Chapter 2 Functions, Linear Equations, and Models Exercise Set 2.1 1. correspondence 2. exactly 3. domain 4. range 5. horizontal 6. vertical 7. “f of 3,” “f at 3,” or “the value of f at 3” 8. vertical 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 than 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 not a function because one member of the domain, 2008, corresponds to three musicians and another, 2009, corresponds to two musicians. 14. The correspondence is a function because each celebrity corresponds to just one birthday. 15. The correspondence is a function because each predator corresponds to just one prey. 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 USB flash drive would have only one storage capacity. 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. 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) The domain is the set of all x-values. It is { } 3, 2, 0, 4 − − . b) The range is the set of all y-values. It is { } 10,3,5,9 − . c) The correspondence is a function. 22. a) The domain is the set of all x-values. It is { } 0,1,2,5 . b) The range is the set of all y-values. It is { } 1,3 − . c) The correspondence is a function. 23. a) The domain is the set of all x-values. It is { } 1,2,3,4,5 . b) The range is the set of all y-values. It is { } 1. c) The correspondence is a function. 24. a) The domain is the set of all x-values. It is { } 1. b) The range is the set of all y-values. It is { } 1,2,3,4,5 . c) The correspondence is not a function. 25. a) The domain is the set of all x-values. It is { } 2,3,4 − .
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Chapter 2
Functions, Linear Equations, and Models Exercise Set 2.1 1. correspondence 2. exactly 3. domain 4. range 5. horizontal 6. vertical 7. “f of 3,” “f at 3,” or “the value of f at 3” 8. vertical 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 than 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 not a function because one member of the domain, 2008, corresponds to three musicians and another, 2009, corresponds to two musicians. 14. The correspondence is a function because each celebrity corresponds to just one birthday. 15. The correspondence is a function because each predator corresponds to just one prey. 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 USB flash drive would have only one storage capacity. 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. 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) The domain is the set of all x-values. It is { }3, 2,0,4− − .
b) The range is the set of all y-values. It is { }10,3,5,9− .
c) The correspondence is a function. 22. a) The domain is the set of all x-values. It is { }0,1,2,5 .
b) The range is the set of all y-values. It is { }1,3− .
c) The correspondence is a function. 23. a) The domain is the set of all x-values. It is { }1,2,3,4,5 .
b) The range is the set of all y-values. It is { }1 .
c) The correspondence is a function. 24. a) The domain is the set of all x-values. It is { }1 .
b) The range is the set of all y-values. It is { }1,2,3,4,5 .
c) The correspondence is not a function. 25. a) The domain is the set of all x-values. It is { }2,3,4− .
78 Chapter 2: Intermediate Algebra: Graphs and Models
b) The range is the set of all y-values. It is { }8, 2,4,5− − .
c) The correspondence is not a function. 26. a) The domain is the set of all x-values. It is { }0,4,7,8 .
b) The range is the set of all y-values. It is { }0,4,7,8 .
c) The correspondence is a function. 27. 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− ≤ ≤ .
28. 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 1f = − .
b) The domain is the set of all x-values in the graph. It is { }| 4 3x 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 –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− ≤ ≤ .
29. 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 { }| 4 2x 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 –2. Thus, the x-value for which ( ) 2f x = is –2.
d) The range is the set of all y-values in the graph. It is { }| 3 3y y− ≤ ≤ .
30. 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 = .
b) The domain is the set of all x-values in the graph. It is { }| 4 3x 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 0. 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− ≤ ≤ .
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, 3. Thus, ( )1 3f = .
b) The domain is the set of all x-values in the graph. It is { }| 4 3x 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 –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− ≤ ≤ .
32. 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 = .
b) The domain is the set of all x-values in the graph. It is { }| 3 4x 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 –1. Thus, the x-value for which ( ) 2f x = is –1.
d) The range is the set of all y-values in the graph. It is { }| 0 5y y≤ ≤ .
33. 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, 1. Thus, ( )1 1f = .
b) The domain is the set of all x-values in the graph. It is { }| 3, 1,1,3,5x − − .
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 { }| 1,0,1,2,3y − .
34. 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 = .
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 .
35. 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 = .
b) The domain is the set of all x-values in the graph. It is { }| 3 4x 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. There are two such points. They are ( )1,2−
and ( )3,2 . Thus, the x-values for which
( ) 2f x = are –1 and 3.
d) The range is the set of all y-values in the graph. It is { }| 4 5y y− ≤ ≤ .
36. 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 { }| 5 2x x− ≤ ≤ .
80 Chapter 2: Intermediate Algebra: Graphs and Models
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 ( )5,2−
and ( )1,2 . Thus, the x-values for which
( ) 2f x = are –5 and 1.
d) The range is the set of all y-values in the graph. It is { }| 3 5y y− ≤ ≤ .
37. 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, 1. Thus, ( )1 1f = .
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 − .
38. 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 { }| 4 4x 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 { }| 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 .
45. Domain: { }| is a real number 2x x and x ≠ − ;
range: { }| is a real number 4y y and y ≠ −
46. Domain: { }| is a real number 5x x and x ≠ ;
range: { }| is a real number 2y y and y ≠
47. Domain: { }| 0x x ≥ , or [ )0,∞ ;
range: { }| 0y y ≥ , or [ )0,∞
48. Domain: { }| 3x x ≤ , or ( ],3−∞ ;
range: { }| 0y y ≥ , or [ )0,∞
49. We can use the vertical line test.
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.
It is possible for a vertical line to intersect the graph more than once. Thus this is not the graph of a function. 51. We can use the vertical line test.
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. 52. 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.
53. 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. 54. We can use the vertical line test.
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. 55. 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.
82 Chapter 2: Intermediate Algebra: Graphs and Models
56. We can use the vertical line test.
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. 57. ( ) 2 3g x x= +
84 Chapter 2: Intermediate Algebra: Graphs and Models
75. ( ) 2 5
5 2 5
0 2
0
f x x
x
x
x
= −− = −
==
76. ( )
( ) ( )2 5
4 2 4 5
8 5
13
f x x
f
= −
− = − −= − −= −
77. ( )
( )( )
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
= += +
− =− =− =
− =
− =
78. ( )
( )( )
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
= +− = +
− − =− − =
− =
− =− =
79. ( )
( ) ( )13
1 1 12 3 2
16
251 246 6 6
4
4
4
f x x
f
= +
= += += + =
80. ( )
( ) ( )13
1 1 13 3 3
361 19 9 9
359
4
4
4
f x x
f
= +
− = − +
= − + = − +=
81. ( ) 4
7 4
4 7 3
f x x
x
x
= −= −= − = −
So, ( )3 7f − = .
82. ( )
( )( )
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 − =
83. ( ) 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 − = − .
