � 0.55. Find P(A and B).

5. Suppose that P(A and B) � 0.25, P(A) � 0.4, and P(B) � 0.6.

a. Find P( ). b. Find P(A|B).

6. Suppose a survey of 1500 accidents produces the data in the above table.The table shows the cause of the accident and the gender of the driver ofthe car. Answer these questions based on these data.

a. If an accident is due to poor judgement, what is the probability that the driver is male?

b. If the driver of a car involved in an accident is female, what is the probability that the cause is mechanical failure?

7. There are 6 blue and 4 red marbles in a bag. If you draw two marbles atrandom, what is the probability that they are both blue

a. if you replace the first marble before you draw again?

b. if you do not replace the first marble before you draw again?

8. a. As a review problem, find .

b. The answer to part a should be the same as the answer to part b ofExercise 7. Explain why.

6C2

10C2

B�A

B�A

Copyright ©

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ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


Mechanical failure Intoxication Poor judgement

Male 415 168 255

Female 360 74 228

5. a. 0b. 0.76

6. a. 0.375b. 0.875c. 0.125d. 0.125e. 0.375f. 0.5

7. a. � 1.8%b. � 39.6%

8. a. 15%b. 35%c. 15%d. 35%

Lesson 10.6

1. 0.15 2. 0.7

3. a. � 0.38b. 0.625c. 0.79

4. 0.1375

5. a. 0.625b. � 0.417

6. a. � 0.528b. � 0.544

7. a. 0.36

b.

8. a.


Lesson 10.7

1.


3.




Lesson 11.1

1. �3, �2, �1, 0, 1, 2

2. 5, 10, 15, 20, 25, 30

3. 2, 5, 8, 11, 14, 17

4. 2, 5, 10, 17, 26, 37

5.

6. 4, 9, 16, 25, 36, 49

7. 1, �3, �7, �11

8. 4, 14, 24, 34

9. �2, �4, �8, �16

10. 5, �5, 5, �5

�12

, �112

, �212

, �312

, �412

, �512

0001–2418

2419–5774

5775–8588

8589–9769

9770–10,000

001–198

199–259

260–279

280–311

312–429

430–693

694–880

881–1000

13

13

ANSWERSC

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◆Skill Using a simulation to estimate the experimental probability of an event

Recall Simulations are often used when direct experimentation is very expensive or evenimpossible.

◆ ExampleThe probabilities that a real-estate agent will sell0 to 5 properties in one week are given in thetable at right. a. Distribute 2-digit random integers from 01 to

100 among these values so that thecorresponding random integers can be used to simulate the number of houses thereal-estate agent sells per week.

b. Use this distribution to simulate the agent’sweekly sales for 26 consecutive weeks.

◆ Solutiona. Add a “random numbers” column

to the table.

A probability of 0.12 means thatthere are 12 chances out of 100that this event will happen. Soassign the numbers 1–12 for theevent of selling 0 houses.To follow this pattern, assign thenext 21 numbers for theprobability of 0.21. This will bethe numbers 13–33.Continue so that all numbers from 1 to 100 are assigned.

b. Use your calculator to generate 6 randomnumbers between 1 and 100.

[Your calculator might use a differentcommand from the one shown here,such as Int(rand*100)+1.]

The screen at right shows a practice runfor the first 6 random numbers. Matcheach of these numbers to the appropriate row of the “random numbers” column of the table.

For example, since 46 fits in the 34–59 row, this implies that 2 houses weresold that week.

Verify that the next 5 random numbers shown would imply weekly sales of 4, 4, 1, 3, and 2 houses.

To complete this problem you would need to generate 20 more randomnumbers and match them to the appropriate rows of the table.

Reteaching

10.7 Experimental Probability and Simulation

Number Probability

0 0.12

1 0.21

2 0.26

3 0.19

4 0.12

5 0.10

Total 1.00

Properties sold

Number Probability Randomnumbers

0 0.12 01–12

1 0.21 13–33

2 0.26 34–59

3 0.19 60–78

4 0.12 79–90

5 0.10 91–100

Use a random number generator to complete the followingexercises.

1. A baseball player’s statistics are shown in this table.Complete the last column by assigning randomnumbers from 001 to 1000.

2. Simulate the player’s activity for 10 times at batand record your results.

The probabilities that there are 0 to 4 significant leaksof toxins into a large bay on any one day aregiven in the table at right.

3. Distribute random integers from 0001 to 10,000among these 5 values so that the correspondingrandom integers can be used to simulate the numberof significant toxic spills into the bay each day.

4. Use the distribution you constructed in Exercise 3 to simulate the numberof spills into the bay for 30 consecutive days.

5. How do the probabilities from your simulation compare with theprobabilities given?

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


Outcome Probability Randomnumbers

Single 0.198

Double 0.061

Triple 0.020

Home run 0.032

Walk 0.118

Ground out 0.264

Fly out 0.187

Strike out 0.120

Total 1.000

Batting Statistics

Number Probability Randomof leaks numbers

0 0.2418

1 0.3356

2 0.2814

3 0.1181

4 0.0231

Total 1.0000

Toxins

5. a. 0b. 0.76

6. a. 0.375b. 0.875c. 0.125d. 0.125e. 0.375f. 0.5

7. a. � 1.8%b. � 39.6%

8. a. 15%b. 35%c. 15%d. 35%

Lesson 10.6

1. 0.15 2. 0.7

3. a. � 0.38b. 0.625c. 0.79

4. 0.1375

5. a. 0.625b. � 0.417

6. a. � 0.528b. � 0.544

7. a. 0.36

b.

8. a.


Lesson 10.7

1.


3.




Lesson 11.1

1. �3, �2, �1, 0, 1, 2

2. 5, 10, 15, 20, 25, 30

3. 2, 5, 8, 11, 14, 17

4. 2, 5, 10, 17, 26, 37

5.

6. 4, 9, 16, 25, 36, 49

7. 1, �3, �7, �11

8. 4, 14, 24, 34

9. �2, �4, �8, �16

10. 5, �5, 5, �5

�12

, �112

, �212

, �312

, �412

, �512

0001–2418

2419–5774

5775–8588

8589–9769

9770–10,000

001–198

199–259

260–279

280–311

312–429

430–693

694–880

881–1000

13

13

ANSWERSC

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◆Skill A Writing the terms of a sequence when given an explicit or recursive formula

Recall tn represents the nth term, where n is a natural number.

For example, t5 represents the fifth term of a sequence.

◆ Example 1Write the first 5 terms of the sequence defined by the explicit formula tn � �2n � 5.

◆ SolutionIf n � 1, then t1 � �2(1) � 5 � 3.t2 � �2(2) � 5 � 1 t3 � �2(3) � 5 � �1 t4 � �2(4) � 5 � �3 t5 � �2(5) � 5 � �5

The first 5 terms are 3, 1, �1, �3, and �5.

Recall tn � 1 represents the term just previous to the term tn.

For example, if tn is the third term, then tn � 1 is the second term.

◆ Example 2Write the first 4 terms of the sequence defined by t1 � 5 and tn � 2tn � 1 � 3.

◆ Solutiont1 � 5 t3 � 2t2 � 3 � 2(13) � 3 � 29t2 � 2t1 � 3 � 2(5) � 3 � 13 t4 � 2t3 � 3 � 2(29) � 3 � 61

The first 4 terms are 5, 13, 29, and 61.

Write the first 6 terms of each sequence defined by the givenexplicit formula.

1. tn � n � 4 2. tn � 5n

3. tn � 3n � 1 4. tn � n2 � 1

5. 6. tn � (n � 1)2

Write the first 4 terms of each sequence defined by the givenrecursive formula.

7. t1 � 1 and tn � tn � 1 � 4 8. t1 � 4 and tn � tn � 1 � 10

9. t1 � �2 and tn � 2tn � 1 10. t1 � 5 and tn � �tn � 1

11. 12. t1 � 2 and tn � 5tn � 1 �2t1 � 1 and tn �12

tn�1

tn � �n �12

Reteaching

11.1 Sequences and Series

Evaluate the sum.

13. 14. 15.

16. 17. 18.

19. 20. 21. �200

k�1k�

80

k�1k�

50

k�1k

�2

k�1(k2 � 3k)�

3

j�1(5j � 6)�

6

i�1(i � 3)

�5

k�12k�

3

k�1k2�

4

k�1(k � 1)

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Using -notation to define and evaluate a series

Recall represents the sumof the 5 terms where k takes on each integer value from 3 to 7.

◆ Example 1Write the terms of and then find the sum.

◆ Solutionk � 1; first term: 2(1) � 1 � 1k � 2; second term: 2(2) � 1 � 3k � 3; third term: 2(3) � 1 � 5k � 4; fourth term: 2(4) � 1 � 7

The sum is 1 � 3 � 5 � 7 � 16.

◆ Example 2

Evaluate .

◆ SolutionIt would take a long time to write out and find the sum of 100 terms. Here’s a shortcut.

Let tk be the kth term and notice the following pattern.t1 � 1, t100 � 100 and so t1 � t100 � 101t2 � 2, t99 � 99 and so t2 � t99 � 101t3 � 3, t98 � 98 and so t3 � t98 � 101…t50 � 50, t51 � 51 and so t50 � t51 � 101

Since each pair of terms has a sum of 101 and there are 50 of these pairs, the total must be 101 · 50 or 5050.

There is a formula that can be used for this type of problem.

In this example, .�100

k�1k �

100(101)2

� 5050

�n

k�1k �

n(n � 1)2

�100

k�1k

�4

k�1(2k � 1)

�7

k�32k

�

5. a. 0b. 0.76

6. a. 0.375b. 0.875c. 0.125d. 0.125e. 0.375f. 0.5

7. a. � 1.8%b. � 39.6%

8. a. 15%b. 35%c. 15%d. 35%

Lesson 10.6

1. 0.15 2. 0.7

3. a. � 0.38b. 0.625c. 0.79

4. 0.1375

5. a. 0.625b. � 0.417

6. a. � 0.528b. � 0.544

7. a. 0.36

b.

8. a.


Lesson 10.7

1.


3.




Lesson 11.1

1. �3, �2, �1, 0, 1, 2

2. 5, 10, 15, 20, 25, 30

3. 2, 5, 8, 11, 14, 17

4. 2, 5, 10, 17, 26, 37

5.

6. 4, 9, 16, 25, 36, 49

7. 1, �3, �7, �11

8. 4, 14, 24, 34

9. �2, �4, �8, �16

10. 5, �5, 5, �5

�12

, �112

, �212

, �312

, �412

, �512

0001–2418

2419–5774

5775–8588

8589–9769

9770–10,000

001–198

199–259

260–279

280–311

312–429

430–693

694–880

881–1000

13

13

ANSWERSC

opyright ©by H

olt, Rinehart and W



11.

12. 2, 8, 38, 188

13. 14 14. 14 15. 30 16. 3 17. 48

18. 14 19. 1275 20. 3240 21. 20,100

Lesson 11.2

1. arithmetic; d � 6

2. not arithmetic

3. arithmetic; d � �1




7. 62 8. 49 9. �83 10. 9.3 11. 38

12. 92 13. �91 14. 200 15. 16, 21

16. �2, 0, 2, 4, 6

17. 34

18. �5, �7, �9, �11, �13, �15, �17, �19,�21, �23

19. 51.5, 53, 54.5

20. 1.2, 1.9

Lesson 11.3

1. 171 2. �210 3. 572 4. 1085

5. 1800 6. 630 7. 195 8. 210

9. 5050 10. 25,250

11. ; 3775

12. ; 7650

13. ; �2600

14. 60 15. 115 16. �15 17. 91

18. 996 19. 160

Lesson 11.4

1. geometric; r � 3

2. not geometric


4. geometric; r � �0.25

5. not geometric

6. geometric; r � �2

7. 448 8. 0.125

9. 729 10. 12,500

11. 1701 or �1701

12. 4, 16

13. 15 or �15

14. 10, 20, 40 or �10, 20 ,�40

15. � 14.14

Lesson 11.5

1. 42 2. 10,922 3. 2,796,202

4. 6.25 5. 6.64 6. 6.67

7. ; 363

8. ; 2550

9. ; 1.875t1 � 1; r �12

; n � 4

t1 � 10; r � 2; n � 8

t1 � 3; r � 3; n � 5

tn � �2n � 1

tn � 6n

tn � 3n � 1

1, 12

, 14

, 18

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Recognizing arithmetic sequences and finding the common difference

Recall An arithmetic sequence is a sequence in which each pair of successive terms differby the same number d, called the common difference.

◆ Example Based on the terms given, determine whether each sequence is arithmetic. If so,identify the common difference, d.a. 1, 4, 9, 16, 25, . . . b. 8, 5, 2, �1, �4, . . .

◆ Solutiona. 4 � 1 � 3 9 � 4 � 5 6 � 9 � 7 25 � 16 � 9

Since the differences are not the same, this is not an arithmetic sequence.

b. 5 � 8 � �3 2 � 5 � �3 �1 � 2 � �3 �4 � (�1) � �3Since all the differences are the same this is an arithmetic sequence.The common difference is �3; d � �3.

◆Skill B Finding the nth term of an arithmetic sequence

Recall The nth term, tn, of an arithmetic sequence is given by .

◆ Example 1Find the fifteenth term of the arithmetic sequence that begins with 5, 13, 21, 29,37, . . .

◆ SolutionSince the first term is 5, .

Since 13 � 5 � 8 and 21 � 13 � 8, the common difference is 8, d � 8

The fifteenth term is 117.t15 � 5 � (15 � 1)(8) � 117

t1 � 5

tn � t1 � (n � 1)d

Based on the terms given, state whether or not each sequence isarithmetic. If it is, identify the common difference, d.

1. 4, 10, 16, 22, 28, . . . 2. 2, 4, 8, 16, 32, . . . 3. �2, �3, �4, �5, �6, . . .

4. 7, 0, �7, �14, �21, . . . 5. 1, 3, 5, 7, 9, . . . 6. 3, 3, 3, 3, 3, . . .

Reteaching

11.2 Arithmetic Sequences

Find the twentieth term of each arithmetic sequence.

7. 5, 8, 11, 14, . . . 8. 30, 31, 32, 33, . . .

9. 12, 7, 2, �3, . . . 10. 1.7, 2.1, 2.5, 2.9, . . .

Find the indicated number of arithmetic means between thetwo given numbers.

15. two between 11 and 26 16. five between �4 and 8

17. one between 28 and 40 18. ten between �3 and �25

19. three between 50 and 56 20. two between 0.5 and 2.6

Find the indicated term given two other terms.

11. Find t10 if t3 � 10 and t5 � 18. 12. Find t15 if t5 � �8 and t8 � 22.

13. Find t25 if t1 � 5 and t4 � �7. 14. Find t100 if t4 � 8 and t10 � 20.

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆ Example 2Find the fifteenth term of an arithmetic sequence if and .

◆ SolutionStep 1: Find d.

The table indicates that 25 � 3d � 58, so d � 11.

Step 2: Find t1. Since ,

Step 3: Find t15.

� 146� �8 � 14 � 11

t15 � t1 � (15 � 1)d

t1 � �825 � t1 � 3 � 11t4 � t1 � (4 � 1)d

t7 � 58t4 � 25

◆Skill C Finding the arithmetic means between two numbers

Recall The terms between any two nonconsecutive terms of an arithmetic sequence arecalled the arithmetic means.

◆ Example Find three arithmetic means between 8 and �20.

◆ Solution8, , , , �20

d d d d This indicates that 8 � 4d � �20. So d � �7.

Add �7 to each term to find the next term and fill in the blanks.

8, , , , �20 The three arithmetic means are 1, �6, and �13.�13�6 1

n 4 5 6 7

tn 25 ? ? 58

d d d

11.

12. 2, 8, 38, 188

13. 14 14. 14 15. 30 16. 3 17. 48

18. 14 19. 1275 20. 3240 21. 20,100

Lesson 11.2


2. not arithmetic





7. 62 8. 49 9. �83 10. 9.3 11. 38

12. 92 13. �91 14. 200 15. 16, 21

16. �2, 0, 2, 4, 6

17. 34

18. �5, �7, �9, �11, �13, �15, �17, �19,�21, �23

19. 51.5, 53, 54.5

20. 1.2, 1.9

Lesson 11.3

1. 171 2. �210 3. 572 4. 1085

5. 1800 6. 630 7. 195 8. 210

9. 5050 10. 25,250

11. ; 3775

12. ; 7650

13. ; �2600

14. 60 15. 115 16. �15 17. 91

18. 996 19. 160

Lesson 11.4


2. not geometric



5. not geometric


7. 448 8. 0.125

9. 729 10. 12,500

11. 1701 or �1701

12. 4, 16

13. 15 or �15

14. 10, 20, 40 or �10, 20 ,�40

15. � 14.14

Lesson 11.5

1. 42 2. 10,922 3. 2,796,202

4. 6.25 5. 6.64 6. 6.67

7. ; 363

8. ; 2550

9. ; 1.875t1 � 1; r �12

; n � 4

t1 � 10; r � 2; n � 8

t1 � 3; r � 3; n � 5

tn � �2n � 1

tn � 6n

tn � 3n � 1

1, 12

, 14

, 18

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◆Skill A Finding the sum of the first n terms of an arithmetic series

Recall The sum of the first n terms of an arithmetic series is given by .

