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Chapter 2

Functions, Linear Equations, and Models Exercise Set 2.1 1. domain 2. range 3. exactly 4. more 5. horizontal 6. vertical 7. “f of 3,” “f at 3,” or “the value of f at 3” 8. function 9. The correspondence is a function because each member of the domain corresponds to just one member of the range. 10. The correspondence is not a function because a member of the domain (6) corresponds to more then one member of the range. 11. The correspondence is a function because each girl’s age corresponds to just one weight. 12. The correspondence is a function because each boy’s age corresponds to just one weight. 13. The correspondence is a function because each celebrity corresponds to just one birthday. 14. The correspondence is a function because each predator corresponds to just one prey. 15. The correspondence is not a function because one member of the domain, July 24, corresponds to three celebrities and another, July 18, corresponds to three celebrities. 16. The correspondence is not a function because one member of the domain, Texas, corresponds to four members of the range.

17. The correspondence is a function because each pumpkin corresponds to just one price. 18. The correspondence is not a function, since it is reasonable to assume that at least one member of the rock band plays more than one instrument. The correspondence is a relation, since it is reasonable to assume that each member of a rock band plays at least one instrument. 19. The correspondence is a function because each team member would have only one number on his or her uniform. 20. The correspondence is a function because each triangle would have only one area. 21. a) Locate 1 on the horizontal axis and then find the point on the graph for which 1 is the first coordinate. From that point, look to the vertical axis to find the corresponding y- coordinate, –2. Thus, ( )1 2f = − .

b) The domain is the set of all x-values in the graph. It is { }| 2 5x x− ≤ ≤ .

c) To determine which member(s) of the domain are paired with 2, locate 2 on the vertical axis. From there look left and right on the graph to find any points for which 2 is the second coordinate. One such point exists. Its first coordinate is 4. Thus, the x-value for which ( ) 2f x = is

4. d) The range is the set of all y-values in the graph. It is { }| 3 4y y− ≤ ≤ .

22. a) Locate 1 on the horizontal axis and then find the point on the graph for which 1 is the first coordinate. From that point, look to the vertical axis to find the


70 Chapter 2: Intermediate Algebra: Graphs and Models

corresponding y- coordinate, –1. Thus, ( )1 1f = − .


c) To determine which member(s) of the domain are paired with 2, locate 2 on the vertical axis. From there look left and right on the graph to find any points for which 2 is the second coordinate. One such point exists. Its first coordinate is –3. Thus, the x-value for which ( ) 2f x = is –3.

d) The range is the set of all y-values in the graph. It is { }| 2 5y y− ≤ ≤ .

23. a) Locate 1 on the horizontal axis and then find the point on the graph for which 1 is the first coordinate. From that point, look to the vertical axis to find the corresponding y- coordinate, 3. Thus, ( )1 3f = .



3. d) The range is the set of all y-values in the graph. It is { }|1 4y y≤ ≤ .




d) The range is the set of all y-values in the graph. It is { }| 0 4y y≤ ≤ .

25. a) Locate 1 on the horizontal axis and then find the point on the graph for which 1 is the first coordinate. From that point, look to the vertical axis to find the corresponding y- coordinate, –2. Thus, ( )1 2f = − .






c) To determine which member(s) of the domain are paired with 2, locate 2 on the vertical axis. From there look left and right on the graph to find any points for which 2 is the second coordinate. One such point exists. Its first coordinate is 0.


Exercise Set 2.1 71

Thus, the x-value for which ( ) 2f x = is

0. d) The range is the set of all y-values in the graph. It is { }| 5 4y y− ≤ ≤ .








d) The range is the set of all y-values in the graph. It is { }| 0 5y y≤ ≤ .


b) The domain is the set of all x-values in the graph. It is { }| 3, 1,1,3,5x − − .


3. d) The range is the set of all y-values in the graph. It is { }| 1,0,1,2,3y − .


b) The domain is the set of all x-values in the graph. It is { }| 4, 3, 2, 1,0,1, 2x − − − − .

c) To determine which member(s) of the domain are paired with 2, locate 2 on the vertical axis. From there look left and right on the graph to find any points for which 2 is the second coordinate. There are two such points. They are ( )2,2−

and ( )0, 2 . Thus, the x-values for which

( ) 2f x = are –2 and 0..

d) The range is the set of all y-values in the graph. It is { }|1, 2,3, 4y .

31. a) Locate 1 on the horizontal axis and then find the point on the graph for which 1 is the first coordinate. From that point, look to the vertical axis to find the



corresponding y- coordinate, 4. Thus, ( )1 4f = .



and ( )3, 2 . Thus, the x-values for which

( ) 2f x = are –1 and 3.





and ( )1,2 . Thus, the x-values for which

( ) 2f x = are –5 and 1.



b) The domain is the set of all x-values in the graph. It is { }| 4 5x x− < ≤ .

c) To determine which member(s) of the domain are paired with 2, locate 2 on the vertical axis. From there look left and right on the graph to find any points for which 2 is the second coordinate. All points in the set { }| 2 5x x< ≤ satisfy

this condition. These are the x-values for which ( ) 2f x = .

d) The range is the set of all y-values in the graph. It is { }| 1,1,2y − .



c) To determine which member(s) of the domain are paired with 2, locate 2 on the vertical axis. From there look left and right on the graph to find any points for which 2 is the second coordinate. All points in the set { }| 0 2x x< ≤ satisfy

this condition. These are the x-values for which ( ) 2f x = .

d) The range is the set of all y-values in the graph. It is { }|1, 2,3, 4y .

35. We can use the vertical line test.


Exercise Set 2.1 73

Visualize moving the vertical line across the graph. No vertical line will intersect the graph more than once. Thus the graph is the graph of a function. 36. We can use the vertical line test.

It is possible for a vertical line to intersect the graph more than once. Thus this is not the graph of a function. 37. We can use the vertical line test.


It is possible for a vertical line to intersect the graph more than once. Thus this is not the graph of a function.


It is possible for a vertical line to intersect the graph more than once. Thus this is not the graph of a function. 40. We can use the vertical line test.


It is possible for a vertical line to intersect the graph more than once. Thus this is not the graph of a function.




Visualize moving the vertical line across the graph. No vertical line will intersect the graph more than once. Thus the graph is the graph of a function. 43. ( ) 2 3g x x= +

a) ( )0 2 0 3 0 3 3g = ⋅ + = + =

b) ( ) ( )4 2 4 3 8 3 5g − = − + = − + = −

c) ( ) ( )7 2 7 3 14 3 11g − = − + = − + = −

d) ( )8 2 8 3 16 3 19g = ⋅ + = + =

e) ( ) ( )2 2 2 3

2 4 3 2 7

g a a

a a

+ = + += + + = +

f) ( ) ( )2 2 3 2 2 5g a a a+ = + + = +

44. ( ) 3 2h x x= −

a) ( )4 3 4 2 12 2 10h = ⋅ − = − =

b) ( )8 3 8 2 24 2 22h = ⋅ − = − =

c) ( ) ( )3 3 3 2 9 2 11h − = − − = − − = −

d) ( ) ( )4 3 4 2 12 2 14h − = − − = − − = −

e) ( ) ( )1 3 1 2

3 3 2 3 5

h a a

a a

− = − −= − − = −

f) ( ) ( )1 3 2 1

3 2 1 3 3

h a a

a a

− = − −= − − = −

45. ( ) 25 4f n n n= +

a) ( ) 20 5 0 4 0 0 0 0f = ⋅ + ⋅ = + =

b) ( ) ( ) ( )21 5 1 4 1 5 4 1f − = − + − = − =

c) ( ) 23 5 3 4 3 45 12 57f = ⋅ + ⋅ = + =

d) ( ) 25 4f t t t= +

e) ( ) ( )2

2 2

2 5 2 4 2

5 4 8 20 8

f a a a

a a a a

= + ⋅

= ⋅ + = +

f) ( ) ( )2 22 2 5 4 10 8f a a a a a⋅ = + = +

46. ( ) 23 2g n n n= −

a) ( ) 20 3 0 2 0 0 0 0g = ⋅ − ⋅ = − =

b) ( ) ( ) ( )21 3 1 2 1 3 2 5g − = − − − = + =

c) ( ) 23 3 3 2 3 27 6 21g = ⋅ − ⋅ = − =

d) ( ) 23 2g t t t= −

e) ( ) ( )2

2 2

2 3 2 2 2

3 4 4 12 4

g a a a

a a a a

= − ⋅

= ⋅ − = −

f) ( ) ( )2 22 2 3 2 6 4g a a a a a⋅ = − = −

47. ( ) 3

2 5

xf x

x

−=

−

a) ( ) 0 3 3 3 30

2 0 5 0 5 5 5f

− − −= = = =

⋅ − − −

b) ( ) 4 3 1 14

2 4 5 8 5 3f

−= = =

⋅ − −

c) ( ) ( )1 3 4 4 4

12 1 5 2 5 7 7

f− − − −

− = = = =− − − − −

d) ( ) 3 3 0 03 0

2 3 5 6 5 1f

−= = = =

⋅ − −

e) ( ) ( )2 3

22 2 5

1 1

2 4 5 2 1

xf x

x

x x

x x

+ −+ =

+ −− −= =+ − −

48. ( ) 3 4

2 5

xs x

x

−=

+

a) ( ) 3 10 4 30 4 2610

2 10 5 20 5 25s

⋅ − −= = =

⋅ + +

b) ( ) 3 2 4 6 4 22

2 2 5 4 5 9s

⋅ − −= = =

⋅ + +

c)

1 33 4 41 2 2

12 1 52 52

3 8 55 1 52 2 2

6 6 2 6 12

s⋅ − − = = +⋅ +

− −= = = − ⋅ = −

d) ( ) ( )( )

3 1 4 3 4 7 71 , or

2 1 5 2 5 3 3s

− − − − −− = = = −

− + − +


Exercise Set 2.1 75

e) ( ) ( )( )

3 3 43

2 3 5

3 9 4 3 5

2 6 5 2 11

xs x

x

x x

x x

+ −+ =

+ ++ − += =+ + +

49. ( ) 2 3

4A s s=

( ) 2 34 4 4 3 6.93

4A = = ≈

The area is 2 24 3 cm 6.93 cm≈ .

50. ( ) 2 3

4A s s=

( ) 2 3 36 6 36 9 3 15.59

4 4A = = = ≈

The area is 2 29 3 in 15.59 in≈ . 51. ( ) 24V r rπ=

( ) ( )23 4 3 36V π π= =

The area is 2 236 in 113.10 inπ ≈ . 52. ( ) 24V r rπ=

( ) ( )25 4 5 100V π π= =

The area is 2 2100 cm 314.16 cmπ ≈ .

53. ( ) 133

dP d = +

( ) 20 2020 1 1

33 33P = + =

The pressure at 20 ft is 20

133

atm.

( ) 30 1030 1 1

33 11P = + =


111

atm.

( ) 100 1 1100 1 1 3 4

33 33 33P = + = + =


433

atm.

54. ( ) 0.112W d d=

( ) ( )16 0.112 16 1.792W = =

The amount of water resulting from snow melting at a depth of 16 cm is 1.792 cm. ( ) ( )25 0.112 25 2.8W = =

The amount of water resulting from snow melting at a depth of 25 cm is 2.8 cm. ( ) ( )100 0.112 100 11.2W = =

The amount of water resulting from snow melting at a depth of 100 cm is 11.2 cm. 55. ( ) ( )2 5 2 8 5 16 5 11f x x= − = − = − =

56. ( ) 2 5

13 2 5

18 2

9

f x x

x

x

x

= −= −==

57. ( ) 2 5

5 2 5

0 2

0

f x x

x

x

x

= −− = −

==

58. ( )

( ) ( )2 5

4 2 4 5

8 5

13

f x x

f

= −

− = − −= − −= −

59. ( )

( )( )

13

1 12 3

1 12 3

81 12 2 3

7 12 3

7 32 1

212

4

4

4

f x x

x

x

x

x

x

x

= +

= +− =− =− =

− =

− =

60. ( )

( )( )

13

1 13 3

1 13 3

1 12 13 3 3

13 13 3

13 33 1

4

4

4

13

f x x

x

x

x

x

x

x

= +

− = +− − =− − =

− =

− =

− =



61. ( )( ) ( )

13

1 1 12 3 2

16

251 246 6 6

4

4

4

f x x

f

= +

= +

= += + =

62. ( )

( ) ( )13

1 1 13 3 3

361 19 9 9

359

4

4

4

f x x

f

= +

− = − +

= − + = − +=

63. ( ) 4

7 4

4 7 3

f x x

x

x

= −= −= − = −

So, ( )3 7f − = .

64. ( )

( )( )

12

12

12

1 12 5

110

5 1

5 1

1 5

5

f x x

x

x

x

x

x

= += +

− =− =

− =

− =

So, ( )1 110 2f − =

65. ( ) 0.1 0.5

3 0.1 0.5

3 0.5 0.1

2.5 0.1

2.5

0.125

f x x

x

x

x

x

x

= −− = −

− + =− =−

=

− =

So, ( )25 3f − = − .

66. ( ) 2.3 1.5

10 2.3 1.5

1.5 2.3 10

1.5 7.7

7.7 77, or 5.13

1.5 15

f x x

x

x

x

x

= −= −= −= −−

= = − −

67. The graph crosses the x-axis at only one point whose coordinate is ( )2,0− , so –2 is the zero

of the function. 68. The graph crosses the x-axis at two points whose coordinates are ( )1,0− and ( )3,0 , so

the zeros of the function are –1 and 3. 69. The graph does not cross the x-axis at all so there are no zeros of this function. 70. The graph crosses the x-axis at only one point whose coordinates are ( )4,0 , so 4 is the zero

of this function. 71. The graph crosses the x-axis at two points whose coordinates are ( )2,0− and ( )2,0 , so

the zeros of this function are –2 and 2. 72. The graph does not cross the x-axis at all so there are no zeros of this function. 73. We want to find any x-values for which ( ) 0f x = , so we substitute 0 for ( )f x and

solve: ( ) 5

0 5

5

f x x

x

x

= −= −=

The zero of this function is 5. 74. We want to find any x-values for which ( ) 0f x = , so we substitute 0 for ( )f x and

solve: ( ) 3

0 3

3

f x x

x

x

= += +

− =

The zero of this function is –3. 75. We want to find any x-values for which ( ) 0f x = , so we substitute 0 for ( )f x and

solve: ( )

( )( )

12

12

12

10

0 10

10

10 2

20

f x x

x

x

x

x

= +

= +− =

− =− =

The zero of this function is –20.


Exercise Set 2.1 77

76. We want to find any x-values for which ( ) 0f x = , so we substitute 0 for ( )f x and

solve: ( )

( )

23

23

23

32

6

0 6

6

6

9

f x x

x

x

x

x

= −

= −=

=

=

The zero of this function is 9. 77. We want to find any x-values for which ( ) 0f x = , so we substitute 0 for ( )f x and

solve: ( ) 2.7

0 2.7

2.7

f x x

x

x

= −= −=

The zero of this function is 2.7. 78. We want to find any x-values for which ( ) 0f x = , so we substitute 0 for ( )f x and

solve: ( ) 0.5

0 0.5

0.5

f x x

x

x

= −= −=

The zero of this function is 0.5. 79. We want to find any x-values for which ( ) 0f x = , so we substitute 0 for ( )f x and

solve: ( )

73

3 7

0 3 7

7 3

f x x

x

x

x

= += +

− =− =

The zero of this function is 73− .

80. We want to find any x-values for which ( ) 0f x = , so we substitute 0 for ( )f x and

solve: ( )

85

5 8

0 5 8

8 5

f x x

x

x

x

= −= −==

The zero of this function is 85 .

81. ( )0f refers to the output from the function f

when the input is 0. That is, we substitute the value 0 for the independent variable and determine the value of the function. When we find the zeros of a function, we are trying to determine those input values which result in an output value of 0. That is, we are trying to find any x-values for which ( ) 0f x = , so

we substitute 0 for ( )f x and solve:

82. When we evaluate ( ) 2 5 7f x x x= − + for a

specific input value, say x a= , we substitute the value a for each instance of the variable x. Then we evaluate the resulting expression. For example, after substituting a for x, we would have the expression 2 5 7a a− + ; so, we would square a, then subtract 5 times a, and then subtract 7. The result is the output for the function which is equal to ( )f a .

83. 84. 85.



86. 87. 88. 89. 90. 91. Writing Exercise. The independent variable is z. The notation ( )n z indicates that this is

the case. 92. Writing Exercise. Every function is a relation, because each member of the domain corresponds to a member of the range. Not every relation is a function, because in some relations one or more members of the domain can correspond to more than one member of the range.

93. To find ( )( )4f g − , we first find ( )4g − :

( ) ( )4 2 4 5 8 5 3g − = − + = − + = − .

Then ( )( ) ( ) ( )24 3 3 3 1

3 9 1 27 1 26.

f g f− = − = − −

= ⋅ − = − =

To find ( )( )4g f − , we first find ( )4f − :

( ) ( )24 3 4 1 3 16 1 48 1 47f − = − − = ⋅ − = − =

Then ( )( ) ( )4 47 2 47 5

94 5 99.

g f g− = = ⋅ +

= + =

94. To find ( )( )1f g − , we first find ( )1g − .

( ) ( )1 2 1 5 2 5 3g − = − + = − + = .

Then ( )( ) ( ) 21 3 3 3 1

27 1 26

f g f− = = ⋅ −

= − =

To find ( )( )1g f − , we first find ( )1f − .

( ) ( )21 3 1 1 3 1 1 2f − = − − = ⋅ − =

Then ( )( ) ( )1 2 2 2 5 9g f g− = = ⋅ + = .

95. To find ( )( )( )( )f f f f tiger , we start with

the innermost function and work our way out. Since ( )f tiger dog= , we have

( )( )( )( ) ( )( )( )f f f f tiger f f f dog= .

Since ( )f dog cat= , we have

( )( )f f cat .

Since ( )f cat fish= , we have

( )f fish .

Finally, ( )f fish worm= . So,

( )( )( )( )f f f f tiger worm= .