84. ( ) 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
= −= −= −= −−= = − −
85. The graph crosses the x-axis at only one point whose coordinate is ( )2,0− , so –2 is the zero
of the function. 86. 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. 87. The graph does not cross the x-axis at all so there are no zeros of this function. 88. The graph crosses the x-axis at only one point whose coordinates are ( )4,0 , so 4 is the zero
of this function. 89. 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. 90. The graph does not cross the x-axis at all so there are no zeros of this function.
88 Chapter 2: Intermediate Algebra: Graphs and Models
b) ( )10f
Since 10 10 10− ≤ ≤ , ( )f x x2= . Thus,
( ) 210 10 100f = = .
c) ( )11f
Since 11 10> , ( ) 2 10f x x= + . Thus,
( ) 211 11 10 121 10 131f = + = + = .
118. ( )2
2
2 3, if 2
, if 2 4
5 7, if 4
x x
f x x x
x x
⎧ − ≤⎪= < <⎨⎪ − ≥⎩
a) ( )0f
Since 0 2≤ , ( ) 22 3f x x= − . Thus,
( ) ( )20 2 0 3 0 3 3f = = − = − = − .
b) ( )3f
Since 2 3 4< < , ( ) 2f x x= . Thus,
( ) 23 3 9f = = .
c) ( )6f
Since 6 4≥ , ( ) 5 7f x x= − . Thus,
( )6 5 6 7 30 7 23f = ⋅ − = − = .
119. Thinking and Writing Exercise. The
expression 3
2
x + is defined for all real
numbers, but the expression 2
3x + is not
defined for 3x = − . 120. Thinking and Writing Exercise. The notation ( )n z implies that n is a function of z, or that
the value of n depends on the variable z. Thus, z is the independent variable.
121. 6 3 3 1
2 7 9 3
− = − = −− −
122. ( )2 4 2
5 8 3
− − −= −
−
123. ( )
( )5 5 0
03 10 13
− − −= =
− −
124. ( )2 3 5
13 2 5
− −= − = −
− −
125. 2 8
2 8
2 8
x y
y x
y x
− =− = − +
= −
126. 5 5 10
5 5 10
2
x y
y x
y x
+ == − += − +
127. 2 3 6
3 2 6
22
3
x y
y x
y x
+ == − +
= − +
128. 5 4 8
4 5 8
52
4
x y
y x
y x
− =− = − +
= −
129. Thinking and Writing Exercise. Jaylan should choose the number of fish as the independent variable, since the amount of food depends on the number of fish. Therefore, the amount of food is the dependent variable, and the number of fish is the independent variable. 130. Thinking and Writing Exercise ( )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.
131. To find ( )( )4f g − , we first find ( )4g − :
132. 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− = = ⋅ + = .
133. 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= .
134. 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. 135. 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 largest contraction was approximately 2 min 50 sec into the test. 136. 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.
137. 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. 138. 139. 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.
96 Chapter 2: Intermediate Algebra: Graphs and Models
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 3 hr, 200 ml was dripped. This is a rate of 200/3, or
2366 ml/hr. Then 100 ml was dripped in
the next hour, a rate of 100 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.
68. The marathoner’s speed is given by
change in distancechange in time
. Note that the runner
reaches the 22-mi point 56 min after the 15-mi point was reached or after 2 hr, 56 min. We will express time in hours: 2 hr, 56 min = 14
152
hr. Then change in distance 22 15 7
change in time 14 142 215 1515 157 , or 7.5 mph14 2
−= =−
= ⋅ =
The marathoner’s speed is 7.5 mph. 69. The skier’s speed is given by
change in distancechange in time
. Note that the skier
reaches the 12-km mark 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. 70. The rate at which the number of recycling
groups increased is given by change in number of groups
.change in time
change in number of groups 4224 2936
change in time 28 months
128846
28
−=
= =
The number of recycling groups is increasing at a rate of 46 groups per month.
71. The work rate is given by change in portion of house painted
change in time⋅
change in portion of house paintedchange in time
2 1 53 4 5 512 18 0 8 12 8 96
−= = = ⋅ =−
The painter’s work rate is 596
of the house per
hour. 72. The average rate of descent is given by
change in altitudechange in time
. We will express time in
minutes:
60 min311 hr hr 90 min
2 2 1 hr= ⋅ =
2 hr 10 min, = 2 hr + 10 min
60 min2 hr 10 min 120 min 10 min1 hr
130 min
= ⋅ + = +
=
Then
change in altitude 0 12 000
change in time 130 90
12 000 300.40
− ,= −− ,= = −
The average rate of descent is 300 ft/min. 73. The rate at which the number of hits is
increasing is given by change in number of hits
change in time.
change in number of hits 430 000 80 000change in time 2009 2007
350 000175 000
2
, − ,= −,= = ,
The number of hits is increasing at a rate of 175,000 hits/yr.
74. a) Graph III is appropriate, because it shows
that the rate before January 1 is $3000/month while it is $2000/month after January 1.
b) Graph IV is appropriate, because it shows 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 1 and $2000− /month after January 1.
75. ( ) 0.75 30C d d= +
0.75 signifies the cost per mile is $0.75; 30 signifies that the minimum cost to rent a truck is $30.
76. ( ) 0 05 200P x x= . +
0.05 signifies that a salesperson earns a 5% commission on sales. 200 indicates that a salesperson’s base salary is $200 per week.
77. 1( ) 52
L t t= +
12
signifies that Lauren’s hair grows 12
in. per
month. 5 signifies that her hair is 5 in long immediately after she gets it cut.
78. ( ) 1913439
5D t t= +
191
5 signifies that the demand increases
191
5
billion kWh per year, for years after 2000. 3439 signifies that the demand was 3439 billion kWh in 2000.
79. ( ) 175.5
8A t t= +
1
8 signifies that the life expectancy of
American females increases 1
8 of a year, per
year, for years after 1970. 75.5 signifies that the life expectancy for a female born in 1970 was 75.5 years.
80. ( ) 12
8G t t= +
1
8 signifies that the grass grows
1
8in. per day.
2 signifies that the grass is 2 in. long immediately after is it cut.
81. ( ) 0.89 16.63P t t= + 0.89 signifies that the average price of a ticket increases by $0.89 per year, for years after 2000. 16.63 signifies that the cost of a ticket is $16.63 in 2000.
82. ( ) 2 2.5C d d= +
2 signifies that the cost per mile of a taxi ride is $2. 2.5 signifies that the minimum cost of a taxi ride is $2.50.
83. ( ) 849 5960C t t= +
849 signifies that the number of acres of organic cotton increases by 849 acres per year, for years after 2006. 5960 signifies that 5960 acres were planted with organic cotton in 2006.
84. ( ) 25 75C x x= +
25 signifies that the cost per person is $25. 75 signifies that the setup cost for the party is $75.
85. ( ) 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 18x t| ≤ ≤ .
86. ( ) 2000 15 000V 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) 0 2000 15 000
2000 15 000
7 5
t
t
t
= − + ,= ,= .
It will take 7.5 yr for the machine to depreciate completely.