◆ Example 1Find the sum 95 � 96 � 97 � 98 � 99 � 100 � 101 � 102 � 103 � 104 � 105.

◆ Solution

Since there are 11 terms, and .

The sum of these 11 terms is 1100.

Recall The nth term of an arithmetic sequence is given by .

◆ Example 2Find S20 for the arithmetic sequence 2, 10, 18, 26, 34, . . .

◆ Solution

S20 � 20(2 � 1542 ) � 1560

t20 � 2 � (20 � 1)8 � 154

d � 8t1 � 2

tn � t1 � (n � 1)d

tn � t1 � (n � 1)d

S11 � 11(95 � 1052 ) � 1100

t11 � 105

tn � t11n � 11

t1 � 95

Sn � n(t1 � tn

2 )

Use the formula for an arithmetic series to find each sum.

1. 3 � 7 � 11 � 15 � 19 � 23 � 27 � 31 � 35

2. 10 � 5 � 0 � 5 � 10 � 15 � 20 � 25 � 30 � 35 � 40 � 45

3. 7 � 16 � 25 � 34 � . . . � 97

4. 20 � 21 � 22 � 23 � . . . � 50

For each arithmetic series, find S30.

5. 2, 6, 10, 14, . . . 6. �8, �6, �4, �2, . . .

7. 50, 47, 44, 41, . . . 8. 7, 7, 7, 7, . . .

9. Find the sum of the first 100 natural numbers.

10. Find the sum of the multiples of 5 from 5 to 500.

Reteaching

11.3 Arithmetic Series

Evaluate.

14. 15. 16.

17. 18. 19. �20

k�1(50 � 4k)�

12

k�1(10k � 18)�

7

k�1(3k � 1)

�10

k�1(15 � 3k)�

5

k�1(k � 20)�

8

k�1(12 � k)

Find an explicit formula for each arithmetic sequence and use agraphics calculator to find S50.

11. 2, 5, 8, 11, . . . 12. 6, 12, 18, 24, . . . 13. �3, �5, �7, �9, . . .

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Using a graphics calculator to evaluate an arithmetic series

Recall An arithmetic series is the indicated sum of the terms of an arithmetic sequence.

◆ ExampleUse a graphics calculator to find S80 for 5, 15, 25, 35, 45, . . .

◆ SolutionYou must first find an explicit formula for thearithmetic sequence.

Use where and .

Do not replace n with 80 yet.

Note: The commands shown may be different on your calculator.

S80 � 32,000

tn � 10n � 5

tn � 5 � (n � 1)10

d � 10t1 � 5tn � t1 � (n � 1)d

◆Skill C Evaluating an arithmetic series expressed in -notation

Recall indicates the sum of 15 terms where the first value of k is 1 and the last

value of k is 15.

◆ ExampleFind the sum indicated by .

◆ Solution

using

or �20

n�1(3n � 2) � 590S20 � 590

Sn � n(t1 � tn

2 )S20 � 20(1 � 582 )

t20 � 3 � 20 � 2 � 58

t � 3 � 1 � 2 � 1

�20

n�1(3n � 2)

�15

k�1(5k � 3)

�

11.

12. 2, 8, 38, 188

13. 14 14. 14 15. 30 16. 3 17. 48

18. 14 19. 1275 20. 3240 21. 20,100

Lesson 11.2


2. not arithmetic





7. 62 8. 49 9. �83 10. 9.3 11. 38

12. 92 13. �91 14. 200 15. 16, 21

16. �2, 0, 2, 4, 6

17. 34

18. �5, �7, �9, �11, �13, �15, �17, �19,�21, �23

19. 51.5, 53, 54.5

20. 1.2, 1.9

Lesson 11.3

1. 171 2. �210 3. 572 4. 1085

5. 1800 6. 630 7. 195 8. 210

9. 5050 10. 25,250

11. ; 3775

12. ; 7650

13. ; �2600

14. 60 15. 115 16. �15 17. 91

18. 996 19. 160

Lesson 11.4


2. not geometric



5. not geometric


7. 448 8. 0.125

9. 729 10. 12,500

11. 1701 or �1701

12. 4, 16

13. 15 or �15

14. 10, 20, 40 or �10, 20 ,�40

15. � 14.14

Lesson 11.5

1. 42 2. 10,922 3. 2,796,202

4. 6.25 5. 6.64 6. 6.67

7. ; 363

8. ; 2550

9. ; 1.875t1 � 1; r �12

; n � 4

t1 � 10; r � 2; n � 8

t1 � 3; r � 3; n � 5

tn � �2n � 1

tn � 6n

tn � 3n � 1

1, 12

, 14

, 18

ANSWERSC

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Recognizing geometric sequences and finding the common ratio

Recall A geometric sequence is a sequence in which the ratio of each pair of successiveterms is the same number r, called the constant ratio.

◆ ExampleBased on the terms given, determine whether each sequence is geometric. If so,identify the common ratio, r.a. 3, 6, 12, 24, . . . b. 2, 4, 12, 48, . . . c. 125, 25, 5, 1, . . .

◆ Solution

a.

Since the ratios are the same, this is a geometric sequence.The common ratio is 2; r � 2.

b.

Since the ratios are not the same, this is not a geometric sequence.

c.

Since the ratios are the same, this is a geometric sequence.

The common ratio is .15

; r �15

15

525

�15

25125

�15

4812

� 4124

� 342

� 2

2412

� 2126

� 263

� 2

Determine whether each sequence is a geometric sequence. If so,identify the common ratio, r.

1. 4, 12, 36, 108, . . . 2. 5, 10, 15, 20, . . . 3. 30, 60, 120, 240, . . .

4. 40, �10, 2.5, �0.625, . . . 5. 60, 30, 10, 2.5, . . . 6. 2, −4, 8, −16, . . .

Reteaching

11.4 Geometric Sequences

◆Skill B Finding the nth term of a geometric sequence

Recall The nth term, tn, of a geometric sequence is given by .

◆ Example 1Find the 9th term of the geometric sequence that begins 2, 10, 50, 250, . . .

◆ SolutionSince the first term is 2, .

Since , the common ratio is 5; r � 5.

According to the formula, .t9 � 2(5)9�1 � 781,250

102

� 5

t1 � 2

tn � t1rn�1

Find the indicated number of geometric means between the twogiven numbers.

12. two between 1 and 64 13. one between 5 and 45

14. three between 5 and 80 15. one between 10 and 20

Find the seventh term of each geometric sequence.

7. 7, 14, 28, . . . 8. 8, −4, 2, . . . 9. 64, 96, 144, . . .

Find the indicated term of each geometric sequence when giventwo other terms.

10. Find t6 if and . 11. Find t8 if and . t5 � 63t3 � 7t4 � 500t1 � 4

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆ Example 2Find the eighth term of a geometric sequence if and .

◆ SolutionStep 1: Find r.

Step 2: Find t1. Since ,

or

or

Notice that using for t1 would give the same two values for t8.�34

t1 � �34

t1 �34

6 � t1(�2)36 � t1(2)3

t4 � t1r3

t6 � 24t4 � 6

n 4 5 6

tn 6 ? 24

r r

The table indicates that

So .r � 2r2 � 4

6r2 � 24

◆Skill C Finding the geometric means between two numbers

Recall The terms between any two nonconsecutive terms of a geometric sequence arecalled the geometric means.

◆ ExampleFind two geometric means between 6 and 20.25.

◆ Solution

6, , , 20.25 This indicates that

r r r

Multiply each term by 1.5 to find the next term and fill in the blanks.

6, , , 20.25 The two geometric means are 9 and 13.5.13.59

r � (3.375)13 � 1.5

6r3 � 20.25

Step 3: Find t8.

or

or t8 � �96t8 � 96

t8 �34

(�2)7t8 �34

(2)7

t8 � t1r7

11.

12. 2, 8, 38, 188

13. 14 14. 14 15. 30 16. 3 17. 48

18. 14 19. 1275 20. 3240 21. 20,100

Lesson 11.2


2. not arithmetic





7. 62 8. 49 9. �83 10. 9.3 11. 38

12. 92 13. �91 14. 200 15. 16, 21

16. �2, 0, 2, 4, 6

17. 34

18. �5, �7, �9, �11, �13, �15, �17, �19,�21, �23

19. 51.5, 53, 54.5

20. 1.2, 1.9

Lesson 11.3

1. 171 2. �210 3. 572 4. 1085

5. 1800 6. 630 7. 195 8. 210

9. 5050 10. 25,250

11. ; 3775

12. ; 7650

13. ; �2600

14. 60 15. 115 16. �15 17. 91

18. 996 19. 160

Lesson 11.4


2. not geometric



5. not geometric


7. 448 8. 0.125

9. 729 10. 12,500

11. 1701 or �1701

12. 4, 16

13. 15 or �15

14. 10, 20, 40 or �10, 20 ,�40

15. � 14.14

Lesson 11.5

1. 42 2. 10,922 3. 2,796,202

4. 6.25 5. 6.64 6. 6.67

7. ; 363

8. ; 2550

9. ; 1.875t1 � 1; r �12

; n � 4

t1 � 10; r � 2; n � 8

t1 � 3; r � 3; n � 5

tn � �2n � 1

tn � 6n

tn � 3n � 1

1, 12

, 14

, 18

ANSWERSC

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Finding the sum of the first n terms of a geometric series

Recall The sum of the first n terms of a geometric series is given by .

◆ Example Find each sum for the geometric series 4 � 12 � 36 � 108 � . . .a. b.

◆ Solution

Since , the common ratio is 3.

a. b.

13,120 118,096��

S10 � 4(1 � 310

1 � 3 )S8 � 4(1 � 38

1 � 3 )124

� 3

S10S8

Sn � t1(1 � rn

1 � r )

Find each indicated sum of the geometric series 2 � 8 � 32 � 128 � . . .

1. 2. 3.

Find each sum of the geometric series 10 � (�5) � 2.5 � (�1.25) � . . .Round answers to nearest hundredth, if necessary.

4. 5. 6. S11S8S4

S11S7S3

Reteaching

11.5 Geometric Series and Mathematical Induction

◆Skill B Evaluating a geometric series expressed in -notation

Recall indicates the sum of 5 terms where the first value of k is 1 and the last

value of k is 5.

◆ Example

For a. identify t1, r, and n. b. find the indicated sum.

◆ Solutiona. , and ; b.

� 2340n � 4�7515

� 5r �t2

t1

S4 � 15(1 � 54

1 � 5 )t2 � 3 � 52 � 75t1 � 3 � 51 � 15

�4

k�1(3 � 5k)

�15

k�1(2 � 3k)

�

Evaluate t1, r, and n, and evaluate the sum of each series.

7. 8. 9. �4

k�1(12)k�1

�8

k�1(5 � 2k)�

5

k�13k

10. Prove by mathematical induction that .1 � 4 � 7 � . . . � (3n � 2) �n(3n � 1)

2

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill C Using mathematical induction to prove a statement

Recall Mathematical induction is used to prove that a statement in terms of n, where n is anatural number, is true for all values of n.

◆ Example Prove by mathematical induction that .

◆ SolutionStep 1: Show that this statement is true when .

Step 2: Let k be some particular value of n and assume that the following is true:

Replace n with k.

Step 3: Write a statement for k � 1 terms. This is the statement to be proved.

Rewrite the right-hand side of this last statement as

Step 4: Recall what was assumed to be true in Step 2, and add 3(k � 1) to eachside of that equation.

common denominator

By showing that the results from steps 3 and 4 (boxed expressions) are the same,you have proved that if the formula is true for k (what you assumed to be true),it is also true for k � 1. Since k is just some value of n, let k � n � 1. Step 1showed that this formula is true when n � 1, and step 2 proved that it must alsobe true for n � 2. If it is true when n � 2, it must also be true for n � 3. Followingthe same logic, it must be true for the next value of n, and eventually for everyvalue of n.

�

�3k2 � 9k � 6

2

�3k(k � 1)

2�

6(k � 1)2

3 � 6 � 9 � . . . � 3k � 3(k � 1) �3(k � 1)

2� 3(k � 1)

3 � 6 � 9 � . . . � 3k � 3(k � 1) �3(k � 1)(k � 1 � 1)

2

3 � 6 � 9 � . . . � 3k �3k(k � 1)

2

3 � 3

3(1) � 3(1)(1 � 1)

2

n � 1

3 � 6 � 9 � . . . � 3n �3n(n � 1)

2

3(k � 1)(k � 2)2

3(k � 1)(k � 2)2

11.

12. 2, 8, 38, 188

13. 14 14. 14 15. 30 16. 3 17. 48

18. 14 19. 1275 20. 3240 21. 20,100

Lesson 11.2


2. not arithmetic





7. 62 8. 49 9. �83 10. 9.3 11. 38

12. 92 13. �91 14. 200 15. 16, 21

16. �2, 0, 2, 4, 6

17. 34

18. �5, �7, �9, �11, �13, �15, �17, �19,�21, �23

19. 51.5, 53, 54.5

20. 1.2, 1.9

Lesson 11.3

1. 171 2. �210 3. 572 4. 1085

5. 1800 6. 630 7. 195 8. 210

9. 5050 10. 25,250

11. ; 3775

12. ; 7650

13. ; �2600

14. 60 15. 115 16. �15 17. 91

18. 996 19. 160

Lesson 11.4


2. not geometric



5. not geometric


7. 448 8. 0.125

9. 729 10. 12,500

11. 1701 or �1701

12. 4, 16

13. 15 or �15

14. 10, 20, 40 or �10, 20 ,�40

15. � 14.14

Lesson 11.5

1. 42 2. 10,922 3. 2,796,202

4. 6.25 5. 6.64 6. 6.67

7. ; 363

8. ; 2550

9. ; 1.875t1 � 1; r �12

; n � 4

t1 � 10; r � 2; n � 8

t1 � 3; r � 3; n � 5

tn � �2n � 1

tn � 6n

tn � 3n � 1

1, 12

, 14

, 18

ANSWERSC

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10. Step 1:

Step 2:

Step 3:

Step 4:

Lesson 11.6

1. geometric; convergent

2. not geometric


4. geometric; divergent

5. 6. 7. 1 8. 9. 2 10.

11. 12. 13. 14. 15. 1

16.

Lesson 11.7

1. 1, 6, 15, 20, 15, 6, 1

2. 1, 7, 21, 35, 35, 21, 7, 1

3. 1, 8, 28, 56, 70, 56, 28, 8, 1

4. 220 5. 5005 6. 70 7. 18

8. 18 9. 2,042,975 10. � 0.22

11. � 0.33 12. � 0.33 13. � 0.36

14. � 0.63 15. � 0.004 16. � 0.14

17. � 0.04

Lesson 11.8

1.

2. x4 � 4x3y � 6x2y2 � 4xy3 � y4

3. x7 � 7x6y � 21x5y2 � 35x4y3 � 35x3y4 �21x2y5 � 7xy6 � y7

4. r5 � 10r4s � 40r3s2 � 80r2s3 � 80r s4 � 32r5

5. 81a4 � 540a3b � 1350a2b2 � 1500ab3 �625b4

6. 7. 8.

9. 10. 11.

12. � 0.06 13. � 0.25 14. � 0.78

15. � 0.06


Lesson 12.1

1. 5.65; 6; 8

2. 4.52; 3.85; 3.1

3. $814.50; $15, $10

4. 69.13; 61; 24

5. 81.5; 80; 75 and 80

6.

28 192

mean: 6.86

5 5 25

6 8 48

7 5 35

8 6 48

9 4 36

495x8y4924x6y6495x4y8

�20xy9�8064x5y511,520x8y2

6ab5 � b6a6 � 6a5b � 15a4b2 � 20a3b3 � 15a2b4 �

137

4399

311

13

79

�37

14

23

23

�(k � 1)(3k � 2)

2

�3k2 � 5k � 2

2

�k(3k � 1)

2� [3(k � 1) � 2]

[3(k � 1) � 2]1 � 4 � 7 � . . . � (3k � 2) �

�(k � 1)(3k � 2)

2

�(k � 1)[3(k � 1) � 1]

2

[3(k � 1) �2]1 � 4 � 7 � . . . � (3k � 2) �

1 � 4 � 7 � . . . � (3k � 2) �k(3k � 1)

2

1 �1(3 � 1 � 1)

2

ANSWERSC

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Identifying convergent and divergent geometric series

Recall In a geometric series, the ratio, r, of any two consecutive terms is a constant.A geometric series converges if ; it diverges if .

◆ Example For each series, determine if it is geometric. If so, determine whether it isconvergent or divergent.

a. . . . b. . . . c. . . .

◆ Solution

a.

This series is geometric with . Since , this series converges.

b.

This is not a geometric series because there is no common ratio.

c.

This series is geometric with . Since , this series diverges.��2� � 1r � �2

�24 12 � �212 (�6) � �2�6 3 � �2

34

23

�98

23

12

�43

�23� � 1r �23

827

49

�23

49

23

�23

23

1 �23

3 � 6 � 12 � 24 �12

�23

�34

�45

�1 �23

�49

�8

27�

�r � � 1�r � � 1

Determine whether each series is geometric; if so, tell whether itconverges or diverges.