96. To find the strength of the largest contraction, locate the highest point on the graph, and find the corresponding pressure on the y-axis, which is the second coordinate of the point. The largest contraction was approximately 22 mm of mercury. 97. To find the time during the test when the largest contraction occurred, locate the highest point on the graph, and find the corresponding time on the x-axis, which is the first coordinate of the point. The time of the


Exercise Set 2.1 79

largest contraction was approximately 2 min 50 sec into the test. 98. Writing Exercise. About 12 mm; we would expect the contraction at 7 min to be about the same size as the contraction at 4 min since the largest contractions occurred about 3 min apart. 99. The two largest contractions occurred at about 2 minutes, 50 seconds and 5 minute, 40 seconds. The difference in these times is 2 minutes 50 seconds, so the frequency is about 1 every 3 minutes. 100. 101. We know that ( )1, 7− − and ( )3,8 are both

solutions of ( )g x mx b= + . Substituting, we

have

( )7 1m b− = − + , or 7 m b− = − + ,

and ( )8 3m b= + , or 8 3m b= + .

Solve the first equation for b and substitute that expression into the second equation.

( )

7 First equation

7 Solving for

8 3 Second equation

8 3 7 Substituting

8 3 7

8 4 7

15 4

15

4

m b

m b b

m b

m m

m m

m

m

m

− = − +− == += + −= + −= −=

=

We know that 7m b− = , so 15

74

b− = , or

13

4b− = . We have

15

4m = and

13

4b = − , so

( ) 15 13

4 4g x x= − .

102. For walking, we have the two points ( )1

22 ,210 and ( )343 ,300 , where the first

coordinate represents mph and the second calories per hour. Writing this as a function ( )w x mx b= + , we have

( )12210 2m b= + and ( )3

4300 3m b= + .

Solving the first equation for b and substituting that expression into the second equation. 210 2.5 First equation

210 2.5 Solving for

300 3.75 Second equation

300 3.75 210 2.5 Substituting

300 1.25 210

90 1.25

72

m b

b m b

m b

m m

m

m

m

= += −= += + −= +==

We know that 210 2.5m b= + , so

( )210 2.5 72

210 180

30 .

b

b

b

= += +=

.

So, we have ( ) 72 30w x x= + . If someone

walks 124 mph, then they would burn

( ) ( )4.5 72 4.5 30 354w = + = calories in one

hour, or 708 calories in two hours. Similarly, for bicycling we have the two points ( )1

25 ,210 and ( )13,660 , where the

first coordinate represents mph and the second, calories per hour. Writing this as a function, ( )c x mx b= + , we have

( )12210 5m b= + and ( )660 13m b= + .

Solving the first equation for b and substituting that expression into the second equation. 210 5.5 First equation

210 5.5 Solving for

660 13 Second equation

660 13 210 5.5 Substituting

450 7.5

60

m b

b m b

m b

m m

m

m

= += −= += + −==

We know that 210 5.5m b= + , so

( )210 5.5 60

210 330

120

b

b

b

= += +

− =



We now have ( ) 60 120c x x= − . If someone

rides their bicycle at 14 mph, they would burn ( ) ( )14 60 14 120 840 120 720c = − = − =

calories in hour. So bicycling 14 mph for one hour will burn more calories than walking 1

24 mph for two

hours. Exercise Set 2.2 1. e 2. d 3. c 4. f 5. a 6. b 7. Graph ( ) 2 7f x x= −

We make a table of values. Then we plot the corresponding points and connect them.

( )1 5

2 3

3 1

5 3

x f x

−−−

8. Graph ( ) 3 7g x x= −


( )0 7

1 4

2 1

3 2

x g x

−−−

9. Graph ( ) 1

3 2g x x= − +


( )0 2

3 1

6 0

9 1

x g x

−

10. Graph ( ) 1

2 5f x x= − −


( )10 0

8 1

4 3

0 5

x f x

−− −− −

−

11. Graph ( ) 2

5 4h x x= −


( )0 4

5 2

10 0

x h x

−−

12. Graph ( ) 4

5 2h x x= +


( )0 2

5 6

10 10

x h x

13. 3 4y x= +

The y-intercept is ( )0, 4 , or simply 4.

14. 4 9y x= −

The y-intercept is ( )0, 9− , or simply –9.


Exercise Set 2.2 81

15. ( ) 4 3g x x= − −


16. ( ) 5 7g x x= − +

The y-intercept is ( )0,7 , or simply 7.

17. 3

8 4.5y x= − −

The y-intercept is ( )0, 4.5− , or simply –4.5.

18. 15

7 2.2y x= +

The y-intercept is ( )0,2.2 , or simply 2.2.

19. ( ) 2.9 9f x x= −


20. ( ) 3.1 5f x x= − +


21. 37 204y x= +


22. 52 700y x= − +


23. Slope = Change in 5 9 4

2Change in 4 6 2

y

x

− −= = =

− −

24. Slope = Change in 1 7 8 4

Change in 2 8 6 3

y

x

− − −= = =

− −

25. Slope = Change in 4 8 12

2Change in 9 3 6

y

x

− − −= = = −

−

26. Slope = ( )15 12Change in

Change in 9 17

15 12 3 3

9 17 26 26

y

x

− − −=

− −− + −= = =− − −

27. Slope = ( )

( )( )

8 54 15 2 10 10

4 2 4 25 3 5 3

3 310 10

1012 215 15 15

3 15 910 2 4

Change in

Change in

y

x

− −= =− − − − +

= =− + −

= − = −

28. Slope = ( )

( )( )

1 24 5

313 4

5 8 320 20 20 3 912

20 5 259 5412 12 12

Change in

Change in

y

x

− − −= =

−− +

= = − = −− −

29. Slope = ( )Change in 43.6 43.6

Change in 4.5 9.7

0 00

4.5 9.7 14.2

y

x

−= =

− −

= =+

30. Slope = ( )( )

2.6 3.1Change in

Change in 1.8 2.8

2.6 3.1 0.5 1

1.8 2.8 1 2

y

x

− − −=− − −− += = =− +

31. a) The graph of 3 5y x= − has a positive

slope, 3, and the y-intercept is ( )0, 5− .

Thus, graph II matches this equation. b) The graph of 0.7 1y x= + has a positive

slope, 0.7, and the y-intercept is ( )0,1 .

Thus graph IV matches this equation. c) The graph of 0.25 3y x= − − has a

negative slope, –0.25, and the y-intercept is ( )0. 3− . Thus graph III matches this

equation. d) The graph of 4 2y x= − + has a negative

slope, –4, and the y-intercept is ( )0, 2 .

Thus graph I matches this equation. 32. a) The graph of 1

2 5y x= − has a positive

slope, 12 , and the y-intercept is ( )0, 5− .

Thus graph II matches this equation. b) The graph of 2 3y x= + has a positive

slope, 2, and the y-intercept is ( )0,3 .

Thus graph IV matches this equation. c) The graph of 3 1y x= − + has negative

slope, –3, and the y-intercept is ( )0,1 .

Thus graph I matches this equation.



d) The graph of 34 2y x= − − has a negative

slope, 34− , and the y-intercept is ( )0, 2− .

Thus graph III matches this equation. 33. 5

2 3y x= +

Slope is 52 ; y-intercept is ( )0,3 .

From the y-intercept, we go up 5 units and to the right 2 units. This gives us the point ( )2,8 . We can now draw the graph.

As a check, we can rename the slope and find another point.

5 5 1 5

2 2 1 2

− −= ⋅ =

− −

From the y-intercept, we go down 5 units and to the left 2 units. This gives us the point ( )2, 2− − . Since ( )2, 2− − is on the line, we

have a check. 34. 2

5 4y x= +

Slope is 25 ; y-intercept is ( )0, 4 .



12

12

2 2 1

5 5 2.5

− −= ⋅ =

− −

From the y-intercept, we go down 1 unit and to the left 2.5 units. This gives us the point ( )2.5,3− . Since ( )2.5,3− is on the line, we

have a check.

35. ( ) 52 2f x x= − +

Slope is 52− ; y-intercept is ( )0, 2 .

From the y-intercept, we go down 5 units and to the right 2 units. This gives us the point ( )2, 3− . We can now draw the graph.


12

12

5 5 2.5

2 2 1

−− −= ⋅ =

− −

From the y-intercept, we go up 2.5 units and to the left 1 unit. This gives us the point ( )1,4.5− . Since ( )1,4.5− is on the line, we

have a check. 36. ( ) 2

5 3f x x= − +

Slope is 25− ; y-intercept is ( )0,3 .

From the y-intercept, we go down 2 units and to the right 5 units. This gives us the point ( )5,1 . We can now draw the graph.


12

12

2 2 1

5 5 2.5

− − −= ⋅ =

From the y-intercept, we go down 1 unit and to the right 2.5 units. This gives us the point ( )2.5,2 . Since ( )2.5,2 is on the line, we

have a check. 37. 2 5

2 5

2 5

x y

y x

y x

− =− = − +

= −

Slope is 2, or 2

1; y-intercept is ( )0, 5− .


Exercise Set 2.2 83

From the y-intercept, we go up 2 units and to the right 1 unit. This gives us the point ( )1, 3− . We can now draw the graph.


2 2 4 8

1 1 4 4= ⋅ =

From the y-intercept, we go up 8 units and to the right 4 unit. This gives us the point ( )4,3 . Since ( )4,3 is on the line, we

have a check. 38. 2 4

2 4

x y

y x

+ == − +

Slope is –2, or 2

1

−; y-intercept is ( )0, 4 .

From the y-intercept, we go down 2 units and to the right 1 unit. This gives us the point ( )1,2 . We can now draw the graph.


2 2 3 6

1 1 3 3

− − −= ⋅ =

From the y-intercept, we go down 6 units and to the right 3 unit. This gives us the point ( )3, 2− . Since ( )3, 2− is on the line, we


3 2F x x= +

Slope is 13 ; y-intercept is ( )0, 2 .

From the y-intercept, we go up 1 unit and to the right 3 units. This gives us the point ( )3,3 . We can now draw the graph.


1 1 2 2

3 3 2 6= ⋅ =

From the y-intercept, we go up 2 units and to the right 6 unit. This gives us the point ( )6, 4 . Since ( )6, 4 is one the line we have a

check. 40. ( ) 3 6g x x= − +

Slope is –3, or 3

1

−; y-intercept is ( )0,6 .

From the y-intercept, we go down 3 units and to the right 1 unit. This gives us the point ( )1,3 . We can now draw the graph.


3 3 2 6

1 1 2 2

− − −= ⋅ =

From the y-intercept, we go down 6 units and to the right 2 units. This gives us the point ( )2,0 . Since ( )2,0 is one the line we have a

check. 41. 6 4 6

6 4 6

3 2 3

21

3

y x

y x

y x

y x

+ == − += − +−

= +

Slope is 2

3


From the y-intercept, we go down 2 units and to the right 3 unit. This gives us the point ( )3, 1− . We can now draw the graph.




2 2 1 2

3 3 1 3

− − −= ⋅ =

− −

From the y-intercept, we go up 2 units and to the left 3 units. This gives us the point ( )3,3− . Since ( )3,3− is one the line we have

a check. 42. 4 20

4 20

15

4

y x

y x

y x

+ == −

= −

Slope is 1


From the y-intercept, we go up 1 unit and to the right 4 unit. This gives us the point ( )4, 4− . We can now draw the graph.


1 1 1 1

4 4 1 4

− −= ⋅ =

− −

From the y-intercept, we go down 1 unit and to the left 4 units. This gives us the point ( )4, 6− − . Since ( )4, 6− − is one the line we

have a check. 43. ( ) 0.25g x x= −

Slope is 0.25− , or 1

4


From the y-intercept, we go down 1 unit and to the right 4 units. This gives us the point ( )4, 1− . We can now draw the graph.


1 1 1 1

4 4 1 4

− − −= ⋅ =

− −

From the y-intercept, we go up 1 unit and to the left 4 units. This gives us the point ( )4,1− . Since ( )4,1− is one the line we

have a check. 44. ( ) 1.5 3F x x= −

Slope is 1.5 , or 3




3 3 2 6

2 2 2 4= ⋅ =

From the y-intercept, we go up 6 units and to the right 4 units. This gives us the point ( )4,3 . Since ( )4,3 is one the line we

have a check. 45. 4 5 10

5 4 10

42

5

x y

y x

y x

− =− = +

= −

Slope is 4




Exercise Set 2.2 85


12

12

4 4 2

5 5 2.5= ⋅ =

From the y-intercept, we go up 2 units and to the right 2.5 units. This gives us the point ( )2.5,0 . Since ( )2.5,0 is one the line we

have a check. 46. 5 4 4

4 5 4

51

4

x y

y x

y x

+ == − +−

= +

Slope is 5

4




5 5 1 5

4 4 1 4

− − −= ⋅ =

− −

From the y-intercept, we go up 5 units and to the left 4 units. This gives us the point ( )4,6− . Since ( )4,6− is one the line we


4 2f x x= −

Slope is 5




5 5 1 5

4 4 1 4

− −= ⋅ =

− −

From the y-intercept, we go down 5 units and to the left 4 units. This gives us the point ( )4, 7− − . Since ( )4, 7− − is one the line we


3 2f x x= +

Slope is 4

3, or

4

3

−−

; y-intercept is ( )0, 2 .

From the y-intercept, we go down 4 units and to the left 3 units. This gives us the point ( )3, 2− − . We can now draw the graph.

As a check, we can rename the slope and find another point. (Use slope as given.)

4

3

From the y-intercept, we go up 4 units and to the right 3 units. This gives us the point ( )3,6 . Since ( )3,6 is one the line we have a

check. 49. ( )

( )( )

( )

12 4 3

12 4 3

4 3 12

33

4

f x x

f x x

f x x

f x x

− =

− + = −

= − +−= +

Slope is 3

4




From the y-intercept, we go down 3 units and to the right 4 units. This gives us the point ( )4,0 . We can now draw the graph.


3 3 1 3

4 4 1 4

− − −= ⋅ =

− −

From the y-intercept, we go up 3 units and to the left 4 units. This gives us the point ( )4,6− . Since ( )4,6− is one the line we

have a check. 50. ( )

( )

( )

15 5 2

5 2 15

23

5

f x x

f x x

f x x

+ = −

= − −−

= −

Slope is 2

5

−; y-intercept is ( )0, 3− .



12

12

2 2 1

5 5 2.5

−− −= ⋅ =

− −

From the y-intercept, we go up 1 unit and to the left 2.5 units. This gives us the point ( )2.5, 2− − . Since ( )2.5, 2− − is one the line

we have a check.

51. ( )( )

4.5

0 4.5

g x

g x x

=

= +

Slope is 0 ; y-intercept is ( )0,4.5 .

From the y-intercept, we go up or down any number of units and any number of units to the left or right. Any point on the graph will lie on a horizontal line 4.5 units above the x- axis. We can now draw the graph.

52. ( )( )

34

34 0

g x x

g x x

=

= +

Slope is 3

4; y-intercept is ( )0,0 .



3 3 1 3

4 4 1 4

− −= ⋅ =

− −

From the y-intercept, we go down 3 units and to the left 4 units. This gives us the point ( )4, 3− − . Since ( )4, 3− − is one the line we

have a check. 53. Use the slope-intercept equation, ( )f x mx b= + , with 2

3m = and 9b = − .

( )( ) 2

3 9

f x mx b

f x x

= +

= −

54. Use the slope-intercept equation, ( )f x mx b= + , with 3

4m = − and 12b = .

( )( ) 3

4 12

f x mx b

f x x

= +

= − +


Exercise Set 2.2 87

55. Use the slope-intercept equation, ( )f x mx b= + , with 6m = − and 2b = .

( )( ) 6 2

f x mx b

f x x

= +

= − +

56. Use the slope-intercept equation, ( )f x mx b= + , with 2m = and 1b = − .

( )( ) 2 1

f x mx b

f x x

= +

= −


9m = − and 5b = .

( )( ) 7

9 5

f x mx b

f x x

= +

= − +


11m = − and 9b = .

( )( ) 4

11 9

f x mx b

f x x

= +

= − +

59. Use the slope-intercept equation, ( )f x mx b= + , with 5m = and 1

2b = .

( )( ) 1

25

f x mx b

f x x

= +

= +

60. Use the slope-intercept equation, ( )f x mx b= + , with 6m = and 2

3b = .

( )( ) 2

36

f x mx b

f x x

= +

= +

61. We can use the coordinates of any two points on the line. We’ll use ( )0,100 and ( )9,40 .

change in 40 100 60

Slope = change in 9 0 9

20 2, or 6 .

3 3

y

x

− −= =

−

= − −

The distance from the finish line is decreasing

at a rate of 2

63

meters per second.

62. We can use the coordinates of any two points on the line. We’ll use ( )0,30 and ( )3,3 .

change in 3 30 27


9.

y

x

− −= =

−= −

The value is decreasing at a rate of $900 per year. 63. We can use the coordinates of any two points on the line. We’ll use ( )1,4 and ( )3,10 .

change in 10 4 6


3.

y

x

−= =

−=

The distance is increasing at a rate of 3 miles per hour. 64. We can use the coordinates of any two points on the line. We’ll use ( )1,1 and ( )5,3 .

change in 3 1 2


0.5.

y

x

−= =

−=

The weight is increasing at a rate of 0.5 lb per bag. 65. We can use the coordinates of any two points on the line. We’ll use ( )0,50 and ( )2,200 .

change in 200 50 150


75.

y

x

−= =

−=

The number of pages read is increasing at a rate of 75 pages per day. 66. We can use the coordinates of any two points on the line. We’ll use ( )0,5 and ( )4,6 .

change in 6 5 1

Slope = .change in 4 0 4

y

x

−= =

−

The distance is increasing at a rate of 1

4 km

per minute. 67. We can use the coordinates of any two points on the line. We’ll use ( )15,470 and

( )55,510 .


Slope = 1.change in 55 15 40

y

x

−= = =

−

The average SAT math score is increasing at a rate of 1 point per thousand dollars of family income.



68. We can use the coordinates of any two points on the line. We’ll use ( )35, 495 and

( )65,525 .


Slope = 1.change in 65 35 30

y

x

−= = =

−

The average SAT verbal score is increasing at a rate of 1 point per thousand dollars of family income. 69. The marathoner’s speed is given by

change in distance

change in time. Note that the

marathoner reaches the 22-mi point 56 min after the 15- mi point. We will express

change in time in hours: 56 14

60 15= hr. Then

change in distance 22 15

change in time 14 /15

15 157 7.5.

14 2

−=

= ⋅ = =

The speed is 7.5 mph. 70. The skier’s speed is given by

change in distance

change in time. Note that the skier

reaches the 12-km 45 min after the 3-km mark was reached or after 15 + 45, or 60 min. We will express time in hours: 15 min = 0.25 hr and 60 min = 1 hr. Then

change in distance 12 3 9

12.change in time 1 0.25 0.75

−= = =

−

The speed is 12 km/h. 71. The average rate of descent is given by

change in altitude

change in time. We will express time in

minutes: 1 3 60 min

1 hr hr 90 min2 2 1 hr

= ⋅ =

2 hr, 10 min = 2 hr + 10 min =

60 min

2 hr 10min 120min 10min1 hr

130min

⋅ + = +

=

Then

change in altitude 0 12,000 12,000

change in time 130 90 40

300.

