CHAPTER 1 Functions Graphs, and Models; Linear Functions. Algebra Toolbox Exercises. 1. {1,2,3,4,5,6,7,8} and

CHAPTER Algebra Toolbox CHAPTER Functions Graphs, and Models; Linear Functions Algebra Toolbox Exercises. {,,,4,,6,7,8} and { xx< 9, x } Remember that x means that x is a natural number.. Yes.. Yes. Every element of B is also in A. 4. No. {,,,4,...}. Therefore,.. Yes. Every integer can be written as a fraction with the denominator equal to. 6. Yes. Irrational numbers are by definition numbers that are not rational.. (,7] 4. (,7 ]. (,4) 6. 4 0 4 7. Note that > x implies x<, therefore 8. 9. 4 0 4 4 0 4 (, ) 7. Integers. Note this set of integers could also be considered a set of rational numbers. See question. 8. Rational numbers 9. Irrational numbers 0. 0. x >. x (4, ). x Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Functions, Graphs, and Models; Linear Functions. ( 4,) 0. a( b+ 8c) ab 8ac. 6 6. 7 4 4 4. The x term has a coefficient of. The x term has a coefficient of 4. The constant term is 8.. The x 4 term has a coefficient of. The x term has a coefficient of 7. The constant term is.. 4( x y) ( x+ y) 4x 4y x y x 6y x 6y. 4 ( x y) + 4xy ( y xy) ( x 4y) 8x 4y+ 4xy y+ xy x+ 4y ( 8x x) ( 4y y 4y) ( 4xy xy) + + + + 6x y+ 9xy. x( 4yz 4) ( xyz x) 8xyz 8x xyz + x ( 8xyz xyz) ( 8x x) + + xyz x 4 4 6. ( z z + 0z 6) + ( z + 4z z ) 4 4 ( z + z ) + ( 4z ) + ( z z ) + ( 0z) + ( 6 ) + + 4 z 4z 7z 0z x y y y y 4 4 4 + + + 7. ( x x) + ( 9 + 0) 4 4 7 9 x y + y y x 8. 4( p + d ) 4p + 4d 9. ( x 7y) ( i x) + ( i 7y) 6x+ 4y 4. x 6 x x. 6 x 6 8 x x 8 6. x + 6 x + 6 x Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Algebra Toolbox 7. 4x 6+ x 4x x 6+ x x x 6 x + 6+ x 9 x 9 x 8. x 4 7x x + 7x 4 7x+ 7x 0x 4 0x + 4 + 0x 6 0x 6 0 0 6 x 0 x 9. x 4 x 4 4 4 x 48 x 6 40. x 8 + 4x x 4x 8 + 4x 4x x 8+ 8 + 8 x 0 x 0 x 0 Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

4 CHAPTER Functions, Graphs, and Models; Linear Functions Section. Skills Check. Using Table A a. is an x-value and therefore is an input into the function f ( x ). b. f ( ) represents an output from the function.. No. In the given table, x is not a function of y. If y is considered the input variable, one input will correspond with more than one output. Specifically, if y 9, then x or x 7. 6. Yes. Each input y produces exactly one output x. c. The domain is the set of all inputs. D:{ 9, 7,, 6,,7, 0}. The range is the set of all outputs. R:{ 4,,6,7,9,0 } d. Every input x into the function f yields exactly one output y f ( x).. Using Table B a. is an x-value and therefore is an input into the function f ( x ). b. g ( 7) represents an output from the function c. The domain is the set of all inputs. D:{ 4,, 0,,, 7,}. The range is the set of all outputs. R:{,,7,8,9,0, } d. Every input x into the function f yields y g x. exactly one output 7. a. f () b. f () 0 () 0 (4) 0 c. f () 8. a. f ( ) b. f ( ) 8 f ( ) + ( ) + 8 + 8 6 c. 9. Recall that R( x) x+ 8. a. R( ) ( ) + 8 + 8 7 b. R( ) ( ) + 8 + 8. f ( 9) f 7 9 c. R () () + 8 0 + 8 8 4. g( 4) g() 8 0. Recall that Cs () 6 s. a. C () 6 () 6 (9) 6 8 Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section. b. C( ) 6 ( ) 6 (4) 6 8 8 8. a. Yes. Every input yields exactly one output. b. Not a function. If x, then y 4 or y 6. c. C( ) 6 ( ) 6 () 6 4 9. a. Not a function. If x, then y or y 4. b. Function. Every input yields exactly one output.. Yes. Every input corresponds with exactly, 0,,,. one output. The domain is { } The range is { 8,,,,7 }.. No. Every input x does not match with exactly one output y. Specifically, if x then y or y 4.. No. The graph fails the vertical line test. Every input does not match with exactly one output. 4. Yes. The graph passes the vertical line test. Every input matches with exactly one output.. No. If x, then y or y 7. One input yields two outputs. The relation is not a function. 6. Yes. Every input x yields exactly one output y. 7. a. Not a function. If x 4, then y or y 8. b. Yes. Every input yields exactly one output. 0. a. Function. Every input yields exactly one output. b. Not a function. If x, then y or y.. No. If x 0, then (0) + y 4 y 4 y ±. So, one input of 0 corresponds with outputs of and. Therefore the equation is not a function.. Yes. Every input for x corresponds with exactly one output for y.. C π r, where C is the circumference and r is the radius. 4. D is found by squaring E, multiplying the result by, and subtracting.. A function is a correspondence that assigns to each element of the domain exactly one element of the range. 6. The domain of a function is the set of all possible inputs into the function. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