1. . . . 2. . . .

3. . . . 4. . . .32

�94

�278

�8116

�1 �12

�14

�18

�

12

�23

�34

�45

�13

�16

�1

12�

124

�

Reteaching

11.6 Infinite Geometric Series

◆Skill B Finding the sum of an infinite geometric series

Recall The sum of a convergent infinite geometric series is equal to .

◆ Example 1Find the sum of the infinite geometric series

◆ Solution

and , so .

You can check your result by using the general formula for the sum of a geometric

series, . In this case, graph

. Notice what happens to

the value of y as x gets larger.

y � 1(1 �

1 �

(23)x

23

)Sn � t1(1 � rn

1 � r )

S �t1

1 � r�

11 � 2

3

� 3r �23

t1 � 1

1 �23

�49

�8

27� . . .

t1

1 � r

10

0 100

Write each decimal as a fraction in simplest form.

11. 12. 13.

14. 15. 16. 0.0270.990.43

0.270.30.7

Find the sum of each infinite geometric series, if it exists.

5. . . . 6. . . . 7. . . .

Evaluate.

8. 9. 10. ��

k�1(�

34)k

��

k�1(23)k

��

k�1(15)k

12

�14

�18

�1 �12

�14

�18

�13

�16

�1

12�

Copyright ©

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ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


Recall The symbol for infinity is �; indicates the sum of an infinite number of terms.

◆ Example 2

Evaluate .

◆ Solution

Let k � 1; . Let k � 2; .

��

k�1( 25k) �

121 �

25

15

�12

S �t1

1 � r�

�15

22525

r �t2

t1�

t2 �225

t1 �25

��

k�1( 25k)

��

k�1

◆Skill C Changing a repeating decimal to a fraction

Recall A repeating decimal represents a rational number and can be written in the form ,where a and b are integers.

◆ Example 1Write as a fraction in lowest terms.

◆ Solution. . .

and

Thus, .

Check with your calculator. . . .7 11 � 0.6363630.63 �7

11

S �t1

1 � r�

0.631 � 0.01

�0.630.99

�6399

�7

11

r �t2

t1�

0.00630.63

� 0.01

t2 � 0.0063t1 � 0.630.63 � 0.63 � 0.0063 � 0.000063 � 0.00000063 �

0.63

ab

10. Step 1:

Step 2:

Step 3:

Step 4:

Lesson 11.6


2. not geometric



5. 6. 7. 1 8. 9. 2 10.

11. 12. 13. 14. 15. 1

16.

Lesson 11.7

1. 1, 6, 15, 20, 15, 6, 1

2. 1, 7, 21, 35, 35, 21, 7, 1

3. 1, 8, 28, 56, 70, 56, 28, 8, 1

4. 220 5. 5005 6. 70 7. 18

8. 18 9. 2,042,975 10. � 0.22

11. � 0.33 12. � 0.33 13. � 0.36

14. � 0.63 15. � 0.004 16. � 0.14

17. � 0.04

Lesson 11.8

1.

2. x4 � 4x3y � 6x2y2 � 4xy3 � y4

3. x7 � 7x6y � 21x5y2 � 35x4y3 � 35x3y4 �21x2y5 � 7xy6 � y7

4. r5 � 10r4s � 40r3s2 � 80r2s3 � 80r s4 � 32r5

5. 81a4 � 540a3b � 1350a2b2 � 1500ab3 �625b4

6. 7. 8.

9. 10. 11.

12. � 0.06 13. � 0.25 14. � 0.78

15. � 0.06


Lesson 12.1

1. 5.65; 6; 8

2. 4.52; 3.85; 3.1

3. $814.50; $15, $10

4. 69.13; 61; 24

5. 81.5; 80; 75 and 80

6.

28 192

mean: 6.86

5 5 25

6 8 48

7 5 35

8 6 48

9 4 36

495x8y4924x6y6495x4y8

�20xy9�8064x5y511,520x8y2

6ab5 � b6a6 � 6a5b � 15a4b2 � 20a3b3 � 15a2b4 �

137

4399

311

13

79

�37

14

23

23

�(k � 1)(3k � 2)

2

�3k2 � 5k � 2

2

�k(3k � 1)

2� [3(k � 1) � 2]

[3(k � 1) � 2]1 � 4 � 7 � . . . � (3k � 2) �

�(k � 1)(3k � 2)

2

�(k � 1)[3(k � 1) � 1]

2

[3(k � 1) �2]1 � 4 � 7 � . . . � (3k � 2) �

1 � 4 � 7 � . . . � (3k � 2) �k(3k � 1)

2

1 �1(3 � 1 � 1)

2

ANSWERSC

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olt, Rinehart and W



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◆Skill A Finding entries in Pascal’s triangle

Recall Each entry in Pascal’s triangle is the sum of the two entries directly above it.

◆ Example 1Without looking in the textbook, complete the fifth row of Pascal’s triangle.

1 ← Row 01 1

1 2 11 3 3 1

1 4 6 4 1 ← Row 4

◆ SolutionAdding adjacent numbers in row 4 gives 1, 5, 10, 10, 5, 1 for Row 5.

Recall To find the combination of n things taken r at a time, .

◆ Example 2Find the sixth entry in the fifteenth row of Pascal’s triangle.

◆ SolutionInstead of writing out all these rows of the triangle, consider the following.The first entry in any row is 1, which is . To find the rest of the entries, use

.

To find the sixth entry:

The sixth entry in the fifteenth row of Pascal’s triangle is 3003.nCk�1 � 15C6�1 � 15C5 �

15!5!(15 � 5)!

�15!

5!10!� 3003.

nCk�1

nC0

nCr �n!

r!(n � r)!

Complete the following rows for Pascal’s triangle.

1. Row 6:

2. Row 7:

3. Row 8:

Find the value for each entry in Pascal’s triangle.

4. fourth entry in the twelfth row 5. tenth entry in the fifteenth row

6. fifth entry in the eighth row 7. second entry in the eighteenth row

8. eighteenth entry in the eighteenth row 9. tenth entry in the twenty-fifth row

Reteaching

11.7 Pascal’s Triangle

A fair coin is tossed 8 times. Find the following probabilities.

10. P(exactly 3 heads) 11. P(5 or 6 heads)

12. P(5 or 6 tails) 13. P(more than 4 heads)

14. P(at least 4 heads) 15. P(8 tails)

16. P(more than 5 tails) 17. P(7 or 8 heads)

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Using Pascal’s triangle to find probabilities

Recall If a probability experiment that has 2 equally likely outcomes is repeated for n

trials, the probability of either outcome occurring exactly k times is .

◆ Example 1A lawyer has two phone lines, A and B. Suppose an incoming call is equally likelyto be on line A or line B. Use Pascal’s triangle to find the probability that 0, 1, 2,3, 4, and 5 out of the next 5 calls will be on line A.

◆ SolutionRow 5: 5C0 5C1 5C2 5C3 5C4 5C5

1 5 10 10 5 1

total: 1 � 5 � 10 � 10 � 5 � 1 � 32

Notice that 32 is 25. (the total of the 5th row entries)

◆ Example 2For the lawyer in Example 1, find the probability that at least 3 of the next 7 callsare on line B.

◆ SolutionAt least 3 means 3, 4, 5, 6, or 7 calls.

The probability is approximately 0.77.

� 35 � 35 � 21 � 7 � 1128

� 0.77

7C3

27 �7C4

27 �7C5

27 �7C6

27 �7C7

27

nCk

2n

number of calls on line A 0 1 2 3 4 5

probability 132

532

1032

1032

532

132

10. Step 1:

Step 2:

Step 3:

Step 4:

Lesson 11.6


2. not geometric



5. 6. 7. 1 8. 9. 2 10.

11. 12. 13. 14. 15. 1

16.

Lesson 11.7

1. 1, 6, 15, 20, 15, 6, 1

2. 1, 7, 21, 35, 35, 21, 7, 1

3. 1, 8, 28, 56, 70, 56, 28, 8, 1

4. 220 5. 5005 6. 70 7. 18

8. 18 9. 2,042,975 10. � 0.22

11. � 0.33 12. � 0.33 13. � 0.36

14. � 0.63 15. � 0.004 16. � 0.14

17. � 0.04

Lesson 11.8

1.

2. x4 � 4x3y � 6x2y2 � 4xy3 � y4

3. x7 � 7x6y � 21x5y2 � 35x4y3 � 35x3y4 �21x2y5 � 7xy6 � y7

4. r5 � 10r4s � 40r3s2 � 80r2s3 � 80r s4 � 32r5

5. 81a4 � 540a3b � 1350a2b2 � 1500ab3 �625b4

6. 7. 8.

9. 10. 11.

12. � 0.06 13. � 0.25 14. � 0.78

15. � 0.06


Lesson 12.1

1. 5.65; 6; 8

2. 4.52; 3.85; 3.1

3. $814.50; $15, $10

4. 69.13; 61; 24

5. 81.5; 80; 75 and 80

6.

28 192

mean: 6.86

5 5 25

6 8 48

7 5 35

8 6 48

9 4 36

495x8y4924x6y6495x4y8

�20xy9�8064x5y511,520x8y2

6ab5 � b6a6 � 6a5b � 15a4b2 � 20a3b3 � 15a2b4 �

137

4399

311

13

79

�37

14

23

23

�(k � 1)(3k � 2)

2

�3k2 � 5k � 2

2

�k(3k � 1)

2� [3(k � 1) � 2]

[3(k � 1) � 2]1 � 4 � 7 � . . . � (3k � 2) �

�(k � 1)(3k � 2)

2

�(k � 1)[3(k � 1) � 1]

2

[3(k � 1) �2]1 � 4 � 7 � . . . � (3k � 2) �

1 � 4 � 7 � . . . � (3k � 2) �k(3k � 1)

2

1 �1(3 � 1 � 1)

2

ANSWERSC

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◆Skill A Using the Binomial Theorem to expand binomials raised to a power

Recall ,where

◆ Example 1Write in expanded form.

◆ Solution

Remember that means 5C0. 5C0 � 1

Notice that the powers of a go from 5 to 0 and the powers of b go from 0 to 5.Also, the coefficients are the fifth row of Pascal’s triangle.

◆ Example 2Write 5 in expanded form.

◆ SolutionThis can be done in the same way as Example 1 by replacing a with 2x and b with �y. However, the 2 in the 2x will cause the coefficients to be differentfrom those above.

Notice the alternating positive and negative terms.(2x � y)5 � 32x5 � 80x4y � 80x3y2 � 40x2y3 � 10xy4 � y5

� (55)(2x)0(�y)5� (54)(2x)1(�y)4 �

(53)(2x)2(�y)3( 5

2)(2x)3(�y)2 �(51)(2x)4(�y)1 �(2x � y)5 � (50)(2x)5(�y)0 �

(2x � y)

(a � b)5 � a5 � 5a4b � 10a3b2 � 10a2b3 � 5ab4 � b5

(50)(a � b)5 � (50)a5b0 � (51)a4b1 � (52)a3b2 � (53)a2b3 � (54)a1b4 � (55)a0b5

(a � b)5

(nk) � nCk(x � y)n � �n

k�0(nk)xn�kyk

Expand each binomial raised to a power.

1.

2.

3.

4.

5. (3a � 5b)4

(r � 2s)5

(x � y)7

(x � y)4

(a � b)6

Reteaching

11.8 The Binomial Theorem

For the expansion of (2x � y)10, find the indicated term.

6. 3rd term 7. 6th term 8. 10th term

For the expansion of (x � y)12, find the term that contains eachmonomial.

9. x4y8 10. x6y6 11. x8y4

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Writing a particular term of a binomial expansion

Recall Since the first term is found with k � 0, the 6th term is found with k � 6 � 1 � 5.The value of k is always 1 less than the number of the term.

◆ Example 1Write the fourth term of .

◆ Solutionn � 8 and k � 4 � 1 � 3

◆ Example 2Write the term in the expansion of that contains a4b5.

◆ Solutionn � 9 and k � 5 (exponent of b factor) (9

5)a4b5 � 126a4b5

(a � b)9

(83)x5(�3y)3 � �1512x5y3

(x � 3y)8

◆Skill C Using the Binomial Theorem to determine probabilities

Recall If a probability experiment has two possible outcomes that are not equally likely,the sum of the two probabilities must still equal 1.

◆ ExampleThe probability that a movie is showing on any one of a block of 5 premiumcable channels is 80%. What is the probability that at any given time 4 of the 5channels are showing a movie?

◆ SolutionFor each channel, P(showing movie) � 0.8 and P(not showing movie) � 0.2.

There is approximately a 41% chance that exactly 4 movies are showing.

(51)(0.8)4(0.2)1 � 0.41

Ten students assume that the chance of getting an A on aparticular chemistry test is 75%. If this is true, what is theprobability that

12. exactly 5 of these students will get an A?

13. exactly 7 of these students will get an A?

14. 7 or more of these students will get an A?

15. all 10 students will get an A?

10. Step 1:

Step 2:

Step 3:

Step 4:

Lesson 11.6


2. not geometric



5. 6. 7. 1 8. 9. 2 10.

11. 12. 13. 14. 15. 1

16.

Lesson 11.7

1. 1, 6, 15, 20, 15, 6, 1

2. 1, 7, 21, 35, 35, 21, 7, 1

3. 1, 8, 28, 56, 70, 56, 28, 8, 1

4. 220 5. 5005 6. 70 7. 18

8. 18 9. 2,042,975 10. � 0.22

11. � 0.33 12. � 0.33 13. � 0.36

14. � 0.63 15. � 0.004 16. � 0.14

17. � 0.04

Lesson 11.8

1.

2. x4 � 4x3y � 6x2y2 � 4xy3 � y4

3. x7 � 7x6y � 21x5y2 � 35x4y3 � 35x3y4 �21x2y5 � 7xy6 � y7

4. r5 � 10r4s � 40r3s2 � 80r2s3 � 80r s4 � 32r5

5. 81a4 � 540a3b � 1350a2b2 � 1500ab3 �625b4

6. 7. 8.

9. 10. 11.

12. � 0.06 13. � 0.25 14. � 0.78

15. � 0.06


Lesson 12.1

1. 5.65; 6; 8

2. 4.52; 3.85; 3.1

3. $814.50; $15, $10

4. 69.13; 61; 24

5. 81.5; 80; 75 and 80

6.

28 192

mean: 6.86

5 5 25

6 8 48

7 5 35

8 6 48

9 4 36

495x8y4924x6y6495x4y8

�20xy9�8064x5y511,520x8y2

6ab5 � b6a6 � 6a5b � 15a4b2 � 20a3b3 � 15a2b4 �

137

4399

311

13

79

�37

14

23

23

�(k � 1)(3k � 2)

2

�3k2 � 5k � 2

2

�k(3k � 1)

2� [3(k � 1) � 2]

[3(k � 1) � 2]1 � 4 � 7 � . . . � (3k � 2) �

�(k � 1)(3k � 2)

2

�(k � 1)[3(k � 1) � 1]

2

[3(k � 1) �2]1 � 4 � 7 � . . . � (3k � 2) �

1 � 4 � 7 � . . . � (3k � 2) �k(3k � 1)

2

1 �1(3 � 1 � 1)

2

ANSWERSC

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◆Skill A Finding the mean, median, and mode

Recall The mean is the arithmetic average, denoted .The median is the numerical middle value.The mode is the most often repeated data value.

◆ ExampleThe students in a defensive driving course admitted to the following number oftraffic tickets each.

3 2 2 5 4 2 1 0 2 82 4 3 1 2 1 1 5 2 2

Find the mean, median, and mode for this set of data.

◆ SolutionThe total number of tickets is 52.The total number of students is 20.

The mean is 2.6.

Arrange the data in order from least to greatest.

0 1 1 1 1 2 2 2 2 22 2 2 3 3 4 4 5 5 8

The average of the two middle values (tenth and eleventh values) is 2. Themedian is 2.

The number 2 occurs most often, so the mode is 2.

x �5220

� 2.6

x

Find the mean, median, and mode of each data set. Give answersto the nearest hundredth, when necessary.

1. 4, 5, 8, 6, 7, 8, 9, 1, 5, 4, 6, 8, 9, 6, 4, 3, 3, 2, 8, 7

mean: median: mode:

2. 1.7, 3.1, 5.5, 6.4, 7.8, 9.3, 2.7, 3.1, 6.7, 0.3, 4.6, 1.2, 8.0, 2.9

mean: median: mode:

3. $5, $8000, $10, $20, $15, $10, $10, $35, $25, $15

mean: median: mode:

4. number of CDs in students’ collections: 38, 52, 24, 95, 118, 70, 24, 132

mean: median: mode:

5. grades on an exam: 95, 100, 75, 90, 80, 80, 75, 70, 85, 65

mean: median: mode:

Reteaching

12.1 Measures of Central Tendency

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Finding the mean from a frequency table

Recall Using a frequency table is a shortcut for finding the mean of a large sets of data.

◆ Example 1The following grades were recorded on a class quiz.