− −= =

−= −

The average rage of descent is 300 ft/min.

72. The rate at which the painter is working is

given by change in work

change in time. Then

2 1 53 4 12change in work 5

.change in time 8 0 8 96

−= = =

−

The painter is painting the house at a rate of 5

96 of the house per hour.

73. The rate at which the average length of a cell- phone call was increasing is given by

change in length of average call


change in length of average call

change in time

3.00 2.15 0.850.085.

2005 1995 10

−= = =−

The rate of increase in average length of cell- phone call is 0.085 minutes per year. 74. The rate at which the number of local exchange phone calls was decreasing is given

by change in number of calls


change in number of calls 426 544

change in time 2003 1998

11823.6

5

−=

−−= = −

The rate of local exchange calls is decreasing at a rate of 23.6 billion calls per year. 75. a) Graph II indicated that 200 ml of fluid was dripped in the first 3 hr, a rate of

200

3 ml/hr. It also indicates that 400 ml

of fluid was dripped in the next 3 hr, a

rate of 400

3 ml/hr, and that this rate

continues until the end of the time period

shown. Since the rate of 400

3 ml/hr is

double the rate of 200

3 ml/hr, this graph

is appropriate for the given situation. b) Graph IV indicates that 300 ml of fluid was dripped in the first 2 hr, a rate of 300/2, or 150 ml/hr. In the next 2 hr, 200 ml was dripped. This is a rate of 200/2,


Exercise Set 2.2 89

or 100 ml/hr. Then 100 ml was dripped in the next 3 hr, a rate of 100/3, or 1

333

ml/hr. Finally, in the remaining 2 hr, 0 ml of fluid was dripped, a rate of 0/2, or 0 ml/hr. Since the rate at which the fluid was given decreased as time progressed and eventually became 0, this graph is appropriate for the given situation. c) Graph I is the only graph that shows a constant rate for 5 hours, in this case from 3 PM to 8 PM. Thus, it is appropriate for the given situation. d) Graph III indicates that 100 ml of fluid was dripped in the first 4 hr, a rate of 100/4, or 25 ml/hr. In the next 4 hr, 300 ml was dripped. This is a rate of 300/4, or 75 ml/hr. In the last hour 200 ml was dripped, a rate of 200 ml/hr. Since the rate at which the fluid was given gradually increased, this graph is appropriate for the given situation. 76. a) Graph III indicates sales increased $3000 during the month before January 1, or $3000/month while sales increased by $2000 during the month after January 1, or $2000/month. b) Graph IV is appropriate because it show that the rate before January 1 is $3000/month while it is –$4000/month after January 1. c) Graph I is appropriate because it shows that the rate is $1000/month before January 1 and $2000/month after January 1. d) Graph II is appropriate because it shows that the rate is $4000/month before January and –$2000/month after January 1. 77. ( ) 25 75C x x= +

25 signifies that the cost per person is $25; 75 signifies that the setup cost for the party is $75.

78. ( ) 0.05 200P x x= +

0.05 signifies that a salesperson earns 5% commission on sales; 200 indicates that a salesperson’s base salary is $200 per week.

79. ( ) 11

2L t t= +

1

2 signifies that Ty’s hair grows

1

2 in. per

month; 1 signifies that his hair is 1 in. long immediately after he gets it cut. 80. ( ) 9

25 22D t t= +

925 signifies that the demand increases 9

25

quadrillion Btu’s per year for years after 2000; 22 signifies that the demand was 22 quadrillion Btu’s in 2000. 81. ( ) 1

7 75.5A t t= +

17 signifies that the life expectancy of

American women increases 17 yr per year, for

years after 1970; 75.5 signifies that the life expectancy in 1970 was 75.5 yr. 82. ( ) 1

8 2G t t= +

18 signifies that the grass grows 1

8 in. per day;

2 signifies that the grass is 2 in. long when cut. 83. ( ) 0.227 4.29P t t= +

0.227 signifies that the price increases $0.227 per year, for years since 1995; 4.29 signifies that the average cost of a movie ticket in 1995 was $4.29. 84. ( ) 0.5 7f t t= − +

–0.5 signifies that receipts decrease $0.5 billion per year, for years after 1995; 7 signifies that receipts in 1995 were $7 billion. 85. ( ) 2 2.5C d d== +

2 signifies that the cost per mile of taxi ride is $2; 2.5 signifies that the minimum cost of a taxi ride is $2.50



86. ( ) 0.3 20C d d= +

0.3 signifies that the cost per mile of renting the truck is $0.30; 20 signifies that the minimum cost is $20. 87. ( ) 5000 90,000F t t= − +

a) –5000 signifies that the truck’s value depreciates $5000 per year; 90,000 signifies that the original value of the truck was $90,000.

b) We find the value of t for which ( ) 0F t = .

0 5000 90,000

5000 90,000

18

t

t

t

= − +==

It will take 18 yr for the truck to depreciate completely.

c) The truck’s value goes from $90,000 when 0t = to $0 when 18t = , so the

domain of F is { }| 0 18t t≤ ≤ .

88. ( ) 2000 15000V t t= − +

a) –2000 signifies that the color separator’s value depreciates $2000 per year; 15,000 signifies that the original value of the separator was $15,000.

b) We find the value of t for which ( ) 0V t = .

0 2000 15,000

2000 15,000

7.5

t

t

t

= − +==

It will take 7.5 yr for the machine to depreciate completely. c) The machine’s value goes from $15,000 when 0t = to $0 when 7.5t = , so the

domain of V is { }| 0 7.5t t≤ ≤ .

89. ( ) 150 900v n n= − +

a) –150 signifies that the snowblower’s value depreciates $150 per winter of use; 900 signifies that the original value of the snowblower was $900.

b) We find the value of n for which ( ) 300v n = .

300 150 900

600 150

4

n

n

n

= − +− = −

=

The snowblower’s trade-in value will be $300 after 4 winters of use.

c) First we find the value of n for which ( ) 0v n = .

0 150 900

900 150

6

n

n

n

= − +− = −

=

The value of the snowblower goes from $900 when 0n = to $0 when 6n = . Since the snowblower is used only in the winter, we express the domain of v as { }| 0,1,2,3,4,5,6n .

90. ( ) 300 2400T x x= − +

a) –300 signifies that the mower’s value depreciates $300 per summer of use; 2400 signifies that the original value of the mower was $2400.

b) We find the value of x for which ( ) 1200T x = .

1200 300 2400

1200 300

4

x

x

x

= − +− = −

=

The mower’s value will be $1200 after 4 summers of use.

c) First we find the value of x for which ( ) 0T x = .

0 300 2400

2400 300

8

x

x

x

= − +− = −

=

The value of the mower goes from $2400 when 0x = to $0 when 8x = . Since the mower is used only in the summer, we express the domain of T as { }0,1,2,3,4,5,6,7,8 .

91. Writing Exercise. Find the slope-intercept form of the equation.

4 5 12

5 4 12

4 12

5 5

x y

y x

y x

+ == − +

= − +

This form of the equation indicates that the line has a negative slope and thus should slant


Exercise Set 2.2 91

down from left to right. The student

apparently graphed 4 12

5 5y x= + .

92. Writing Exercise. Using algebra, we find that the slope-intercept form of the equation is

5 3

2 2y x= − . This indicates that the y-

intercept is 3

0,2

− , so a mistake has been

made. It appears that the student graphed

5 3

2 2h x= + .

93. ( ){ }( ){ }( ){ }

[ ]{ }{ }{ }

09 2 3 5 2 3 2

9 2 3 5 2 3 1 2

9 2 3 5 2 3 1

9 2 3 5 6 2

9 2 15 18 6

9 5 6

45 54

x x x y

x x x

x x x

x x x

x x x

x

x

− + − + −

= − + − + −

= − + − − = − − −

= − + +

= += +

94. ( ) ( ) ( ) ( )3 3 332 3 2 3 6 95 5 125a b a b a b− = − = −

95. ( )( ) ( )2 3 5 2 5 3 1

7 4

13 2 13 2

26

m n m n m n

m n

+ +− = − ⋅ ⋅

= −

96. ( )[ ][ ]

2 3 2 3 10

2 3 2 6 10

2 3 2 4

2 3 6 12

8 3 12

x y x

x y x

x y x

x y x

x y

− − + + = − − − +

= − − += − + −= − −

97. ( )( ) ( )( )02 3 5 2 3 2 34 3 4 1 4x y x y x y x y− = =

98. ( ) ( ) ( ) ( )( )( ) ( )( )

2 2 225 7 5 7

25 2 7 2

10 1410 14

2 2

1

2

1

4 4

x y x y

x y

x yx y

− − −−− − − −

− − − −

=

=

= =

99. Writing Exercise. You would want m to be negative because this would signify that the debt is decreasing. 100. a) b) The profit increases each year but not as much as in the previous year. 101. Writing Exercise. m must be measured in trillions of dollars per month. 102. We first solve for y. rx py s

py rx s

r sy x

p p

+ == − +

= − +

The slope is r

p− , and the y-intercept is

0,s

p

.

103. We first solve for y.

( )

rx py s ry

ry py rx s

y r p rx s

r sy x

r p r p

+ = −+ = − ++ = − +

= − ++ +

The slope is r

r p−

+, and the y-intercept is

0,s

r p

+

.

104. Since ( )1 1,x y and ( )2 2,x y are two points on

the graph of y mx b= + , then 1 1y mx b= +

and 2 2y mx b= + . Using the definition of



slope, we have

( ) ( )

( )

2 12 1

2 1 2 1

2 1

2 1

Slope =

.

mx b mx by y

x x x x

m x xm

x x

+ − +−=

− −−

= =−

105. ( ) ( ) ( )f c d f c f d+ = +

Let 1c = and 2d = . Then

( ) ( ) ( )1 2 3 3f c d f f m b+ = + = = + , but

( ) ( ) ( ) ( )2 3 2f c f d m b m b m b+ = + + + = + .

The given statement is false. 106. ( ) ( ) ( )f cd f c f d=


( ) ( ) ( )2 3 6 6 6f cd f f m b m b= ⋅ = = ⋅ + = + ,

but ( ) ( ) ( ) ( ) ( )( )

2 2

2 3 2 3

6 5 .

f c f d f f m b m b

m mb b

= = ⋅ + ⋅ +

= + +

Thus, the given statement is false. 107. ( ) ( )f kx kf x=

Let 2k = . Then ( ) ( )2 2f kx f x mx b= = + ,

but ( ) ( )2 2 2kf x mx b mx b= + = + . The given

statement is false. 108. ( ) ( ) ( )f c d f c f d− = −


( ) ( )( )5 2

3 3 3

f c d f

f m b m b

− = −

= = ⋅ + = + , but

( ) ( ) ( ) ( )( ) ( )

5 2

5 2

5 2

3

f c f d f f

m b m b

m b m b

m

− = −

= ⋅ + − ⋅ += + − −=

Thus, the given statement is false. 109. Observe that parallel lines rise or fall at the same rate. Thus, their slopes are the same. For the line containing ( )3,k− and ( )4,8 ,

8 8

slope = .3 4 7

k k− −=

− − −

For the line containing ( )5,3 and ( )1, 6− ,

6 3 9

slope = 1 5 4.

− −=

−

Then we have

8 9

7 44 32 63

4 31

31

4

k

k

k

k

−=

−− = −

= −

= −

110. a) Graph III indicates that the first 2 mi and the last 3 mi were traveled in approximately the same length of time and at a fairly rapid rate. The mile following the first two miles was traveled at a much slower rate. This could indicate that the first two miles were driven, the next mile was swum and the last three miles were driven, so this graph is most appropriate for the given situation.

b) The slope in Graph IV decreases at 2 mi and again at 3 mi. This could indicate that the first two miles were traveled by bicycle, the next mile was run, and the last 3 miles were walked, so this graph is most appropriate for the given situation.

c). The slope in Graph I decreases at 2 mi and then increases at 3 mi. This could indicate that the first two miles were traveled by bicycle, the next mile was hiked, and the last three miles were traveled by bus, so this graph is most appropriate for the given situation.

d) The slope in Graph II increases at 2 mi and again at 3 mi. This could indicate that the first two miles were hiked, the next mile was run, and the last three miles were traveled by bus, so this graph is most appropriate for the given situation.

111. a) ( )6 5 5

slope = 5 4 4

c c c c

b b b b

− − −= = −

− −

b) ( )

slope = 0

d e d e

b b

+ −=

−

Since we cannot divide by 0, the slope is undefined.


Exercise Set 2.3 93

c) ( ) ( )( ) ( )

( )

slope =

2 2

2

2

2

a d a d

c f c f

a d a d

c f c f

a d

f

a d a d

f f

− − − +− − +

− − − −=− − −

− −=

−− + += =−

112. ( ) 3 17.5,C n n= + for 1 10n≤ ≤ ,

( ) 6 17.5,C n n= + for 11 20n≤ ≤ ,

( ) 7 17.5,C n n= + for 21 30n≤ ≤ ,

( ) 8 17.5,C n n= + for 31n ≥ .

113. Exercise Set 2.3 1. horizontal 2. undefined 3. vertical 4. y-axis 5. 0; x 6. 0; y 7. parallel

8. standard 9. linear 10. –1 11. 9 3

12

y

y

− ==

The graph of 12y = is a horizontal line.

Since 9 3y − = is equivalent to 12y = , the

slope of the line 9 3y − = is 0.

12. 1 7

6

x

x

+ ==

The graph of 6x = is a vertical line. Since 1 7x + = is equivalent to 6x = , the slope of the line 1 7x + = is undefined. 13. 8 6

6

83

4

x

x

x

=

=

=

The graph of 3

4x = is a vertical line. Since

8 6x = is equivalent to 3

4x = , the slope of

the line 8 6x = is undefined. 14. 3 5

8

y

y

− ==

The graph of 8y = is a horizontal line.

Since 3 5y − = is equivalent to 8y = , the

slope of the line 3 5y − = is 0.

15. 3 28

28

3

y

y

=

=

The graph of 28

3y = is a horizontal line.

Since 3 28y = is equivalent to 28

3y = , the

slope of the line 3 28y = is 0.

16. 19 6

19

6

y

y

= −

− =



The graph of 19

6y = − is a horizontal line.

Since 19 6y= − is equivalent to 19

6y = − ,

the slope of the line 19 6y= − is 0.

17. 9 12

3

x

x

+ ==

The graph 3x = is a vertical line. Since 9 12x+ = is equivalent to 3x = , the slope of the line 9 12x+ = is undefined. 18. 2 18

9

x

x

==

The graph of 9= is a vertical line. Since 2 18x = is equivalent to 9x = , the slope of the line 2 18x = is undefined. 19. 2 4 3

2 7

7

2

x

x

x

− ==

=

The graph of 7

2x = is a vertical line. Since

2 4 3x − = is equivalent to 7

2x = , the slope

of the line 2 4 3x − = is undefined. 20. 5 1 16

5 17

17

5

y

y

y

− ==

=

The graph of 17


Since 5 1 16y − = is equivalent to 17

5y = , the

slope of the line 5 1 16y − = is 0.

21. 5 4 35

5 39

39

5

y

y

y

− ==

=

The graph of 39


Since 5 4 35y − = is equivalent to 39

5y = ,

the slope of the line 5 4 35y − = is 0.

22. 2 17 3

2 20

10

x

x

x

− ===

The graph of 10x = is a vertical line. Since 2 17 3x − = is equivalent to 10x = , the slope of the line 2 17 3x − = is undefined. 23. 3 3 2

2

y x y

x

+ = +=

The graph of 2x = is a vertical line. Since 3 3 2y x y+ = + is equivalent to 2x = , the

slope of the line 3 3 2y x y+ = + is undefined.

24. 4 12 4

12

x y y

x

− = −=

The graph of 12x = is a vertical line. Since 4 12 4x y y− = − is equivalent to 12x = , the

slope of the line 4 12 4x y y− = − is

undefined. 25. 5 2 2 7

5 2 5

5

3

x x

x x

x

− = −= −

= −

The graph of 5

3x = − is a vertical line. Since

5 2 2 7x x− = − is equivalent to 5

3x = − , the

slope of the line 5 2 2 7x x− = − is undefined. 26. 5 3 9

4 6

3

2

y y

y

y

+ = +=

=

The graph of 3


Since 5 3 9y y+ = + is equivalent to 3

2y = ,

the slope of the line 5 3 9y y+ = + is 0.


Exercise Set 2.3 95

27. 2

53

y x= − +

The equation is written in slope-intercept

form. We see that the slope is 2

3− .

28. 3

42

y x= − +

The equation is written in slope-intercept

form. We see that the slope is 3

2− .

29. Graph 5y = .

This is a horizontal line that crosses the y-axis at ( )0,5 . If we find some ordered pairs, note

that, for any x-value chosen, y must be 5.

2 5

0 5

3 5

x y

−

30. Graph 1x = − . This is a vertical line that crosses the x-axis at ( )1,0− . If we find some ordered pairs,

note that, for any y-value chosen, x must be –1.

1 5

1 0

1 3

x y

−−− −

31. Graph 3x = . This is a vertical line that crosses the x-axis at ( )3,0 . If we find some ordered pairs,

note that, for any y-value chosen, x must be 3.

3 5

3 0

3 3

x y

−

32. Graph 2y = .

This is a horizontal line that crosses the y-axis at ( )0, 2 . If we find some ordered pairs, note

that, for any x-value chosen, y must be 2.

2 2

0 2

3 2

x y

−

33. Graph ( )4 20f x⋅ =

First solve for ( )f x .

( )( )

4 20

5

f x

f x

⋅ =

=

This is a horizontal line that crosses the vertical axis at ( )0,5 .

34. Graph ( )6 12g x⋅ = .

First solve for ( )g x .

( )( )

6 12

2

g x

g x

⋅ =

=

This is a horizontal line that crosses the y-axis at ( )0, 2 .

35. Graph 3 15x = − . Since y does not appear, we solve for x. 3 15

5

x

x

= −= −

This is a vertical line that crosses the x-axis at ( )5,0− .



36. Graph 2 10x = . Since y does not appear, we solve for x. 2 10

5

x

x

==

This is a vertical line that crosses the x-axis at ( )5,0 .

