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College Algebra with Corequisite Support

Chapter 4

College Algebra with Corequisite SupportChapter 4

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4.1 Linear Functions

1.

m= 43 02 = 1 2 = 1 2 ; m= 43 02 = 1 2 = 1 2 ; decreasing because m<0. m<0.

2.
m= 1,8681,442 2,0122,009 = 426 3 =142 people per year m= 1,8681,442 2,0122,009 = 426 3 =142 people per year
3.

y=7x+3 y=7x+3

4.

H( x )=0.5x+12.5 H( x )=0.5x+12.5

5.
This graph shows a decreasing function graphed on an x y coordinate plane. The x-axis runs from negative 8 to 10 and the y-axis runs from negative 8 to 8. The function passes through the points (0,6), (4,3) and (8,0).
6.

Possible answers include (3,7), (3,7), (6,9), (6,9), or (9,11). (9,11).

7.
This graph shows three functions on an x, y coordinate plane. One shows an increasing function y = x that passes through points (0, 0) and (2, 2).  A second shows an increasing function y = 2 times x that passes through the points (0, 0) and (2, 4).  The third is an increasing function y = 2 times x plus 4 and passes through the points (0, 4) and (2, 8).
8.

( 16,0 ) ( 16,0 )

9.
  1. f(x)=2x; f(x)=2x;
  2. g(x)= 1 2 x g(x)= 1 2 x
10.

y= 1 3 x+6 y= 1 3 x+6

4.2 Modeling with Linear Functions

1.

C( x )=0.25x+25,000 C(x)=0.25x+25,000
The y-intercept is (0,25,000) (0,25,000). If the company does not produce a single doughnut, they still incur a cost of $25,000.

2.

41,100 2020

3.

21.57 miles

4.3 Fitting Linear Models to Data

2.

54°F 54°F

3.

150.871 billion gallons; extrapolation

4.1 Section Exercises

1.

Terry starts at an elevation of 3000 feet and descends 70 feet per second.

3.

d( t )=10010t d( t )=10010t

5.

The point of intersection is ( a,a ). ( a,a ). This is because for the horizontal line, all of the y y coordinates are a a and for the vertical line, all of the x x coordinates are a. a. The point of intersection is on both lines and therefore will have these two characteristics.

7.

Yes

9.

Yes

11.

No

13.

Yes

15.

Increasing

17.

Decreasing

19.

Decreasing

21.

Increasing

23.

Decreasing

25.

2

27.

–2

29.

y= 3 5 x1 y= 3 5 x1

31.

y=3x2 y=3x2

33.

y= 1 3 x+ 11 3 y= 1 3 x+ 11 3

35.

y=1.5x3 y=1.5x3

37.

perpendicular

39.

parallel

41.

f(0)=(0)+2 f(0)=2 yint:(0,2) 0=x+2 xint:(2,0) f(0)=(0)+2 f(0)=2 yint:(0,2) 0=x+2 xint:(2,0)

43.

h(0)=3(0)5 h(0)=5 yint:(0,5) 0=3x5 xint:( 5 3 ,0 ) h(0)=3(0)5 h(0)=5 yint:(0,5) 0=3x5 xint:( 5 3 ,0 )

45.

2x+5y=20 2(0)+5y=20 5y=20 y=4 yint:(0,4) 2x+5(0)=20 x=10 xint:(10,0) 2x+5y=20 2(0)+5y=20 5y=20 y=4 yint:(0,4) 2x+5(0)=20 x=10 xint:(10,0)

47.

Line 1: m = –10 Line 2: m = –10 Parallel

49.

Line 1: m = –2 Line 2: m = 1 Neither

51.

Line 1: m=2   Line 2: m=2   Parallel Line 1: m=2   Line 2: m=2   Parallel

53.

y=3x3 y=3x3

55.

y= 1 3 t+2 y= 1 3 t+2

57.

0

59.

y= 5 4 x+5 y= 5 4 x+5

61.

y=3x1 y=3x1

63.

y=2.5 y=2.5

65.

F

67.

C

69.