98. Thinking and Writing Exercise. a) Answers will vary. b) The profit increases each year but not as
much as in the previous year. 99. 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.
100. Look for the section of the graph that has the
greatest positive slope. Ponte sul Pesa to Panzano is the steepest part of the trip.
101. The longest uphill climb is the widest rising
line. It is the trip from Sienna to Castellina in Chianti.
102. From Le Bolle to Passo dei Pecorai and from
Passo dei Pecorai to Strada in Chianti the road’s grade is about the same. The grade of the road from Strada in Chianti to Poggio Ugolino is about 1
10 of the previous two
sections. Brittany begin her ride in Le Bolle.
103. Reading from the graph the trip from Castellina in Chianti to Ponte sul Pesa is downhill, then to Panzano is uphill and then to Creve in Chianti is downhill. All sections are about the same grade. So the trip began at Castellina in Chianti.
104. From Ponte sul Pesa to Panzano the elevation
is about 4% as shown.
change in elevationgrade
change in horizontal distance
500 300 m 200 m35 30 km 5 km200 m 2 0.04 4%
5000 m 50
=
−= =−
= = = =
105.
( )
rx py s ry
ry py rx s
y r p rx s
sry xr p r p
+ = −+ = − ++ = − +
= − ++ +
The slope is rr p
− + , and the y-intercept
is ( )0 .sr p
, +
106. We first solve for y .
rx py s
py rx s
sry xp p
+ == − +
= − +
The slope is rp
− , and the y-intercept is
( )0 .sp
,
107. Since 1 1( , )x y and 2 2( , )x y are two points on
112 Chapter 2: Intermediate Algebra: Graphs and Models
76. 19x y− =
We rewrite the second equation in slope- intercept form: 19
19
x y
y x
− == −
a) The slope of this line is 1, so the slope of a parallel line is also 1.
b) The reciprocal of this slope is 1
11= , and
the opposite of this number is −1, so the slope of a perpendicular line is −1. 77. 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= + .
78. 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= − − .
79. 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= − − .
80. First we find the slope of the given line. 3 10
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= + .
81. 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− .
The y-intercept is 1
0,3
⎛ ⎞−⎜ ⎟⎝ ⎠, so the equation of
the desired function is ( ) 2 1
5 3f x x= − − .
82. 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
− =− =
= −
= −
The slope of the given line is 1
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= + .
83. First we find the slope of the given line. 3 12
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 = − .
84. 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 = .
85. 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 4y x= − + .
114 Chapter 2: Intermediate Algebra: Graphs and Models
94. 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. 95. 10xy =
The equation cannot be written in standard form for a linear equation, Ax By C+ = ,
since there is an xy term. Thus, the equation is not linear.
96. 10
yx
=
10xy =
When trying to convert the equation into standard linear form, multiply both sides by x. The equation cannot be written in standard form for a linear equation, since there is an xy term. The equation is not linear. 97. ( )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.
98. ( )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− .
99. ( ) 1g x
x=
Replacing ( )g x with y and attempt to write
the equation in standard form.
1
1
yx
xy
=
=
The equation is not linear because it has an xy-term. 100. ( ) 3f x x=
Replace ( )f x with y and attempt to write the
equation in standard form. 3
3 0
y x
x y
=
− + =
The equation is not linear because it has an 3x -term.
101. ( ) 2
5
f xx=
Replace ( )f x with y and attempt to write the
equation in standard form.
2
2
2
5
5
5 0
yx
y x
x y
=
=
− + =
The equation is not linear because it has an 2x -term.
102. ( )
32
g xx= +
Replace ( )g x with y and attempt to write the
equation in standard form.
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.
103. Thinking and Writing Exercise. 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.
104. Thinking and 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.
123. 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. The point-slope form of a line is:
incandescent wattage as the first coordinate and the corresponding CFL wattage as the second coordinate.
To predict the CFL wattage that creates light equivalent to a 75-watt incandescent bulb, locate the point on the line directly above 75. Then move horizontally to the vertical axis and read the CFL wattage value. The wattage is about 19 watts. To predict the CFL wattage that creates light equivalent to a 120-watt incandescent bulb, locate the point on the line directly above 120. Then move horizontally to the vertical axis and read the CFL wattage value. The wattage is about 30 watts.
46. The data values were plotted as shown.
To predict the CFL wattage that creates light
equivalent to a 40-watt incandescent bulb, locate the point on the line directly above 40. Then move horizontally to the vertical axis and read the CFL wattage value. The wattage
is about 9 watts. To predict the CFL wattage that creates light equivalent to a 150-watt incandescent bulb, locate the point on the line directly above 150. Then move horizontally to the vertical axis and read the CFL wattage value. The wattage is about 38 watts.
47. Plot the given data values using body weight
as the first coordinate and the corresponding number of drinks as the second coordinate.
To estimate the number of drinks that a 140-lb person would have to drink to be considered intoxicated, locate the point on the line directly above 140. Then move horizontally to the vertical axis and read the corresponding number of drinks. The estimated number of drinks is 3.5. To estimate the number of drinks that a 230-lb person would have to drink to be considered intoxicated, locate the point on the line directly above 230. Then move horizontally to the vertical axis and read the corresponding number of drinks. The estimated number of drinks is 5.75.
48. The data values were plotted as shown.
To estimate the number of drinks that a 120-lb person would have to drink to be considered intoxicated, locate the point on the line directly above 120. Then move horizontally to the vertical axis and read the corresponding number of drinks. The estimated number of drinks is 3. To estimate the number of drinks that a 250-lb person would have to drink to be considered intoxicated, locate the point on the line directly above 250. Then move horizontally to the vertical axis and read the
124 Chapter 2: Intermediate Algebra: Graphs and Models
corresponding number of drinks. The estimated number of drinks is 6.25.
49. a) The problem says to let t = the number of
years after 2000, and a(t) = the world production capacity, in millions of vehicles, for the year t. Then the production capacity of 84 million vehicles in 2008, and of 97 million in 2015 correspond to data points: (2008 2000,84) (8,84)− = and
(2015 2000,97) (15,97).− =
Find the slope of the function that fits the data:
(15) (8)
15 897 84 13
7 7
a am
−=−
−= =
Use the value of m, and either data point, and substitute them into the point-slope equation for a line.
( )13( ) 84 8
713 104
( ) 847 7
13 104 588( )
7 7 713 484
( ) , 7 7
13 484or ( )
7
a t t
a t t
a t t
a t t
ta t
− = −
= − +
= − +
= +
+=
b) In 2013, t = 2013 – 2000 = 13.
( ) ( )13 13 484 169 48413 =
7 7653 2
93 93.37 7
a+ +=
= = ≈
The model predicts the production capacity in 2013 will be about 93.3 million vehicles.
c) Set ( ) 100a t = and solve for t. 13 484
1007
13 484 700
13 700 484 216
21616.6
132000 16.6 2016.6
t
t
t
t
+ =
+ == − =
= ≈
+ =
The model predicts that the production
capacity will reach 100 million vehicles in 2016.