37. Graph ( )4 3 12 3g x x x⋅ + = + .

First solve for ( )g x .

( )( )( )

4 3 12 3

4 12

3

g x x x

g x

g x

⋅ + = +

⋅ =

=

This is a horizontal line that crosses the vertical axis at ( )0,3 .

38. Graph ( )3 2f x− = .

First solve for ( )f x .

( )( )

3 2

1

f x

f x

− =

=

This is a horizontal line that crosses the vertical axis at 0,1.

39. Graph 4x y+ = .

To find the y-intercept, let 0x = and solve for y. 4

0 4

4

x y

y

y

+ =+ =

=

The y-intercept is ( )0, 4 .

To find the x-intercept, let 0y = and solve

for x. 4

0 4

4

x y

x

x

+ =+ =

=

The x-intercept is ( )4,0 .

Plot these points and draw a line. A third point could be used as a check.

40. Graph 5x y+ = .

To find the y-intercept, let 0x = and solve for y. 5

0 5

5

x y

y

y

+ =+ =

=

The y-intercept is ( )0,5 .


for x. 5

0 5

5

x y

x

x

+ =+ =

=




Exercise Set 2.3 97

41. Graph 2 6y x= + .

To find the y-intercept, let 0x = and solve for y. 2 6

2 0 6

6

y x

y

y

= += ⋅ +=



for x. 2 6

0 2 6

6 2

3

y x

x

x

x

= += +

− =− =

The x-intercept is ( )3,0− .

Plot these points and draw a line. A third point could be used as a check. 42. Graph 3 9y x= + .


3 0 9

9

y x

y

y

= += ⋅ +=



for x. 3 9

0 3 9

9 3

3

y x

x

x

x

= += +

− =− =


Plot these points and draw a line. A third point to be used as a check.

43. Graph 3 5 15x y+ = − .

To find the y-intercept, let 0x = and solve for y. 3 5 15

3 0 5 15

5 15

3

x y

y

y

y

+ = −⋅ + = −

= −= −

The y-intercept is ( )0, 3− .


for x. 3 5 15

3 5 0 15

3 15

5

x y

x

x

x

+ = −+ ⋅ = −

= −= −



44. Graph 5 4 20x y− = .


5 0 4 20

4 20

5

x y

y

y

y

− =⋅ − =

− == −



for x. 5 4 20

5 4 0 20

5 20

4

x y

x

x

x

− =− ⋅ =

==





45. Graph 2 3 18x y− = .


2 0 3 18

3 18

6

x y

y

y

y

− =⋅ − =

− == −



for x. 2 3 18

2 3 0 18

2 18

9

x y

x

x

x

− =− ⋅ =

==



46. Graph 3 2 18x y+ = − .


3 0 2 18

2 18

9

x y

y

y

y

+ = −⋅ + = −

= −= −



for x.

3 2 18

3 2 0 18

3 18

6

x y

x

x

x

+ = −+ ⋅ = −

= −= −



47. Graph 3 6y x= .


3 6 0

3 0

0

y x

y

y

y

== ⋅==



for x. 3 6

3 0 6

0 6

0

y x

x

x

x

=⋅ =

==


Since the x- and y-intercepts are the same point, find one additional point. Letting 2x = ,

3 6

3 6 2

3 12

4

y x

y

y

y

== ⋅== ,

we obtain the point ( )2,4



Exercise Set 2.3 99

48. Graph 5 15y x= .


5 15 0

5 0

0

y x

y

y

y

== ⋅==



for x. 5 15

5 0 15

0 15

0

y x

x

x

x

=⋅ =

==

The x-intercept is ( )0,0 . Since the x- and y-

intercepts are the same, we obtain one additional point. Letting 2x = ,

5 15

5 15 2

5 30

6

y x

y

y

y

== ⋅== ,

we obtain the point ( )2,6 .


49. Graph ( ) 3 7f x x= − .

To find the y-intercept, let 0x = and solve

for ( )f x .

( )( )( )( )

3 7

3 0 7

0 7

7

f x x

f x

f x

f x

= −

= ⋅ −

= −

= −


To find the x-intercept, let ( ) 0f x = and

solve for x.

( ) 3 7

0 3 7

7 3

7

3

f x x

x

x

x

= −= −=

=

The x-intercept is 7

,03

.


50. Graph ( ) 2 9g x x= − .


for ( )g x .

( )( )( )( )

2 9

2 0 9

0 9

9

g x x

g x

g x

g x

= −

= ⋅ −

= −

= −


To find the x-intercept, let ( ) 0g x = and

solve for x. ( ) 2 9

0 2 9

9 2

9

2

g x x

x

x

x

= −= −=

=


,02

.




51. Graph 1.4 3.5 9.8y x− = − .

To find the y-intercept, let 0x = and solve for y.

( )1.4 3.5 9.8

1.4 3.5 0 9.8

1.4 0 9.8

1.4 9.8

7

y x

y

y

y

y

− = −− = −

− = −= −= −



for x.

( )1.4 3.5 9.8

1.4 0 3.5 9.8

0 3.5 9.8

3.5 9.8

2.8

y x

x

x

x

x

− = −− = −− = −− = −

=

The x-intercept is ( )2.8,0 .


52. Graph 3.6 2.1 22.68x y− = .

To find the y-intercept, let 0x = and solve for y.

( )3.6 2.1 22.68

3.6 0 2.1 22.68

0 2.1 22.68

10.9

x y

y

y

y

− =− =− =

= −

The y-intercept is ( )0, 10.9− .


for x.

( )3.6 2.1 22.68

3.6 2.1 0 22.68

3.6 0 22.68

3.6 22.68

6.3

x y

x

x

x

x

− =− =

− ===

The x-intercept is ( )6.3,0 .


53. Graph ( )5 2 7x g x+ = .


for ( )g x .

( )( )( )( )

( )

5 2 7

5 0 2 7

0 2 7

2 7

7

2

x g x

g x

g x

g x

g x

+ =

⋅ + =

+ =

=

=

The y-intercept is 7

0,2

.


solve for x.

( )5 2 7

5 2 0 7

5 0 7

5 7

7

5

x g x

x

x

x

x

+ =+ ⋅ =+ =

=

=


,05

.


54. Graph ( )3 4 11x f x− = .


for ( )f x .


Exercise Set 2.3 101

( )( )( )( )

( )

3 4 11

3 0 4 11

0 4 11

4 11

11

4

x f x

f x

f x

f x

f x

− =

⋅ − =

− =

− =

= −


0,4

− .


solve for x. ( )3 4 11

3 4 0 11

3 0 11

3 11

11

3

x f x

x

x

x

x

− =− ⋅ =− =

=

=


,03

.


55. ( ) 20 4f x x= −


for ( )f x .

( )( )( )( )

20 4

20 4 0

20 0

20

f x x

f x

f x

f x

= −

= − ⋅

= −

=



solve for x.

( ) 20 4

0 20 4

4 20

5

f x x

x

x

x

= −= −==


Choice (c) with window [ ]10,10, 10,30− −

will display both intercepts. 56. ( ) 3 7g x x= +


for ( )g x .

( )( )( )( )

3 7

3 0 7

0 7

7

g x x

g x

g x

g x

= +

= ⋅ +

= +

=



solve for x. ( ) 3 7

0 3 7

7 3

7

3

g x x

x

x

x

= += +

− =

− =


,03

− .

Choice (a) with window [ ]10,10, 10,10− −

will display both intercepts. 57. ( ) 35 7000p x x= − +


for ( )p x .

( )( )( )( )

35 7000

35 0 7000

0 7000

7000

p x x

p x

p x

p x

= − +

= − ⋅ +

= +

=


To find the x-intercept, let ( ) 0p x = and

solve for x. ( ) 35 7000

0 35 7000

35 7000

200

p x x

x

x

x

= − += − +==


Choice (d) with window [ ]0,500,0, 10,000

will display both intercepts.



58. ( ) 0.2 0.01r x x= −


for ( )r x .

( )( )( )( )

0.2 0.01

0.2 0.01 0

0.2 0

0.2

r x x

r x

r x

r x

= −

= − ⋅

= −

=

The y-intercept is ( )0,0.2 .

To find the x-intercept, let ( ) 0r x = and

solve for x. ( ) 0.2 0.01

0 0.2 0.01

0.01 0.2

0.220

0.01

r x x

x

x

x

= −= −=

= =


Choice (b) with window [ ]5,30, 1,1− − will

display both intercepts. 59. We first solve for y and determine the slope of each line. 8

8

x y

y x

+ == +

The slope of 8y x= + is 1.

5

5

y x

y x

− = −= −

The slope of 5y x= − is 1.

The slopes are the same; the lines are parallel. 60. We first solve for y and determine the slope of each line. 2 3

2 3

x y

y x

− == −

The slope of 2 3x y− = is 2.

2 9

2 9

y x

y x

− == +

The slope of 2 9y x− = is 2.


3 9

y x

y x

+ == −

The slope of 9 3y x+ = is 3.

3 2

3 2

3 2

x y

x y

y x

− = −+ =

= +

The slope of 3 2x y− = − is 3.


6 8

y x

y x

+ = −= − −

The slope of 8 6y x+ = − is –6.

2 5

2 5

x y

y x

− + == +

The slope of 2 5x y− + = is 2.

The slopes are not the same; the lines are not parallel. 63. We determine the slope of each line. ( ) 3 9f x x= +

The slope of ( ) 3 9f x x= + is 3.

2 6 2

3 1

y x

y x

= − −= − −

The slope of 2 6 2y x= − − is –3.

The slopes are not the same; the lines are not parallel. 64. We determine the slope of each line. ( ) 7 9f x x= − −

The slope of ( ) 7 9f x x= − − is –7.

3 21 7

77

3

y x

y x

− = +

= − −

The slope of 7

73

y x= − − is –7.

The slopes are the same; the lines are parallel. 65. We determine the slope of each line. The slope of ( ) 4 3f x x= − is 4.

4 7

4 7

1 7

4 4

y x

y x

y x

= −= − +

= − +

The slope of 4 7y x= − is 1

4− .



The product of their slopes is 1

44

− , or –1;

the lines are perpendicular. 66. We determine the slope of each line. 2 5 3

2 3 5

2 3

5 52 3

5 5

x y

x y

x y

y x

− = −+ =

+ =

= +

The slope of 2 5 3x y− = − is 2

5.

2 5 4

5 2 4

2 4

5 5

x y

y x

y x

+ == − +

= − +

The slope of 2 5 4x y+ = is 2

5− .

The product of their slopes is

2 2 4

15 5 25 − = − ≠ −

, so the lines are not

perpendicular. 67. We determine the slope of each line. 2 7

2 7

1 7

2 2

x y

y x

y x

+ == − +

= − +

The slope of 2 7x y+ = is 1

2− .

2 4 4

4 2 4

2 4

4 41

12

x y

y x

y x

y x

+ == − +

= − +

= − +

The slope of 2 4 4x y+ = is 1

2− .

The product of their slopes is 1 1

2 2 − −

, or

1

4, so the lines are not perpendicular.

68. We determine the slope of each line. The slope of 7y x= − + is 1− .

The slope of ( ) 3f x x= + is 1.

The product of their slopes is ( )1 1 1− = − , so

the lines are perpendicular. 69. The slope of the given line is 3. Therefore, the slope of a line parallel to it is also 3. The y-intercept is ( )0,9 , so the equation of the

desired function is ( ) 3 9f x x= + .

70. The slope of the given line is –5. Therefore, the slope of a line parallel to it is also –5. The y-intercept is ( )0, 2− , so the equation of the

desired function is ( ) 5 2f x x= − − .

71. First we find the slope of the given line. 2 3

2 3

x y

y x

+ == − +

The slope of the given line is –2. Therefore, the slope of a line parallel to it is also –2. The y-intercept is ( )0, 5− , so the equation of the

desired function is ( ) 2 5f x x= − − .


3 10

3 10

x y

x y

y x

= +− =

= −

The slope of the given line is 3. Therefore, the slope of a line parallel to it is also 3. The y-intercept is ( )0,1 , so the equation of the

desired function is ( ) 3 1f x x= + .

73. First we find the slope of the given line. 2 5 8

5 2 8

2 8

5 5

x y

y x

y x

+ == − +

= − +

The slope of the given line is 2

5− . Therefore,

the slope of a line parallel to it is also 2

5− .


0,3

− , so the equation of

the desired function is ( ) 2 1

5 3f x x= − − .



74. First we find the slope of the given line. 3 6 4

3 4 6

6 3 4

1 2

2 3

x y

x y

y x

y x

− =+ =

= +

= +


2. Therefore,

the slope of a line parallel to it is also 1

2. The

y-intercept is 4

0,5

, so the equation of the

desired function is ( ) 1 4

2 5f x x= + .


0 4

y

y x

== +

The slope of the given line is 0. Therefore, the slope of a line parallel to it is also 0. The y-intercept is ( )0, 5− , so the equation of the

desired function is ( ) 5f x = − .

76. First we find the slope of the given line.

5 10

10 5

10

2

y

y

y x

==

= +

The slope of the given line is 0. Therefore, the slope of a line parallel to it is also 0. The y-intercept is ( )0,12 , so the equation of the

desired function is ( ) 12f x = .

77. The slope of the given line is 1. The slope of a line perpendicular to it is the opposite of the reciprocal of 1, or –1. The y-intercept is ( )0, 4 , so we have 1 4y x= − ⋅ + , or

4y x= − + .

78. The slope of the given line is 2. The slope of a line perpendicular to it is the opposite of the

reciprocal of 2, or 1

2− . The y-intercept is

( )0, 3− , so we have 1

32

y x= − ⋅ − , or

1

32

y x= − − .

79. First find the slope of the given line.

2 3 6

3 2 6

22

3

x y

y x

y x

+ == − +

= − +


3− . The slope

of a line perpendicular to it is the opposite of

the reciprocal of 2

3− , or

3

2. The y-intercept

is ( )0, 4− , so we have 3

42

y x= ⋅ − , or

3

42

y x= − .

80. First find the slope of the given line. 4 2 8

2 4 8

2 4

x y

y x

y x

+ == − += − +

The slope of the given line is 2− . The slope of a line perpendicular to it is the opposite of

the reciprocal of 2− , or 1

2. The y-intercept

is ( )0,8 , so we have 1

82

y x= ⋅ + , or

1

82

y x= + .

81. First find the slope of the given line. 5 13

5 13

5 13

x y

x y

y x

− =− =

= −

The slope of the given line is 5. The slope of a line perpendicular to it is the opposite of

the reciprocal of 5, or 1

5− . The y-intercept

is 1

0,5

, so we have 1 1

5 5y x= − ⋅ + , or

1 1

5 5y x= − + .



82. First find the slope of the given line. 2 5 7

2 7 5

5 2 7

2 7

5 5

x y

x y

y x

y x

− =− =

= −

= −


5. The slope

of a line perpendicular to it is the opposite of

the reciprocal of 2

5, or

5

2− . The y-intercept

is 1

0,8

− , so we have

5 1

2 8y x= − ⋅ − , or

5 1

2 8y x= − − .

83. This equation is in the standard form for a linear equation, Ax By C+ = , with 5A = ,

3B = − , and 15C = . Thus, it is a linear equation. Solve for y to find the slope. 5 3 15

3 5 15

55

3

x y

y x

y x

− =− = − +

= −

The slope is 5

3.

84. We write 3 5 15 0x y+ + = in standard form

for a linear equation, Ax By C+ = , as

3 5 15x y+ = − with 3A = , 5B = , and

15C = − . Thus, it is a linear equation. Solve for y to find the slope. 3 5 15

5 3 15

33

5

x y

y x

y x

+ = −= − −

= − −

The slope is 3

5− .

85. We write 16 4 10y+ = in standard form

for a linear equation, Ax By C+ = , as

0 4 6x y+ = − with 0A = , 4B = , and

6C = − . Thus, it is a linear equation. Solve for y to find the slope.

4 6

30

2

y

y x

= −

= + −

The slope is 0 . 86. We write 3 12 0x − = in standard form for a linear equation, Ax By C+ = , as

3 0 12x y+ = with 3A = , 0B = , and

12C = . Thus, it is a linear equation and the line is vertical. 87. ( ) 23 6g x x=

Replace ( )g x with y and attempt to write the

equation in standard form. 2

2

3 6

6 3 0

y x

x y

=− + =

The equation is not linear, because it has an 2x term. 88. ( )2 4 8x f x+ =

Replacing ( )f x with y, we have 2 4 8x y+ =

which is a linear equation. Solve for y to find the slope. 2 4 8

4 2 8

12

2

x y

y x

y x

+ == − +

= − +

The slope is 1

2− .

89. ( )3 7 2 4y x= −

The equation can be written in standard form for a linear equation, Ax By C+ =

( )3 7 2 4

3 14 28

14 3 28

y x

y x

x y

= −= −

− + = −

with 14A = − , 3B = , and 28C = − . Thus, it is a linear equation. Solve for y to find the slope. 14 3 28

3 14 28

14 28

3 3

x y

y x

y x

− + = −= −

= −

The slope is 14

3.



90. ( )2 5 3 5

10 6 5

10 5 6

6 5 10

x y

x y

y x

x y

− =− =

= ++ =

This is a linear equation since it is in the form Ax By C+ = , with 6A = , 5B = , and

10C = . Solve for y to find the slope. 6 5 10

5 6 10

62

5

x y

y x

y x

+ == − +

= − +

The slope is 6

5− .

91. ( ) 10g x

x− =

Replacing ( )g x with y and attempt to write

the equation in standard form.

1

0

1 0

1

yx

xy

xy

− =

− ==

The equation is not linear because it has an xy-term.

92. ( ) 10f x

x+ =

Replace ( )f x with y and attempt to write the

equation in standard form.

1

0

1 0

1

yx

xy

xy

+ =

+ == −

The equation is not linear because it has an xy-term.

93. ( ) 2

5

f xx=

Replace ( )f x with y and attempt to write the


2

2

2

5

5

5 0

yx

y x

x y

=

=− + =

The equation is not linear because it has an 2x -term.

94. ( )

32

g xx= +

Replace ( )g x with y and attempt to write the


32

6 2

2 6

yx

y x

x y

= +

= +− + =

The equation is linear. From the next-to-last step above, we see that the slope is 2. 95. For each 10 km/hr increase, the corresponding force of resistance does not remain constant but increases. The slope is not constant so a linear function does not give an approximate fit. 96. Writing Exercise. For each 5° temperature decrease, the corresponding decrease in wind chill temperature is 6° or 7°. Thus, the slope is nearly constant, so a linear function will give an approximate fit. 97. 1

2 3y x= +

The slope is 12 ; y-intercept is ( )0,3 . From

the y-intercept, we go up 1 unit and to the right 2 units. This give us the point ( )2,4 .