90 70 80 90 90 100 100 90 70 80 100 100 90 80 70 90

Make a frequency table and use it to find the mean.

◆ SolutionEach frequency is the number of students that earn a particular grade.

The mean grade is

approximately 86.9.

◆ Example 2In the table shown at right, each class represents a range of data values. Estimate the mean for this grouped frequency table.

◆ SolutionFirst find the mean of each class.

Find the product of each classmean and frequency.

Divide the total of the productsby the total of the frequencies.

The estimated mean is 19.6.x �43122

� 19.6

x �1390

16� 86.9

grade frequency product

70 3 210

80 3 240

90 6 540

100 4 400

Total 16 1390

class frequency

1–10 8

11–20 2

21–30 7

31–40 5

class class mean frequency product

1–10 5.5 8 44

11–20 15.5 2 31

21–30 25.5 7 178.5

31–40 35.5 5 177.5

22 431

On your own paper, make a frequency table for each data set,and find the mean to the nearest hundredth.

6. 7, 6, 6, 7, 6, 5, 8, 6, 5, 9, 8, 5, 6, 8, 7. 500, 100, 500, 400, 300, 800, 800, 100, 600,9, 5, 8, 8, 6, 8, 7, 5, 6, 9, 7, 7, 9, 6 500, 700, 400, 400, 500, 600, 900, 800, 300

On your own paper, make a grouped frequency table for each setof data and estimate the mean as a whole number.

8. 75, 78, 99, 88, 74, 89, 61, 91, 9. 320, 480, 560, 750, 460, 380,84, 78, 75, 92, 88, 67, 82, 85 720, 790, 380, 590, 510, 630

10. Step 1:

Step 2:

Step 3:

Step 4:

Lesson 11.6


2. not geometric



5. 6. 7. 1 8. 9. 2 10.

11. 12. 13. 14. 15. 1

16.

Lesson 11.7

1. 1, 6, 15, 20, 15, 6, 1

2. 1, 7, 21, 35, 35, 21, 7, 1

3. 1, 8, 28, 56, 70, 56, 28, 8, 1

4. 220 5. 5005 6. 70 7. 18

8. 18 9. 2,042,975 10. � 0.22

11. � 0.33 12. � 0.33 13. � 0.36

14. � 0.63 15. � 0.004 16. � 0.14

17. � 0.04

Lesson 11.8

1.

2. x4 � 4x3y � 6x2y2 � 4xy3 � y4

3. x7 � 7x6y � 21x5y2 � 35x4y3 � 35x3y4 �21x2y5 � 7xy6 � y7

4. r5 � 10r4s � 40r3s2 � 80r2s3 � 80r s4 � 32r5

5. 81a4 � 540a3b � 1350a2b2 � 1500ab3 �625b4

6. 7. 8.

9. 10. 11.

12. � 0.06 13. � 0.25 14. � 0.78

15. � 0.06


Lesson 12.1

1. 5.65; 6; 8

2. 4.52; 3.85; 3.1

3. $814.50; $15, $10

4. 69.13; 61; 24

5. 81.5; 80; 75 and 80

6.

28 192

mean: 6.86

5 5 25

6 8 48

7 5 35

8 6 48

9 4 36

495x8y4924x6y6495x4y8

�20xy9�8064x5y511,520x8y2

6ab5 � b6a6 � 6a5b � 15a4b2 � 20a3b3 � 15a2b4 �

137

4399

311

13

79

�37

14

23

23

�(k � 1)(3k � 2)

2

�3k2 � 5k � 2

2

�k(3k � 1)

2� [3(k � 1) � 2]

[3(k � 1) � 2]1 � 4 � 7 � . . . � (3k � 2) �

�(k � 1)(3k � 2)

2

�(k � 1)[3(k � 1) � 1]

2

[3(k � 1) �2]1 � 4 � 7 � . . . � (3k � 2) �

1 � 4 � 7 � . . . � (3k � 2) �k(3k � 1)

2

1 �1(3 � 1 � 1)

2

ANSWERSC

opyright ©by H

olt, Rinehart and W



7.

18 9200

mean: 511.11

8.

16 1292

estimated mean: 81

9.

12 6494

estimated mean: 541

Lesson 12.2

1. Stem Leaf

1 2, 32 0, 43 0, 3, 7, 7, 74 1, 1, 3, 4, 5, 6, 85 2, 3, 4, 5, 7, 86 6, 87 0, 2, 7, 8, 98 0, 9, 99 1, 1, 7

52.5; 37

2. Stem Leaf

10 2, 5, 6, 811 412 4, 513 0, 114 1, 2, 5, 615 3, 5, 916 8, 8, 917 8, 818 4, 519 4

145.5; 168 and 178

3.

ounces8 9 10 11 12 13 14 15 16

5

3

1num

ber

of

can

s

4

2

100 2 200

200 0 0

300 2 600

400 3 1200

500 4 2000

600 2 1200

700 1 700

800 3 2400

900 1 900

ANSWERSC

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60–69 64.5 2 129

70–79 74.5 5 372.5

80–89 84.5 6 507

90–99 94.5 3 283.5

300–399 349.5 3 1048.5

400–499 449.5 2 899

500–599 549.5 3 1648.5

600–699 649.5 1 649.5

700–799 749.5 3 2248.5

8 4

9 0

10 3

11 1

12 1

13 0

14 2

15 4

16 3

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Making and using a stem-and-leaf plot

Recall The leaves in a stem-and-leaf plot are usually single digits.

◆ ExampleFor a class of students, the following data represent the number of hours eachstudent spent during the week watching television.

10 43 5 27 25 14 26 42 18 8 7 10 16 30 35 22

a. Make a stem-and-leaf plot. b. Find the median and mode.

◆ Solutiona. stem leaf

0 5, 7, 81 0, 0, 4, 6, 82 2, 5, 6, 73 0, 54 2, 3

Use your own paper to make a stem-and-leaf plot for each dataset. Then find the median and the mode.

1. 91, 89, 58, 41, 68, 97, 44, 55, 80, 77, 37, 48, 2. 130, 114, 155, 124, 178, 178, 106, 168, 145, 43, 70, 89, 78, 12, 30, 33, 79, 46, 45, 53, 72, 169,168, 125, 131, 105, 102, 108, 146, 153, 41, 20, 37, 37, 57, 66, 54, 24, 91, 13, 52 141, 184, 159, 194, 142, 185

median: mode: median: mode:

Reteaching

12.2 Stem-and-Leaf Plots, Histograms, and Circle Graphs

b. You should list the leaves for each stemfrom least to greatest. Because there are 16data items, the median is the average ofthe eighth and ninth items counting fromthe least value.

The median is 20.

The mode is 10 because this numberoccurs most often.

18 � 222

� 20

◆Skill B Making a histogram

Recall The height of each vertical bar in a histogram represents the frequency of the valueor values marked below it on the horizontal axis.

◆ ExampleThis set of data represents shoe sizes for 9 9.5 10 8.5 9 11a group of boys. 9 9.5 10.5 10 11 8

Make a frequency table and a histogram for this set of data.

◆ Solution

shoe sizes8 8.5 9 9.5 10 10.5 11

3

2

1

num

ber

of

bo

ys

size 8 8.5 9 9.5 10 10.5 11

frequency 1 1 3 2 2 1 2

On your own paper, make a frequency table and histogram foreach data set.

3. ounces in canned goods: 11, 15, 16, 16, 14, 4. lengths of jeans in inches: 33, 34, 32, 34, 30, 15, 10, 8, 8, 12, 15, 16, 8, 15, 8, 14, 10, 10 34, 32, 34, 30, 34, 36, 33, 32, 32, 35, 36, 35,

31, 34, 36, 36

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill C Making a circle graph

Recall Each sector of a circle graph is some fraction or percent of 360°.

◆ ExampleMake a circle graph for the data about shoe sizes shown in the Skill B example.

◆ SolutionYou must find the total number of boys and then find fractions of 360� for thenumber of boys having each shoe size.

There are 12 boys and 1 of them wears size 8. So, of them have shoe size 8.112

Make a circle graph for the following data. Round degreemeasures to the nearest whole degree.

5. Students were asked about their favorite type of music. The responses of 500 students are recorded in the following table.

size frequency sector of circle

8 1

8.5 1

9 3

9.5 2

10 2

10.5 1

11 2 212

� 360� � 60�

112

� 360� � 30�

212

� 360� � 60�

212

� 360� � 60�

312

� 360� � 90�

112

� 360� � 30�

112

� 360� � 30�8.5

8

11

10.510

9.5

9

Classical 43

Rock 115

Rhythm & Blues 78

Country 181

Jazz 57

Other 26

7.

18 9200

mean: 511.11

8.

16 1292

estimated mean: 81

9.

12 6494

estimated mean: 541

Lesson 12.2

1. Stem Leaf

1 2, 32 0, 43 0, 3, 7, 7, 74 1, 1, 3, 4, 5, 6, 85 2, 3, 4, 5, 7, 86 6, 87 0, 2, 7, 8, 98 0, 9, 99 1, 1, 7

52.5; 37

2. Stem Leaf

10 2, 5, 6, 811 412 4, 513 0, 114 1, 2, 5, 615 3, 5, 916 8, 8, 917 8, 818 4, 519 4

145.5; 168 and 178

3.

ounces8 9 10 11 12 13 14 15 16

5

3

1num

ber

of

can

s

4

2

100 2 200

200 0 0

300 2 600

400 3 1200

500 4 2000

600 2 1200

700 1 700

800 3 2400

900 1 900

ANSWERSC

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.


60–69 64.5 2 129

70–79 74.5 5 372.5

80–89 84.5 6 507

90–99 94.5 3 283.5

300–399 349.5 3 1048.5

400–499 449.5 2 899

500–599 549.5 3 1648.5

600–699 649.5 1 649.5

700–799 749.5 3 2248.5

8 4

9 0

10 3

11 1

12 1

13 0

14 2

15 4

16 3

4.

5.

Lesson 12.3

1. 51; 74; 85; 92; 18

2. 81; 10; 33.5; 51; 41

3. 85; 31; 51; 79; 48

4.

5.

6.

Lesson 12.4

1. 23; 7.2

2. 6; 1.2

3. 1.67; 1.29

4. 150; 12.25

5. 39.6; 6.29

6. 5.25; 2.29

Lesson 12.5

1. 23.0% 2. 34.6% 3. 7.7% 4. 1.0%

5. 11.5% 6. 31.6% 7. 77.2% 8. 77.5%

9. 22.5% 10. 17.2% 11. 82.8%

0 5 10 15

2 7.5 9.5 12 15

0 10 20 30 40 50 60 70 80

2 10 33.5 51 83

2.5 3.0 3.5 4.0

2.5 2.8 3.05 3.4 3.8

Classical

Other

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olt, Rinehart and W



30 2

31 1

32 4

33 2

34 6

35 2

36 4

Classical 31°

Rock 83°

Rhythm & Blues 56°

Country 130°

Jazz 41°

Other 19°

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Finding the range, quartiles, and interquartile range

Recall The range is the difference between the maximum and minimum values.The quartiles divide the data into 4 sets where Q2 is the median.The interquartile range is given by .

◆ Example Earthquake intensities are measured on the Richter scale. For the earthquake intensities listed at right, find the following:a. rangeb. the quartilesc. the interquartile range

◆ Solutiona. Arrange the data in order from least to greatest.

6.4 6.5 6.6 6.8 6.8 6.8 6.9 7.0 7.0 7.0 7.1 7.2 7.2 7.3 8.2

The range is 8.2 � 6.4 � 1.8.

b. Find the median of all the data values and label it Q2.Find the median of the lower half of the values and label it Q1.Find the median of the upper half of the values and label it Q3.

6.4 6.5 6.6 6.8 6.8 6.8 6.9 7.0 7.0 7.0 7.1 7.2 7.2 7.3 8.2↑ ↑ ↑Q1 Q2 Q3

The quartiles are

c. The interquartile range is , or . 7.2 � 6.8 � 0.4Q3 � Q1

Q1 � 6.8, Q2 � 7.0, and Q3 � 7.2.

Q3 � Q1

Find the range, quartiles, and interquartile range for each data set.

1. 75, 90, 53, 85, 75, 83, 73, 80, 46, 89, 91, 93, 85, 95, 68, 88, 97, 70, 96, 93, 86

range: Q1: Q2: Q3: interquartile range:

2. 4, 7, 9, 31, 34, 2, 11, 33, 36, 2, 8, 13, 35, 37, 24, 34, 31, 50, 52, 57, 60, 69, 78, 83


3. 82, 65, 11, 31, 50, 95, 33, 88, 79, 10, 15, 45, 51, 66, 53


Reteaching

12.3 Box-and-Whisker Plots

7.2 6.8 8.26.8 6.8 7.06.5 7.2 7.07.3 6.9 7.16.4 7.0 6.6

Make a box-and-whisker plot for each data set.

4. The lengths of several samples of beetles in centimeters are: 2.5, 2.8, 3.1,3.6, 3.4, 3.8, 3.0, 2.8, 3.5, 3.3, 2.6, 3.0, 2.9, 2.7, 3.4, 3.2, 3.7, 2.5, 3.1, 2.9,2.5, 3.1, 3.8, 2.9

5. The ages of the members of the Richards family reunion are: 4, 7, 9, 31,34, 2, 11, 33, 36, 2, 8, 13, 35, 37, 24, 34, 31, 50, 52, 57, 60, 69, 78, 83.

6. Number of students over 4 weeks needing before-school tutoring inalgebra: 7, 12, 10, 8, 7, 12, 15, 10, 10, 7, 2, 9, 12, 14, 6, 10, 12, 9, 8, 8.

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Making a box-and-whisker plot

Recall The box in a box-and-whisker plot always represents the middle half of the values.

◆ Example A class quiz has a maximum score of 10. Make a box-and-whisker plot torepresent the following quiz scores.

4 10 8 9 6 10 8 9 6 9 7 10 9 6 8 9 6 8

◆ SolutionArrange the data in order and find the quartiles.

4 6 6 6 6 7 8 8 8 8 9 9 9 9 9 10 10 10↑ ↑ ↑

Q1 Q2 Q3

Draw a number line that contains the minimum and maximum values.

Make a box that extends from Q1 to Q3 and draw “whiskers” from the box thatextend to the minimum and maximum values.

Draw a vertical line within the box at Q2, the median.

1 2 3 4 5 6 7 8 9 10 11 12

1 2 3 4 5 6 7 8 9 10 11 12

4.

5.

Lesson 12.3

1. 51; 74; 85; 92; 18

2. 81; 10; 33.5; 51; 41

3. 85; 31; 51; 79; 48

4.

5.

6.

Lesson 12.4

1. 23; 7.2

2. 6; 1.2

3. 1.67; 1.29

4. 150; 12.25

5. 39.6; 6.29

6. 5.25; 2.29

Lesson 12.5

1. 23.0% 2. 34.6% 3. 7.7% 4. 1.0%

5. 11.5% 6. 31.6% 7. 77.2% 8. 77.5%

9. 22.5% 10. 17.2% 11. 82.8%

0 5 10 15

2 7.5 9.5 12 15

0 10 20 30 40 50 60 70 80

2 10 33.5 51 83

2.5 3.0 3.5 4.0

2.5 2.8 3.05 3.4 3.8

Classical

Other

Jazz

RockRhythm &

Blues

Country

inches30 31 32 33 34 35 36

5

3

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ber

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4

2

ANSWERSC

opyright ©by H

olt, Rinehart and W



30 2

31 1

32 4

33 2

34 6

35 2

36 4

Classical 31°

Rock 83°

Rhythm & Blues 56°

Country 130°

Jazz 41°

Other 19°

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Finding the range and the mean deviation

Recall The sum 1 � 2 � 3 � 4 can be represented in summation notation as .

gives the mean deviation of a set of data.

◆ ExamplePaul’s test grades are 78, 80, 75, 85, and 82; Marie’s test grades are 60, 83, 73, 100,and 84. Compare the range and mean deviation for these two sets of test grades.

◆ SolutionPaul’s range of test grades is 85 � 75 � 10; Marie’s range is 100 � 60 � 40.The range, which measures the spread of values, is much greater for Marie’s grades.

To find the mean deviation, you must first find the mean.

Paul: Marie:

Then find the absolute value of the difference between the mean, , and each test grade.

Paul: Marie:

Although Paul and Marie have the same mean grade, the mean deviation showsthat the dispersion of Marie’s grades is much greater than Paul’s.

� 10.8� 2.8

15 �

5

i�1�xi � 80� �

15

� 5415 �

5

i�1�xi � 80� �

15

� 14

�5

i�1�xi � 80� � 54�

5

i�1�xi � 80� � 14

x

x �60 � 83 � 73 � 100 � 84

5� 80x �

78 � 80 � 75 � 85 � 825

� 80

1n �

n

i�1�xi � x�

�4

i�1i

Find the range and mean deviation for each data set.