50. a) The problem says to let t = the number of
years after 2000, and v(t) = the number of Las Vegas convention attendees, in millions, for the year t. Then the 4.6 million attendees in 2002, and 6.1 million in 2006 correspond to data points: (2002 2000,4.6) (2,4.6)− = and
(2006 2000,6.1) (6,6.1).− =
Find the slope of the function that fits the data:
(6) (2)
6 26.1 4.6 1.5
0.3754 4
v vm
−=−−= = =
Use the value of m, and either data point, and substitute them into the point-slope equation for a line.
( )( ) 4.6 0.375 2
( ) 0.375 0.75 4.6
( ) 0.375 3.85
v t t
v t t
v t t
− = −= − += +
b) In 2011, t = 2011 – 2000 = 11. ( ) ( )11 0.375 11 3.85
4.125 3.85
7.975
v = += +=
The model predicts there will be 7.975 million attendees in 2011.
c) Set ( ) 8v t = and solve for t. 0.375 3.85 8
0.375 8 3.85 4.15
4.1511.1
0.3752000 11.1 2011.1
t
t
t
+ == − =
= ≈
+ =
The model predicts that the number of attendees will reach 8 million in 2011.
51. a) The problem says to let t = the number of
years after 1990, and E(t) = the life expectancy, in years, for females born in the year t. Then the expected value of 79.0 years in 1994, and 80.2 years in 2006 correspond to data points: (1994 1990,79.0) (4,79.0)− = and
(2006 1990,80.2) (16,80.2).− =
Find the slope of the function that fits the data:
Use the value of m, and either data point, and substitute them into the point-slope equation for a line.
( )( ) 79 0.1 4
( ) 0.1 0.4 79
( ) 0.1 78.6
E t t
E t t
E t t
− = −= − += +
b) In 2012, t = 2012 – 1990 = 22. ( )(22) 0.1 22 78.6
2.2 78.6 80.8
E = += + =
The model predicts that the life expectancy for a female born in 2012 will be 80.8 years.
52. a) The problem says to let t = the number of
years after 1990, and E(t) = the life expectancy, in years, for males born in the year t. Then the expected value of 72.4 years in 1994, and 75.1 years in 2006 correspond to data points: (1994 1990,72.4) (4,72.4)− = and
(2006 1990,75.1) (16,75.1).− =
Find the slope of the function that fits the data:
(16) (4)
16 475.1 72.4 2.7
0.22512 12
E Em
−=−
−= = =
Use the value of m, and either data point, and substitute them into the point-slope equation for a line.
( )( ) 72.4 0.225 4
( ) 0.225 0.9 72.4
( ) 0.225 71.5
E t t
E t t
E t t
− = −= − += +
b) In 2012, t = 2012 – 1990 = 22. ( )(22) 0.225 22 71.5
4.95 71.5 76.45
E = += + =
The model predicts that the life expectancy for a male born in 2012 will be 76.45 years.
53. a) The problem says to let t = the number of
years after 2000, and N(t) = the amount of solid waste recycled, in millions of tons, in the year t. Then the 53 million tons recycled in 2000, and the 61 million
tons recycled in 2008 correspond to data points: (2000 2000,53) (0,53)− = and
(2008 2000,61) (8,61).− =
Find the slope of the function that fits the data:
(8) (0)
8 061 53 8
18 8
N Nm
−=−
−= = =
Use the value of m, and either data point, and substitute them into the point-slope equation for a line.
( )( ) 53 1 0
( ) 53
N t t
N t t
− = −= +
b) In 2012, t = 2012 – 2000 = 12. (12) 12 53
65
N = +=
The model predicts that 65 million tons of solid waste will be recycled in 2012.
54. a) The problem says to let t = the number of
years after 2000, and A(t) = the contributions by PACs, in millions of dollars, to federal candidates in the year t. Then the $282 million contributed in 2002, and the $412.8 million contributed in 2008 correspond to data points: (2002 2000,282) (2,282)− = and
(2008 2000,412.8) (8,412.8).− =
Find the slope of the function that fits the data:
( ) ( )8 2
8 2412.8 282 130.8
21.86 6
A Am
−=
−−= = =
Use the value of m, and either data point, and substitute them into the point-slope equation for a line.
( )( ) 282 21.8 2
( ) 21.8 43.6 282
( ) 21.8 238.4
A t t
A t t
A t t
− = −= − += +
b) In 2012, t = 2012 – 2000 = 12. ( )(12) 21.8 12 238.4
261.6 238.4 500
A = += + =
The model predicts that PACs will donate $500 million to federal candidates in 2012.
126 Chapter 2: Intermediate Algebra: Graphs and Models
55. a) The problem says to let t = the number of years after 2006, and C(t) = the percentage of Americans familiar with the term "carbon footprint" in the year t. Then the 38% of Americans familiar with the term in 2007, and the 57% familiar with the term in 2009 correspond to data points: (2007 2006,38) (1,38)− = and
(2009 2006,57) (3,57).− =
Find the slope of the function that fits the data:
( ) ( )3 1
3 157 38 19
9.52 2
C Cm
−=
−−= = =
Use the value of m, and either data point, and substitute them into the point-slope equation for a line.
( )( ) 38 9.5 1
( ) 9.5 9.5 38
( ) 9.5 28.5
C t t
C t t
C t t
− = −= − += +
b) In 2012, t = 2012 – 2006 = 6. ( )(6) 9.5 6 28.5
57 28.5 85.5
C = += + =
The model predicts that 85.5% of Americans will be familiar with the term "carbon footprint" in 2012.
c) Set ( ) 100C t = and solve for t.
9.5 28.5 100
9.5 100 28.5
9.5 71.5
71.57.5
9.52006 7.5 2013.5
t
t
t
t
+ == −=
= ≈
+ =
The model predicts all Americans will be familiar with the term “carbon footprint” in 2013.
56. a) The problem says to let t = the number of
years after 2000, and M(t) = the Medicaid long-term care expenses, in billions of dollars, in the year t. Then the $92 billion in expenses in 2002, and the $109 billion in expenses in 2006 correspond to data points: (2002 2000,92) (2,92)− = and
(2006 2000,109) (6,109).− =
Find the slope of the function that fits the
data: ( ) ( )6 2
6 2109 92 17
4.254 4
M Mm
−=
−−= = =
Use the value of m, and either data point, and substitute them into the point-slope equation for a line.
( )( ) 92 4.25 2
( ) 4.25 8.5 92
( ) 4.25 83.5
M t t
M t t
M t t
− = −= − += +
b) In 2010, t = 2010 – 2000 = 10. ( )(10) 4.25 10 83.5
42.5 83.5 126
M = += + =
The model predicts that Medicaid's long-term medical expenses will be $126 billion in 2010.
c) Set ( ) 150M t = and solve for t.