We can now draw the graph.



98. ( ) 1f x x= − +

The slope is 1− , or 1

1


From the y-intercept, we go down 1 unit and to the right 1 unit. This gives us the point ( )1,0 . We can now draw the graph.

99. 3 3x y− =


3 0 3

0 3.

3

x y

y

y

y

− =⋅ − =− =

= −



for x. 3 3

3 0 3

3 3

1

x y

x

x

x

− =− =

==


Plot these points and draw the line.

100. 2 3 6x y+ =


2 0 3 6

0 3 6

2

x y

y

y

y

+ =⋅ + =+ =

=



for x.

2 3 6

2 3 0 6

2 6

3

x y

x

x

x

+ =+ ⋅ =

==


Plot these points and draw the line.

101. 102. 103. 104. 105.



106. 107. Writing Exercise. Consider the equation Ax By C+ = . Writing this equation in slope-

intercept form we have A C

y xB B

= − + . If we

choose a value for x for which is not a multiple of B, the corresponding y-value will be a fraction. Similarly, if Ax− is a multiple of B but C is not, then the corresponding y- value will be a fraction. Under these conditions, Jim will avoid fractions if he graphs using intercepts. 108. Writing Exercise. A line’s x- and y-intercepts coincide only when the line passes through the origin. The equation for such a line is of the form y mx= .

109. The line contains the points ( )5,0 and

( )0, 4− . We use the points to find the slope.

4 0 4 4

Slope = 0 5 5 5

− − −= =

− −

Then the slope-intercept equation is

4

45

y x= − . We rewrite this equation in

standard form.

4

45

5 4 20

4 5 20

y x

y x

x y

= −

= −− + = −

This equation can also be written as 4 5 20x y− = .

110. ( )0y mx b m= + ≠

To find the x-intercept, let 0y = .

0 mx b

b mx

bx

m

= +− =

− =

Thus, the x-intercept is ,0b

m −

.

111. 23rx y p s+ = −

The equation is in standard form with A r= , 3B = , and 2C p s= − . It is linear.

112. 2 9py sx r y= − −

Try to put the equation in standard form.

( )

2

2

2

9

9

9

py sx r y

sx py r y

sx p r y

= − −− + + = −

− + + = −

The equation is in standard form with A s= − , 2B p r= + , and 9C = − . It is

linear. 113. 2 5r x py= +

Try to put the equation in standard form.

2

2

5

5

r x py

r x py

= +− =

The equation is in standard form with 2A r= , B p= − , and 5C = . It is linear.

114. 17x

pyr− =

The equation is in standard form with 1

Ar

= ,

B p= − , and 17C = . It is linear.

115. Let equation A have intercepts ( ),0a and

( )0,b . Then equation B has intercepts

( )2 ,0a and ( )0,b .

Slope of A = 0

0

b b

a a

−= −

−

Slope of B = 0 1

0 2 2 2

b b b

a a a

− = − = − −

The slope of equation B is 1

2 the slope of

equation A. 116.

( ) ( )3 5 8

5 3 8

ax y x by

a x b y

+ = − +− + + =

If the graph is a horizontal line, then the coefficient of x is 0.

5 0

5

a

a

− ==



Then we have ( )3 8b y+ = .

If the graph passes through ( )0, 4 , we have

( )3 4 8

3 2

1

b

b

b

+ =+ == −

117. First write the equation in standard form.

( ) ( )

3 5 8

5 3 8

5 3 8

ax y x by

ax x y by

a x b y

+ = − +− + + =

− + + =

If the graph is a vertical line, then the coefficient of y is 0.

3 0

3

b

b

+ == −

Then we have ( )5 8a x− =

If the line passes through ( )4,0 , we have:

( )5 4 8 Substituting 4 for

5 2

7

a x

a

a

− =− =

=

118. a) b) c) 3 5 7 2

7 4

7

4

x x

x

x

− = +− =

− =

d) ( )2 5 10

2 10 10

20

x x

x x

x

− = +− = +

=

119. a) Solve each equation for y, enter each on the equation-editor screen, and then examine a table of values for the two functions. Since the difference between the y-values is the same for all x-values, the lines are parallel. b) Solve each equation for y, enter each on the equation-editor screen, and then examine a table of values for the two functions. Since the difference between the y-values is not the same for all x- values, the lines are not parallel. Exercise Set 2.4 1. True 2. False 3. False 4. True 5. True 6. True 7. True 8. True 9. False 10. True 11. ( )1 1y y m x x− = − Point-slope equation

( )4 2 1y x− = − − Substitute –2 for m, 1 for

1x , and 4 for 1y



To graph the equation, we count off a slope of

2

1

−, starting at ( )1,4 , and draw the line.

12. ( )1 1y y m x x− = − Point-slope equation

( )1 5 3y x− = − Substitute 5 for m, 3 for

1x , and 1 for 1y


5

1, or

5

1

−−

starting at ( )3,1 , and draw the line.



1x , and 2 for 1y


3

1, or

3

1

−−

starting at ( )5,2 , and draw the

line.



1x , and 3 for 1y


2

1, or

2

1

−−

starting at ( )7,3 , and draw the

line.


( ) ( )( )14 2

2y x− − = − − Substitute

1

2 for m,

–2 for 1x , and –4 for 1y


1

2, starting at ( )2, 4− − , and draw the line.


( ) ( )( )7 1 5y x− − = − − Substitute 1 for m,

–5 for 1x , and –7 for

1y


1

1, starting at ( )5, 7− − , and draw the line.


( )0 1 8y x− = − − Substitute –1 for m, 8 for

1x , and 0 for 1y


1

1

−, starting at ( )8,0 , and draw the line.




( )( )0 3 2y x− = − − − Substitute –3 for m, –

2 for 1x , and 0 for 1y


3

1

−, starting at ( )2,0− , and draw the line.

19. ( )

( )27

1 1

9 8y x

y y m x x

− = −

− = −

Slope: 2

7, 1 8x = , and 1 4y = , so the slope is

2

7 and a point ( )1 1,x y on the graph is ( )8,9 .

20. ( )

( )1 1

3 9 2y x

y y m x x

− = −

− = −

Slope: 9 , 1 2x = , and 1 3y = , so the slope is


21. ( )

( ) ( )( )1 1

2 5 7

2 5 7

y x

y x

y y m x x

+ = − −

− − = − −

− = −

Slope:–5, 1 7x = , and 1 2y = − , so the slope is

5− and a point ( )1 1,x y on the graph is

( )7, 2− .

22. ( )( )( )

( )

29

29

1 1

4 5

4 5

y x

y x

y y m x x

− = − +

− = − − −

− = −

Slope: 2

9− , 1 5x = − , and 1 4y = , so the slope

is 2


( )5,4− .

23. ( )

( )( )( )

53

53

1 1

4 2

4 2

y x

y x

y y m x x

− = − +

− = − − −

− = −

Slope: 5

3− , 1 2x = − , and 1 4y = , so the slope

is 5


( )2,4− .

24. ( )

( ) ( )( )1 1

7 4 9

7 4 9

y x

y x

y y m x x

+ = − −

− − = − −

− = −

Slope: –4, 1 9x = , and 1 7y = − , so the slope

is –4 and a point ( )1 1,x y on the graph is

( )9, 7− .

25.

( )( )

47

47

1 1

0 0

y x

y x

y y m x x

=

− = −

− = −

Slope: 4

7, 1 0x = , and 1 0y = , so the slope is

4




( )( ) ( )( )

( )

1 1

1

1

Point-slope equation

Substituting 0.6 4 0.6 3for , 3 for ,

and 4 for

4 0.6 1.8 Simplifying

0.6 5.8 Simplifying

Function notation0.6 5.8

y y m x x

y xm x

y

y x

y x

f x x

− = −−− − = − − −

−−

+ = − −= − −= − −

26.

( )( )1 1

3

0 3 0

y x

y x

y y m x x

=− = −

− = −

Slope: 3, 1 0x = , and 1 0y = , so the slope is


27. ( )

( ) ( )

( )

1 1

1 1


Substituting 4 for ,3 4 2

2 for , and 3 for

3 4 8 Simplifying

4 11 Simplifying

Function notation4 11

y y m x x

my x

x y

y x

y x

f x x

− = −

− − = −−

+ = −= −= −

28. ( )

( )( )

( )

1 1

1

1


Substituting 4 for 5 4 1, 1 for , and 5

for

5 4 4 Simplifying

4 1 Simplifying

Function notation4 1

y y m x x

y xm x

y

y x

y x

f x x

− = −−− = − − −

−

− = − −= − += − +

29. ( )( )( )

( )

1 1

3355

1

1

3 125 5

3 285 5

3 285 5


Substituting for 8 4

, 4 for , and 8

for

Simplifying8

Simplifying

Function notation

y y m x x

y x

m x

y

y x

y x

f x x

− = −

−− = − − −−

− = − −= − += − +

30. ( )

( )( )

( )

1 1

1155

1

1

1 25 5

315 5

315 5



, 2 for , and 1

for

Simplifying1

Simplifying

Function notation

y y m x x

y x

m x

y

y x

y x

f x x

− = −

−− = − − −−

− = − −= − += − +

31.



32. ( )

( ) ( )

( )

1 1

1

1


Substituting 2.3 for 5 2.3 4

, 4 for , and 5

for

5 2.3 9.2 Simplifying

2.3 14.2 Simplifying

Function notation2.3 14.2

y y m x x

y x

m x

y

y x

y x

f x x

− = −

− − = −−

+ = −= −= −

33. ( )

( ) ( )

( )

1 1

2277

1

1

27

27

27



, 0 for , and 6

for

Simplifying6 0

Simplifying6

Function notation6

y y m x x

y x

m x

y

y x

y x

f x x

− = −

− − = −−

+ = −= −= −

34. ( )( )

( )

1 1

1144

1

1

14

14

14



, 0 for , and 3

for

Simplifying3 0

Simplifying3

Function notation3

y y m x x

y x

m x

y

y x

y x

f x x

− = −

− = −

− = −= += +

35. ( )

( )( )

( )

1 1

3355

1

1

3 125 5

3 425 5

3 425 5



, 4 for , and 6

for

Simplifying6

Simplifying

Function notation

y y m x x

y x

m x

y

y x

y x

f x x

− = −

− = − −−

− = += += +

36. ( )

( ) ( )

( )

1 1

2277

1

1

2 127 7

2327 7

2327 7



, 6 for , and 5

for

Simplifying5

Simplifying

Function notation

y y m x x

y x

m x

y

y x

y x

f x x

− = −

−− − = − −−

+ = − += − −= − −



37. First find the slope of the line

6 4 2 1

5 1 4 2m

−= = =

−

Use the point-slope equation with 1

2m = and

( ) ( )1 11,4 ,x y= . (We could let ( ) ( )1 15,6 ,x y=

instead and obtain an equivalent equation.)

( )

( )

14 1

21 1

42 21 7

2 21 7

Using function notation2 2

y x

y x

y x

f x x

− = −

− = −

= +

= +


1 6 5

4 2 2m

−= = −

−


2m = −

and ( ) ( )1 12,6 ,x y= . (We could let

( ) ( )1 14,1 ,x y= instead and obtain an

equivalent equation.)

( )

( )

56 2

25

6 525

1125

11 Using function notation2

y x

y x

y x

f x x

− = − −

− = − +

= − +

= − +

39. First find the slope of the line:

( )3 3 3 3 6

1.56.5 2.5 4 4

m− − +

= = = =−

Use the point-slope equation with 1.5m =

and ( ) ( )1 16.5,3 ,x y= .

( )

( )

3 1.5 6.5

3 1.5 9.75

1.5 6.75

1.5 6.75 Using function notation

y x

y x

y x

f x x

− = −− = −

= −= −


( )1.7 1.3

0.67 2

m− −

= =−

Use the point-slope equation with 0.6m =

and ( ) ( )1 17,1.7 ,x y= . (We could let

( ) ( )1 12, 1.3 ,x y− = instead and obtain an

equivalent equation.) ( )

( )

1.7 0.6 7

1.7 0.6 4.2

0.6 4.2 1.7

0.6 2.5

0.6 2.5 Using function notation

y x

y x

y x

y x

f x x

− = −− = −

= − += −= −

41. First find the slope of the line:

2 3 5

50 1 1

m− − −

= = =− −

Observe that the y-intercept is ( )0, 2− . The

using the slope-intercept equation we have 5 2y x= − .

We could also use the point-slope equation with 5m = and ( ) ( )1 11,3 ,x y= .

( )

( )

3 5 1

3 5 5

5 2

5 2

y x

y x

y x

f x x

− = −− = −

= −= −


( )4 0 4

0 3 3m

− −= = −

− −




3m = − .

and ( ) ( )1 13,0 ,x y− = .

( )

( )

( )

40 3

34

334

434

43

y x

y x

y x

f x x

− = − − −

= − +

= − −

= − −


( )( )

6 3 6 3 3 3

4 2 4 2 2 2m

− − − − + −= = = =− − − − + −


2m = .

and ( ) ( )1 12, 3 ,x y− − = .

( ) ( )

( )

( )

33 2

23

3 223

3 323

23

2

y x

y x

y x

y x

f x x

− − = − −

+ = +

+ = +

=

=


( )( )

1 7 1 7 63

2 4 2 4 2m

− − − − += = = =− − − − +

Use the point-slope equation with 3m = .

and ( ) ( )1 12, 1 ,x y− − = .

( ) ( )( )

( )

1 3 2

1 3 2

3 5

3 5

y x

y x

y x

f x x

− − = − − + = +

= += +

45.

To estimate the unemployment for Indiana for 2001, locate the point directly above the year 2001. We then estimate its second coordinate by moving horizontally from that point to the y-axis. The unemployment in 2001 is approximately 5.5%. Using the same approach for year 2004, we determine that the unemployment is approximately 8.7%. 46. To estimate the number of calories burned by a walker weighing 150 lb, we locate the point directly above 150 lb. We then estimate the second coordinate by moving horizontally from that point to the y-axis. The number of calories burned is approximately 88 calories per hour. Following the same procedure for a runner weighing 200 lb, we determine the number of calories burned per hour to be approximately 120. 47. To estimate the number of calories burned by a person walking 4.0 mph, we locate the point directly above 4.0 mph. We then estimate the second coordinate by moving horizontally from that point to the y-axis. The number of calories burned is approximately 90 calories per hour. Following the same procedure for a person walking at a speed of 5.5 mph, we determine the number of calories burned per hour to be approximately 120. 48.



To estimate the amount spent per person on reading material in 2001, we locate the point directly above 2001. We then estimate the second coordinate by moving horizontally from that point to the y-axis. The amount spent is approximately $142 per year. Following the same procedure, we determine the approximate amount spent per person in 2004 to be about $132. 49. (a) We form pairs of the type ( ),t C where t

is the number of years since 1971 and C is the calories consumed per day. We have two pairs, ( )0,1542 and ( )29,1877 .

These are two points on the graph of the linear function we are seeking. We use the point-slope form to write an equation relating C and t:

1877 1542 335

29 0 29m

−= =

−

( )

( )

3351542 0

29335

154229

3351542

29335

1542 29

C t

C t

C t

C t t

− = −

− =

= +

= +

(b) 2009 is 38 years since 1971, so to predict the number of calories an American woman will consume per day in 2009, we find ( )38C :

( ) ( )33538 38 1542

2957, 448

291981

C = +

=

≈

The predicted calorie consumption in 2009 will be approximately 1981 calories per day.

(c) Substitute 2000 for ( )C t and solve for t:

335

2000 154229

3352000 1542

29335

45829

13,282

33540

t

t

t

t

t

= +

− =

=

=

≈

An American woman will consume 2000 calories per day about 40 years since 1971, or in 2011. 50. (a) We form pairs of the type ( ),t C where t

is the number of years since 1971 and C is the calories consumed per day. We have two pairs, ( )0, 2450 and

( )29, 2618 .

These are two points on the graph of the linear function we are seeking. We use the point-slope form to write an equation relating C and t:

2618 2450 168

29 0 29m

−= =

−

( )

( )

1682450 0

29168

245029

1682450

29168

2450 29

C t

C t

C t

C t t

− = −

− =

= +

= +

(b) 2008 is 37 years since 1971, so to predict the number of calories an American man will consume per day in 2008, we find ( )37C :

( ) ( )16837 37 2450

2977,266

292664

C = +

=

≈

The predicted calorie consumption in 2008 will be approximately 2664 calories per day.



(c) Substitute 2750 for ( )C t and solve for t:

168

2750 245029

1682750 2450

29168

30029

725

1452

t

t

t

t

t

= +

− =

=

=

≈

An American man will consume 2750 calories per day about 52 years since 1971, or in 2023. 51. (a) We form pairs of the type ( ),t E where t

is the number of years since 1993 and E is the life expectancy for a male in the United States.. We have two pairs, ( )0,72.2 and ( )10,74.8 .

These are two points on the graph of the linear function we are seeking. We use the point-slope form to write an equation relating E and t:

74.8 72.2 2.6

0.2610 0 10

m−

= = =−

( )

( )

72.2 0.26 0

72.2 0.26

0.26 72.2

0.26 72.2

E t

E t

E t

E t t

− = −− =

= += +

(b) 2009 is 16 years since 1993, so to predict the life expectancy of a male in 2009 find ( )16E :

( ) ( )16 0.26 16 72.2

76.36

E = +=

The predicted life expectancy of a male in 2009 is 76.36 years. 52. (a) We form pairs of the type ( ),t E where t

is the number of years since 1993 and E is the life expectancy of a female in the United States.. We have two pairs, ( )0,78.8 and ( )10,80.1 .

These are two points on the graph of the linear function we are seeking. We use

the point-slope form to write an equation relating E and t:

80.1 78.8 1.3

0.1310 0 10

m−

= = =−

( )

( )

78.8 0.13 0

78.8 0.13

0.13 78.8

0.13 78.8

E t

E t

E t

E t t

− = −− =

= += +

(b) 2010 is 17 years since 1993, so to predict the life expectancy of a female in 2010 find ( )17E :

( ) ( )17 0.13 17 78.8

81.01

E = +=

The predicted life expectancy of a female in 2010 is 81.01 years. 53. (a) We form pairs of the type ( ),t A where t

is the number of years since 1994 and A is amount, in millions of dollars, contributed to congressional candidates by PACs. We have two pairs, ( )0,189.6

and ( )10,310.5 .