6 CHAPTER Functions, Graphs, and Models; Linear Functions 7. The range of a function is the set of all possible outputs from the function. 8. The vertical-line test says that if no vertical line intersects the graph of an equation in more than one point, then the equation is a function. Section. Exercises 9. a. No. Every input (x, given day) would correspond with multiple outputs (p, stock prices). Stock prices fluctuate throughout the trading day. b. Yes. Every input (x, given day) would correspond with exactly one output (p, the stock price at the end of the trading day). 0. a. Yes. Every input (stepping on the scale) corresponds with exactly one output (the man s weight). b. No. Every input corresponds with multiple outputs. The man s weight will fluctuate throughout the given year, x.. Yes. Every input (month) corresponds with exactly one output (cents per pound).. a. Yes. Every input (age in years) corresponds with exactly one output (life insurance premium). b. No. One input of $.8 corresponds with six outputs.. Yes. Every input (education level) corresponds with exactly one output (average income). 4. Yes. The graph of the equation passes the vertical line test.. Yes. Yes. Every input (depth) corresponds with exactly one output (pressure). The graph of the equation passes the vertical line test. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section. 7 6. Yes. The graph of the equation passes the vertical line test. 7. a. Yes. Every input (day of the month) corresponds with exactly one output (weight). b. The domain is {,,, 4,,,4}. c. The range is { 7,7,7,74,7,76,77,78 }. d. The highest weights were on May and May. e. The lowest weight was May 4. f. Three days from May until May 4. 8. a. No. One input of 7 matches with two outputs of 70 and 8. b. Yes. Every input (average score on the final exam) matches with exactly one output (average score on the math placement test). 9. a. B () 6, b. (),047 B represents the balance owed by the couple at the end of two years. B. c. Year. d. t 4 40. a. The couple must make payments for 0 years. 0 f 0,000 b. f ( 89,000). It will take the couple years to payoff an $89,000 mortgage at 7.%. c. f ( 0,000) 0 d. f ( i ) f if ( 40,000) i 40,000 0,000 0 The expressions are not equal. 4. a. When t 00, the ratio is approximately 4. b. f (00) 4. For year 00 the projected ratio of working-age population to the elderly is 4. c. The domain is the set of all possible inputs. In this example, the domain consists of all the years, t, represented in the figure. Specifically, the domain is {99,000,00,00,0, 00,0,00}. d. As the years, t, increase, the projected ratio of the working-age population to the elderly decreases. Notice that the bars in the figure grow smaller as the time increases. 4. a. Approximately million b. f ( 890) 4. Approximately 4 million women were in the work force in 890. c. { 890,900,90,90,940, 90,960,970,980,990 d. Increasing. Note that as the year increases, the number of women in the work force also increases. 4. a. f (990) 49,67 b. The domain is the set of all possible inputs. In this example, the domain is all the years, t, represented in the table. Specifically, the domain is {98,986,987,,997,998}. } Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

8 CHAPTER Functions, Graphs, and Models; Linear Functions c. The maximum number of firearms is 8,697, occurring in year 99. Note that f (99) 8,687. 44. a. The domain is { 0,,0,,8 }. b. The range is {.0,.06,.0,.6,.48 }. c. When the input is 0, the output is.0. In 990,.0 billion people in the U.S. were admitted to movies. d. As the years past 980 increase, the movie admissions also increase. The table represents an increasing function. 4. a. Yes. Every year, t, corresponds with exactly one percentage, p. b. f (840) 68.6. f (840) represents the percentage of U.S. workers in a farm occupation in the year 840. c. If f( t ) 7, then t 90. d. f (960) 6. implies that in 960, 6.% of U.S. workers were employed in a farm occupation. e. As the time, t, increases, the percentage, p, of U.S. workers in farm occupations decreases. Note that the graph is sloping down if it is read from left to right. increase rapidly because of the fast growth in Internet usage in the U.S. 47. a. f ( 990).4. In 990 there are.4 workers for each retiree. b. 00 c. As the years increase, the number of workers available to support retirees decreases. Therefore, funding for social security into the future is problematic. Workers will need to pay larger portion of their salaries to fund payments to retirees. 48. a. When the input is 99 the output is approximately 0. This implies that the pregnancy rate per 000 girls in 99 was approximately 0. b. The rate was in 989 and 99. c. The rate increased from 984-8 and again from 987-99. d. 99. In 99, the pregnancy rate per 000 girls is approximately 7. 49. a. R ( 00) ( 00) 6400. The revenue generated from selling 00 golf hats is $6400. b. R ( 00) ( 00 ) $80,000 46. a. In 99, 9.4 million homes used the Internet. b. f ( 997).8. In 997,.8 million U.S. homes used the Internet. c. 998 0. a. C ( 00) 4000 + ( 00) 6400. The production cost of manufacturing 00 golf hats is $6400. b. C ( 00) 4000 + ( 00) $4,000 d. The function is increasing very rapidly. Beyond 998, the function continues to Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section. 9. a. P 00 40(00) 0.(00) 000,000,000 000 98,000 The profit generated from selling 00 ipod players is $98,000. b. P( 4000),800,000,600,000 000 $98,000 40(4000) 0.(4000) 000 h 6+ 96 6 6+ 96 6 86 The height of the ball after one second is 86 feet.. a. () () () h 6+ 96 6 6 + 88 44 0 After three seconds the ball is 0 feet high. b. P 00 0(00) 4000 4000 4000 0 The profit generated from selling 00 golf hats is $0.. a. b. P 00 0 00 4000 0,000 4000 $46, 000 f 000 0.0 000 +.80 0 +.80 0.80 The monthly charge for using 000 kilowatt hours is $0.80.. a. c. Test t. ( ) 6+ 96( ) 6( ) h 6 + 480 400 86 After five seconds the ball is 86 feet high. The ball does eventually fall, since the height at t is lower than the height at t. Considering the following table of values for the function, it seems reasonable to estimate that the ball stops climbing at t. f 00 0.0 00 +.80 7. +.80 $6.0 b. 4. a. P 00 00 0. 00 000 00 000 000 00 The daily profit for producing 00 Blue Chief bicycles is $00. b. P 60 60 0. 60 000 0 60 000 $60 6. a. f f 99 6.6 999 66. b. g ( 99) 48.0. In 99 48.0% of Hispanic males have completed at least some college. 98 4.0 999. c. h h Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

