 College Algebra 2e

Chapter 4

4.1Linear Functions

1 .

$m= 4−3 0−2 = 1 −2 =− 1 2 ; m= 4−3 0−2 = 1 −2 =− 1 2 ;$ decreasing because $m<0. m<0.$

2 .
$m= 1,868−1,442 2,012−2,009 = 426 3 =142 people per year m= 1,868−1,442 2,012−2,009 = 426 3 =142 people per year$
3 .

$y=−7x+3 y=−7x+3$

4 .

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

5 . 6 .

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

7 . 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 .

$y=– 1 3 x+6 y=– 1 3 x+6$

4.2Modeling 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 4.3Fitting Linear Models to Data 1 . $54°F 54°F$ 2 . 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 . $d( t )=100−10t d( t )=100−10t$ 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 . $y= 3 5 x−1 y= 3 5 x−1$ 31 . $y=3x−2 y=3x−2$ 33 . $y=− 1 3 x+ 11 3 y=− 1 3 x+ 11 3$ 35 . $y=−1.5x−3 y=−1.5x−3$ 37 . perpendicular 39 . parallel 41 . $f(0)=−(0)+2 f(0)=2 y−int:(0,2) 0=−x+2 x−int:(2,0) f(0)=−(0)+2 f(0)=2 y−int:(0,2) 0=−x+2 x−int:(2,0)$ 43 . $h(0)=3(0)−5 h(0)=−5 y−int:(0,−5) 0=3x−5 x−int:( 5 3 ,0 ) h(0)=3(0)−5 h(0)=−5 y−int:(0,−5) 0=3x−5 x−int:( 5 3 ,0 )$ 45 . $−2x+5y=20 −2(0)+5y=20 5y=20 y=4 y−int:(0,4) −2x+5(0)=20 x=−10 x−int:(−10,0) −2x+5y=20 −2(0)+5y=20 5y=20 y=4 y−int:(0,4) −2x+5(0)=20 x=−10 x−int:(−10,0)$ 47 . Line 1: m = –10 Line 2: m = –10 Parallel 49 . Line 1: m = –2 Line 2: m = 1 Neither 51 . 53 . $y=3x−3 y=3x−3$ 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 . $y=3x−1 y=3x−1$ 63 . $y=−2.5 y=−2.5$ 65 . F 67 . C 69 . A 71 . 73 . 75 . 77 . 79 . 81 . 83 . 85 . $y=3 y=3$ 87 . $x=−3 x=−3$ 89 . Linear, $g(x)=−3x+5 g(x)=−3x+5$ 91 . Linear, $f(x)=5x−5 f(x)=5x−5$ 93 . Linear, $g(x)=− 25 2 x+6 g(x)=− 25 2 x+6$ 95 . Linear, $f(x)=10x−24 f(x)=10x−24$ 97 . $f(x)=−58x+17.3 f(x)=−58x+17.3$ 99 . 101 . 1. $a=11,900, b=1000.1a=11,900, b=1000.1$ 2. $q(p)=1000p –100q(p)=1000p –100$ 103 . 105 . $y=− 16 3 y=− 16 3$ 107 . $x=ax=a$ 109 . $y= d c–a x– ad c–a y= d c–a x– ad c–a$ 111 . $y=100x–98y=100x–98$ 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 .

$C( t )=12,025−205t C( t )=12,025−205t$

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 .

$y=30t−300 y=30t−300$

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. 5.5 billion cubic feet
2. 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 . No.

11 . No.

13 . Interpolation. About $60°F. 60°F.$

15 .

17 .

19 . 21 . 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 .

29 .

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

31 .

$y=0.121x−38.841,r=0.998 y=0.121x−38.841,r=0.998$

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 .

$y=20.25x−671.5 y=20.25x−671.5$

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 .

$y=2x−2 y=2x−2$

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 . 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 37 . 39 .

2023

41 .

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