These are two points on the graph of the linear function we are seeking. We use the point-slope form to write an equation relating E and t:

310.5 189.6 120.9

12.0910 0 10

m−

= = =−

( )

( )

189.6 12.09 0

189.6 12.09

12.09 189.6

12.09 189.6

A t

A t

A t

A t t

− = −− =

= += +

(b) 2008 is 14 years since 1994, so to predict the amount of PAC contributions in 2008 find ( )14A :

( ) ( )14 12.09 14 189.6

358.86

A = +=

The predicted PAC9 contribution for 2008 is $358.86 million.



54. (a) We form pairs of the type ( ),p A where p

is the price per pound of coffee and A is the amount of coffee, in millions of lb, sold at price p. We have two pairs, ( )8,6.5 and ( )9,4.0 .

These are two points on the graph of the linear function we are seeking. We use the point-slope form to write an equation relating A and p:

4.0 6.5 2.5

2.59 8 1

m− −

= = = −−

( )

( )

6.5 2.5 8

6.5 2.5 20

2.5 26.5

2.5 26.5

A p

A p

A p

A p p

− = − −− = − +

= − += − +

(b) To predict how much coffee consumers would buy at a price of $6 per pound, find ( )6A :

( ) ( )6 2.5 6 26.5

11.5

A = − +=

If coffee were selling at $6 per pound, consumers would be willing to buy 11.5 million lb. 55. (a) We form pairs of the type ( ),t N where t

is the number of years since 1996 and N is the number of millions of tons of recycled solid waste. We have two pairs, ( )0,57.3 and ( )7,72.3 .

These are two points on the graph of the linear function we are seeking. We use the point-slope form to write an equation relating N and t:

72.3 57.3 15

7 0 7m

−= =

−

( )

( )

1557.3 0

715

57.37

1557.3

715

57.3 7

N t

N t

N t

N t t

− = −

− =

= +

= +

(b) 2010 is 14 years since 1996, so to predict the amount of recycled waste in 2010, find ( )14N :

( ) 1514 14 57.3

787.3

N = ⋅ +

=

In 2010, Americans are expected to recycle 87.3 million tons of solid waste. 56. (a) We form pairs of the type ( ),p A where p

is the price per pound of coffee and A is the amount of coffee, in millions of lb, supplied at price p. We have two pairs, ( )8,5.0 and ( )9,7.0 .

These are two points on the graph of the linear function we are seeking. We use the point-slope form to write an equation relating A and p:

7 5 2

29 8 1

m−

= = =−

( )

( )

5 2 8

5 2 16

2 11

2 11

A p

A p

A p

A p p

− = −− = −

= −= −

(b) To predict how much coffee suppliers would be willing to sell at a price of $6 per pound, find ( )6A :

( )6 2 6 11

12 11

1

A = ⋅ −= −=

If coffee were selling at $6 per pound, suppliers would be willing to supply 1 million lb. 57. (a) We form pairs of the type ( ),t N where t

is the number of years since 1999 and N is the number of Americans who used the Internet to find travel information. We have two pairs, ( )0, 48 and ( )7,79 .

These are two points on the graph of the linear function we are seeking. We use the point-slope form to write an equation relating N and t:

79 48 31

7 0 7m

−= =

−



( )

( )

3148 0

731

48731

48731

48 7

N t

N t

N t

N t t

− = −

− =

= +

= +

(b) 2009 is 10 years since 1999, so to predict the number of Americans who used the Internet to find travel information in 2009, find ( )10N :

( ) 3110 10 48

7646

792

N = ⋅ +

=

≈

In 2009, approximately 92 million Americans will use the Internet to find travel information. (c) Substitute 110 for ( )N t and solve for t:

31

110 48731

110 48731

627

14

t

t

t

t

= +

− =

=

=

110 Americans will find travel information on the Internet 14 years after 1999, or in 2013. 58. (a) We form pairs of the type ( ),t R where t

is the number of years since 1991 and R is the time for the 100-meter run. We have two pairs, ( )0,9.86 and ( )14,9.77 .

These are two points on the graph of the linear function we are seeking. We use the point-slope form to write an equation relating R and t:

9.77 9.86 .09 9

14 0 14 1400m

− −= = = −

−

( )

( )

99.86 0

14009

9.861400

99.86

14009

9.86 1400

R t

R t

R t

R t t

− = − −

− = −

= − +

= − +

(b) 2008 is 17 years since 1991, so to predict the time for 2008, find ( )17R :

( ) 917 17 9.86

14009.75

R = − ⋅ +

≈

2015 is 24 years since 1991, so to predict the time for 2015, find ( )24R :

( ) 924 24 9.86

14009.71

R = − ⋅ +

≈

The predicted record times for the 100- meter run in 2008 and 2015 are 9.75 sec and 9.71 sec, respectively. (c) Substitute 9.6 for ( )R t and solve for t:

9

9.6 9.861400

99.86 9.6

14009

0.261400

40.4444

40

t

t

t

t

t

= − +

= −

=

=≈

About 40 years after 1991, or in 2031, the new record will be 9.6 sec. 59. (a) We form pairs of the type ( ),t A where t

is the number of acres, in millions, in the National Park system. We have two pairs, ( )0,78.2 and ( )4,79.0 .

These are two points on the graph of the linear function we are seeking. We use the point-slope form to write an equation relating A and t:

79.0 78.2 0.8

0.24 0 4

m−

= = =−



( )

( )

78.2 0.2 0

78.2 0.2

0.2 78.2

0.2 78.2

A t

A t

A t

A t t

− = −− =

= += +

(b) 2010 is 10 years since 2000, so to predict the amount of land in the National Park system in 2010, find ( )10A :

( )10 0.2 10 78.2

2 78.2

80.2

A = ⋅ += +=

In 2010, the National Park system will contain 80.2 million acres. 60. (a) We form pairs of the type ( ),d P where d

is the depth beneath the ocean’s surface, in ft, and P is the pressure in atm. We have two pairs, ( )100,4 and ( )200,7 .

These are two points on the graph of the linear function we are seeking. We use the point-slope form to write an equation relating P and d:

7 4 3

0.03200 100 100

m−

= = =−

( )

( )

4 0.03 100

4 0.03 3

0.03 1

0.03 1

P d

P d

P d

P d d

− = −− = −

= += +

(b) At 690 ft, in order to find the pressure, we must find ( )690P :

( )690 0.03 690 1

21.7

P = ⋅ +=

The predicted pressure at a depth of 690 ft below the ocean’s surface is 21.7 atm. 61. The points lie approximately on a straight line, so the data appears to be linear. 62. The points lie approximately on a straight line, so the data appears to be linear. 63. The points do not lie on a straight line, so the data is not linear.

64. The points do not lie on a straight line, so the data is not linear. 65. The points lie approximately on a straight line, so the data appears to be linear. 66. The points do not lie on a straight line, so the data is not linear. 67. (a) Enter the data in a graphing calculator, letting x represent the number of years since 1900. Then use the linear regression feature to find the desired function. We have ( ) 0.1673799884 63.2346443W x x= + .

(b) 2010 – 1900 = 110, so we use one of the methods discussed earlier in the text to find ( )110W . We predict that the life

expectancy in 2010 will be about 81.6 yr. This estimate is higher than the one found in Exercise 52. 68. (a) Enter the data in a graphing calculator, letting x represent the number of years since 1900. Then use the linear regression feature to find the desired function. We have ( ) 0.2050086755 53.5967727M x x= + .

(b) 2009 – 1900 = 109, so we use one of the methods discussed earlier in the text to find ( )109M . We predict that the life

expectancy in 2009 will be about 75.9 yr. This estimate is lower than the one found in Exercise 51 69. (a) Enter the data in a graphing calculator, letting x represent the number of years since 1995. Then use the linear regression feature to find the desired function. We have ( ) 885.33 81,693B x x= + .

(b) 2008 – 1995 = 13, so we use one of the methods discussed earlier in the text to find ( )13B . We predict that number of

FDIC-insured financial institutions in 2008 will be about 93,202.



70. (a) Enter the data in a graphing calculator, letting x represent the number of years since 1990. Then use the linear regression feature to find the desired function. We have ( ) 0.1321428571 7.046428571P x x= + .

(b) 2006 – 1990 = 16, so we use one of the methods discussed earlier in the text to find ( )16P . We predict that the number

of near-term births in 2006 is about 9.2%. 71. Writing Exercise. Amy cannot use a linear function to predict her salary for the next five years because the increases in salary are not a constant amount. Each increase is larger than the preceding one because it is calculated on a larger amount than the preceding one. 72. Yes; the slope of the expectancy function for males is greater than the slope for the expectancy for females. Eventually, the line representing the male expectancy function will intersect and be above that of the line for females. 73. The equation is already in slope intercept form. The slope is 3. 74. The equation must first be written in slope- intercept form: ( )10 4 6

10 4 24

4 14

y x

y x

y x

− = − +− = − −

= − −

From the equation, the slope is –4. 75. The equation must first be written in slope- intercept form: 2 6 5

6 2 5

2 5

6 61 5

3 6

x y

y x

y x

y x

+ == − +

= − +

= − +

From the equation, the slope is 1

3− .

76. The equation must first be written in slope- intercept form:

1

21

02

y

y x

=

= ⋅ +

From the equation, the slope is 0. 77. 1

2 5y x= −

The equation is already in slope-intercept form, so the y-intercept is ( )0, 5− . We can

locate the x-intercept by substituting 0 for y and solving for x:

1

0 521

52

10

x

x

x

= −

=

=


78. ( )2 3y x+ = − −

We can determine the y-intercept by substituting 0 for x and solving for y: ( )2 0 3

2 3

1

y

y

y

+ = − −+ =

=


We can determine the x-intercept by substituting 0 for y and solving for x: ( )0 2 3

2 3

1

x

x

x

+ = − −= − +=

The x-intercept is 1,0 . 79. We can determine the y-intercept by substituting 0 for x and solving for y: 4 0 5 20

5 20

4

y

y

y

⋅ + ===


We can determine the x-intercept by substituting 0 for y and solving for x: 4 5 0 20

4 20

5

x

x

x

+ ⋅ ===




80. 4y = −

This is a horizontal line through the point ( )0, 4− . So, the y-intercept is ( )0, 4− and

there is no x-intercept. 81. The function was ( ) 0.2 78.2A t t= + . The

first coefficient means that 0.2 million acres are being added to the National Park system every year. The second coefficient means that there were 78.2 million acres in the National Park system in year 2000 (when 0t = ). 82. In 1999, there were 48 million Americans using the Internet to find travel information. According to our model (answer to Exercise 57b), there will be 92 million Americans using the Internet for travel information in 2009. We would expect that in 2004, the number of users would be about half way between 48 million and 92 million, or 70 million. This might lead us to believe that our model predicts the number of users to be higher than it actually is. 83. Familiarize. The value of C, in degrees Celsius, for any value F, in degrees Fahrenheit can be modeled by a line that contains the points ( )32,0 and ( )212,100 .

Translate.

100 0 100 5

212 32 180 9m

−= = =

−

( )

( )

50 32

95

329

C F

C F

− = −

= −

Carry out. Using function notation, we have

( ) ( )532

9C F F= − . To find out what Celsius

temperature corresponds to 70° F, we find ( )70C :

( ) ( )

( )

570 70 32

95

38921.1

C = −

=

≈

.

Check. We can repeat our calculations. We could also graph the function and determine that ( )70,21.1 is on (or close to) the graph.

State. 70° F corresponds to about 21.1°C. 84. Familiarize. The value C of the computer, in dollars, after t months can be modeled by a line that contains the points ( )6,900 and

( )8,750 .

Translate. We find an equation relating C and t.

750 900 150

758 6 2

m− −

= = = −−

( )900 75 6

900 75 450

75 1350

C t

C t

C t

− = − −− = − +

= − +

Carry out. Using function notation, we have ( ) 75 1350C t t= − + . To find how much the

computer cost we find ( )0C :

( )0 75 0 1350

1350

C = − ⋅ +=

Check. We can repeat our calculations. We could also graph the function and determine that ( )0,1350 is on the graph.

State. The computer cost $1350. 85. Familiarize. The total cost C of the phone, in dollars, after t months, can be modeled by a line that contains the points ( )5,230 and

( )9,390 .


390 230 160

409 5 4

m−

= = =−

( )230 40 5

230 40 200

40 30

C t

C t

C t

− = −− = −

= +

Carry out. Using function notation, we have ( ) 40 30C t t= + . To find the costs already

incurred when the service began we find ( )0C :

( )0 40 0 30 30C = ⋅ + =

Check. We can repeat the calculations. We could also graph the function and determine that ( )0,30 is on the graph.



State. Mel had already incurred $30 in costs when his service just began. 86. Familiarize. The total cost C of operating the restaurant, in dollars, after t months, can be modeled by a line that contains the points ( )4,7500 and ( )7,9250 .


9250 7500 1750

7 4 3m

−= =

−

( )17507500 4

31750 7000

75003 3

1750 22,500 7000

3 3 31750 15,500

3 3

C t

C t

C t

C t

− = −

− = −

= + −

= +

Carry out. Using function notation, we have

( ) 1750 15,500

3 3C t t= + . To find the costs to

operate the restaurant after 10 months, we find ( )10C :

( ) 1750 15,50010 10

3 311,000

C = ⋅ +

=

Check. We can repeat the calculations. We could also graph the function and determine that ( )10,11000 is on the graph.

State. The cost to operate the restaurant after 10 months will be $11,000. 87. From previous work (Exercise 54), the amount of coffee A, in millions of lb, consumers would demand at price p, is given by: ( ) 2.5 26.5 A p p= − + .

Also, from previous work Exercise 56), the amount of coffee A, in millions of pounds, suppliers are willing to sell at price p, is given by: ( ) 2 11A p p= −

To determine the price at which supply equals demand, we must solve: 2.5 26.5 2 11

37.5 4.5

8.33

p p

p

p

− + = −=≈

So, at a price of about $8.33 per pound, supply will equal demand.

88. First solve the equation for y and determine the slope of the given line. 2 6 Given line

2 6

1 13

2 2

x y

y x

y x m

+ == − +

= − + = −


2− .

The slope of every line parallel to the given

line must also be 1

2− . We find the equation

of the line with slope 1

2− and containing the

point ( )3,7 .

( )

( )1 1 Point-slope equation

17 3

21 3

72 21 17

2 2

y y m x x

y x

y x

y x

− = −

− = − −

− = − +

= − +


3 7 3

x y

y x m

− == − =

The slope of the given line is 3. The slope of every line parallel to the given line must also be 3. We find the equation of the line with slope 3 and containing the point ( )1,4− .

( )( )( )

( )

1 1 Point-slope equation

4 3 1

4 3 1

4 3 3

3 7

y y m x x

y x

y x

y x

y x

− = −

− = − −

− = +− = +

= +


2 3 2

x y

y x m

+ = −= − − = −

The slope of the given line is –2. The slope of a perpendicular line is given by

the opposite of the reciprocal of –2, or 1

2.



We find the equation of the line with slope 1

2

containing the point ( )2,5 .

( )15 2 Substituting

21

5 121

42

y x

y x

y x

− = −

− = −

= +


3

1 1

3 3

x y

x y

x y m

− ==

= =


3.

The slope of a perpendicular line is given by

the opposite of the reciprocal of 1

3, or –3.

We find the equation of the line with slope –3 containing the point ( )4,0 .

( )0 3 4 Substituting

3 12

y x

y x

− = − −= − +

92. The price must be a positive number, so we have 0p > . Furthermore, the amount of

coffee sold must be a non-negative number. We have: 2.5 26.5 0

2.5 26.5

10.6

p

p

p

− + ≥− ≥ −

≤

Thus, the domain is { }| 0 10.6p p< ≤ .

93. The price must be a positive number, so we have 0p > . Furthermore, the amount of

coffee supplied must be a nonzero number. We have: 2 11 0

2 11

5.5

p

p

p

− ≥≥≥

Thus, the domain is { }| 5.5p p ≥ .

94. (a) We have two pairs, ( )3, 5− and ( )7, 1− .

Use the point-slope form:

( )1 5 1 5 4

17 3 4 4

m− − − − +

= = = =−

( ) ( )

( )

5 1 3

5 3

88 Using function

notation

y x

y x

y xg x x

− − = −+ = −

= −= −

(b) ( )2 2 8 10g − = − − = −

(c) ( ) 8g a a= −

If ( ) 75g a = , we have

8 75

83.

a

a

− ==

Exercise Set 2.5 1. domain 2. subtract 3. evaluate 4. domains 5. excluding 6. sum 7. Since ( )2 3 2 1 5f = − ⋅ + = − , and

( ) 22 2 2 6g = + = , we have

( ) ( )2 2 5 6 1f g+ = − + = .

8. Since ( ) ( )1 3 1 1 4f − = − − + = , and

( ) ( )21 1 2 3g − = − + = , we have

( ) ( )1 1 4 3 7f g− + − = + = .

9. Since ( )5 3 5 1 14f = − ⋅ + = − and

( ) 25 5 2 27g = + = , we have

( ) ( )5 5 14 27 41f g− = − − = − .

10. Since ( )4 3 4 1 11f = − ⋅ + = − and

( ) 24 4 2 18g = + = , we have

( ) ( )4 4 11 18 29f g− = − − = − .



11. Since ( ) ( )1 3 1 1 4f − = − − + = and

( ) ( )21 1 2 3g − = − + = , we have

( ) ( )1 1 4 3 12f g− ⋅ − = ⋅ = .

12. Since ( ) ( )2 3 2 1 7f − = − − + = and

( ) ( )22 2 2 6g − = − + = , we have

( ) ( )2 2 7 6 42f g− ⋅ − = ⋅ = .

13. Since ( ) ( )4 3 4 1 13f − = − − + = and

( ) ( )24 4 2 18g − = − + = , we have

( ) ( )4 / 4 13/18f g− − = .

14. Since ( )3 3 3 1 8f = − ⋅ + = − and

( ) 23 3 2 11g = + = , we have

( ) ( )3 / 3 8 /11f g = − , or 8

11− .

15. Since ( ) 21 1 2 3g = + = and

( )1 3 1 1 2f = − ⋅ + = − , we have

( ) ( ) ( )1 1 3 2 3 2 5g f− = − − = + = .

16. From Exercise 7, we know that ( )2 6g = and

( )2 5f = − , so ( ) ( ) ( )2 / 2 6 / 5g f = − , or 6

5− .