1. 44, 52, 38, 61, 55 2. 6, 8, 5, 3, 9, 6, 6, 5, 7, 5

range: range:

mean deviation: mean deviation:

Reteaching

12.4 Measures of Dispersion

i xi |xi � 80|1 78 2

2 80 0

3 75 5

4 85 5

5 82 2

i xi |xi � 80|1 60 20

2 83 3

3 73 7

4 100 20

5 84 4

Find the variance and standard deviation for each data set.Round answers to the nearest hundredth, if necessary.

3. number of siblings: 3, 3, 1, 5, 4, 2 4. quiz grades: 80, 100, 90, 80, 70, 90, 60, 70

variance: variance:

standard deviation: standard deviation:

5. football scores: 27, 17, 12, 20, 9 6. high daily temperatures: 92, 94, 97, 91

variance: variance:

standard deviation: standard deviation:

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Finding the variance and standard deviation

Recall For a data set of n values, x1, x2, x3, …, xn, and a mean value of ,

the variance is , and the standard deviation is � which is

(the square root of the variance).

◆ ExampleFind the variance and standard deviation for Marie’s test grades of 60, 83, 73,100, and 84.

◆ SolutionFind the mean.

Make a table to help you evaluate the formula.

The variance is

The standard deviation is or approximately 13.22.� � �174.8

� 174.8

�15

� 874

�2 �1n �

n

i�1(xi � x)2

�n

i�1(xi � x)2 � 874

x �60 � 83 � 73 � 100 � 84

5� 80

��2

�2 �1n �

n

i�1(xi � x)2

x

i xi

1 60 �20 400

2 83 3 9

3 73 �7 49

4 100 20 400

5 84 4 16

(xi � x)2xi � x

4.

5.

Lesson 12.3

1. 51; 74; 85; 92; 18

2. 81; 10; 33.5; 51; 41

3. 85; 31; 51; 79; 48

4.

5.

6.

Lesson 12.4

1. 23; 7.2

2. 6; 1.2

3. 1.67; 1.29

4. 150; 12.25

5. 39.6; 6.29

6. 5.25; 2.29

Lesson 12.5

1. 23.0% 2. 34.6% 3. 7.7% 4. 1.0%

5. 11.5% 6. 31.6% 7. 77.2% 8. 77.5%

9. 22.5% 10. 17.2% 11. 82.8%

0 5 10 15

2 7.5 9.5 12 15

0 10 20 30 40 50 60 70 80

2 10 33.5 51 83

2.5 3.0 3.5 4.0

2.5 2.8 3.05 3.4 3.8

Classical

Other

Jazz

RockRhythm &

Blues

Country

inches30 31 32 33 34 35 36

5

3

1num

ber

of

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s 6

4

2

ANSWERSC

opyright ©by H

olt, Rinehart and W



30 2

31 1

32 4

33 2

34 6

35 2

36 4

Classical 31°

Rock 83°

Rhythm & Blues 56°

Country 130°

Jazz 41°

Other 19°

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Finding the probability of exactly r successes in n trials of a binomial experiment

Recall If the probability of success in a binomial experiment is p, then the probability ofexactly r successes in n trials is given by .

◆ ExampleIf your probability of receiving e-mail is 40% each day, what is the probabilitythat you will receive e-mail exactly 5 out of the next 7 days?

◆ Solutionprobability of success: p � 0.4probability of failure: 1 � p � 0.6

Use a calculator.

The probability of receiving e-mail exactly 5 out of the next 7 days is about 7.7%.

� 0.077� 7C5(0.4)5(0.6)2

P � nCrpr(1 � p)n�r

r � 5n � 7

P � nCrpr(1 � p)n�r

Suppose that the probability of rain is 60% for each of the next 5days. Find the probability that it will rain on

1. exactly 2 days. 2. exactly 3 days.

3. all 5 days. 4. none of these days.

Reteaching

12.5 Binomial Distributions

◆Skill B Finding the probability of exactly r successes or s successes in n trials of a binomialexperiment

Recall Since exactly r successes and exactly s successes are mutually exclusive events, theprobability of r successes or s successes is the sum of these two separate probabilities.

◆ ExampleTheresa has calculated that her probability of getting a strike in any one frame of a bowling game is 30%. What is the probability that she will get a strike in exactly 3 or 4 frames out of 10 in her next game?

◆ Solution

The probability that Theresa will get exactly 3 or 4 strikes is approximately 46.7%.

� 0.467� 10C3(0.3)3(0.7)7 � 10C4(0.3)4(0.7)6

� nCs ps(1 � p)n�sP � nCrpr(1 � p)n�r

s � 4r � 3n � 101 � p � 0.7p � 0.3

The probability that you will have to stop for a red light at thelast intersection before you reach school is 80%. In the next 10times you drive up to this intersection, what is the probabilitythat the light will be red

5. exactly 5 or 6 times? 6. exactly 5, 6, or 7 times?

7. exactly 7, 8, or 9 times?

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill C Finding the probability of “at least” r successes in n trials or “at most” r successes inn trials of a binomial experiment

Recall At most means less than or equal; at least means greater than or equal.

◆ ExampleThe owner of a local bookstore found that 64% of all the books she sold werefiction. Find the probability that out of the next 5 books she sellsa. at least 3 are fiction. b. at most 2 are fiction.

◆ Solution

a. “At least 3” means 3, 4, or 5.

The probability that at least 3 of the next 5 books will be fiction isapproximately 74.9%.

b. “At most 2” means 0, 1, or 2.

The probability that at most 2 of the next 5 books will be fiction isapproximately 25.1%.

Notice that the answers to parts a and b have a sum of 1. This is true because “at least 3” and “at most 2” cover all the possibilities.

� 0.251P � 5C0(0.64)0(0.36)5 � 5C1(0.64)1(0.36)4 � 5C2(0.64)2(0.36)3

� 0.749P � 5C3(0.64)3(0.36)2 � 5C4(0.64)4(0.36)1 � 5C5(0.64)5(0.36)0

On a number cube, the probability of getting a 4 on a given roll

is . What is the probability that in 10 rolls you will get

8. at most two 4’s? 9. at least three 4’s?

On a number cube, the probability of getting an even number

on a given roll is . What is the probability that in 10 rolls you

will get

10. at least seven even numbers? 11. at most six even numbers?

12

16

4.

5.

Lesson 12.3

1. 51; 74; 85; 92; 18

2. 81; 10; 33.5; 51; 41

3. 85; 31; 51; 79; 48

4.

5.

6.

Lesson 12.4

1. 23; 7.2

2. 6; 1.2

3. 1.67; 1.29

4. 150; 12.25

5. 39.6; 6.29

6. 5.25; 2.29

Lesson 12.5

1. 23.0% 2. 34.6% 3. 7.7% 4. 1.0%

5. 11.5% 6. 31.6% 7. 77.2% 8. 77.5%

9. 22.5% 10. 17.2% 11. 82.8%

0 5 10 15

2 7.5 9.5 12 15

0 10 20 30 40 50 60 70 80

2 10 33.5 51 83

2.5 3.0 3.5 4.0

2.5 2.8 3.05 3.4 3.8

Classical

Other

Jazz

RockRhythm &

Blues

Country

inches30 31 32 33 34 35 36

5

3

1num

ber

of

jean

s 6

4

2

ANSWERSC

opyright ©by H

olt, Rinehart and W



30 2

31 1

32 4

33 2

34 6

35 2

36 4

Classical 31°

Rock 83°

Rhythm & Blues 56°

Country 130°

Jazz 41°

Other 19°

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yrig

ht ©

by H

olt,

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Finding the probability of an event using a normal distribution curve

Recall For data that is normally distributedwith mean and standard deviation�, the normal distribution curve atright gives the approximate percent ofdata that will fall within the unitsmarked on the horizontal axis.

◆ ExampleThe scores on a college entrance exam for graduates of a particular high school are normally distributed with a mean of 580 and a standard deviation of 120. Find the probability that a randomly selected student from this high school scoreda. between 460 and 700. b. over 820.

◆ Solutiona. 460 � 580 � 120, so 460 .

According to the curve above, the percent of scores between and is 34%.700 � 580 � 120, so 700 .The percent of scores between and is 34%.

34% � 34% � 68%; the probability of scoring between 460 and 700 is about 68%.

b. 820 � 580 � 2(120), so 820 .

The percent of scores above is 2%.Thus, the probability of scoring over 820 is approximately 2%.

x � 2�

� x � 2�

x � �x� x � �

xx � �

� x � �

x

Use the normal distribution curve shown above to complete thefollowing exercises.

1. The mean for a set of scores is 100 and the standard deviation is 7.5. Findthe probability that a randomly selected score is

a. between 85 and 100. c. less than 100.

b. between 85 and 115.

2. The average age of the employees in a large company is 38 with a standarddeviation of 8. If an employee is selected at random, what is theprobability that the employee’s age is

a. between 38 and 54? c. more than 46?

b. less than 22?

Reteaching

12.6 Normal Distributions

x – 3� x – 2� x – � x x + � x + 2� x + 3�

2%14%

34% 34%14%

0.2

2%

Use a graphics calculator to complete the following exercises.

3. An auto parts manufacturer finds that the useful life of a spark plug in acar is approximately normally distributed with a mean of 9 months and astandard deviation of 2.3 months. What is the probability that arandomly selected spark plug will last

a. more than 1 year? b. less than 7 months?

4. Mortgage statistics collected by a bank indicate that the number of years the average new homeowner will occupy a house before moving or selling is normally distributed with a mean of 6.3 years and a standard deviation of 2.31 years. What is the probability that a randomly selected homeowner

a. will sell the house or move within 3 years?

b. will keep the house from 6 to 8 years?

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Finding the probability of an event using z-scores

Recall z-scores tell how many standard deviations a data value is above or below the mean.

, where a is a given data value.

You will need a graphics calculator for the exercises on this page.

◆ ExampleThe mean shelf life of a particular dairy product is approximately 11 days with astandard deviation of 3 days. If one of these dairy products is randomly selected,find the probability that it will lasta. between 10 and 15 days. b. less than 7 days.

◆ Solutiona. When a � 10, or approximately �0.33.

When a � 15, or approximately 1.33.

Use a calculator to graph

in the window

indicated at right. Find the shadedarea by using the feature(with a lower limit of �0.33 and aupper limit of 1.33) or a similarfeature on the calculator you areusing.

Since the shaded area in this example is approximately 0.54, the probability ofthis dairy product lasting between 10 and 15 days is approximately 54%.

b. When a � 7, , or approximately �1.33.

Use your calculator to find the area from x � �4 to x � �1.33.

The probability of the dairy product lasting less than 7 days is about 0.09, or 9%.

z �7 � 11

3

f(x)dx

f(x) �1

�2πe�x2

2

z �a � x

��

15 � 113

z �a � x

��

10 � 113

z �a � x

�

–4 4

–0.5

1

Lesson 12.6

1. a. 48%b. 96%c. 50%

2. a. 48%b. 2%c. 16%

3. a. 9.7%b. 19.2%

4. a. 7.6%b. 32.2%


Lesson 13.1

1. 2. 3. 4. 5.

6. 7. 8. 9. 10.

11. 53� 12. 35� 13.

14.

15.

16.

Lesson 13.2

1. �330�, 390�

2. �225�, 495�, 855�

3. 30� 4. 68� 5. 30� 6. 55� 7. 30�

8. 60� 9. 40� 10. 40�

11.

12. ,

13.

14.

Lesson 13.3

1.

2.

3.

4. 5. 6. 1 7.

8. 9. 10. �1 11.

12.

13.

14.

Lesson 13.4

1.

2.

3.

4.

5.

6.

7. 45� 8. 270� 9. 150� 10. 300�

11. �540� 12. �330�

13. 10π3

centimeters

�5π3

radians

�π2

radians

7π6

radians

5π4

radians

π4

radians

3π2

radians

(6, �6�3)

(5�3, 5)

(�3�2, 3�2)

��22

12

��32

��32

�32

�32

x � 8�3, y � 8

x � y �8

�2, or 4�2

x � 7�3, r � 14

sec � ��29

5, csc � � �

�292

sin � � �2

�29, cos � �

5�29

, cot � � �52

,

sec � � ��2, csc � � �2

sin � �1

�2, tan � � �1, cot � � �1,

cot � � �5, sec � ��26

5, csc � � ��26

sin � � �1

�26, cos � �

5�26

, tan � � �15

cot � �23

, sec � � ��13

2, csc � � �

�133

sin � � �3

�13, cos � � �

2�13

, tan � �32

,

m∠X � 31�, m∠Y � 59�, XY � 21.1

m∠D � 25�, DF � 21.4, DE � 23.7

m∠A � 49�, AC � 10.5, BC � 12.1

60�

�587

73

3�58

7�58

73

�587

�583

37

7�58

3�58

ANSWERSC

opyr

ight

©by

Hol

t, R

ineh

art a

nd W

inst

on. A

ll rig

hts

rese

rved

.


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by H

olt,

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serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Finding the trigonometric functions of an acute angle

Recall The hypotenuse is the longest side in a right triangle and is opposite the right angle.

◆ ExampleRefer to the triangle shown at right and give values for sin �, cos �, tan �, cot �, sec �, and csc �.

◆ SolutionThe hypotenuse (hyp.) has a length of .The leg opposite (opp.) � has a length of 4.The leg adjacent (adj.) to � has a length of 5.

cot � �adj.opp.

�54

tan � �opp.adj.

�45

sec � �hyp.adj.

��41

5

cos � �adj.hyp.

�5

�41csc � �

hyp.opp.

��41

4sin � �

opp.hyp.

�4

�41

�41

Refer to the triangle at right to find each value. Give exactanswers.

1. sin � 2. cos �

3. tan � 4. csc �

5. sec � 6. cot � 7. sin φ

8. cos φ 9. tan φ 10. csc φ

Reteaching

13.1 Right-Triangle Trigonometry

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ed.

54

√41

�

3√58

�

7

φ

◆Skill B Using inverse trigonometric functions to find the measure of an acute angle

Recall The statements tan � � and � � tan�1 are equivalent.

◆ ExampleIn the triangle shown at right, find the measure of � to the nearest whole degree.

◆ SolutionSince the hypotenuse and the side adjacent to � are given, use the cosine function.

cos � �

Use your calculator in degreemode.� � cos�1(57) � 44°

adj.hyp.

�57

(57)5

7

5�

7

Solve each triangle. Round each angle measure to the nearestdegree and each side length to the nearest tenth.

14. 15. 16.

11

Y

ZX

18

10

FD

65°

E

16 BA

C

41°

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill C Solving a right triangle

Recall Solving a triangle means to use given measures to find the unknown measures of theother sides and angles of the triangle.

◆ Example 1Solve the triangle shown at right.

◆ SolutionUsing the hypotenuse and side opposite ,

Round to the nearest whole degree.

82 � (AC)2 � 122

◆ Example 2Solve the triangle shown at right.

◆ SolutionUsing the side adjacent to ,

, where LM is the hypotenuse.

(LN)2 � (MN)2 � (LM)2

� 16.4 m∠L � 20�LN � �17.52 � 62

m∠L � 70� � 90�

m∠L � m∠M � 90�LM �

6cos 70°

� 17.5

cos 70� �6

LM

∠M

AC � 8.9m∠B � 48�

AC � �122 � 8242� � m∠B � 90�

m∠A � m∠B � 90�

� 42�m∠A � sin�1( 812)

sin ∠A �812

∠A

Copyright ©

by Holt, R


ll rights reserved.

Find � to the nearest degree in each triangle.

11. 12. 13.

10�20

13

�

9

10�

8

812

A C

B

6

N

M

L

70°

Lesson 12.6

1. a. 48%b. 96%c. 50%

2. a. 48%b. 2%c. 16%

3. a. 9.7%b. 19.2%

4. a. 7.6%b. 32.2%


Lesson 13.1

1. 2. 3. 4. 5.

6. 7. 8. 9. 10.

11. 53� 12. 35� 13.

14.

15.

16.

Lesson 13.2

1. �330�, 390�

2. �225�, 495�, 855�

3. 30� 4. 68� 5. 30� 6. 55� 7. 30�

8. 60� 9. 40� 10. 40�

11.

12. ,

13.

14.

Lesson 13.3

1.

2.

3.

4. 5. 6. 1 7.

8. 9. 10. �1 11.

12.

13.

14.

Lesson 13.4

1.

2.

3.

4.

5.

6.

7. 45� 8. 270� 9. 150� 10. 300�

11. �540� 12. �330�

13. 10π3

centimeters

�5π3

radians

�π2

radians

7π6

radians

5π4

radians

π4

radians

3π2

radians

(6, �6�3)

(5�3, 5)

(�3�2, 3�2)

��22

12

��32

��32

�32

�32

x � 8�3, y � 8

x � y �8

�2, or 4�2

x � 7�3, r � 14

sec � ��29

5, csc � � �

�292

sin � � �2

�29, cos � �

5�29

, cot � � �52

,

sec � � ��2, csc � � �2

sin � �1

�2, tan � � �1, cot � � �1,

cot � � �5, sec � ��26

5, csc � � ��26

sin � � �1

�26, cos � �

5�26

, tan � � �15

cot � �23

, sec � � ��13

2, csc � � �

�133

sin � � �3

�13, cos � � �

2�13

, tan � �32

,

m∠X � 31�, m∠Y � 59�, XY � 21.1

m∠D � 25�, DF � 21.4, DE � 23.7

m∠A � 49�, AC � 10.5, BC � 12.1

60�

�587

73

3�58

7�58

73

�587

�583

37

7�58

3�58

ANSWERSC

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©by

Hol

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art a

nd W

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on. A

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.