0 CHAPTER Functions, Graphs, and Models; Linear Functions 7. d. f ( 987).. In 987.% of white males had completed some college. a. Yes. The graph seems to pass the vertical line test. b. Any input into the function must not create a negative number under the radical. Therefore, the radicand, 4s +, must be greater than or equal to zero. 4s + 0 4s + 0 4s s 4 Therefore, based on the equation, the domain is, 4. c. Since s represents wind speed in the given function and wind speed cannot be less than zero, the domain of the function is restricted based on the physical context of the problem. Even though the domain implied by the function is, 4, the actual domain 0,. in the given physical context is [ ) 8. a. 0. + 0.7n 0 0.7n 0. 0.7n 0. 0.7 0.7 n 7 Therefore the domain of R( n ) is all real numbers except or 7,, 7 7. b. In the context of the problem, n represents the factor for increasing the number of questions on a test. Therefore it makes sense that n 0. 9. a. The domain is [ 0,00 ). b. C 7,000 60 60,00 00 60 7,000( 90) C ( 90),,000 00 90 60. a. Considering the square root p + 0 p p Since the denominator can not equal zero, p. Therefore the domain of q is,. b. In the context of the problem, p represents the price of a product. Since the price can not be negative, p 0. The domain is [ 0, ). Also, since q represents the quantity of the product Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section. 6. a. demanded by consumers, q 0. The range is [ 0, ). V V 44(08 48) 44(60) 8640 4(08 7) 4(6),664 () () (08 4()) (8) (8) (08 4(8)) The maximum volume occurs when x 8. Therefore the dimensions that maximize the volume of the box are 8 inches by 8 inches by 6 inches. 6. a. b. S 0 4.9 0 + 98 0 + The initial height of the bullet is meters. b. First, since x represents a side length in the diagram, then x must be greater than zero. Second, to satisfy postal restrictions, the length plus the girth must be less than or equal to 08 inches. Therefore, Length + Girth 08 Length + 4x 08 4x 08 Length 08 Length x 4 Length x 7 4 Since x is greatest if the length is smallest, let the length equal zero to find the largest value for x. 0 x 7 4 x 7 Therefore the conditions on x and the corresponding domain for the function 0 x 7 or x 0,7. V x are [ ] S 9 487. S 0 49 S 487. c. The bullet seems to reach a maximum height at 0 seconds and then begins to fall. See the table in part b) for further verification. c. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Functions, Graphs, and Models; Linear Functions Section. Skills Check. a. b. x y x x, y 7 (, 7) 8 (, 8) (, ) 0 0 (0, 0) (, ) 8 (, 8) 7 (, 7). 4. c. The graphs are the same, although the scale for part b) is smaller than the scale required for part a). [ 4,4] by [ 0,0] c. The graphs are the same... a. b. x y x x, y 9 (, 9) 9 (, 9) (, ) 0 (0, ) (, ) 9 (, 9) 9 (, 9) + 6. [, ] by [, 0] Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section. 7. 0. a. 8. b. [ 0, 0] by [ 0, 0] 9. a. [, ] by [ 0, 0] View b) is better.. a. [ 0, 0] by [ 0, 0] b. [ 0, 0] by [ 0, 0] b. [0, 0] by [ 0, 0] View b) is better. [ 0, 0] by [ 0.0, 0.0] View b) is better. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

4 CHAPTER Functions, Graphs, and Models; Linear Functions. a. [ 0, 0] by [ 0, 0] [ 60, 0] by [ 000, 00] b.. When x 0, y 0. When x 0, y 80. When x 0, y 0. Therefore, [ 0,0] by [ 0,000] is an appropriate viewing window. {Note that answers may vary.} [ 0, 0] by [ 0, 90] View b) is better.. When x or x, y 9. When x 0, y 0, by 0, 70 is an appropriate viewing window. {Note that answers may vary.}. Therefore, [ ] [ ] [ 0,0] by [ 0,000] 6. When x 8, y 0. When x 8, y 7. When x, y 7. Therefore, [,] by [ 0,0] is an appropriate viewing window. {Note that answers may vary.} 4. When x 60, y 0. When x 0, y 0. When x 0, y 870. Therefore, [ 60,0] by [ 000,00] is an appropriate viewing window. {Note that answers may vary.} [,] by [ 0,0] Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section. 7.. [, ] by [ 0, 00] [0, 0] by [0, 600] {Note that answers may vary.} 8.. 9. [, 40] by [ 00, 0] {Note that answers may vary.} t St.t 0..9 6 7.7 8. 4.. a. [ 0, 0] by [0, 80] [, ] by [ 0, 0] 0. q f( q) q q+ 8 8 40 08 4 66 4 40 b. [, ] by [ 0, 0] c. Yes. Yes. Compare the following table of points generated by f( x) x 6 to the given table of points: Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