17. ( )( ) ( ) ( )

( ) ( )2

2

3 1 2

3 3

f g x f x g x

x x

x x

+ = +

= − + + +

= − +

18. ( )( ) ( ) ( )

( ) ( )2

2

2

2 3 1

2 3 1

3 1

g f x g x f x

x x

x x

x x

− = −

= + − − +

= + + −

= + +

19. ( )( ) ( ) ( )

2

2

2 5

3

F G x F x G x

x x

x x

+ = +

= − + −= − +

20. ( )( ) ( ) ( )2

2

2 5

3

F G a F a G a

a a

a a

+ = +

= − + −= − +

21. Using our work from Exercise 20, we have ( )( ) ( ) ( )

( ) ( )2

4 4 4

4 2 5 4

16 2 5 4

23.

F G F G+ − = − + −

= − − + − −= − + +=

22. Using our work from Exercise 20, we have ( )( ) ( ) ( )

( ) ( )2

5 5 5

5 2 5 5

25 2 5 5

33.

F G F G+ − = − + −

= − − + − −= − + +=

23. ( )( ) ( ) ( )

( )2

2

2

2 5

2 5

7

F G x F x G x

x x

x x

x x

− = −

= − − −

= − − +

= + −

Then we have ( )( ) 23 3 3 7

9 3 7

5

F G− = + −= + −=

24. Using our work in Exercise 23, we have ( )( ) 22 2 2 7 1F G− = + − = −

25. ( )( ) ( ) ( )

( )( )2

2 3

2 5

5 10 2

F G x F x G x

x x

x x x

⋅ = ⋅

= − −

= − − +

Then we have

( )( ) ( ) ( ) ( )( )

2 33 5 3 3 10 2 3

5 9 27 10 6

45 27 10 6

56.

F G⋅ − = − − − − + −

= ⋅ − − − −= + − −=

26. Using our work from Exercise 25, we have

( )( ) ( ) ( ) ( )2 34 5 4 4 10 2 4

80 64 10 8 126.

F G⋅ − = − − − − + −= + − − =



27. ( )( ) ( ) ( )2

/ /

2, 5

5

F G x F x G x

xx

x

=

−= ≠−

28. ( )( ) ( ) ( )

( ) ( )2

2

2

5 2

5 2

7

G F x G x F x

x x

x x

x x

− = −

= − − −

= − − +

= − − +

29. Using our work in Exercise 27, we have

( )( ) ( )( )

22 2 4 2 2

/ 25 2 5 2 7

F G− − −

− = = =− − +

30. Using our work in Exercise 27, we have

( )( ) ( )( )

21 2 1 2 1

/ 15 1 5 1 6

F G− − −

− = = = −− − +

31. ( )( ) ( ) ( )2 2 2

26.5 22.5

4

P L P L− = −≈ −≈

We estimate after two years of taking their medication, 4% more of those taking Pravachol suffered heart problems or death than those taking Lipitor. 32. ( )( ) ( ) ( )1 1 1

20 18

2

P L P L− = −≈ −≈

We estimate after one year of taking their medication, 2% more of those taking Pravachol suffered heart problems or death than those taking Lipitor. 33. ( ) ( )( )

( ) ( )2000 2000

2000 2000

1.3 2.2

3.5

N R W

R W

= +

= +≈ +≈

We estimate that 3.5 million U.S. women had children in 2000. 34. ( ) ( )( )

( ) ( )1990 1990

1990 1990

1.3 2.6

3.9

N R W

R W

= +

= +≈ +≈

We estimate that 3.9 million U.S. women had children in 1990. 35. ( )( ) ( ) ( )98 98 98n l n l+ = +

From the middle line of the graph, we can see that ( ) ( )98 98 50n l+ ≈ million. The number

of passengers using the Newark Liberty and LaGuardia airports in 1998 was approximately 50 million. 36. ( )( ) ( ) ( )98 98 98

20 21

41 million

k l k l+ = +≈ +≈

The number of passengers using the Kennedy and LaGuardia airports in 1998 was approximately 41 million. 37. ( ) ( ) ( ) ( )'02 '02 '02 '02F k l n= + +

From the top line of the graph, we can see that ( ) ( ) ( )'02 '02 '02 81k l n+ + ≈ million.

The number of passengers using the Newark Liberty, LaGuardia , and Kennedy airports in 2002 was approximately 81 million. 38. ( ) ( ) ( ) ( )'01 '01 '01 '01F k l n= + +

From the top line of the graph, we can see that ( ) ( ) ( )'01 '01 '01 83k l n+ + ≈ million.

The number of passengers using the Newark Liberty, LaGuardia , and Kennedy airports in 2001 was approximately 83 million. 39. ( )( )

( ) ( )( ) ( ) ( ) ( )

( ) ( )

'02

'02 '02

'02 '02 '02 '02

'02 '02

51 million

F k

F k

k l n k

l n

−

= −

= + + − = +≈

The number of passengers using the Newark Liberty and LaGuardia airports in 2002 was approximately 51 million.



40. ( )( )( ) ( )( ) ( ) ( ) ( )

( ) ( )

'01

'01 '01

'01 '01 '01 '01

'01 '01

54 million

F k

F k

k l n k

l n

−

= −

= + + − = +≈

The number of passengers using the Newark Liberty and LaGuardia airports in 2001 was approximately 54 million.

41. (a) ( ) 5

3f x

x=

−

Since 5

3x − cannot be computed when

the denominator is 0, we find the x-value that causes 3x − to be 0: 3 0

3 Adding 3 to both sides

x

x

− ==

Thus, 3 is not in the domain of f, while all other real numbers are. The domain of f is { }| is a real number 3x x and x ≠ .

(b) ( ) 7

6f x

x=

−

Since 7

6 x− cannot be computed when

the denominator is 0, we find the x-value that causes 6 x− to be 0: 6 0

6 Adding to both sides

x

x x

− ==

Thus, 6 is not in the domain of f, while all other real numbers are. The domain of f is { }| is a real number 6x x and x ≠ .

(c) ( ) 2 1f x x= +

Since we can compute 2 1x + for any real number x, the domain is the set of all real numbers. (d) ( ) 2 3f x x= +

Since we can compute 2 3x + for any real number x, the domain is the set of all real numbers.

(e) ( ) 3

2 5f x

x=

−

Since 3

2 5x − cannot be computed when

the denominator is 0, we find the x-value that causes 2 5x − to be 0: 2 5 0

2 5 Adding 5 to both sides

5 Dividing both sides by 2

2

x

x

x

− ==

=

Thus, 5

2 is not in the domain of f, while

all other real numbers are. The domain

of f is 5

| is a real number 2

x x and x ≠

.

(f) ( ) 3 4f x x= −

Since we can compute 3 4x − for any

real number x, the domain is the set of all real numbers.

42. (a) ( ) 3

1g x

x=

−

Since 3

1x − cannot be computed when

the denominator is 0, we find the x-value that causes 1x − to be 0: 1 0


x

x

− ==

Thus, 1 is not in the domain of g, while all other real numbers are. The domain of g is { }| is a real number 1x x and x ≠ .

(b) ( ) 5g x x= −

Since we can compute 5 x− for any real

number x, the domain is the set of all real numbers.

(c) ( ) 9

3g x

x=

+

Since 9

3x + cannot be computed when

the denominator is 0, we find the x-value that causes 3x + to be 0: 3 0


x

x

+ == − −



Thus, –3 is not in the domain of g, while all other real numbers are. The domain of g is { }| is a real number 3x x and x ≠ − .

(d) ( ) 4

3 4g x

x=

+

Since 4

3 4x + cannot be computed when

the denominator is 0, we find the x-value that causes 3 4x + to be 0: 3 4 0

3 4 Adding 4 to both sides.

4 Dividing both sides by 3

3

x

x

x

+ == − −

= −

Thus, 4

3− is not in the domain of g,

while all other real numbers are. The domain of g is

4


x x and x ≠ −

.

(e) ( ) 3 1g x x= −

Since we can compute 3 1x − for any real number x, the domain is the set of all real numbers. (f) ( ) 7 8g x x= −

Since we can compute 7 8x − for any real number x, the domain is the set of all real numbers. 43. The domain of f and of g is all real numbers. Thus, the Domain of f g+ = Domain of

f g− = Domain of f g⋅ =

{ }| is a real number .x x

44. The domain of f and of g is all real numbers. Thus, the Domain of f g+ = Domain of

f g− = Domain of f g⋅ =

{ }| is a real number .x x

45. Because division by 0 is undefined, we have Domain of { }| is a real number 3f x x and x= ≠ , and

Domain of { }| is a real numberg x x= .

Thus, Domain of f g+ = Domain of f g−

= Domain of f g⋅ = Domain of f.

46. Domain of { }| is any real numberf x= .

Because division by 0 is undefined, we have Domain of { }| is a real number 9g x x and x= ≠ . Thus,

Domain of f g+ = Domain of f g− =

Domain of f g⋅ = Domain of g.

47. Because division by 0 is undefined, we have Domain of { }| is any real number 0f x x and x= ≠ , and

Domain of { }| is any real numberg x x= .


= Domain of f g⋅ = Domain of g.






Domain of { }| is any real numberg x x= .


= Domain of f g⋅ = Domain of f.






Domain of { }| is real number 4g x x and x= ≠ . So,




Domain of f g⋅ =

{ }| is a real number 2 4x x and x and x≠ ≠ .


Domain of { }| is real number 2g x x and x= ≠ . So,


Domain of f g⋅ =


53. Domain of f = Domain of g =

{ }| is a real numberx x . Since ( ) 0g x =

when 3 0x − = , we have ( ) 0g x = when

3x = . We conclude that Domain of /f g =

{ }| is a real number 3x x and x ≠ .



when 5 0x− = , we have ( ) 0g x = when





when 2 8 0x − = , we have ( ) 0g x = when





when 6 2 0x− = , we have ( ) 0g x = when



57. Domain of { }| is a real number 4f x x and x= ≠ .


Since ( ) 0g x = when 5 0x− = , we have

( ) 0g x = when 5x = . We conclude that

Domain of /f g =


58. Domain of { }| is a real number 2f x x and x= ≠ .


Since ( ) 0g x = when 7 0x− = , we have

( ) 0g x = when 7x = . We conclude that

Domain of /f g =


59. ( )( ) ( ) ( )5 5 5 1 3 4F G F G+ = + = + =

( )( ) ( ) ( )7 7 7 1 4 3F G F G+ = + = − + =

60. ( )( ) ( ) ( ) ( )6 6 6 0 3.5 0F G F G⋅ = + = ⋅ =

( )( ) ( ) ( )9 9 9 1 2 2F G F G⋅ = + = ⋅ =

61. ( )( ) ( ) ( )

( )7 7 7

4 1 4 1 5

G F G F− = −

= − − = + =

( )( ) ( ) ( )3 3 3 1 2 1G F G F− = − = − = −

62. ( )( ) ( )( )3 2

/ 3 23 1

FF G

G= = =

( )( ) ( )( )7 1 1

/ 77 4 4

FF G

G

−= = = −

63. From the graph, we see that Domain of { }| 0 9F x x= ≤ ≤ and Domain of

{ }| 3 10G x x= ≤ ≤ . Then Domain of

{ }| 3 9F G x x+ = ≤ ≤ . Since ( )G x is never

0, Domain of { }/ | 3 9F G x x= ≤ ≤ .

64. Using our work in Exercise 63, we have Domain of F G− = { }| 3 9x x≤ ≤ and

Domain of F G⋅ = { }| 3 9x x x≤ ≤ ≤ . Since

( )6 0F = and ( )8 0F = , Domain

/G F = { }| 3 9 6 8x x and x and x≤ ≤ ≠ ≠ .

65.



66. 67. Writing Exercise. The values of ( )( )W L S t+ + decreases as t increases, so

total milk consumption has decreased from 1970 to 2004. 68. Writing Exercise. Answers vary. One possibility is the fear of flying following the Twin Towers attack in New York. 69. 4 7 8

4 7 8 Add 7 to both sides

72 Divide both sides by 4

4

x y

x y y

x y

− == +

= +

70. 3 8 5


3 5Divide both sides by 8

8 83 5

Simpifying8 8

x y

y x x

y x

y x

− =− = − + −

−= + −− −

= −

71. 5 2 3



2 2

x y

y x x

y x

+ = −= − − −

= − −

72. 6 5 2



6 65 1

Simplify6 3

x y

x y y

x y

x y

+ = −= − − −

= − −

= − −

73. Let n represent the number. Five more than twice a number is 49.

5 2 49n↓ ↓ ↓↓ ↓

+ =

74. Let n represent the number.

1

3 572

n − =

75. Let n represent the first integer. ( )1 145x x+ + =

76. Let x represent the number. ( ) 20n n− − =

77. Writing Exercise. The graph of ( )( )y f g x= + will be the graph of ( )y g x=

shifted up c units. 78. Writing Exercise. First draw four graphs with the number of hours after the first dose is taken on the horizontal axis and the amount of Advil absorbed, in mg, on the vertical axis. Each graph would show the absorption of one dose of Advil. Then superimpose the four graphs and, finally, add the amount of Advil absorbed in each graph to create the final graph. 79. Domain of

5


f x x and x = ≠ −

;

domain of { }| is a real number 3g x x and x= ≠ − ;

( ) 0g x = when 4 1 0x − = , or when 1x = or

1x = − . Then domain of

5

/ { | is a real number 2

3 1 1}.

f g x x and x

and x and x and x

= ≠ −

≠ − ≠ ≠ −

80. Domain of { }| is a real number 4F x x and x= ≠ ;

domain of { }| is a real number 3G x x and x= ≠ ;

( ) 0g x = when 2 4 0x − = , or when 2x = −

or 2x = . Then domain of

/ { | is a real number 4

3 2 2}.

F G x x and x

and x and x and x

= ≠≠ ≠ − ≠


Chapter 2 Review Exercises 131

81. Answers may vary.

82. The domain of each function is the set of first coordinates for that function. Domain of { }2, 1,0,1,2f = − − and

Domain of { }4, 3, 2, 1,0,1g = − − − − .

Domain of Domain of f g f g+ = − =

Domain of { }2, 1,0,1f g⋅ = − − .

Since ( )1 0g − = , we conclude that Domain

of { }/ 2,0,1f g = − .

83. Domain of { }| 1 5m x x= − < < .

Domain of { }| is any real numbern x x= .

Since ( ) 0n x = when 2 3 0x − = , we have

( ) 0n x = when 3

2x = . We conclude that

Domain of /m n =

32

{ | is a real number 1 5

}

x x and x

and x

− < <≠

.

84. ( )( ) ( ) ( )2 2 2 1 4 5f g f g+ − = − + − = + =

( )( ) ( ) ( )0 0 0 3 5 15f g f g⋅ = ⋅ = ⋅ =

( )( ) ( ) ( )/ 1 1 / 1 4 / 6 2 / 3f g f g= = =

85. Answers may vary. ( ) 1

2f x

x=

+ and

( ) 1

5g x

x=

−.

86. Because 2 0y = when 3x = , the domain of

{ }3 | is a real number 3y x x and x= ≠ .

Since the graph produced using Connected mode contains the line 3x = , it does not represent 3y accurately. The domain of the

graph produced using Dot mode does not include 3, so it represents 3y more

accurately.

87. Left to the student. 88. Think of adding, subtracting, multiplying, or dividing the y-values for various x-values. (a) IV (b) I (c) II (d) III Chapter 2 Review Exercises 1. True 2. False 3. False 4. False 5. False 6. True 7. True 8. True 9. True 10. True 11. (a) ( )2 3f =

(b) Domain of { }| 2 4f x x= − ≤ ≤ .

(c) ( ) 2f x = when 1x = − .

(d) Range of { }|1 5f y y= ≤ ≤ .

12. ( ) 0.233 5.87A t t= +

Substituting 20 for t, we have ( ) ( )20 0.233 20 5.87

4.66 5.87

10.53

A = += +=

Thus, the median age of cars in 2010 will be 10.53 yr. 13. It is not possible for a vertical line to cross the graph at more than one point, so this is the graph of a function.



14. It is possible for a vertical line to cross the graph at more than one point so this is not the graph of a function. 15. ( ) ( )1 3 1 1

3 3 1

3 2

g a a

a

a

+ = + −= + −= +

16. ( ) 1 3 1 1

3

g a a

a

+ = − +=

17. ( ) ( )0 3 0 1

0 1 1

g = −= − = −

18. We must find x such that ( ) 10g x = .

( ) 3 1 10

3 11

11

3

g x x

x

x

= − ==

=

So ( )g x is 10 when 11

3x = .

19. We must find all values of x such that ( ) 0g x = .

( ) 3 1 0

3 1

1

3

g x x

x

x

= − ==

=

So ( )g x is zero when 1

3x = .

20. ( ) 4 9g x x= − −

The equation is already in slope-intercept form. The slope is –4 and the y-intercept is ( )0, 9− .

21. 6 2 14y x− + =

Writing the equation in slope-intercept form we have

6 2 14



6 61 7

Simplify3 3

y x

y x x

y x

y x

− + =− = − + −

−= + −− −

= −

The slope is 1

3 and the y-intercept is

7

0,3

− .

22. To find the rate of change, or slope, of the graph, we select any two points, say ( )2,75

and ( )8,120 .

We then calculate the slope letting ( ) ( )1 1, 2,75x y = and ( ) ( )2 2, 8,120x y =

2 1

2 1

120 75 457.5

8 2 6

y ym

x x

− −= = = =

− −

So, the value of the apartment is increasing at a rate of $7500 per year. 23. Let ( ) ( )1 1, 4,5x y = and ( ) ( )2 2, 3,1x y = − .

2 1

2 1

1 5 4 4

3 4 7 7

y ym

x x

− − −= = = =

− − − −

24. Let ( ) ( )1 1, 16.4, 2.8x y = − and

( ) ( )2 2, 16.4,3.5x y = − .

( )2 1

2 1

3.5 2.8 0.7

16.4 16.4 0

y ym

x x

− −= = =

− − − −

The slope of the line containing the points is undefined since division by zero is undefined. 25. Let x represent the number of months after August 1, 2005, and let y represent the number of Freecycle recycling groups. Then we have two points with coordinates ( )0, 2938 and ( )10,3579 . The rate at which

the recycling groups are being established is



the slope of the line containing these two points.