A

71.
Graph of f with an x-intercept at -4 and y-intercept at -2 which gives us a slope of: 2.
73.
Graph of f with an y-intercept at 3 and a slope of 2/5.
75.
Graph of a line that passes through the points (-3, -4) and (3, 0) which results in a slope of 2/3.
77.
Graph of g(x) = -3x + 2 which goes through the points (0,2) and (1,-1) with a slope of -3.
79.
Graph of k(x) =  .  This line goes through the points (0,-3) and (3,-1) and has a slope of 2/3.
81.
Graph of the line p(t) = 3t -2.  This line goes through the points (0,-2) and (1,1) which has a slope of 3.
83.
Graph of x = -2 which is a line of undefined slope that goes through the point (-2,0).
85.

y=3 y=3

87.

x=3 x=3

89.

Linear, g(x)=3x+5 g(x)=3x+5

91.

Linear, f(x)=5x5 f(x)=5x5

93.

Linear, g(x)= 25 2 x+6 g(x)= 25 2 x+6

95.

Linear, f(x)=10x24 f(x)=10x24

97.

f(x)=58x+17.3 f(x)=58x+17.3

99.
Graph of f(x) = 2500x + 4000
101.
  1. a=11,900, b=1000.1a=11,900, b=1000.1
  2. q(p)=1000p100q(p)=1000p100
103.
graph where the function's slope is 75 and y-intercept is –22.5
105.

y= 16 3 y= 16 3

107.

x=ax=a

109.

y= d ca x ad ca y= d ca x ad ca

111.

y=100x98y=100x98

113.

x< 1999 201 , x> 1999 201 x< 1999 201 , x> 1999 201

115.

$45 per training session.

117.

The rate of change is 0.1. For every additional minute talked, the monthly charge increases by $0.1 or 10 cents. The initial value is 24. When there are no minutes talked, initially the charge is $24.

119.

The slope is –400. this means for every year between 1960 and 1989, the population dropped by 400 per year in the city.

121.

C

4.2 Section Exercises

1.

Determine the independent variable. This is the variable upon which the output depends.

3.

To determine the initial value, find the output when the input is equal to zero.

5.

6 square units

7.

20.01 square units

9.

2,300

11.

64,170

13.

P( t )=75,000+2500t P( t )=75,000+2500t

15.

(–30, 0) Thirty years before the start of this model, the town had no citizens. (0, 75,000) Initially, the town had a population of 75,000.

17.

Ten years after the model began

19.

W(t)=0.5t+7.5 W(t)=0.5t+7.5

21.

( 15,0 ) ( 15,0 ) : The x-intercept is not a plausible set of data for this model because it means the baby weighed 0 pounds 15 months prior to birth. ( 0,7.5 ) ( 0,7.5 ) : The baby weighed 7.5 pounds at birth.

23.

At age 5.8 months

25.

C( t )=12,025205t C( t )=12,025205t

27.

( 58.7,0 ): ( 58.7,0 ): In roughly 59 years, the number of people inflicted with the common cold would be 0. ( 0,12,025 ) ( 0,12,025 ) Initially there were 12,025 people afflicted by the common cold.

29.

2063

31.

y=2t+180 y=2t+180

33.

In 2070, the company’s profit will be zero.

35.

y=30t300 y=30t300

37.

(10, 0) In the year 1990, the company’s profits were zero

39.

Hawaii

41.

During the year 1933

43.

$105,620

45.
  1. 696 people
  2. 4 years
  3. 174 people per year
  4. 305 people
  5. P(t) = 305 + 174t
  6. 2,219 people
47.
  1. C(x) = 0.15x + 10
  2. The flat monthly fee is $10 and there is a $0.15 fee for each additional minute used
  3. $113.05
49.

P(t) = 190t + 4,360

51.
  1. R(t)=2.1t + 16R(t)=2.1t + 16
  2. 5.5 billion cubic feet
  3. During the year 2017
53.

More than 133 minutes

55.

More than $42,857.14 worth of jewelry

57.

More than $66,666.67 in sales

4.3 Section Exercises

1.

When our model no longer applies, after some value in the domain, the model itself doesn’t hold.

3.

We predict a value outside the domain and range of the data.

5.

The closer the number is to 1, the less scattered the data, the closer the number is to 0, the more scattered the data.

7.