4.25 83.5 150
4.25 150 83.5 66.5
66.515.6
4.252000 15.6 2015.6
t
t
t
+ == − =
= ≈
+ =
The model predicts Medicaid long-term care expenses will reach $150 billion in 2015.
57. a) The problem says to let t = the number of
years after 2000, and N(t) = the number of American households, in millions, that conducted some online banking in the year t. Then the 54 million households using online banking in 2009, and the 66 million households using online banking in 2014 correspond to data points: (2009 2000,54) (9,54)− = and
(2014 2000,66) (14,66).− =
Find the slope of the function that fits the data:
(14) (9)
14 966 54 12
2.45 5
N Nm
−=−
−= = =
Use the value of m, and either data point, and substitute them into the point-slope
b) In 2019, t = 2019 – 2000 = 19. ( ) ( )19 2.4 19 32.4
45.6 32.4 78
N = += + =
The model predicts that 78 million American households will use online banking in 2019.
c) Set ( ) 100N t = and solve for t. 2.4 32.4 100
2.4 100 32.4 67.6
67.628.2
2.42000 28.2 2028.2
t
t
t
+ == − =
= ≈
+ =
The model predicts that the number of American households using online banking will reach 100 million in 2028.
58. a) The problem says to let t = the number of
years after 1990, and A(t) = the amount of land, in millions of acres, in the National Park System. Then the 74.9 million acres in 1994, and the 84 million acres in 2010 correspond to data points : (1994 1990,74.9) (4,74.9)− = and
(2010 1990,84) (20,84).− =
Find the slope of the function that fits the data:
( ) ( )20 4
20 484 74.9 9.1
0.5687516 16
A Am
−=
−−= = =
Use the value of m, and either data point, and substitute them into the point-slope equation for a line. ( ) ( )
( )
74.9 0.56875 4
74.9 0.56875 2.275
0.56875 2.275 74.9
0.56875 72.625
A t t
A t
A t t
t
− = −− = −
= − += +
b) In 2014, t = 2014 – 1990 = 24 ( ) ( )24 0.56875 24 72.625
13.65 72.625 86.275
A = += + =
The model predicts that in 2014, there will be 86.275 million acres of land in the National Park system.
59. a) The problem says to let t = the number of years after 1999, and R(t) = the record time, in seconds, in the 100-m run in the year t. Then the record of 9.79 seconds in 1999, and 9.58 seconds in 2009, correspond to data points: (1999 1999,9.79) (0,9.79)− = and
(2009 1999,9.58) (10,9.58).− =
Find the slope of the function that fits the data:
( ) ( )10 0
10 09.58 9.79 0.21
0.02110 10
R Rm
−=
−−= = − = −
Use the value of m, and either data point, and substitute them into the point-slope equation for a line.
( )( ) 9.79 0.021 0
( ) 0.021 9.79
R t t
R t t
− = − −= − +
b) In 2015, t = 2015 – 1999 = 16. ( )(16) 0.021 16 9.79
0.336 9.79 9.454
R = − += − + =
The model predicts that the record in the 100-m run will be 9.454 seconds in 2015. In 2030, t = 2030 – 1999 = 31.
( )(31) 0.021 31 9.79
0.651 9.79 9.139
R = − += − + =
The model predicts that the record in the 100-m run will be 9.139 seconds in 2030.
c) Set ( ) 9.5R t = and solve for t. 0.021 9.79 9.5
0.021 9.5 9.79 0.29
0.2913.8
0.0211999 13.8 2012.8
t
t
t
− + =− = − = −
−= ≈−
+ =
The model predicts that the record in the 100-m run will be 9.5 seconds in 2012.
60. a) The problem says to let d = the depth, in
feet, below the ocean's surface, and P(d) = the pressure, in atms, at a depth d. Then the pressure of 4 atm at 100 feet, and 7 atm at 200 feet, correspond to data points: (4,100) and (7,200).
Find the slope of the function that fits the data:
128 Chapter 2: Intermediate Algebra: Graphs and Models
( ) ( )200 100
200 1007 4 3
0.03100 100
P Pm
−=
−−= = =
Use the value of m, and either data point, and substitute them into the point-slope equation for a line.
( )( ) 4 0.03 100
( ) 0.03 3 4
( ) 0.03 1
P d d
P d d
P d d
− = −= − += +
b) Find P(690). ( )(690) 0.03 690 1
20.7 1 21.7
P = += + =
The model predicts that the pressure at 690 feet below the ocean's surface is 21.7 atmospheres.
61. The points lie approximately in a straight line,
so the graph of this data is linear. 62. The points lie approximately in a straight line,
so the graph of this data is linear. 63. The points do not lie on a straight line, so the
data are not linear. 64. The points do not lie on a straight line, so the
data are not linear. 65. The points lie approximately in a straight line,
so the graph of this data is linear. 66. The points do not lie on a straight line, so the
data are not linear. (They do appear to lie on the absolute value of a linear function.)
67. a) The problem says to let x = the number
of years since 1900, and W = the life expectancy, in years, of a female born in the year x. Enter the data with the number of years since 1900 in 1L and life
expectancy in 2L .
Use the LinReg feature of the graphing calculator to find the equation of the line.
The second screen indicates that the equation is 0.1611 63.6983.W x= +
b) In 2012, 2012 1900 112.x = − = To predict the life expectancy of a female born in 2012, find W(112).
( )(112) 0.1611 112 63.6983
18.0432 63.6983
81.7415 81.7
W = += += ≈
This estimate is 81.7 – 80.8 = 0.9 of a year higher than the previous estimate.
68. a) The problem says to let x = the number
of years since 1900, and M = the life expectancy, in years, of a male born in the year x. Enter the data with the number of years since 1900 in 1L and life
expectancy in 2L .
Use the LinReg feature of the graphing calculator to find the equation of the line:
The second screen indicates that the equation is 0.2008 53.8983.M x= +
b) In 2012, 2012 1900 112.x = − = To predict the life expectancy of a male born in 2012, find M(112).
( ) ( )112 0.2008 112 53.8983.
22.4896 53.8983
76.3879 76.4
M = += += ≈
This estimate is 76.4 – 76.45 = 0.05,− or 0.05 of a year lower, than the previous estimate.
69. a) The problem says to let t = the number of years since 2000, and N = the number of registered nurses, in millions, employed in the year t. Enter the data with the number of years since 2000 in 1L and the
number of registered nurses in 2L .
Use the LinReg feature of the graphing calculator to find the equation of the line.
The second screen indicates that the equation is 0.0433 2.1678.N t= +
b) In 2012, 2012 2000 12.t = − = To estimate the number of registered nurses in 2012, find N(12).
( ) ( )12 0.0433 12 2.1678
0.5196 2.1678
2.6874 2.69
N = += += ≈
The model predicts there will be approximately 2.69 million registered nurses in the U.S. in 2112.
70. a) The problem says to let x = the number
of years since 2005, and A = the amount spent, in millions of dollars, on Cyber Monday in the year x. Enter the data with the number of years since 2005 in 1L and the amount spent in 2L .