6 CHAPTER Functions, Graphs, and Models; Linear Functions. a. f ( 0) ( 0) ( 0) 400 00 00 b. x 0 implies 0 years after 000. Therefore the answer to part a) yields the millions of dollars earned in 00. 4. a. 6. a. f ( 0) 00( 0) ( 0) 0,000 0 990 b. In 00, x 0. Therefore, 990 thousands of units or 9,90,000 units are produced in 00. [, ] by [ 0, 0] b. [, ] by [ 0, 0] c. Yes. Yes. Compare the following table of points generated by f( x) x to the given table of points: d. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section. 7 Section. Exercises 7. a. x Year 990 For 994, x 994 990 4 For 998, x 998 990 8 b. y (8) 07(8) + 06 70 70 represents the number of welfare cases in Niagara, Canada in 998. c. For 99, x. Therefore, y () 07() +,06,7 There were,7 welfare cases in Niagara, Canada in 99. 0. a. t Year 980 For 98, t 98 980 For 988, t 988 980 8 For 000, t 000 980 0 b. P f(4) 4 + 740 4 + 07 477 477 represents the cost of prizes and expenses in millions of dollars for state lotteries in 984. x x c. min max 980 980 0 997 980 7 8. a. y 7 44 9.88 44 + 409.9, 874.7 + 409.9,086.7 In 944 there were,086.7 thousand women in the workforce.. S 00 + 64t 6t a. b. In 980, x 80. y 7( 80) 9.88( 80) + 409.9 44,800 90.4 + 409.9 4,68.89 In 980 there were 4,68.89 thousand women in the workforce. b. [0, 6] by [0, 00] 9. a. t Year 99 For 996, t 996 99 For 004, t 004 99 9 b. P f (8) 6.0 8 +..69..96 represents the percentage of households with Internet access in 00. x x c. min max 99 99 0 00 99 0 Considering the table, S 48 feet when x is or when x is. The height is the same for two different times because the height of the ball increases, reaches a maximum height, and then decreases. c. The maximum height is 64 feet, occurring seconds into the flight of the ball. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

8 CHAPTER Functions, Graphs, and Models; Linear Functions b.. V 600,000 0,000x a. When x 6, y.4. Therefore, when the median male salary is $6,000, the median female salary is $,40. 4. S.x+.6 a. [0, 0] by [0, 600,000] b. [0, ] by [0, 60] b. [0, 0] by [0, 600,000] When x 0, y 40,000.. F 0.M +.78 a. [0, ] by [0, 60] In 00, federal spending on education is approximately $9.68 billion. [0, 8] by [0, 6] Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section. 9. S t t+ a. 0.07 4.8 8.9 b. b. [0, 7] by [0, 00] c. The cost in 996 is approximately $,087. million. 7. B( t) 0.7 +.8t a. 6. [0, 7] by [0, 00] When t, S. c. 99 corresponds to t 99 980. When t, S 0.07() 4.8() + 8.9. See the graph in part b above. The estimated number of osteopathic students in 99 is,. L t + t+ a.. 740. 07. [0, 0] by [0, 00] b. The tax burden increased. Reading the graph from left to right, as t increases so does B(t). f x 0.07x +.69x+. 8. a. [0, 7] by [00,,000] [0, 0] by [0, 00] b. The graph shows years 90 through 000. c. The graph is increasing between 90 and 000. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

0 CHAPTER Functions, Graphs, and Models; Linear Functions d. P x 00x 0.0x 000 4. In 960, the juvenile arrest rate is 0. per 00,000 people. [0, 000] by [,000, 00,000] P x 00x 8000 0.0x 4. In 998 the juvenile arrest rate is 6.06 per 00,000 people. C x,000 + 00x+ 0.x 9. [0, 00] by [0,,000,000] 4. f ( t) 98.06t+,90.77 a. Since the base year is 990, 990-00 correspond to values of t between 0 and inclusive. [0, 0] by [0,000,,000] R x x 0.x 40. b. For 990: f 0 98.06 0 +,90.77,90.77 For 00: f 98.06 +,90.77 47,64.67 [0, 00] by [0, 000] Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section. c. [0, ] by [0,000, 0,000] {Note that answers may vary.} In 990, approximately 9.% of high school seniors had used cocaine. 44. y x x + x+ 0.0094 0.6. 8. a. Since the base year is 97, 97-996 correspond to values of x between 0 and. b. Since percentages are between 0 and 00, y must correspond to values between 0 and 00. c. 4. x (years since 990) y (number of near-hits) 0 8 4 9 86 4 00 40 6 7 7 9 8 9 0 4 d. [0, 0] by [0, 00] y ( number of near-hits) 40 400 0 00 0 00 0 00 0 0 0 4 6 8 0 x ( years since 990) [0, 0] by [0, 0] e. 990 corresponds to x 990 97. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Functions, Graphs, and Models; Linear Functions 46. a. 99.9 million or 99,900,000 b. b. Years after 000 Population (millions) 0 7. 0 99.9 0 4.9 0. 40 77.4 0 40.7 60 4 [, 6] by [60, 80] Yes. The fit is reasonable but not perfect. c. 48. a. [ 0, 70] by [0, 00] 47. a. [, 6] by [00, 600] b. [, 6] by [60, 80] [, 6] by [00, 600] Yes. The fit is reasonable but not perfect. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section. 49. a. In 00 the unemployment rate was.%. b. [ 0, 0] by [0, 0] c. c. [, 0] by [, 6] [ 0, 0] by [0, 0] Yes. The fit is reasonable but not perfect. [, 0] by [, 0] Yes. The fit is reasonable but not perfect. 0. a. The dropout rate in 004 is.6%. b. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