61.966 years

9.
Scatter plot with a collection of points appearing at (1,46); (2,50); (3,59); (4,75); (5, 100); and (6,136); they do not appear linear

No.

11.
Scatterplot with a collection of points at (1,1); (3,9); (5,28); (7,65); (9,125); and (11,216); they do not appear linear

No.

13.
Scatterplot with a collection of points at (16,46); (18,50); (20,54); (25,55); and (30,62); they appear nonlinear

Interpolation. About 60°F. 60°F.

15.

This value of r indicates a strong negative correlation or slope, so C This value of r indicates a strong negative correlation or slope, so C

17.

This value of r indicates a weak negative correlation, so B This value of r indicates a weak negative correlation, so B

19.
Scatter plot with domain 0 to 10 and a range from -1 to 4 with the line of best fit drawn going through the points: (0,1.5); (1.5, -0.1); (2.1,1.9); (3.4, 1.5); (4.5,2.5); (5.8,2.2); (6.8,3.8); (7.8,3.6); (8.8,2); and (10,2.4).
21.
Scatter plot with a domain of 0 to 10 and a range of 2 to 6 and the line of best fit going through the points: (0,2.1); (1,3.9); (2.1,3.6); (3.6,3.9); (4.4,4); (5.6,4.2); (6.8,5); (7.8,5); (9,5.6); and (10,6)
23.

Yes, trend appears linear because r=0.985 r=0.985 and will exceed 12,000 near midyear, 2016, 24.6 years since 1992.

25.

y=1.640x+13.800, y=1.640x+13.800, r=0.987 r=0.987

27.

y=0.962x+26.86,   r=0.965 y=0.962x+26.86,   r=0.965

29.

y=1.981x+60.197; y=1.981x+60.197; r=0.998 r=0.998

31.

y=0.121x38.841,r=0.998 y=0.121x38.841,r=0.998

33.

(−2,−6),(1,−12),(5,−20),(6,−22),(9,−28); (−2,−6),(1,−12),(5,−20),(6,−22),(9,−28); Yes, the function is a good fit.

35.

(189.8,0) (189.8,0) If 18,980 units are sold, the company will have a profit of zero dollars.

37.

y=0.00587x+1985.41 y=0.00587x+1985.41

39.

y=20.25x671.5 y=20.25x671.5

41.

y=10.75x+742.50 y=10.75x+742.50

Review Exercises

1.

Yes

3.

Increasing

5.

y=3x+26 y=3x+26

7.

3

9.

y=2x2 y=2x2

11.

Not linear.

13.

parallel

15.

(–9,0);(0,–7) (–9,0);(0,–7)

17.

Line 1: m=2; m=2; Line 2: m=2; m=2; Parallel

19.

y=0.2x+21 y=0.2x+21

21.


This is a graph of f of t = 2 times t minus 5 on a x, y coordinate plane.  The x-axis ranges from -4 to 6 and the y-axis ranges from -6 to 6. The curve is an increasing linear function that goes through the points (0,-5) and (2.5,0).
23.

More than 250

25.

118,000

27.

y=300x+11,500 y=300x+11,500

29.
  1. 800
  2. 100 students per year
  3. P( t )=100t+1700 P( t )=100t+1700
31.

18,500

33.

$91,625

35.

Extrapolation


Scatter plot with the points (1990,5600); (1995,5950); (2000,6300); (2005,6600); and (2010,6900).
37.


Scatter plot of: (2,78); (4,81); (6,85); (8,90); and (10,99) and the line of best fit running through these points.  The line of best fit goes through most of the points.
39.

2023

41.

y=1.294x+49.412; r=0.974 y=1.294x+49.412; r=0.974

43.

2027

45.

7,660

Practice Test

1.

Yes

3.

Increasing

5.

y = −1.5x − 6

7.

y = −2x − 1

9.

No

11.

Perpendicular

13.

(−7, 0); (0, −2)

15.

y = −0.25x + 12

17.

Slope = −1 and y-intercept = 6

19.

150

21.

165,000

23.

y = 875x + 10,625

25.
  1. 375
  2. dropped an average of 46.875, or about 47 people per year
  3. y = −46.875t + 1250
27.
29.

In early 2018

31.

y = 0.00455x + 1979.5

33.

r = 0.999

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