◆Skill B Finding the reference angle

Recall A reference angle is the positive acute angle between the terminal side of a given angle and the x-axis.

◆ Example Find the reference angle for each given angle.a. 115° b. 210° c. �135°

◆ Solutiona. b. c.

180° � 115° � 65°. 210° � 180° � 30°. �135° � 360° � 225°

The reference angle is 65°. The reference angle is 30°. The reference angle is 45°.

225� � 180� � 45�

115°

x

y

ref.

x

y

60°

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Finding coterminal angles

Recall An angle is in standard position if its vertex is at the origin and its initial side isalong the positive x-axis. Angles in standard position are coterminal if they havethe same terminal side.

◆ Example Which of 120°, 420°, and �300°are coterminal with 60°?

◆ SolutionThe measures of coterminal angles differ by a multiple of 360°.120° � 60° � 60° not a multiple of 360°420° � 60° � 360° is a multiple of 360°�300° � 60° � �360° is a multiple of 360°The angles coterminal with 60° are 420° and �300°.

For � in standard position, select the listed angles that arecoterminal with �.

1. � � 30°; �690°, �390°, �330°, �150°, 150°, 390°, 730°

2. � � 135°; �360°, �225°, �45°, 45°, 315°, 495°, 855°

Reteaching

13.2 Angles of Rotation

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Rin

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t and

Win

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. All

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s re

serv

ed.

210°

x

y

ref.

225°

x

y

ref.

–135°

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill C Finding the values of all 6 trigonometric functions of an angle when given a pointon the terminal side of the angle

Recall If P(x, y) is a point on the terminal side of � in standard position, then

and

◆ Example 1Find the exact values of the 6trigonometric functions of � given that(�3, 4) is on the terminal side of �.

◆ SolutionSince .

Recall The distance r is always positive.

◆ Example 2The terminal side of � lies in Quadrant III,

and . Find the values of the

other 5 trigonometric functions of �.◆ Solution

Therefore, y � �1, r � 2, and .

cot � � ��3sec � �2

�3csc � � �2tan � � �

1�3

cos � ��32

sin � � �12

x � �22 � (�1)2 � �3

sin � �yr, where r � 0

sin � � �12

cot � � �34

sec � � �53

csc � �54

tan � � �43

cos � � �35

sin � �45

r � �(�3)2 � 42, r � 5

cot � �xy

sec � �rx

csc � �ry

tan � �yx

cos � �xr

sin � �yr

r � �x2 � y2

Copyright ©

by Holt, R


ll rights reserved.

Find the reference angle for each given angle.

3. 150° 4. 248° 5. 330° 6. 55°

7. �150° 8. �240° 9. 400° 10. 500°

x

y

�

ry

x

P(–3, 4)

x

y

�

–12

√3

Find the exact values of the 6 trigonometric functions of � giveneach point on the terminal side of � in standard position.

11. P(�2, �3) 12. P(5, �1)

Find the exact values of the other 5 trigonometric functions of �given the following information.

13. 14. IV; tan � � �25

II; cos � � �1

�2

Lesson 12.6

1. a. 48%b. 96%c. 50%

2. a. 48%b. 2%c. 16%

3. a. 9.7%b. 19.2%

4. a. 7.6%b. 32.2%


Lesson 13.1

1. 2. 3. 4. 5.

6. 7. 8. 9. 10.

11. 53� 12. 35� 13.

14.

15.

16.

Lesson 13.2

1. �330�, 390�

2. �225�, 495�, 855�

3. 30� 4. 68� 5. 30� 6. 55� 7. 30�

8. 60� 9. 40� 10. 40�

11.

12. ,

13.

14.

Lesson 13.3

1.

2.

3.

4. 5. 6. 1 7.

8. 9. 10. �1 11.

12.

13.

14.

Lesson 13.4

1.

2.

3.

4.

5.

6.

7. 45� 8. 270� 9. 150� 10. 300�

11. �540� 12. �330�

13. 10π3

centimeters

�5π3

radians

�π2

radians

7π6

radians

5π4

radians

π4

radians

3π2

radians

(6, �6�3)

(5�3, 5)

(�3�2, 3�2)

��22

12

��32

��32

�32

�32

x � 8�3, y � 8

x � y �8

�2, or 4�2

x � 7�3, r � 14

sec � ��29

5, csc � � �

�292

sin � � �2

�29, cos � �

5�29

, cot � � �52

,

sec � � ��2, csc � � �2

sin � �1

�2, tan � � �1, cot � � �1,

cot � � �5, sec � ��26

5, csc � � ��26

sin � � �1

�26, cos � �

5�26

, tan � � �15

cot � �23

, sec � � ��13

2, csc � � �

�133

sin � � �3

�13, cos � � �

2�13

, tan � �32

,

m∠X � 31�, m∠Y � 59�, XY � 21.1

m∠D � 25�, DF � 21.4, DE � 23.7

m∠A � 49�, AC � 10.5, BC � 12.1

60�

�587

73

3�58

7�58

73

�587

�583

37

7�58

3�58

ANSWERSC

opyr

ight

©by

Hol

t, R

ineh

art a

nd W

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on. A

ll rig

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rese

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.


◆Skill B Finding exact values of the trigonometric functions for an angle whose referenceangle is 30°, 45°, or 60°

Recall

30° 2

60°1

√31

1

45°

45°

√2

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Solving 45-45-90 and 30-60-90 triangles

Recall

◆ Example Find the lengths of the other 2 sides in each right triangle.a. b.

◆ Solutiona. a � 5 b.

x � a � 5

r � 2a �12�3

(or 4�3)r � a�2 � 5�2

y � a �6

�3 (or 2�3)

a�3 � 6 → a �6

�3 (or 2�3)

30°6

yr 60°

r

x45°

45°5

a

a

45°

45°

a√2

30°

60°

a√3

a2a

Find the missing side lengths in each right triangle.

1. 2. 3.30°

16y

x

60°y

x45°

45° 830°

7r

x

60°

Reteaching

13.3 Trigonometric Functions of Any Angle

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ed.

� 30° 45° 60°

sin �

cos �

tan � 1 �31�3

12

1�2

�32

�32

1�2

12

Find each trigonometric value. Give exact answers.

4. sin 120° 5. cos 330° 6. tan 225° 7. cos 150°

8. sin 240° 9. sin 150° 10. tan 315° 11. cos 225°

NAME _________________________________________________ CLASS _______________ DATE ______________


A reference angle is the positive acute anglebetween the terminal side of a given angleand the x-axis.

One mnemonic for remembering whichfunctions are positive in each quadrant is“All students take calculus.”

◆ Example Find each exact value.a. sin 315° b. cos 240° c. tan 210°

◆ Solutiona. 315° is in Quadrant IV, b. 240° is in Quadrant III, c. 210° is in Quadrant III,

where sine is negative. where cosine is where tangent is The reference angle is negative. The reference positive. The reference45°. angle is 60°. angle is 30°.sin 315° � �sin 45° cos 240° � �cos 60° tan 210° � tan 30°

�1

�3� �

12

� �1

�2

Copyright ©

by Holt, R


ll rights reserved.

x

y

sin & cscare “+”

all 6are “+”

tan & cotare “+”

cos & secare “+”

II

III

I

IV

◆Skill C Finding the coordinates of a point P on a circle

Recall If P(x, y) lies at the intersection of the terminal side of � in standard position and acircle centered at the origin with radius r, then P(x, y) � P(r cos �, r sin �).

◆ Example Find the coordinates of point P shown inthe figure at right.

◆ Solutionand

and

The coordinates of point P are .(�2�3, � 2)

r sin � � 4(�12)r cos � � 4(�

�32 )

sin 210� � �sin 30� � �12

cos 210� � �cos 30� � ��32

x

y

210°

r = 4

P(x, y)

ref.

Point P is located at the intersection of a circle centered at the originwith a radius of r and the terminal side of angle � in standard position.Find the exact coordinates of point P.

12. � � 135°, r � 6 13. � � 30°, r � 10 14. � � 300°, r � 12

Lesson 12.6

1. a. 48%b. 96%c. 50%

2. a. 48%b. 2%c. 16%

3. a. 9.7%b. 19.2%

4. a. 7.6%b. 32.2%


Lesson 13.1

1. 2. 3. 4. 5.

6. 7. 8. 9. 10.

11. 53� 12. 35� 13.

14.

15.

16.

Lesson 13.2

1. �330�, 390�

2. �225�, 495�, 855�

3. 30� 4. 68� 5. 30� 6. 55� 7. 30�

8. 60� 9. 40� 10. 40�

11.

12. ,

13.

14.

Lesson 13.3

1.

2.

3.

4. 5. 6. 1 7.

8. 9. 10. �1 11.

12.

13.

14.

Lesson 13.4

1.

2.

3.

4.

5.

6.

7. 45� 8. 270� 9. 150� 10. 300�

11. �540� 12. �330�

13. 10π3

centimeters

�5π3

radians

�π2

radians

7π6

radians

5π4

radians

π4

radians

3π2

radians

(6, �6�3)

(5�3, 5)

(�3�2, 3�2)

��22

12

��32

��32

�32

�32

x � 8�3, y � 8

x � y �8

�2, or 4�2

x � 7�3, r � 14

sec � ��29

5, csc � � �

�292

sin � � �2

�29, cos � �

5�29

, cot � � �52

,

sec � � ��2, csc � � �2

sin � �1

�2, tan � � �1, cot � � �1,

cot � � �5, sec � ��26

5, csc � � ��26

sin � � �1

�26, cos � �

5�26

, tan � � �15

cot � �23

, sec � � ��13

2, csc � � �

�133

sin � � �3

�13, cos � � �

2�13

, tan � �32

,

m∠X � 31�, m∠Y � 59�, XY � 21.1

m∠D � 25�, DF � 21.4, DE � 23.7

m∠A � 49�, AC � 10.5, BC � 12.1

60�

�587

73

3�58

7�58

73

�587

�583

37

7�58

3�58

ANSWERSC

opyr

ight

©by

Hol

t, R

ineh

art a

nd W

inst

on. A

ll rig

hts

rese

rved

.


NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Converting between degree and radian measure

Recall π radians � 180°, so .

◆ Example 1Convert the following to radian measure.a. 50° b. �150°

◆ Solution

a. b.

180x � 50π 180x � �150π

radians radians

◆ Example 2Convert the following to degree measure.

a. radians b. radians

◆ Solution

a. b.

x � 120° x � �315°

�74

π radians � �315�23

π radians � 120�

πx � �74

π(180�)πx �23

π(180�)

x180�

��

74 ππ

x180�

�

23 ππ

�74

π23

π

�150� � �56

π50� �5

18π

� �56

π�518

π

x � �150π180

x �50π180

�150�180�

�xπ

50�180�

�xπ

degree measure180

�radian measure

π

Convert the following degree measures to radian measures. Give exact answers.

1. 270° 2. 45° 3. 225°

4. 210° 5. �90° 6. �300°

Convert each of the following radian measures to degreemeasures.

7. 8. 9.

10. 11. �3π 12. �11π

65π3

5π6

3π2

π4

Reteaching

13.4 Radian Measure and Arc Length

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olt,

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. All

right

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Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Finding arc length

Recall If � is the measure in radians of a central angle in a circle with radius r, then thelength, s, of the arc intercepted by � is given by s � r�.

◆ Example Find the measure of arc s if

r � 8 centimeters and � .

◆ Solution

The arc length is 6π, about 18.85 centimeters.

s � 8 �3π4

� 6π

�3π4

3π4

8 cm

s

Find the length of the arc intercepted by angle � in a circle withthe given radius.

13. radians, r � 5 cm 14. radians, r � 10 cm 15. radians, r � 12 cm� �5π6

� �5π3

� �2π3

◆Skill C Finding exact values for the trigonometric functions of an angle measured in radians

◆ Example Give the exact value of each expression where the angle measures are in radians.

a. b. c.

◆ SolutionUse reference angles.

a. b. c.

sin(�270�) � sin 90� � 1tan 240� � tan 60� � �3cos 150� � �cos 30� � ��32

�3π2

radians � �270°4π3

radians � 240�5π6

radians � 150�

x

y

3π2

–

x

y

4π3

x

y

5π6

sin(�3π2 )tan 4π

3cos 5π

6

Evaluate each expression. Give exact answers.

16. 17. 18.

19. 20. 21. sin πtan(�π4)cos(�

7π6 )

tan 5π6

cos 2π3

sin 3π4

Lesson 12.6

1. a. 48%b. 96%c. 50%

2. a. 48%b. 2%c. 16%

3. a. 9.7%b. 19.2%

4. a. 7.6%b. 32.2%


Lesson 13.1

1. 2. 3. 4. 5.

6. 7. 8. 9. 10.

11. 53� 12. 35� 13.

14.

15.

16.

Lesson 13.2

1. �330�, 390�

2. �225�, 495�, 855�

3. 30� 4. 68� 5. 30� 6. 55� 7. 30�

8. 60� 9. 40� 10. 40�

11.

12. ,

13.

14.

Lesson 13.3

1.

2.

3.

4. 5. 6. 1 7.

8. 9. 10. �1 11.

12.

13.

14.

Lesson 13.4

1.

2.

3.

4.

5.

6.

7. 45� 8. 270� 9. 150� 10. 300�

11. �540� 12. �330�

13. 10π3

centimeters

�5π3

radians

�π2

radians

7π6

radians

5π4

radians

π4

radians

3π2

radians

(6, �6�3)

(5�3, 5)

(�3�2, 3�2)

��22

12

��32

��32

�32

�32

x � 8�3, y � 8

x � y �8

�2, or 4�2

x � 7�3, r � 14

sec � ��29

5, csc � � �

�292

sin � � �2

�29, cos � �

5�29

, cot � � �52

,

sec � � ��2, csc � � �2

sin � �1

�2, tan � � �1, cot � � �1,

cot � � �5, sec � ��26

5, csc � � ��26

sin � � �1

�26, cos � �

5�26

, tan � � �15

cot � �23

, sec � � ��13

2, csc � � �

�133

sin � � �3

�13, cos � � �

2�13

, tan � �32

,

m∠X � 31�, m∠Y � 59�, XY � 21.1

m∠D � 25�, DF � 21.4, DE � 23.7

m∠A � 49�, AC � 10.5, BC � 12.1

60�

�587

73

3�58

7�58

73

�587

�583

37

7�58

3�58

ANSWERSC

opyr

ight

©by

Hol

t, R

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art a

nd W

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.


14.

15. 10π centimeters

16. 17. 18. 19.

20. �1 21. 0

Lesson 13.5

1.

2.

3.

4.

5.

6.

7.

8.

Lesson 13.6

1. 30�, or

2. 135�, or

3. 60�, or

4. �90�, or

5. 30�, or

6. �45�, or �π4

π6

�π2

π3

3π4

π6

y

x–1

1

y = cos x

y = cos x + 1

π2

π2

3π2

– π 2π

y

x–1

1

y = cos x

y = cos(x – )π2

π2

π2

3π2

– π 2π

y

�90° 180° 270° 360°–1

1

y = tan �

y = tan �12

y

�90° 180° 270° 360°–1

1

y = tan �y = –tan �

y

�90° 180° 270° 360°–1

1

y = cos �y = cos 3�

y

�90° 180° 270° 360°–1

1

y = cos �y = cos �1

2

��32

�1

�3�

12

��22

50π3

centimeters

ANSWERSC

opyright ©by H

olt, Rinehart and W



y

x–1

1

y = tan x

y = tan x – 1

π2

π2

3π2

– π 2π

y

x–1

1

y = tan x

y = –tan(x + )π4

π2

π2

3π2

– π 2π

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Text

Recall Text

◆ ExampleSBHCT

Sketch each pair of functions on the same set of axes. Use 0° � � � 360°.

1. 2.

3. y � tan �, y � �tan � 4. y � tan �, y � tan 12

�

y � cos �, y � cos 3�y � cos �, y �12

cos �

Reteaching

13.5 Graphing Trigonometric Functions

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by H

olt,

Rin

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t and

Win

ston

. All

right

s re

serv

ed.

◆Skill A Graphing functions of the form y � a sin b�, y � a cos b�, and y � a tan b�

Recall The sine and cosine are periodic functions with a period of 360° or 2π radians.The tangent function has a period of 180° or π radians.

◆ Example 1Graph y � sin �, y � 2 sin �, andy � sin 2� on the same set of axes.

◆ SolutionThe graph of y � a sin b� has anamplitude (height above x-axis) of

and period of .

Check your results with a graphics calculator.

◆ Example 2Graph y � sin � and y � �sin � onthe same set of axes.

◆ SolutionNotice that the graph of y � �sin �is the reflection of y � sin � acrossthe (horizontal) �-axis.