4 CHAPTER Functions, Graphs, and Models; Linear Functions Section. Skills Check b.. Recall that linear functions must be in the form f ( x) ax+ b. a. Not linear. The equation has a nd degree (squared) term. b. Linear. c. Not linear. The x-term is in the denominator of a fraction. y y 6 6 8 4 4. m x x [ 0, 0] by [ 0, 0] 6. a. x-intercept: Let y 0 and solve for x. x + (0) 7 x 7 y-intercept: Let x 0 and solve for y. y y m x x 4 ( 0) 8 8 4 0 undefined Zero in the denominator creates an undefined expression.. b. 0+ y 7 y 7 7 y y.4 x-intercept: (7, 0), y-intercept: (0,.4) y y 0 m 0 x x ( 6) 4 4.. a. x-intercept: Let y 0 and solve for x. x (0) x x [, 0] by [ 0, 0] 7. a. x-intercept: Let y 0 and solve for x. y-intercept: Let x 0 and solve for y. (0) y y y x-intercept: (, 0), y-intercept: (0, ) Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section. (0) 9 6x 0 9 6x 0 9 9 9 6x 6x 9 9 x 6 x. b. [, ] by [ 0, 0] y-intercept: Let x 0 and solve for y. y 9 6(0) y 9 y x-intercept: (., 0), y-intercept: (0, ) 9. Horizontal lines have a slope of zero. Vertical lines have an undefined slope. 0. a. Positive b. Negative b. c. Undefined d. Zero. m 4, b 8 [ 0, 0] by [ 0, 0] 8. a. x-intercept: Let y 0 and solve for x. 0 9x x 0 y-intercept: Let x 0 and solve for y. y 9(0) y 0 x-intercept: (0, 0), y-intercept: (0, 0). Note that the origin, (0, 0), is both an x- and y-intercept.. x+ y 7 y x+ 7 x + 7 y 7 y x+ 7 m, b. y y,horizontal line m 0, b 4. x 6, vertical line undefined slope, no y-intercept Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

6 CHAPTER Functions, Graphs, and Models; Linear Functions. a. m 4, b b. Rising. The slope is positive c. exercise 6. Exercise displays the greatest steepness. 9. For a linear function, the rate of change is equal to the slope. m 4. 0. For a linear function, the rate of change is equal to the slope. m. [, 0] by [, 0] 6. a. m 0.00, b 0.0 b. Rising. The slope is positive. c.. For a linear function, the rate of change is equal to the slope. m.. For a linear function, the rate of change is equal to the slope. 00 m.. For a linear function, the rate of change is equal to the slope. y y 7 0 m. x x 4 [ 00, 00] by [ 0.0, 0.0] 7. a. m 00, b 0,000 b. Falling. Slope is negative. c. 4. For a linear function, the rate of change is equal to the slope. y y m. x x 6 4. The lines are perpendicular. The slopes are negative reciprocals of one another. 6. For line : y y m x x 8 7 [0, 00] by [0, 0,000] 8. Steepness refers to the rise of the line as the graph is read from left to right. Therefore, exercise 7 is the least steep, followed by Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section. 7 For line : x 7y 7y x+ 7y x+ 7 7 y x 7 m 7 Since the slopes are equal, the lines are parallel. Section. Exercises. Linear. Rising the slope is positive. 0. m.. Non-linear. The function does not fit the form f ( x) mx+ b.. Linear. Falling the slope is negative. 0.76 m. 7. a. The identity function is y x. Graph ii represents the identity function. b. The constant function is y k, where k is a real number. In this case, k. Graph i represents a constant function. 8. The slope of the identity function is one ( m ). 9. a. The slope of a constant function is zero ( m 0). b. The rate of change of a constant function equals the slope, which is zero. 0. The rate of change of the identity function equals the slope, which is one. 4. Linear. Falling the slope is negative. 0.6 m.. a. x-intercept: Let p 0 and solve for x. 0 p 9x 0 0(0) 9x 0 9x 0 0 x 9 0 The x-intercept is,0 9. b. p-intercept: Let x 0 and solve for p. 0 p 9x 0 0 p 9(0) 0 0 p 0 p The y-intercept is ( 0, ). In 990, the percentage of high school students using marijuana daily is %. c. x 0 corresponds to 990, x corresponds 99, etc. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