3579 2938 641

64.110 0 10

m−

= = =−

The number of Frecycling groups was increasing at a rate of 64.1 groups per month. 26. ( ) 645 9800C t t= +

The first coefficient, 645, indicates that the tuition is increasing at a rate of $645 per year. The second coefficient, $9800, indicates the amount of tuition costs in 1997. 27. Substituting the values for slope and y- intercept in point-slope equation of a line produces the equation 2

7 6y x= − , which

written in function form is ( ) 27 6f x x= − .

28. 2 4 8x y− + =

We determine the y-intercept by substituting 0 for x and solving for y. 2 0 4 8

4 8

2

y

y

y

− ⋅ + ===

We determine the x-intercept by substituting 0 for y and solving for x. 2 4 0 8

2 8

4

x

x

x

− + ⋅ =− =

= −

The x- and y-intercepts are ( )4,0− and ( )0, 2 ,

respectively.

29.

30. 31. 32. 33. 34. The window should show both intercepts: ( )98,0 and ( )0,14 . One possibility is

[ ]10,120, 5,15− − , with Xscl =10.

35. ( )

( )5 1

2 2

y x

x y

+ = −− =

Determine the slope of Equation (1): 5

5

y x

y x

+ = −= − −

The slope is –1. Determine the slope of Equation (2):



2

2

2

x y

y x

y x

− =− = − +

= −

The slope is 1. Since the product of the slopes is 1 1 1− ⋅ = − , the lines are perpendicular. 36. ( )

( )3 5 7 1

7 3 7 2

x y

y x

− =− =

Determine the slope of Equation (1): 3 5 7

3 5

7 7

x y

x y

− =

− =

The slope is 3

7.

Determine the slope of Equation (2): 7 3 7

7 3 7

31

7

y x

y x

y x

− == +

= +

The slope is 3

7 7.

Both lines have the same slope so they are parallel. 37. First find the slope of the line 7y x= + . The

line is already in slope-intercept form and has slope 1m = . We must determine the equation of a line with slope 1/1 1m = − = − and y-intercept of 0, 3− . The line has the equation 3y x= − − .

38. The equation 2 7 0x − = can be written in the form 2 0 7x y+ = so it is a linear equation.

since it is in the form Ax By C+ = where A

and B are not both 0. 39. First we substitute y for ( )f x to obtain

3 8 7x y− = . Since this is in the form

Ax By C+ = where A and B are not both 0, it

is a linear equation. 40. Since one of the variables is squared, this is not a linear equation.

41. This equation can also be written in the form 12 7 1p q−− = . The variable q is written to

the –1 power; so this is not a linear equation. 42. The point-slope form of the equation of a line is ( )1 1y y m x x− = − . Let 2m = − and

( ) ( )1 1, 3,4x y = − . Substituting these values in

the point-slope form gives

( )( )4 2 3y x− = − − − or ( )4 2 3y x− = − + .

43. The slope-intercept form of the equation of a line is y mx b= + . First find the slope of the

line containing the points ( )2,5 and ( )4, 3− − .

3 5 8 4

4 2 6 3m

− − −= = =− − −

Use either point to determine the value of b. We choose ( )2,5 .

4

34

5 238

53

15 8

3 37

3

y x b

b

b

b

b

= +

= ⋅ +

= +

− =

=

The line has the equation 4 7

3 3y x= + or

written in function notation ( ) 4 7

3 3f x x= + .

44. Locate the point directly above the year 1985. Then estimate its second coordinate by moving horizontally from that point to the y- axis. The minimum hourly wage in 1985 is about $3.50.



45. Locate the point directly above the year 2009. Then estimate its second coordinate by moving horizontally from that point to the y- axis. The minimum hourly wage in 2009 is about $5.50. 46. (a) We form pairs of the type ( ),t R where t

is the number of years since 1983 and R is the time for the 200-meter run. We have two pairs, ( )0,19.75 and

( )20,19.32 .

These are two points on the graph of the linear function we are seeking. We use the point-slope form to write an equation relating R and t:

19.32 19.75 .43

0.021520 0 20

m− −

= = = −−

( )

( )

19.75 0.0215 0

19.75 0.0215

0.0215 19.75

0.0215 19.75

R t

R t

R t

R t t

− = − −− = −

= − += − +

(b) 2008 is 25 years since 1983, so to predict the time for 2008, find ( )25R :

( )25 0.0215 25 19.75

19.2125

R = − ⋅ +≈

2013 is 30 years since 1983, so to predict the time for 2015, find ( )30R :

( )30 0.0215 30 19.75

19.105

R = − ⋅ +≈

The predicted record times for the 200- meter run in 2008 and 2013 are 19.21 sec and 19.11 sec, respectively. 47. The points lie approximately on a straight line, so the data appear to be linear. 48. Entering the data in a graphing calculator and using the linear regression feature, we have ( ) 3692.377049 80935.5082B t t= + .

49. 2009 is 19 years after 1990. Use one of the methods discussed in the text to find ( )19B .

We estimate that the number of twin births in 2009 will be 151,091.

50. ( ) 5

3f x

x=

+

Since 5

3x + cannot be computed when

the denominator is 0, we find the x-value that causes 3x + to be 0: 3 0


x

x

+ == − −

Thus, –3 is not in the domain of f, while all other real numbers are. The domain of f is { }| is a real number 3x x and x ≠ − .

51. ( ) 3

5

xf x

+=

Since we can compute 3

5

x + for any

real number x, the domain is the set of all real numbers. 52. ( )( ) ( ) ( )

( )( )( )( )

2

4 4 4

3 4 6 4 1

12 6 16 1

6 17

102

g h g h⋅ = ⋅

= ⋅ − +

= − += ⋅=

53. ( )( ) ( ) ( )

( ) ( )[ ] [ ]( )

2

2 2 2

3 2 6 2 1

6 6 4 1

12 5

17

g h g h− − = − − −

= − − − − + = − − − +

= − −= −

54. ( )( ) ( ) ( )

( )( )2

/ 1 1 / 1

3 1 6

1 1

3 6

1 19 9

2 2

g h g h− = − −

− −=

− +− −=+

−= = −



55. ( )( ) ( ) ( )2

2

2

3 6 1

3 6 1

3 5

g h x x x

x x

x x

+ = − + +

= − + += + −

56. Domain of { }| is any real numberg x x= , and

Domain of { }| is any real numberh x x= .

Thus, Domain of g h+ = Domain of g h⋅ =

. 57. Domain of g = Domain of h=


when 3 6 0x − = , we have ( ) 0g x = when

2x = . We conclude that Domain of /h g =


58. Writing Exercise. For a function, every member of the domain corresponds to exactly one member of the range. Thus, for any function, each member of the domain corresponds to exactly one member of the range. Therefore, a function is a relation. In a relation, every member of the domain corresponds to at least one, but not necessarily exactly one, member of the range. Therefore, a relation may or may not be a function. 59. The slope of a line is the rise between two points on the line divided by the run between those points. For a vertical line, there is no run between any two points, and division by 0 is undefined. Therefore, the slope is undefined. For a horizontal line, there is no rise between any two points, so the slope is 0/run, or 0. 60. The y-intercept is found by substituting 0 for x and solving for y. First replace ( )f x with

y.

( ) ( )( )( )

2

2

02

0

3 0.17 5 2 7

3 0.17 5 2 7

3 0.17 0 5 2 0 7

3 0 5 7

3 1 7

9

x

x

f x x x

y x x

y

y

y

y

+ = + − −

+ = + − −

+ = ⋅ + − ⋅ −

+ = + −+ = −

= −


61. Begin by writing each equation in slope- intercept form.

3 4 12

3 12 4

33

4

x y

x y

x y

− =− =

− =

6 9

9 6

9

6 6

ax y

ax y

ax y

+ = −+ = −

+ =− −

For the lines to be parallel, their slopes must

be equal. We set both slopes, 3

4 and

6

a

−

equal and solve for a.

3

4 64 18

18 9

4 2

a

a

a

=−= −

= − = −

62. Each jar costs $2.49. Also, there is a $0.60 charge per jar for shipping and handling. This makes a total charge of $2.49 + $0.60, or $3.09 per jar. There is a flat fee of $3.75 for shipping and handling no matter how many jars are purchased. Let x = the number of jars purchased and let ( )f x represent the total

cost. Then ( ) 3.09 3.75f x x= + .

63. (a) The slope of the line represents distance covered per unit time, or speed. So, the steeper the line, the greater the speed. Graph III matches this situation since a slower (walking) speed is followed by a faster (train) speed, followed by a slower (walking) speed. (b) Graph IV matches this situation since biking would be faster than running, and running would be faster than walking. (c) Graph I matches this situation since sitting in one spot to fish would correspond to 0 speed. (d) Graph II matches this situation. Waiting corresponds to 0 speed, followed by a fast speed for the train ride, followed by a slower speed for running the rest of the way to work.


Chapter 2 Test 137

Chapter 2 Test 1. (a) To determine ( )2f − locate the point

directly above –2 on the x-axis and then estimate the second coordinate by moving horizontally to the y-axis. From the graph, ( )2 1f − = .

(b) The domain is the set of all values for which f is defined. From the graph it appears that the domain of f = { }| 3 4x x− ≤ ≤ .

(c) To determine any x-value(s) for which ( ) 1

2f x = , locate the point 12 on the y-

axis and move left or right to locate any points on the graph. Then estimate the first coordinate by moving to the x-axis. From the graph ( ) 1

23f = .

(d) The range of f is the set of all outputs or second values of the function. From the graph it appears that the range of f = { }| 1 2y y− ≤ ≤ .

2. (a) 2009 is 9 years after the year 2000. To predict the number of book sales in 2009, find ( )9S .

( ) ( )9 1.24 9 34.5

11.16 34.5

45.66

S = += +=

The predicted sales is $45.66 billion. (b) 1.24 signifies that the rate of increase in U.S. book sales was $1.24 billion per year. 34.5 signifies that U.S. book sales were $34.5 billion in 2000. 3.

Locate the point directly above the year 2002. Estimate the second coordinate of this point by moving horizontally to the y-axis. The number of international visitors to the United States in 2002 was approximately 47.1 million. 4. ( ) 3

5 12f x x= − +

The equation is already in slope-intercept

form. The slope is 3

5− and the y-intercept is

( )0,12 .

5. 5 2 7y x− − =

Put the equation in slope-intercept form. 5 2 7

5 2 7

2 7

5 5

y x

y x

y x

− − =− = +

= − −

The slope is 2

5− and the y-intercept is

7

0,5

− .

6. Let ( ) ( )1 1, 2, 2x y = − − and ( ) ( )2 2, 6,3x y = .

( )( )

2 1

2 1

3 2 3 2 5

6 2 6 2 8

y ym

x x

− −− += = = =

− − − +

7. Let ( ) ( )1 1, 3.1,5.2x y = − and

( ) ( )2 2, 4.4,5.2x y = − .

( )2 1

2 1

5.2 5.2 00

4.4 3.1 1.3

y ym

x x

− −= = = =

− − − − −

8. Using the two labeled points, ( ) ( )1 1, 60,150x y = and ( ) ( )2 2, 120,300x y = ,

we determine the change in y per change in x, or the slope of the line containing the two points.

2 1

2 1

300 150 150 75

120 60 60 30

y ym

x x

− −= = = =

− −

The rate of change is 75 calories per 30 minutes, or 2.5 calories per minute. 9. If the slope is 5m = − and the y-intercept is

( )0, 1− , using the slope-intercept form we can



write the equation as 5 1y x= − − , or in

function notation as ( ) 5 1f x x= − − .

10. 11. 12. 13. 14. Given the function ( ) 2 9f x x= + , or

2 9y x= + , the y-intercept is ( )0,9 . We can

determine the x-intercept by substituting 0 for y and solving for x.

2 9

0 2 9

2 9

9

2

y x

x

x

x

= += +

− =

= −


,02

− , or ( )4.5,0− .

Since the standard viewing window is [–10, 10, –10, 10], the intercepts will be visible. 15. ( )

( )4 2 3 1

3 4 12 2

y x

x y

+ =− + = −

Determine the slope of Equation (1):

4 2 3

4 3 2

3 2

4 43 1

4 2

y x

y x

y x

y x

+ == −

= −

= −

Determine the slope of Equation (2)

3 4 12

4 3 12

33

4

x y

y x

y x

− + = −= −

= −

Since both lines have slope of 3

4, the lines

are parallel. 16. ( )

( )2 5 1

2 6 2

y x

y x

= − +− =

The first equation is already in slope-intercept form, so its slope is –2. Determine the slope of Equation (2):

2 6

2 6

13

2

y x

y x

y x

− == +

= +

The slope of Equation (2) is 1

2.

The product of the slopes of the two lines is

1

2 12

− ⋅ = − , so the lines are perpendicular.


Chapter 2 Test 139

17. (a) We can write 8 7 0x − = in the form 8 0 7x y+ = , so it is a linear equation.

(b) The x-variable is squared so this is not a linear equation.

(c) 2 5 3x y− = is already in Ax By C+ =

form, so it is a linear equation. 18. We are given 4m = and ( ) ( )1 1, 2, 4x y = − − .

The point-slope form of the equation of this

line is ( ) ( )( )4 4 2y x− − = − − , or

( )4 4 2y x+ = + .

19. Let ( ) ( )1 1, 3, 1x y = − and ( ) ( )2 2, 4, 2x y = − .

Determine the slope of the line containing the two points.

( )2 1

2 1

2 1 2 1 11

4 3 4 3 1

y ym

x x

− − −− − + −= = = = = −

− − −

Determine the value for b: y mx b= +

Using either point, we substitute the values for x and y and find b. We choose point ( )3, 1− .

1 1 3

1 3

2

y mx b

b

b

b

= +− = − ⋅ +− = − +

=

The equation is 2y x= − + , or ( ) 2f x x= − +

using function notation. 20. (a) Let m = the number of miles driven and let C = the cost to drive m miles. We have two points of the form ( ),m C ,

( )250,100 and ( )300,115 .

Determine the slope of the line:

115 100 15 3

0.3300 250 50 10

m−

= = = =−

Determine the b value:

( )

3

103

100 25010

100 75

25

C m b

b

b

b

= +

= +

= +=

The equation is 0.3 25C m= + , or

( ) 0.3 25C m m= + using function

notation.

(b) To determine the cost to drive the van 500 mi, we must find ( )500C .

( ) ( )500 0.3 500 25

150 25

175

C = += +=

To drive the van 500 mi, it will cost $175. 21. (a) Enter the data in a graphing calculator, letting x represent the number of years since 1900. Then use the linear regression feature to find the desired function. We have ( ) 0.5375687433 89.50755151A x x= − + .

(b) To predict the number of accidental deaths in 2010, we must compute ( )110A since 2010 – 1900 = 110. We

predict the number of accidental deaths to be about 30.4 deaths per 100,000 population.

22. ( ) 6

2 1

xg x

x

−=

+

Since 6

2 1

x

x

−+

cannot be computed when

the denominator is 0, we find the x-value that causes 2 1x + to be 0: 2 1 0

2 1

1

2

x

x

x

+ == −

= −

Thus, 1

2− is not in the domain of g, while

all other real numbers are. The domain

of g is 1


x x and x ≠ −

.

23. ( ) 3 4g x x= − − and ( ) 2 1h x x= +

(a) ( ) ( )22 2 1

4 1

5

h − = − += +



(b) ( )0 3 0 4

0 4

4

g = − ⋅ −= −= −

(c) ( ) ( )5 3 4 5

3 1

g t t

t

+ = − − += − +

(d) ( )( ) ( ) ( )

( ) ( )( ) ( )( ) ( )

2

3 3 3

3 3 4 3 1

9 4 9 1

13 10

130

g h g h⋅ = ⋅

= − ⋅ − ⋅ +

= − − ⋅ +

= − ⋅= −

(e) To find the zeros of ( )g x , we set

( ) 0g x = and solve for x.

3 4 0

3 4

4

3

x

x

x

− − =− =

= −

(f) Domain of h = Domain of g =


when 3 4 0x− − = , we have ( ) 0g x =

when 4

3x = − . We conclude that

Domain of /f g =

4


x x and x ≠ −

.

24. (a) 1 hr and 40 min is equal to

40 2 5

1 160 3 3

= = hr. We must compute

5

3f

.

5 5

5 153 3

5 25

30

f = + ⋅

= +=

The cyclist will be 30 mi from the starting point 1 hr and 40 min after passing the 5-mi marker. (b) From the equation ( ) 5 15f t t= + , we see

that the cyclist is advancing 15 mi for every hr he travels, so the rate is 15 mph.

25. The line in question must be parallel to the line having equation 2 5 8x y− = . Writing

this equation in slope-intercept form, we have 2 5 8

5 2 8

2 8

5 5

x y

y x

y x

− =− = − +

= −

The slope must be 2

5. The line must contain

the point ( )3,2− . Using the point-slope form

we can determine the equation:

( )( )

( )

22 3

52

2 35

y x

y x

− = − −

− = +

We can rewrite this equation in slope- intercept form.

( )22 3

52 6

25 52 6

25 52 16

5 5

y x

y x

y x

y x

− = +

− = +

= + +

= +

26. The line in question must be perpendicular to the line having equation 2 5 8x y− = .

Writing this equation in slope-intercept form, we have 2 5 8

5 2 8

2 8

5 5

x y

y x

y x

− =− = − +

= −

The slope must be ( )25

51/

2− = − . The line

must contain the point ( )3,2− . Using the

point-slope form we can determine the equation:

( )( )

( )

52 3

25

2 32

y x

y x

− = − − −

− = − +

We can rewrite this equation in slope- intercept form.


Chapter 2 Test 141

( )52 3

25 15

22 25 15

22 25 11

2 2

y x

y x

y x

y x

− = − +

− = − −

= − + −

= − −

27. If the graph is to be parallel to the line 3 2 7x y− = , then we must determine the

slope: 3 2 7

2 3 7

3 7 3

2 2 2

x y

y x

y x m

− =− = − +

= − =

The line must contain the two points ( ),3r

and ( )7, s . The slope of this line must be 3

2.

( ) ( )

3 3

2 73 7 2 3

21 3 2 6

21 3 6 2

27 3 2

2 3 27

3 27 27 3

2 2 2

s

rr s

r s

r s

r s

s r

rs r

−=−

− = −− = −− + =− == − +

−= − + =

28. Answers vary. In order to have the restriction

on the domain of / /f g h that 3

4x ≠ and

2

7x ≠ , we need to find some function ( )h x

such that ( ) 0h x = when 2

7x = .

2

77 2

7 2 0

x

x

x

=

=− =

One possible answer is ( ) 7 2h x x= − .

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