Use the LinReg feature of the graphing calculator to find the equation of the line.
The second screen indicates that the equation is 102.6 504.2A x= +
b) In 2011, 2011 2005 6.x = − = To estimate the amount spent on Cyber Monday in 2011, find A(6). ( ) ( )6 102.6 6 504.2
615.6 504.2
1119.8
A = += +=
The model predicts that approximately $1119.8 million, or $1.1198 billion, was spent on Cyber Monday 2111.
71. Thinking and Writing Exercise. If the
y-intercept is given, then it is easy to use the slope-intercept form of a line. Whether the point-slope form of a line is used, or the slope-intercept form is used, the two given points must be used to calculate the slope of the line. But once that is done, one can immediately use the slope and the y-intercept to determine the equation of the line.
72. Thinking and Writing Exercise. The equation
for the life expectancy for males has a greater positive slope than the equation for females. Therefore, the lines are not parallel and will intersect at some point. Beyond the time corresponding to that intersection point, the life expectancy for males will be higher than for females.
130 Chapter 2: Intermediate Algebra: Graphs and Models
77. ( )3
xf x
x=
−
3
x
x − is undefined when its denominator is 0.
Find the value(s) of x that make 3 0.x − = 3 0
3
x
x
− ==
3 is not in the domain of f(x). The domain of
f(x) is { } is a real number and 3x x x ≠ . 78. ( ) 2 1g x x= −
2 1x − is well-defined for any real number x. The domain is the set of all real numbers.
79. ( ) 6 11g x x= +
6 11x + is well-defined for any real number x. The domain is the set of all real numbers.
80. ( ) 7
2
xf x
x
−=
7
2
x
x
− is undefined when its denominator is 0.
Therefore, find the value(s) of x that make 2 0.x = 2 0
00
2
x
x
=
= =
0 is not in the domain of f(x). The domain of
f(x) is { } is a real number and 0 .x x x ≠ 81. Thinking and Writing Exercise. The slope-
intercept form is more useful when using a graphing calculator because to graph using the calculator requires the equation be solved for y. However, the point-slope form of a line is easily solved for y by adding 1y to both
sides of the equation:
( )( )
1 1
1 1.
y y m x x
y m x x y
− = −
= − +
82. Thinking and Writing Exercise. Any
nonvertical line contains an infinite number of points, any one of which can be used in the point-slope form of the line. But any nonvertical line only passes through the y-axis in one place. Therefore, there is only one value of b that can be used in the y mx b= + (slope-intercept) form of the line.
83. First simplify the equation. ( )3 0 52
3 0
3
y x
y
y
− = −− =
=
y c= is the form of a horizontal line that
passes through the y-axis at ( )0, .c In this
case, 3,c = so the graph is:
84. First simplify the equation.
( )4 0 93
4 0
4
y x
y
y
+ = ++ =
= −
y c= is the form of a horizontal line that
passes through the y-axis at ( )0, .c In this
case, 4,c = − so the graph is:
85. Two points with coordinates ( )4, 1− and
( )3, 3− are shown on the graph. First
determine the slope of the line. ( )1 3 1 3 2
24 3 4 3 1
m− − − − +
= = = =− −
Use the point-slope equation with 2m = and
( ) ( )1 1, 4, 1x y = − (or ( ) ( )1 1, 3, 3x y = − ) to
are shown on the graph. First determine the slope of the line.
1 5 4 4
5 2 3 3m
− −= = = −
−
Use the point-slope equation with4
3m = −
and ( ) ( )1 1, 2,5x y = (or ( ) ( )1 1, 5,1x y = ) to
determine the required equation. ( ) ( )4
3
843 3
843 3
8 1543 3 3
2343 3
5 2
5
5
y x
y x
y x
y x
y x
− = − −
− = − += − + += − + += − +
87. 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
+ == − +
= − + = −
The slope of the given line is 1
2− . Every line
parallel to the given line must also have a
slope of 1
2− . Find the equation of the line
with a slope of 1
,2
− 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
− = −
− = − −
− = − +
= − +
88. First solve the equation for y and determine
the slope of the given line. 3 7 Given line
3 7 3
x y
y x m
− == − =
The slope of the given line is 3. Every line parallel to the given line must also have a slope of 3. Find the equation of the line with a slope of 3, 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
− = −
− = − −
− = +− = +
= +
89. First solve the equation for y and determine
the slope of the given line. 2 3 Given line
2 3 2
x y
y x m
+ = −= − − = −
The slope of the given line is 2.− Every line perpendicular to the given line must have a
71. Thinking and Writing Exercise. To determine the values where S M W J> + + one must estimate where the sum on the right is equal to the value of S. This appears to be the case around 1985.
Check:
(1985) 41,
(1985) (1985) (1985)
27 8 5 40
S
M W J
≈+ +≈ + + =
The values are close. Similarly, it appears to be the case around 2003.
Check: (2003) 52
(2003) (2003) (2003)
21 21 9 51
S
M W J
≈+ +≈ + + =
Again, the values are close. Since it is difficult to do better than a rough estimate
with a graph of this type. Any answer near (1985, 2003) would be acceptable.
72. Thinking and Writing Exercise. The total
number of births, found by adding the values of C and B, appear to have increased from about 3.7 million to 4.1 million from 1970 to 2004. However, there were some time periods where the number of births remained approximately constant, and some periods where the number of births even decreased (such as from 1970 to 1975 or 1990 to 1995). The percent of Caesarean section births is determined by dividing the number of Caesarean section births by the total number of births. This value definitely increased from 1970 to 2004 as a result of the overall increase in C and decrease in B.
73. 6 3
6 3
1 1
6 2
x y
y x
y x
− =− = − +
= −
74. 3 8 5
8 3 5
3 5
8 8
x y
y x
y x
− =− = − +
= −
75. 5 2 3
2 5 3
5 3
2 2
x y
y x
y x
+ = −= − −
= − −
76. 8 4
8 4
1 4
8 81 1
8 2
x y
y x
y x
y x
+ == − +
= − +
= − +
77. Let n represent the number. 2 5 49.n + =
78. Let n represent the number. 1
3 57.2
n − =
79. Let n represent the smaller integer. ( )1 145.n n+ + =
140 Chapter 2: Intermediate Algebra: Graphs and Models
80. Let n represent the number. ( ) 20n n− − =
81. Thinking and Writing Exercise. We would
change the vertical axis to correspond to absorption of Advil and graph the points accordingly. If taken 4 times a day, we would expect to see 4 peaks over the course of 1 day.
82. Thinking and Writing Exercise. The graph of
( )( )f g x+ will have the same shape as the
graph of g, but will be shifted upward (because c is positive) by c units.