8 CHAPTER Functions, Graphs, and Models; Linear Functions d. For a linear function the rate of change is equal to the slope. 0 m. 8. a. Eating Asparagus [ 0, 0] by [, ] 6. a. y-intercept: Let x 0 and solve for y. y 88,000 00(0) 88,000 Initially the value of the building is $88,000. b. x-intercept: Let y 0 and solve for x. c. 0 88,000 00x 00x 88,000 88,000 x 60 00 The value of the building is zero (the building is completely depreciated) after 60 months or 0 years. Asparagus Consumption 0.7 0.6 0. 0.4 0. 0. 0. 0 98 990 99 000 Year b. The data can be modeled by a constant function. c. y 0.6 d. Eating Asparagus Asparagus Consumption 0.8 0.6 0.4 0. y 0.6 0 98 990 99 000 Year 9. a. m 6. [0, 60] by [0,,000,000] 7. a. The data can be modeled by a constant function. Every input x yields the same output y. b. y.8 c. A constant function has a slope equal to zero. b. Each year, the percent of Fortune Global 00 firms recruiting via the Internet increased by 6.%. 40. a. For a linear function, the rate of change is equal to the slope. m 0.7069. The slope is negative. b. The percentage is decreasing. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section. 9 4. a. For a linear function, the rate of change is equal to the slope. m. The 7 slope is positive. b. For each one degree increase in temperature, there is a increase in the 7 number of cricket chirps. 4. a. m.84 b. The rate of growth is.84 hundred dollars per year. 4. a. Yes, it is linear. b. m 0.99 c. For each one dollar increase in white median annual salaries, there is a 0.99 dollar increase in minority median annual salaries. 44. a. m 0. For each one unit increase in x, the number of years since 90, there is a 0. decrease in y, the percent of voters voting in presidential elections. b. The rate of decrease is 0. percent per year. 4. a. To determine the slope, rewrite the equation in the form f ( x) ax+ b or y mx+ b. 0 p 9x 0 0 p 9x+ 0 0 p 9x+ 0 0 0 9 p x+ 0 9 m.6 0 b. Each year, the percentage of high school seniors using marijuana daily increases by approximately 0.6%. 46. a. p 8d 496 Solving for p : p 8d + 496 8d + 496 p 8 496 p d + 6 496 p d + 6 Therefore, m b. For every one unit increase in depth, there is a corresponding 6 pound per square foot increase in pressure. 47. x-intercept: Let R 0 and solve for x. R 00 70x 0 00 70x 70x 00 00 x 0 70 0,0. The x-intercept is y-intercept: Let x 0 and solve for R. R 00 70x R 00 70(0) R 00 The y-intercept is ( 0,00 ). Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

0 CHAPTER Functions, Graphs, and Models; Linear Functions 0. a. m. y intercept b 6.0 [0, ] by [0, 00] 48. a. D (0) 0.7(0).09.76 Based on the model, 0,000 ATM transactions correspond to a dollar volume of $.76 billion. b. Fewer than approximately 7,4 ATMs. [0, 0] by [, ] 49. a. m.74 y intercept b 4.6 b. The y-intercept represents the percentage of the population with Internet access in 99. Therefore in 99, 4.6% of the U.S. population had Internet access. c. The slope represents the annual change in the percentage of the population with Internet access. Therefore, the percentage of the population with Internet access increased by.74% each year. b. The y-intercept represents the total amount spent for wireless communications in 99. Therefore in 99, the amount spent on wireless communication in the U.S. was 6.0 billion dollars. c. The slope represents the annual change in the amount spent on wireless communications. Therefore, the amount spent on wireless communications in the U.S. increased by. billion each year.. a. m y y x x 700,000,0,000 0 0 60,000 0 6,000 b. Based on the calculation in part a), the property value decreases by $6,000 each year. The annual rate of change is 6,000.. a. m y y x x 68. 8. 990 890 0.4 00 0.04 b. Based on the calculation in part a), the number of men in the workforce increased by 0.04 million (or 04,000) each year. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.. Marginal profit is the rate of change of the profit function. c. Selling one additional golf ball each month increases revenue by $.60. m y y x x 9000 460 7 00 40 7 8 8. a. m 98 b. The marginal revenue is $98 per unit c. Selling one additional television each month increases revenue by $98. The marginal profit is $8 per unit. 4. Marginal cost is the rate of change of the cost function. m y y x x 0 690 00 00 840 00.8 9. The marginal profit is $9 per unit. Note that 9 m. 60. The marginal profit is $99 per unit. Note that 99 m. The marginal cost is $.80 per unit.. a. m 0.6 b. The marginal cost is $0.6 per unit. c. Manufacturing one additional golf ball each month increases the cost by $0.6 or 6 cents. 6. a. m 98 b. The marginal cost is $98 per unit. c. Manufacturing one additional television each month increases the cost by $98. 7. a. m.60 b. The marginal revenue is $.60 per unit. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Functions, Graphs, and Models; Linear Functions Section.4 Skills Check... 4.. 6. m 4, b. The equation is y 4x+. m, b. The equation is y x+. m, b. The equation is y x+. m, b 8. The equation is y x 8. m, b. The equation is 4 y x+. 4 m, b. The equation is y x+. 7. y y m( x x ) y 4 ( x ( )) y 4 ( x+ ) y 4 x+ y x+ 9 8. y y m( x x ) y ( x ( 4)) y ( x+ 4) y x y x+ 9. y y m( x x ) y ( 6) ( x 4) 4 4 y+ 6 x+ i 4 4 y+ 6 x+ 4 y x 4 0. y y m( x x ) y 6 y 6 ( x+ ) y 6 x y x+ 4 ( x ). y y m( x x ) y 4 0( x ( )) y 4 0 y 4. Since the slope is undefined, the line is vertical. The equation of the line is x a, where a is the x-coordinate of a point on the Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.4 line. Since the line passes through (, 4), the equation is x.. x 9 Equation: y y m( x x ) y 4 ( x ) y 4 x y x+ 4. y 0 y y 7 6. Slope: m x x (4) 6 Equation: y y m( x x) y 7 ( x 4) y 7 x 4 y x+ y y 8 6 6. Slope: m x x Equation: y y m( x x) y 8 ( x ) y 8 x y x 7 y y 6 7. Slope: m x x ( ) Equation: y y m( x x) y 6 ( x ) y 6 x y x+ 4 8. Slope: y y 4 7 m x x 4 ( ) 7 y y 9. Slope: m x x ( 4) Equation: 0. Slope: Equation: y y m( x x) y ( x ( )) y x+ 9 y x+ 4 6 y y m x x y y m( x x) y ( 6 ) ( x ) y+ 6 x+ y x y y. Slope: m undefined x x 9 9 0 The line is vertical. The equation of the line is x 9.. Slope: y y m x x 0 ( ) 8 The line is horizontal. The equation of the line is y. 0. With the given intercepts, the line passes through the points (, 0) and (0, 4). The Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