360�b

�a�

function amplitude period

y � sin � 1 360°

y � 2 sin � 2 360°

y � sin 2� 1 180°

y

�90° 180° 270° 360°–1

1

y = sin �y = 2 sin �

y = sin 2�

y

�90° 180° 270° 360°–1

1

y = sin �y = –sin �

Sketch each pair of functions on the same set of axes. Use .

5. 6.

7. 8.

y

x–1

1

π 2ππ2

π2

3π2

–

y

x–1

1

π 2ππ2

π2

3π2

–

y � tan x, y � tan x � 1y � tan x, y � �tan(x �π4)

y

x–1

1

π 2ππ2

π2

3π2

–

y

x–1

1

π 2ππ2

π2

3π2

–

y � cos x, y � cos x � 1y � cos x, y � cos(x �π2)

�π2 � x � 2π

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Graphing functions of the form , , and

Recall The graph of is a phase shift (horizontal translation) of the graph ofto the right c units.

The graph of is a vertical shift of the graph of up d units.

◆ Example 1Graph and on the same set of axes.

◆ Solution

phase shift of units to the left

◆ Example 2Graph and on the same set of axes.

◆ Solution

vertical shift of 2 units down

y � sin x � 2y � sin x

π4

y � sin(x �π4)y � sin x

y � sin xy � sin x � dy � sin x

y � sin(x � c)

y � tan(x � c) � dy � cos(x � c) � dy � sin(x � c) � d

Copyright ©

by Holt, R


ll rights reserved.

y

x–1

1

π 2ππ2

π2

3π2

–

y

x–1

1

π 2ππ2

π2

3π2

–

y = sin x

y = sin x � �π4

y = sin xy = sin x � 2

14.


16. 17. 18. 19.

20. �1 21. 0

Lesson 13.5

1.

2.

3.

4.

5.

6.

7.

8.

Lesson 13.6

1. 30�, or

2. 135�, or

3. 60�, or

4. �90�, or

5. 30�, or

6. �45�, or �π4

π6

�π2

π3

3π4

π6

y

x–1

1

y = cos x

y = cos x + 1

π2

π2

3π2

– π 2π

y

x–1

1

y = cos x

y = cos(x – )π2

π2

π2

3π2

– π 2π

y

�90° 180° 270° 360°–1

1

y = tan �

y = tan �12

y

�90° 180° 270° 360°–1

1


y

�90° 180° 270° 360°–1

1


y

�90° 180° 270° 360°–1

1


2

��32

�1

�3�

12

��22

50π3

centimeters

ANSWERSC

opyright ©by H

olt, Rinehart and W



y

x–1

1

y = tan x

y = tan x – 1

π2

π2

3π2

– π 2π

y

x–1

1

y = tan x

y = –tan(x + )π4

π2

π2

3π2

– π 2π

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Evaluating inverse trigonometric relations and functions

Recall The domain and range of a function become the range and domain respectively,of the inverse.

◆ Example 1Find each value. Give answers in degrees and radians.

a. b. c.

◆ Solutiona. Since , then or radians.

Notice that although other angles have a sine of , you must choose an angle

that is between �90� and 90� in order to have a value in the appropriate range.

b. Since , then or radians.

c. Since , then or radians.

◆ Example 2Evaluate each expression.

a. b.

◆ Solutiona. Begin inside the parentheses. b.

Therefore,

So, or radiansπ4

� 45�sin(Cos�1(12)) � sin 60� �

�32

Tan�1(sin 90�) � Tan�1(1)Cos�1(12) � 60�

sin 90� � 1

Tan�1(sin 90�)sin(Cos�1(12))

�π4

Tan�1(�1) � �45�tan(�45�) � �1

2π3

Cos�1(�12) � 120�cos 120� � �

12

�32

π3

Sin�1(�32 ) � 60�sin 60� �

�32

Tan�1(�1)Cos�1(�12)Sin�1(�3

2 )

Find each value. Give answers in degrees and in radians. (It may behelpful to review what you learned about 30�-, 45�-, and 60�-angles.)

1. 2. 3.

4. 5. 6.

Evaluate each composite trigonometric expression.

7. 8. 9.

10. 11. 12. Sin�1(sin 90�)Tan�1(sin 0�)Sin�1(cos 0�)

sin(Tan�1(�1

�3))cos(Sin�1( 1�2))tan(Cos�1(�3

2 ))

Tan�1(�1)Cos�1(�32 )Sin�1(�1)

Tan�1(�3 )Cos�1(�1

�2)Sin�1(12)

Reteaching

13.6 Inverses of Trigonometric Functions

Cop

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by H

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Rin

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Win

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. All

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Find the measure of each angle to the nearest whole degree.

13. Find the measure of the smallest angle in a right triangle with sides of 3, 4, and 5 centimeters.

14. What is the angle between the bottom of theladder and the ground as shown at right?

15. Find the angle at the peak of the roof as shownat right.

16. The hypotenuse of a right triangle is 3 times as long as the shorter leg.Find the measure of the angle between the shorter leg and thehypotenuse.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆ Skill B Applying inverse trigonometric functions

Recall

◆ ExampleAt a certain time of the day, the 5 meter flagpoleshown at right casts a shadow that is 3 meters long.What is the angle of elevation of the sun at this time?

◆ SolutionSince 3 meters is the length of the side adjacent to �and 5 meters is the length of the side opposite �, usethe tangent function.

This last equation states that � is the angle that has a tangent of .

Use calculator in degreemode.� � 59�

53

� � tan�1(53)

tan � �53

tan � �opp.adj.

cos � �adj.hyp.

sin � �opp.hyp.

Copyright ©

by Holt, R


ll rights reserved.

5 m

3 m

�

12 ft

3 ft

�

�

45 ft

15 ft

14.


16. 17. 18. 19.

20. �1 21. 0

Lesson 13.5

1.

2.

3.

4.

5.

6.

7.

8.

Lesson 13.6

1. 30�, or

2. 135�, or

3. 60�, or

4. �90�, or

5. 30�, or

6. �45�, or �π4

π6

�π2

π3

3π4

π6

y

x–1

1

y = cos x

y = cos x + 1

π2

π2

3π2

– π 2π

y

x–1

1

y = cos x

y = cos(x – )π2

π2

π2

3π2

– π 2π

y

�90° 180° 270° 360°–1

1

y = tan �

y = tan �12

y

�90° 180° 270° 360°–1

1


y

�90° 180° 270° 360°–1

1


y

�90° 180° 270° 360°–1

1


2

��32

�1

�3�

12

��22

50π3

centimeters

ANSWERSC

opyright ©by H

olt, Rinehart and W



y

x–1

1

y = tan x

y = tan x – 1

π2

π2

3π2

– π 2π

y

x–1

1

y = tan x

y = –tan(x + )π4

π2

π2

3π2

– π 2π

7. 8. 9. 10. 90� 11. 0�

12. 90� 13. 37� 14. 76� 15. 113�

16. 71�


Lesson 14.1

1. 5.4 square centimeters

2. 92.2 square kilometers

3. 14.7 square meters

4. 30 square feet

5.

6.

7.

8. 2 triangles

9. no triangles

10. 2 triangles

11. 1 triangle

Lesson 14.2

1.

2.

3.

4.

Answers may vary due to rounding.

5.

6.

7.

8.

Lesson 14.3

1.

2.

3.

4.

5. sin x cot x � sin x � cos x

6. sin x sec x cot x � sin x � 1

7. cos2 x � sin2 x � 1 � sin2 x � sin2 x �1 � 2 sin2 x

8. (1 � sin x)(1 � sin x) � 1 � sin2 x � cos2 x

1cos x

�cos xsin x

cos xsin x

2π–2π

4

–4

2π–2π

4

–4

2π–2π

4

–4

2π–2π

4

–4

A � 143�, B � 20�, C � 18�

A � 58�, B � 73�, C � 48�

A � 56�, B � 90�, C � 34�

A � 38�, B � 82�, C � 60�

c � 14.0, A � 3�, B � 5�

a � 122.2, B � 38�, C � 28�

b � 49.1, A � 110�, C � 48�

a � 46.7, B � 42�, C � 87�

B � 44�, b � 183.6, c � 249.9

A � 78�, a � 9.8, c � 6.2

C � 97�, b � 5.2, c � 10.1

�12

1�2

1�3

ANSWERSC

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by H

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Rin

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Win

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. All

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serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Finding the area of a triangle

Recall Given SAS, the area is or or .

◆ ExampleFind the area of the triangle shown at right.

◆ SolutionYou must find the measure of the angle between the two given sides.

The sum equals 180�.

area:

The area is 20 square meters.

� 20K �12

(8)(10) sin 30�

m�A � 30�m�A � 80� � 70� � 180�

K �12

ab sin CK �12

ac sin BK �12

bc sin A

Find the area of �ABC to the nearest tenth of a square unit.

1. a � 5 inches, b � 9 inches, 2. b � 13 yards, c � 15 yards,

3. a � 8 meters, c � 4 meters, 4. a � 5 feet, b � 12 feet, m�A � 90�m�A � 67�

m�A � 71�m�C � 14�

Reteaching

14.1 The Law of Sines

c = 8m

b = 10ma

AB

C

80°

70°

◆Skill B Using the law of sines to solve a triangle

Recall can be used to solve triangles when given ASA or SAA.

◆ ExampleSolve �ABC shown at right.

◆ SolutionSince you have c � 24, find �C.

Now use the law of sines twice.Use a proportion involving three of the four values.

a sin 74� 24 sin 48� cross product

Use your calculator in degreemode.� 18.6a �24 sin 48�

sin74�

�

sin 48�a

�sin 74�

24

sin Aa

�sin C

c

� 74�m�C � 180� � (58� � 48�)

sin Aa

�sin B

b�

sin Cc

c = 24

ba

AB

C

48°58°

Solve �ABC. Round your answers to the nearest tenth.

5. 6. 7. A � 27�, C � 109�, a � 120B � 64�, C � 38�, b � 9A � 52�, B � 31�, a � 8

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


Apply the law of sines again to find side b.

� 21.2b �24 sin 58�

sin 74�

sin 58�b

�sin 74�

24

sin Bb

�sin C

c

◆Skill C Analyzing the ambiguous case

Recall When you are given SSA, there may be 0, 1, or 2 possible triangles.

In the right triangle at left, notice that .

There are 5 possible cases.

�A is acute �A is acute �A is acutea � b sin A a � b sin A a � b sin A

no triangle possible one triangle two triangles

�A is obtuse �A is obtusea � b sin A a � b sin A

no triangle possible one triangle

◆ ExampleHow many triangles are determined by , and ?

◆ Solutionis acute and b sin sin 57� 26.8

a � b sin A, because 30 � 26.8 There are two possible triangles.�A � 32�A

a � 30�A � 57�, b � 32

ba

A

b

a

A

b aa

A

ba

A

b a

A

sin A �ab

a � b sin A

ba

A

State how many triangles are determined by each set of given information.

8. 9.

10. 11. A � 130�, a � 20, b � 16A � 30�, b � 15, a � 30

A � 13�, a � 12, b � 60A � 34�, a � 3, b � 4

7. 8. 9. 10. 90� 11. 0�

12. 90� 13. 37� 14. 76� 15. 113�

16. 71�


Lesson 14.1




4. 30 square feet

5.

6.

7.

8. 2 triangles

9. no triangles

10. 2 triangles

11. 1 triangle

Lesson 14.2

1.

2.

3.

4.


5.

6.

7.

8.

Lesson 14.3

1.

2.

3.

4.





1cos x

�cos xsin x

cos xsin x

2π–2π

4

–4

2π–2π

4

–4

2π–2π

4

–4

2π–2π

4

–4

A � 143�, B � 20�, C � 18�

A � 58�, B � 73�, C � 48�

A � 56�, B � 90�, C � 34�

A � 38�, B � 82�, C � 60�

c � 14.0, A � 3�, B � 5�

a � 122.2, B � 38�, C � 28�

b � 49.1, A � 110�, C � 48�

a � 46.7, B � 42�, C � 87�

B � 44�, b � 183.6, c � 249.9

A � 78�, a � 9.8, c � 6.2

C � 97�, b � 5.2, c � 10.1

�12

1�2

1�3

ANSWERSC

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Hol

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Cop

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by H

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Rin

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Win

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. All

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s re

serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Using the law of cosines to solve a triangle when given two sides and the includedangle (SAS)

Recall a2 � b2 � c2 � 2bc cos A

b2 � a2 � c2 � 2ac cos B

c2 � a2 � b2 � 2ab cos C

◆ ExampleSolve the triangle shown at right.

◆ Solutiona2 � b2 � c2 � 2bc cos A

a2 � 52 � 72 � 2(5)(7) cos 39�

a2 � 19.6 Use a calculator.

a � 4.4 Find the square root.

To find the measure of �B, use the law of sines.

m�B � sin�1(0.715)

m�B � 45.7�

sin B �5 sin 39�

4.4

sin 39�4.4

�sin B

5

Solve each triangle.

1. 2.

3. 4.

b = 8a = 6c

AB

C

172°

b = 82 c = 63.2

aCB

A

114°

a = 123

c = 97

bA

B

C

22°

b = 40

c = 60

a

A B

C

51°

Reteaching

14.2 The Law of Cosines

b = 5

c = 7

a

AB

C

39°

Since the sum of all three angles is 180�,

m�C � 180� � (39� � 45.7�).

m�C � 95.3�

Solve each triangle.

5. 6.

7. 8.b = 10

a = 18

c = 9A

BCb = 9

a = 8c = 7

A

B

C

b = 18

a = 15

c = 10

A

B

C

b = 16

a = 10 c = 14

A

B

C

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Using the law of cosines to solve a triangle when given all three sides (SSS)

Recall In geometry, you proved triangles congruent by SSS; however, AAA only givessimilar triangles.

◆ ExampleFind the measures of the angles in �ABC.

◆ Solutiona2 � b2 � c2 � 2bc cos A

72 � 92 � 32 � 2(9)(3) cos A

cos A � 0.759

m�A � cos�1(0.759)

m�A � 41�

b2 � a2 � c2 � 2ac cos B

92 � 72 � 32 � 2(7)(3) cos B

cos B � �0.548

m�B � cos�1(�0.548)

m�B � 123�

m�C � 180� � (41� � 123�)

m�C � 16°

cos B �92 � 72 � 32

�2(7)(3)

cos A �72 � 92 � 32

�2(9)(3)

b = 9

a = 7c = 3

A

B

C

7. 8. 9. 10. 90� 11. 0�

12. 90� 13. 37� 14. 76� 15. 113�

16. 71�


Lesson 14.1




4. 30 square feet

5.

6.

7.

8. 2 triangles

9. no triangles

10. 2 triangles

11. 1 triangle

Lesson 14.2

1.

2.

3.

4.


5.

6.

7.

8.

Lesson 14.3

1.

2.

3.

4.





1cos x

�cos xsin x

cos xsin x

2π–2π

4

–4

2π–2π

4

–4

2π–2π

4

–4

2π–2π

4

–4

A � 143�, B � 20�, C � 18�

A � 58�, B � 73�, C � 48�

A � 56�, B � 90�, C � 34�

A � 38�, B � 82�, C � 60�

c � 14.0, A � 3�, B � 5�

a � 122.2, B � 38�, C � 28�

b � 49.1, A � 110�, C � 48�

a � 46.7, B � 42�, C � 87�

B � 44�, b � 183.6, c � 249.9

A � 78�, a � 9.8, c � 6.2

C � 97�, b � 5.2, c � 10.1

�12

1�2

1�3

ANSWERSC

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NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Verifying the fundamental trigonometric identities

Recall

Also, sin2 � � (sin �)(sin �) or (sin �)2, which is the form you will need to enter intoyour graphics calculator.

◆ ExampleUse a calculator to verify that

.

◆ Solution

Graph (for cot �)

and .

The graphs match exactly.

y2 �cos �sin �

y1 �1

tan �

cot � �cos �sin �

csc � �1

sin �, sec

� �1

cos �, cot � �

1tan �

Verify each identity by graphing each side separately. Sketch thecommon graph.

1. 2.

3. 4. 1 � cot2 � � csc2 �tan2 � � 1 � sec2 �

sin2 � � cos2 � � 1tan � �sin �cos �

Reteaching

14.3 Fundamental Trigonometric Identities

For exercises 5–10, show on your own paper how the firstexpression simplifies to the second expression.

5. sin x cot x to cos x 6. sin x sec x cot x to 1

7. cos2 x � sin2 x to 1 � 2 sin2 x 8. (1 � sin x)(1 � sin x) to cos2 x

9. tan x � cot x to sec x csc x 10. (cos x � sin x)2 to 1 � 2 cos x sin x

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Simplifying expressions by using basic trigonometric identities

Recall Since sin2 � � cos2 � � 1, then sin2 � � 1 � cos2 � and cos2 � � 1 � sin2 �.

◆ Example 1Simplify csc � tan � to sec �.

◆ Solutioncsc � tan � fundamental identities

fundamental identity

◆ Example 2Simplify (sec � � 1)(sec � � 1) to tan2 �.