83. 43 1
( ) ; ( )2 5 3 9
x xf x g x
x x
−= =+ +
: 2 5 0
2 5
5 5, So
2 2
f x
x
x x
+ == −− −= ≠
Domain of { |f x x= is a real number, }52x −≠
: 3 9 0g x + =
3 9
3; So 3
x
x x
= −= − ≠ −
Domain of { |g x x= is a real number,
3}.x ≠ −
( )
( )( )( )
4
2
0
1 0
1 1 1 0
1
g x
x
x x x
x
=
− =
+ + − =
= ±
( )
( )( )4
4
3 3 93 1/
2 5 3 9 2 5 1
x xx xf g
x x x x
+−= =+ + + −
Therefore, the domain of / { |f g x x= is a
real number, 52 , 3, and 1}.x x x−≠ ≠ − ≠ ±
84. 21 4
( ) and G( )4 3
xF x x
x x
−= =− −
Domain of { |F x x= is a real number and
4}x ≠
Domain of { |G x x= is a real number and
3}x ≠
( )
( )( )2
0
4 0
2 2 0
2
G x
x
x x
x
=
− =+ − =
= ±
( )( )
2
2
31 4/
4 3 4 4
xxF G
x x x x
−−= =− − − −
Therefore, the domain of / { |F G x x= is a
real number, 4, 3, and 2}.x x x≠ ≠ ≠ ±
85. Answers may vary. The two functions must
be defined over the intervals [ ]2,3− , but must
not both be defined anywhere else. To remove the value x = 1 from the domain of f / g, make ( )1 0.g =
86. The domain of , , and f g f g f g+ − ⋅ is the
set of numbers common to the domains of f and g:{ 2, 1, 0,1}.− − The domain of /f g is
the same set, excluding any x values ( ) 0.g x = Since ( )1 0,g − = it is excluded
from the set. The domain of /f g is:
{ 2, 0,1}.−
87. The domain of m is { | 1 5}.x x− < <
( ) 0 when 2 3 0
2 3
3
2
n x x
x
x
= − ==
=
We exclude this value from the domain of /m n . The domain of /m n is:
is the intersection of the individual domains of f and g, at least one of the functions needs to be undefined at 2,x = − and at least one needs to be undefined at 5.x = Both functions should be well-defined everywhere else.
1 1
( ) ; ( )2 5
f x g xx x
= =+ −
90. For 1 5y = , 2 2y x= + , and 3 ,y x= the
graph of 1 2y y+ will be the same as the
graph of 2y , but shifted up 5 units.
The graph of 1 3y y+ will be the same as the
graph of 3y but shifted up 5 units. The graph
of 2 3y y+ is similar to the graph of 3y , but it
is shifted up 2 units and rises more steeply as x increases.
91. 1 2.5 1.5y x= + and 2 3y x= −
1
2
2.5 1.5
3
y x
y x
+=−
Domain: { }| is any real number 3x and x ≠
The CONNECTED mode graph contains values that cross the line 3x = , whereas, the DOT mode graph contains no points having 3 as the first coordinate. Thus, the DOT mode graph represents 3y more accurately.
92. Think of adding, subtracting, multiplying, or dividing the y-values for various x-values. a) IV b) I c) II d) III Chapter 2 Study Summary 1. ( ) 2 3
( 1) 2 3( 1) 2 3 5
f x x
f
= −− = − − = + =
2. Use the vertical line test by visualizing a
vertical line moving across the graph. No vertical line will intersect the graph more than once. Thus the graph is the graph of a function.
3. The domain is the set of all real numbers, or
.The range is the set of all y-values in the graph. It is { }| 2y y ≥ − or [ )2,− ∞
4. ( ) 15
4f x x= − is well-defined for all real
numbers. The domain is, therefore, the set of all real numbers.
4. False 5. False 6. True 7. True 8. True 9. False 10. True 11. a) To use the vertical line test visualize a
vertical line moving across the graph. No vertical line will intersect the graph more than once. Thus, the graph is the graph of a function.
b) The domain is the set of all x values: { }| 4 5 .x x− ≤ ≤
The range is the set of all y-values:
{ }| 2 4y y− ≤ ≤ .
12. a) To use the vertical line test visualize a
vertical line moving across the graph. No vertical line will intersect the graph more than once. Thus the graph is the graph of a function.
b) The domain is the set of all x values: ll real numbers or
{ }| is a real number .x x
The range is the set of all y values:
{ }| 1y y ≥ .
13. a) To use the vertical line test visualize a
vertical line moving across the graph. It is possible for a vertical line to intersect the graph more than once. Thus this is not the graph of a function.
14. a) On the graph provided, find the point
with an x value of 3. The point is (3,0) .
Therefore ( )3 0f =
b) Find the point or points with a y value of 2− . The y value is 2− at the point
( 4, 2)− − : ( 4) 2.f − = −
15. a) 4 4
(1)2(1) 1 3
g = =+
b) Since 4
2 1x + is undefined when the
denominator is 0, find the x-value(s) where 2 1 0.x + =
2 1 0
2 1 Adding 1 to both sides.
1 Dividing both sides by 2
2
x
x
x
+ == − −
= −
Thus, 1
2− is not in the domain of g. The
domain of g is: 1
| is a real number 2
x x and x⎧ ⎫≠ −⎨ ⎬⎩ ⎭
.
16. ( )2 , if 0,
3 5, if 0 2
7, if 2
x x
f x x x
x x
⎧ <⎪= − ≤ ≤⎨⎪ + >⎩
a) Since 0 0 2≤ ≤ , ( )0 3 0 5 5f = − = −⋅ .
b) Since 3 2> , ( )3 3 7 10.f = + =
17. The equation ( ) 4 9g x x= − − is in slope-
intercept form. Slope is –4; y-intercept is (0, –9)
144 Chapter 2: Intermediate Algebra: Graphs and Models
19. To find the rate of change, or slope, of the graph, select any two points, say ( )2,75 and
( )8,120 and use the slope formula.
2 1
2 1
120 75 457.5
8 2 6
y ym
x x
− −= = = =− −
The rate of change in the apartment's value is $7500 per year.
20. 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
− − −= = = =− − − −
21. 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. 22. ( ) 11 1542C t t= +
The slope 11 signifies the rate of change in the average number of calories consumed each day. The average number of calories consumed each day has increased by 11 calories per year since 1971. The intercept 1542 signifies the average number of calories consumed each day in 1971.
23. Simplify the equation.
3 7
4
y
y
+ ==
The graph of y = 4 is a horizontal line. Its slope is 0.
24. Simplify the equation.
2 9
9
2
x
x
− =
= −
The graph of 9
2x = − is a vertical line. Its
slope is undefined. 25. 3 2 8x y− =
To find the y-intercept, let x = 0 and solve for y.
3 0 2 8
2 8
4
y
y
y
⋅ − =− =
= −
The y-intercept is (0, –4). To find the x-intercept, let y = 0 and solve for x. 3 2 0 8
3 8
8
3
x
x
x
− ⋅ ==
=
The x-intercept is ( )8 , 03
.