4 CHAPTER Functions, Graphs, and Models; Linear Functions slope of the line is y y 4 0 4 m. x x 0 ( ) Equation: y y m( x x) 4 y 0 4 y ( x+ ) 4 y x+ 4 ( x ) 4. With the given intercepts, the line passes through the points (4, 0) and (0, ). The slope of the line is y y 0 m. x x 0 (4) 4 4 Equation:. Slope: y y m( x x) y 0 ( x 4) 4 y ( x 4) 4 y x 4 8 4 6 y y m x x Equation: y y m( x x) y ( x 4) y x y x+ 6. Slope: y y 7 8 m x x ( 4) 6 Equation: y y m( x x) y 7 ( x ( 4 )) y 7 x y x 7. For a linear function, the rate of change is equal to the slope. Therefore, m. The equation is y y m( x x) y ( x 0) y x y x+. 8. For a linear function, the rate of change is equal to the slope. Therefore, m 8. The equation is y y m( x x) y ( 7) 8( x 0) y+ 7 8x y 8x 7. 9. 0. f( b) f( a) b a f() f( ) ( ) () ( ) 4 The average rate of change between the two points is. f( b) f( a) b a f() f( ) ( ) () ( ) 8 + 9 The average rate of change between the two points is. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.4. a. f ( x+ h) 4 ( x+ h) 4 x h b. c. f( x+ h) f( x) [ ] 4 x h 4 x 4 x h 4 + x h f( x+ h) f( x) h h h. a. f( x+ h) ( x+ h) + x+ h+ b. c.. a. f( x+ h) f( x) [ ] x+ h+ x+ x+ h+ x h f( x+ h) f( x) h h h f x+ h x+ h + 4 ( x xh h ) + + + 4 x + xh+ h + 4 4 4. a. b. c. f x+ h x+ h + ( x xh h ) + + + x + xh+ h + 6 f( x+ h) f( x) x + 6xh+ h + x + x + 6xh+ h + x 6xh + h f ( x+ h) f( x) h 6xh + h h h( 6x+ h) h 6x+ h. a. The difference in the y-coordinates is consistently 0, while the difference in the x-coordinates is consistently 0. Note that 6 8 0, 64 60 0, etc. Considering the scatter plot below, a line fits the data exactly. b. c. f( x+ h) f( x) x + xh+ h + x + x + xh+ h + x 4xh + h 4 4 4 4 4 4 f ( x+ h) f( x) h 4xh + h h h( 4x+ h) h 4x+ h [0, 60] by [00, 800] b. Slope: y y m x x 6 8 0 0 0 0 Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