◆ Solution(sec � � 1)(sec � � 1) � sec2 � � 1

� tan2 � � 1 � 1 fundamental identity� tan2 �

(a � b)(a � b) � a2 � b2

� sec �

sin �sin �

� 1�1

cos �

�1

sin ��

sin �cos �

◆Skill C Finding the values for which a trigonometric expression is defined

Recall Division by zero is undefined. sin � and cos � are defined for all values of �.tan � and sec � are defined for all values of � except odd multiples of 90�.cot � and csc � are defined for all values of � except multiples of 180�.

◆ ExampleFor what values of � is defined?

◆ Solutionsin � � 0, if � is a multiple of 180�.Since these values of � would give a denominator of 0 in the given expression,

is defined for all values � except � � n180� (any multiple of 180�).1 � cos �sin �

1 � cos �sin �

Find the values of � for which each trigonometric expression isdefined.

11. 12. 13. sin �sin2 � � cos2 �

1 � cos �tan �

sin �cos2 � � 1

7. 8. 9. 10. 90� 11. 0�

12. 90� 13. 37� 14. 76� 15. 113�

16. 71�


Lesson 14.1




4. 30 square feet

5.

6.

7.

8. 2 triangles

9. no triangles

10. 2 triangles

11. 1 triangle

Lesson 14.2

1.

2.

3.

4.


5.

6.

7.

8.

Lesson 14.3

1.

2.

3.

4.





1cos x

�cos xsin x

cos xsin x

2π–2π

4

–4

2π–2π

4

–4

2π–2π

4

–4

2π–2π

4

–4

A � 143�, B � 20�, C � 18�

A � 58�, B � 73�, C � 48�

A � 56�, B � 90�, C � 34�

A � 38�, B � 82�, C � 60�

c � 14.0, A � 3�, B � 5�

a � 122.2, B � 38�, C � 28�

b � 49.1, A � 110�, C � 48�

a � 46.7, B � 42�, C � 87�

B � 44�, b � 183.6, c � 249.9

A � 78�, a � 9.8, c � 6.2

C � 97�, b � 5.2, c � 10.1

�12

1�2

1�3

ANSWERSC

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9. tan x � cot x �

sec x csc x

10. (cos x � sin x)2 � cos2 x � 2 cos x sin x �sin2 x � 1 � 2 cos x sin x

11. all � except � � n(180�)

12. all � except � � n(90�)

13. defined for all values of �

Lesson 14.4

1. 2. 3.

4. 5. (�3.33, 4.23) 6. (1, �2)

7. (1.60, �3.23) 8. (2, 2) 9. (4.95, 0.71)

10. (�2.12, �4.95)

Lesson 14.5

1. 2 cot � sin2 � � � sin2 �

� 2 cos � sin �� sin 2�

2.

� cos2 � �

� cos2 � � � cos2 �

� cos2 � � sin2 �� cos 2�

3. 2 cos2 � � 1 � cos2 � � cos2 � � 1� cos2 � � (1 � cos2 �)� cos2 � � sin2 �� cos 2�

4. 5. 0 6. 1 7. 8. 9. 0

10. � 0.259

11. � 0.966

12. � �0.707

13. � �0.383

14. � 0.966

15. � � �0.966

Lesson 14.6

1. or

2.

3. or

4. no solution

5.

6.

7.

8.

9. 0 and 2π

10.

11. 0 and 2π

12.

13.

14. true for all real values of x

π4

and 5π4

π6

, 5π6

, and 3π2

π2

x � 1.57, 3.31, 4.71, 6.12

x � 0.55, 2.12, 3.70, 5.27

x � 0; x � 2.09, 6.28

x � 0; x � 2.09, 4.19

330� � n(360�)� � 0� � n(180�) or 210� � n(360�)

� � 60� � n(180�)

210� � n(360�)� � 150� � n(360�)

�1 ��32

2

�1 ��32

2

��1 ��22

2

��22

�1 ��32

2

�1 ��32

2

�12

12

�32

sin2 �cos2 �

sin2 �cos2 �

1cos2 �

�1

sec2 ��

tan2 �sec2 �

1 � tan2 �1 � tan2 �

�1 � tan2 �

sec2 �

2 cos �sin �

�22

12

�2 � �64

�2 � �64

�1

cos x sin x�

�sin2 x � cos2 x

cos x sin x

sin xcos x

�cos xsin x

ANSWERSC

opyright ©by H

olt, Rinehart and W



Cop

yrig

ht ©

by H

olt,

Rin

ehar

t and

Win

ston

. All

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s re

serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Evaluating expressions using the sum and difference identities

Recall sin(A � B) � sin A cos B � cos A sin Bsin(A � B) � sin A cos B � cos A sin Bcos(A � B) � cos A cos B � sin A sin Bcos(A � B) � cos A cos B � sin A sin B

Note: You may want to review finding the values of trigonometric functions ofcommon angles in Lesson 13.3 before completing the following examplesand exercises.

◆ ExampleFind an exact value for

a. . b. .

◆ Solution

a.

Notice that .

b.

Notice that .π3

�π6

�π2

and cos π2

� 0

� 0

��34

��34

�12

��32

��32

�12

cos(π3

�π6) � cos π

3 cos π

6� sin π

3 sin π

6

sin(π3

�3π4 ) sin π

3� sin 3π

4

��2 � �6

4

� ��64

��24

��32

� ��22

�12

��22

sin(π3

�3π4 ) � sin π

3 cos π

4� cos π

3 sin 3π

4

cos(π3

�π6)sin(π

3�

3π4 )

Find the exact value of each expression.

1. 2.

3. 4. cos(3π4

�π2)cos(2π

3� π)

sin(3π4

�π6)sin(π

4�

π3)

Reteaching

14.4 Sum and Difference Identities

Find the image of each point after the given counterclockwiserotation.

5. (2, 5), 60� 6. (�2, �1), 90�

7. (�3, 2), 150� 8. (�2, �2), 180�

9. (4, �3), 45� 10. (5, 2), 225�

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Using a matrix to find the coordinates of an image point after a rotation of � degrees

Recall If P(x, y) is any point in the plane, then the image of P(x, y) after a rotation of �degrees counterclockwise about the origin is given by the following matrix product.

◆ ExampleFind the image of (3, 4) after a rotation of 120� counterclockwise about the origin.

◆ Solution

�

In your calculator, let and .

The product [A][B] is approximately .

The coordinates of the image point are approximately (�4.96, 0.60).

��4.960.60�

[B] � �34��

12

�32

��32

�12�[A] �

� �12

�32

��32

�12� � �3

4��cos 120�sin 120�

�sin 120�cos 120�� 3

4�

x

y

O

(3, 4)120°

(–4.96 , 0.60)

�cos �sin �

�sin �cos �� x

y�


sec x csc x


11. all � except � � n(180�)

12. all � except � � n(90�)


Lesson 14.4

1. 2. 3.

4. 5. (�3.33, 4.23) 6. (1, �2)

7. (1.60, �3.23) 8. (2, 2) 9. (4.95, 0.71)

10. (�2.12, �4.95)

Lesson 14.5

1. 2 cot � sin2 � � � sin2 �

� 2 cos � sin �� sin 2�

2.

� cos2 � �

� cos2 � � � cos2 �

� cos2 � � sin2 �� cos 2�


4. 5. 0 6. 1 7. 8. 9. 0

10. � 0.259

11. � 0.966

12. � �0.707

13. � �0.383

14. � 0.966

15. � � �0.966

Lesson 14.6

1. or

2.

3. or

4. no solution

5.

6.

7.

8.

9. 0 and 2π

10.

11. 0 and 2π

12.

13.


π4

and 5π4

π6

, 5π6

, and 3π2

π2

x � 1.57, 3.31, 4.71, 6.12

x � 0.55, 2.12, 3.70, 5.27

x � 0; x � 2.09, 6.28

x � 0; x � 2.09, 4.19

330� � n(360�)� � 0� � n(180�) or 210� � n(360�)

� � 60� � n(180�)

210� � n(360�)� � 150� � n(360�)

�1 ��32

2

�1 ��32

2

��1 ��22

2

��22

�1 ��32

2

�1 ��32

2

�12

12

�32

sin2 �cos2 �

sin2 �cos2 �

1cos2 �

�1

sec2 ��

tan2 �sec2 �

1 � tan2 �1 � tan2 �

�1 � tan2 �

sec2 �

2 cos �sin �

�22

12

�2 � �64

�2 � �64

�1

cos x sin x�


cos x sin x

sin xcos x

�cos xsin x

ANSWERSC

opyright ©by H

olt, Rinehart and W



Cop

yrig

ht ©

by H

olt,

Rin

ehar

t and

Win

ston

. All

right

s re

serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Using the double-angle identities to simplify and evaluate expressions

Recall

◆ Example 1Simplify the expression .

◆ Solution

fundamental identities

cosine is reciprocal of secant

double-angle identity

◆ Example 2Use a double-angle identity to find sin 60� using the value of sin 30�.

◆ Solutionsin 60� � sin(2 � 30�)

� 2 sin 30� cos 30� double-angle identity

��32

� 2 �12

��32

� sin 2�

� 2 sin � cos �

�2 sin �cos �

� cos2 �

2 sin �cos �sec2 �

2 tan �1 � tan2 �

�

2 tan �1 � tan2 �

cos 2� � cos2 � � sin2 �sin 2� � 2 sin � cos �

For each exercise, show on your own paper how the firstexpression simplifies to the second expression.

1. 2 cot � sin2 � to sin 2� 2. 3.

Use a double-angle formula to evaluate each of the following.

4. sin 120� 5. sin 180�

6. sin 450� 7. cos 60�

8. cos 600� 9. cos 270�

2 cos2 � � 1 to cos 2�1 � tan2 �1 � tan2 �

to cos 2�

Reteaching

14.5 Double-Angle and Half-Angle Identities

Use half-angle identities to find exact values for each of thefollowing. Check your results with a calculator.

10. sin 15� 11. cos 15�

12. sin 675� 13. cos 112.5�

14. sin 75� 15. cos 165�

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Using the half-angle identities to find exact values of trigonometric functions

Recall and

◆ ExampleUse half-angle formulas to finda. sin 22.5�. b. cos 105�.

◆ Solutiona. The sine of 22.5� will be positive since it is in Quadrant I.

Check on your calculator. and .

b. The cosine of 105� will be negative since it is in Quadrant II.

On your calculator, and .cos 105� � �0.259��1 � �32

2� �0.259

� ��1 � �32

2

� ��1 � cos 210�2

cos 105� � cos 210�2

sin 22.5� � 0.383�1 �1

�22

� 0.383

� �1 �1

�22

� �1 � cos 45�2

sin 22.5� � sin 45�2

cos �2

� �1 � cos �2

sin �2

� �1 � cos �2


sec x csc x


11. all � except � � n(180�)

12. all � except � � n(90�)


Lesson 14.4

1. 2. 3.

4. 5. (�3.33, 4.23) 6. (1, �2)

7. (1.60, �3.23) 8. (2, 2) 9. (4.95, 0.71)

10. (�2.12, �4.95)

Lesson 14.5

1. 2 cot � sin2 � � � sin2 �

� 2 cos � sin �� sin 2�

2.

� cos2 � �

� cos2 � � � cos2 �

� cos2 � � sin2 �� cos 2�


4. 5. 0 6. 1 7. 8. 9. 0

10. � 0.259

11. � 0.966

12. � �0.707

13. � �0.383

14. � 0.966

15. � � �0.966

Lesson 14.6

1. or

2.

3. or

4. no solution

5.

6.

7.

8.

9. 0 and 2π

10.

11. 0 and 2π

12.

13.


π4

and 5π4

π6

, 5π6

, and 3π2

π2

x � 1.57, 3.31, 4.71, 6.12

x � 0.55, 2.12, 3.70, 5.27

x � 0; x � 2.09, 6.28

x � 0; x � 2.09, 4.19

330� � n(360�)� � 0� � n(180�) or 210� � n(360�)

� � 60� � n(180�)

210� � n(360�)� � 150� � n(360�)

�1 ��32

2

�1 ��32

2

��1 ��22

2

��22

�1 ��32

2

�1 ��32

2

�12

12

�32

sin2 �cos2 �

sin2 �cos2 �

1cos2 �

�1

sec2 ��

tan2 �sec2 �

1 � tan2 �1 � tan2 �

�1 � tan2 �

sec2 �

2 cos �sin �

�22

12

�2 � �64

�2 � �64

�1

cos x sin x�


cos x sin x

sin xcos x

�cos xsin x

ANSWERSC

opyright ©by H

olt, Rinehart and W



Cop

yrig

ht ©

by H

olt,

Rin

ehar

t and

Win

ston

. All

right

s re

serv

ed.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill A Solving trigonometric equations algebraically and graphically

Recall Trigonometric equations are true only for certain values of the variable, unlikeidentities which are true for all values of the variable.

◆ Example 1

Find all the solutions for each equation algebraically.

a. b.

◆ Solution

a.

or where n is any integer

◆ Example 2Solve graphically.

◆ Solutionsin x � 2 cos xUse radian mode on your calculator.y1 � sin x and y2 � 2 cos x

These two graphs intersect at and , which

are the solutions of the original equation.

x � 4.25x � 1.11

sin x � 2 cos x for 0 � x � 2π

300� � n(360�)

� � 240� � n(360�)

sin � ��3

2

2 sin � � ��3

sin � � �3 � �sin �

2 cos2 � � 5 cos � � 2 � 0sin � � �3 � �sin �

Find all possible solutions for each equation algebraically.

1. 2.

3. 4.

Solve each equation for graphically.

5. 6.

7. 8. 3 sin 2x � �cos xtan2 x � 1 � tan x

sin x � sin 12

x2 cos2 x � cos x � 1

0 � x � 2π

sin2 � � 5 sin � � 6 � 02 sin2 � � sin � � 0

tan � � �3 � 02 cos � � �3 � 0

Reteaching

14.6 Solving Trigonometric Equations

b.

or

or

(not possible)

or 300� � n(360�)

� � 60� � n(360�)

cos � �12

cos � � 22 cos � � 1

cos � � 2 � 02 cos � � 1 � 0

(2 cos � � 1)(cos � � 2) � 0

2 cos2 � � 5 cos � � 2 � 0

4

–4

Solve each equation for .

9. 10.

11. 12.

13. 14. tan x cot x � 1cos x � sin x

2(1 � cos2 x) � sin x � 1sin2 x � cos x � 1 � 0

csc x sin2 x � 1 � 0sin x cot x � 1

0 � x � 2π

Copyright ©

by Holt, R


ll rights reserved.

NAME _________________________________________________ CLASS _______________ DATE ______________


◆Skill B Solving trigonometric equations containing more than one trigonometric function algebraically

Recall

◆ Example

Solve .

◆ Solution

oror

(not possible)x � 0 or 2πcos x � 2cos x � 1

cos x � 2 � 0cos x � 1 � 0

(cos x � 1)(cos x � 2) � 0

cos2 x � 3 cos x � 2 � 0

�cos2 x � 3 cos x � 2 � 0

1 � cos2 x � 3 cos x � 3 � 0

sin2 x � 3 cos x � 3 � 0

sin2 x � 3 cos x � 3 � 0 for 0 � x � 2π

1 � cot2 � � csc2 �tan2 � � 1 � sec2 �sin2 � � cos2 � � 1

cot � �cos �sin �

tan � �sin �cos �

cot � �1

tan �sec � �

1cos �

csc � �1

sin �


sec x csc x


11. all � except � � n(180�)

12. all � except � � n(90�)


Lesson 14.4

1. 2. 3.

4. 5. (�3.33, 4.23) 6. (1, �2)

7. (1.60, �3.23) 8. (2, 2) 9. (4.95, 0.71)

10. (�2.12, �4.95)

Lesson 14.5

1. 2 cot � sin2 � � � sin2 �

� 2 cos � sin �� sin 2�

2.

� cos2 � �

� cos2 � � � cos2 �

� cos2 � � sin2 �� cos 2�


4. 5. 0 6. 1 7. 8. 9. 0

10. � 0.259

11. � 0.966

12. � �0.707

13. � �0.383

14. � 0.966

15. � � �0.966

Lesson 14.6

1. or

2.

3. or

4. no solution

5.

6.

7.

8.

9. 0 and 2π

10.

11. 0 and 2π

12.

13.


π4

and 5π4

π6

, 5π6

, and 3π2

π2

x � 1.57, 3.31, 4.71, 6.12

x � 0.55, 2.12, 3.70, 5.27

x � 0; x � 2.09, 6.28

x � 0; x � 2.09, 4.19

330� � n(360�)� � 0� � n(180�) or 210� � n(360�)

� � 60� � n(180�)

210� � n(360�)� � 150� � n(360�)

�1 ��32

2

�1 ��32

2

��1 ��22

2

��22

�1 ��32

2

�1 ��32

2

�12

12

�32

sin2 �cos2 �

sin2 �cos2 �

1cos2 �

�1

sec2 ��

tan2 �sec2 �

1 � tan2 �1 � tan2 �

�1 � tan2 �

sec2 �

2 cos �sin �

�22

12

�2 � �64

�2 � �64

�1

cos x sin x�


cos x sin x

sin xcos x

�cos xsin x

ANSWERSC

opyright ©by H

olt, Rinehart and W



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