26. The equation 3 2y x= − + is in slope-intercept
form. The slope is –3 = 3.1− The y-intercept
is (0, 2). From the y-intercept, go down 3 units and to the right 1 unit to (1, –1). Draw the line through the two points.
27. Graph 2 4 8x y− + = using the intercepts.
To find the y-intercept, let x = 0 and solve for y.
2 0 4 8
2
y
y
− ⋅ + ==
The y-intercept is (0, 2). To find the x-intercept, let y = 0 and solve for x.
2 4 0 8
2 8
4
x
x
x
− + ⋅ =− =
= −
The x-intercept is (–4, 0). Plot these points and draw the line.
146 Chapter 2: Intermediate Algebra: Graphs and Models
36. First solve for y and determine the slope of each line. 3 5 7
3 5
7 7
x y
y x
− =
= −
The slope of 3 5 7x y− = is 3
7.
7 3 7
31
7
y x
y x
− =
= +
The slope of 7 3 7y x− = is 3
7.
The slopes are the same, so the lines are parallel.
37. Use the slope-intercept equation
( ) ,f x mx b= + with 2
9m = and b = –4.
( )
( ) 24
9
f x mx b
f x x
= +
= −
38. ( )
( )1 1
1 1
Point-slope equation
10 5 1 Substituting 5 for ,
1 for , and 10 for
y y m x x
y x m
x y
− = −
− = − − −
39. First find the slope using the pair of given
points. 5 6 1 1
2 ( 2) 4 4m
− −= = = −− −
Use the point-slope equation.
( )
( )
15 2
41 1
54 21 11
4 21 11
4 2
y x
y x
y x
f x x
− = − −
− = − +
= − +
= − +
40. The slope in the given equation is 1. The
slope of any perpendicular line is, therefore, 1 1.1
− = − Use the slope and the given
y-intercept (0, 3)− to write the equation in
function form. ( ) 3.f x x= − −
41. Plot and connect the points, using the years as the first coordinate and the corresponding cost per gallon as the second coordinate.
To estimate the cost in 2004, locate the point that is directly above 2004. Then move horizontally from the point to the vertical axis and read the approximate function value. The approximate cost per gallon in 2004 was $1.50.
42. To estimate the cost per gallon in 2008,
extend the graph and extrapolate. The approximate cost per gallon in 2008 was $2.40.
43. a) The equation says to let t = the number of
years since 1980, and R(t) = the record, in seconds, for the 200-m run. Use the given data to form the pairs (3, 19.75) and (27, 19.32). Then find the slope of the function that fits the data.
148 Chapter 2: Intermediate Algebra: Graphs and Models
56. g(x) is well-defined for all real numbers. Its domain is { }| is any real number .x x
57. The domains of g and h are both
{ }| is any real numberx x = . Thus, the
domain of g h+ = .
58. The domains of g and h are
{ }| is a real numberx x . ( ) 0g x = when
3 6 0x − = , when 2x = . Therefore the domain of /h g =
{ }| is a real number 2x x and x ≠ .
59. Thinking and 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 at least one member of the range and the function is, therefore, 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.
60. Thinking and Writing Exercise The slope of a
line is the change in y between any two points on the line divided by the change in x between those points. For a vertical line, there is no change in x between any two points, or, rather, the change in x is 0. Since division by 0 is undefined, the slope is also undefined. For a horizontal line, the change in y between any two points is 0, so the slope is 0/(change in x) = 0.
61. The y-intercept is found by determining
(0).f
( ) ( )( ) ( )( )( )
( )
02
0
0 3 0.17 0 5 2 0 7
0 3 0 5 7
0 3 0 1 7
0 3 6
0 6 3 9
f
f
f
f
f
+ = + − −
+ = + −
+ = + −
+ = −
= − − = −
⋅ ⋅
The y-intercept is ( )0, 9− .
62. Parallel lines have the same slope. 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
+ = −+ = −
+ =− −
The two slopes are 34
and .6a− Solve the
equation 3 .4 6
a= −
34 64 18
18 94 2
a
a
a
= −
= −
= − = −
63. Each package costs $7.99 and costs $2.95 to
ship. This makes a total charge of $7.99 + $2.95 = $10.94 per package. There is also a flat fee of $20 for overnight delivery no matter how many packages are purchased. Let x = the number of packages purchased and ( )f x represent the total cost. Then
( ) cost per pkg # of pkgs + flat fee
10.94 20
f x
x
== +
⋅
64. a) The slope of the line represents the
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 a speed of 0. d) Graph II matches this situation since
waiting corresponds to a speed of 0, and riding a train would be faster than running.
20. Determine the slope of the linear function. ( )2 1 1
14 3 1
m− − − −= = = −
−
Use the slope and the given point in the point-slope equation. ( ) ( ) ( )
( )( )
2 1 4
2 4
2
f x x
f x x
f x x
− − = − −+ = − +
= − +
21. a) The problem says to let t = the number of
years after 1982, and c = the average number of hours commuters sit in traffic in the year t. Use the given information to generate the two points ( )0,16 and
( )25,41 . Determine the slope of the line
containing these points.
41 16 25
125 0 25
m−
= = =−
Use the point-slope equation with 1m =
and ( ) ( )1 1, 0,16t c = to obtain the
equation. ( )16 1 0
16
16
c t
c t
c t
− = −− =
= +
b) 2000 is 2000 – 1982 = 18 years after 1982. Substitute 18 for t and solve for c.
16
18 16
34
c t
c
c
= += +=
Commuters spent an average of 34 hours sitting in traffic in 2000.
c) 2012 is 2012 – 1982 = 30 years after 1982. Substitute 30 for t and solve for c.
16
30 16
46
c t
c
c
= += +=
By the year 2012, the average commuter will spend 46 hours sitting in traffic.
22. Let t = the number of years since 1980 and
B = the number of twins, in thousands, born in the year t. Enter the data with the number of years since 1980 in 1L , and the number of
births, in thousands, in 2L .
Use the LinReg feature of the graphing calculator to find the equation of the line.
The second screen indicates that the equation is 2.6718 65.0212B t= + .
23. 2012 is 2012 – 1980 = 32 years after 1980.
Substitute 32 for t and solve for B.
( )2.6718 65.0212
2.6718 32 65.0212
85.4976 65.0212
150.5188
B t
B
B
B
= += += +=
The predicted number of twin births in 2012 is about 151,000. 24. ( ) ( )5 2 5 1 9h − = − + = −
25. ( )( ) ( ) ( ) 12 1g h x g x h x x
x+ = + = + +
26. The zeros for h are any values of x such that
h(x) = 0. ( ) 2 1
0 2 1
1 2
1
2
h x x
x
x
x
= += +
− =
− =
The only zero of h is x =1
2− .
27. Domain of g:
{ }is a real number and 0x x x ≠ 28. The domain of g is:
{ }is a real number and 0 .x x x ≠
The domain of h is: { } is a real number .x x The domain of g + h is: { }is a real number and 0 .x x x ≠