6 CHAPTER Functions, Graphs, and Models; Linear Functions Equation: y y m( x x) y 8 ( x 0) y 8 x 0 y x+ 6. a. The difference in the y-coordinates is consistently 9, while the difference in the x-coordinates is consistently 6. Note that 7. 8. 9, 6. 7. 9, etc. Considering the scatter plot below, a line fits the data exactly. Section.4 Exercises 7. Let x KWh hours used and let y monthly charge in dollars. Then the equation is y 0.09x+ 8.9. 8. Let x minutes used and let y monthly charge in dollars. Then the equation is y 0.07x+ 4.9. 9. Let t number of years, and let s value of the machinery after t years. Then the equation is s 6,000,600t. [0, 0] by [ 0, 0] 40. Let x age in years, and let y hours of sleep. Then the equation is y 8+ 0. 8 x. or y 8 + 4. 0.x. 0. x. b. Slope: y y m x x 6. 7. 9 9 6 4. a. Let x the number of years since 996, and let P the population of Del Webb s Sun City Hilton Head community. The linear equation modeling the population growth is P 70x+ 98. b. To predict the population in 00, let x 00 996 6. The predicted population is P 70(6) + 98 448. Equation: y y m( x x ) y 6. ( x 9) 7 y 6. x y x 9. + 6. y x+ 4. Let x the number of years past 994, and let y the composite SAT score for the Beaufort County School District. The linear equation modeling the change in SAT score is P 9 + 0.x. 4. a. From year 0 to year, the automobile depreciates from a value of $6,000 to a value of $,000. Therefore, the total depreciation is 6,000 000 or $,000. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.4 7 b. Since the automobile depreciates for years in a straight-line (linear) fashion, each year the value declines by,000 $,000. c. Let t the number of years, and let s the value of the automobile at the end of t years. Then, based on parts a) and b) the linear equation modeling the depreciation is s 000t+ 6,000. P.% 7,000 y 87y where y number of years of service and P annual pension amount in dollars. 44. 4. Notice that the x and y values are always match. That is the number of deputies always equals the number of patrol cars. Therefore the equation is y x, where x represents the number of deputies, and y represents the number of patrol cars. 46. Notice that the y values are always the same, regardless of the x value. That is, the premium is constant. Therefore the equation is y.8, where x represents age, and y represents the premium in dollars. y y 47. m x x 9000 460 7 00 40 8 7 Equation: y y m( x x) y 460 8( x 00) y 460 8x 7,400 y 8x,70 y y 48. m x x 0 680 00 00 80 7 00 6 Equation: y y m( x x) 7 y 680 ( x 00) 6 7 700 y 680 x 6 7 700 8040 y x + 6 7 640 y x 6 y.x. y y 49. m x x 700,000,0,000 0 0 60,000 0 6,000 Equation: y y m( x x) y,90,000 6,000( x 0) y,90,000 6,000x y 6,000x+,90,000 v 6, 000x+,90, 000 Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

8 CHAPTER Functions, Graphs, and Models; Linear Functions 0. a. At t 0, y 860,000. b. ( 0,860,000 ),(,0) y y m x x 0 860,000 0 860,000 4, 400 Equation: y y m( x x) y 0 4,400( x ) y 4, 400x+ 860, 000 y 860,000 4, 400t where t number of years, y value y y. m x x..7 6 6 0.6 0 0.0 Equation: y y m( x x) y.7 0.0( x 6) y.7 0.0x+ 0.8 y 0.0x+.88 p.88 0.0t where t number of years beyond 97, y percentage of cigarette use. Let x median weekly income for whites, and y median weekly income for blacks. The goal is to write y f(x). y y changein y 6.90 0.69 changein 00 m x x x Equation: y y m( x x) y 7 0.69( x 676) y 7 0.69x 48.444 y 0.69x+ 08.6. a. Notice that the change in the x-values is consistently while the change in the y- values is consistently 0.0. Therefore the table represents a linear function. The rate of change is the slope of the linear function. vertical change 0.0 m 0.0 horizontal change b. Let x the number of drinks, and let y the blood alcohol content. Using points (0, 0) and (, 0.0), the slope is y y m x x 0.0 0 0 0.0 0.0. Equation: y y m( x x) y 0 0.0( x 0) y 0.0x 4. a. Notice that the change in the x-values is consistently while the change in the y- values is consistently 0.0. Therefore the table represents a linear function. The rate of change is the slope of the linear function. vertical change 0.0 m 0.0 horizontal change b. Let x the number of drinks, and let y the blood alcohol content. Using points (, 0.) and (0, 0.), the slope is Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.4 9 y y m x x 0. 0. 0 0.0 0.0. Equation: y y m( x x) y 0. 0.0( x ) y 0. 0.0x 0. y 0.0x+ 0.0. a. Let x the year at the beginning of the decade, and let y average number of men in the workforce during the decade. Using points (890, 8.) and (990, 68.) to calculate the slope yields: y y m x x 68. 8. 990 890 0.4 0.04 00 Equation: y y m( x x) y 8. 0.04( x 890) y 8. 0.04x 9.6 y 8.+ 8. 0.04x 9.6 + 8. y 0.04x 94.46 b. Yes. Consider the following table of values based on the equation in comparison to the actual data points. x y (Equation Values) Actual Values 890 8. 8. 900.4.6 90 8.8 7 90. 90 8.6 7 940 4. 40 90 48.4 4.8 960.8 47 970 8.4.6 980 6.46 6.4 990 68. 68. c. It is the same since the points ( 890,8. ) and ( 990,68. ) were used to calculate the slope of the linear model. 6. a. Let t the year, and let p the percentage of workers in farm occupations. Using points (80, 78.) and (994,.6) to calculate the slope yields: y y m x x.6 7.8 994 80 69. 74 0.9770494 0.40 Equation: y y m( x x) y.6 0.9770494( x 994) y.6 0.9770494x+ 79.06099 y 0.9770494x+ 79.66099 y 0.40x+ 79.6 b. The line appears to be a reasonable fit to the data. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

40 CHAPTER Functions, Graphs, and Models; Linear Functions c. Each year between 890 and 994, the percentage of workers in farm-related jobs decreases by 0.40% d. No. The percentage of farm workers would become negative. b. The average rate of change is 08 students per year. It is the same as part a). c. Since enrollment is projected to increase, additional buildings may be necessary. 7. a. f( b) f( a) b a f(00) f(996) 00 996 40. 7..4 y y 9. a. m x x 76 46 0 6.694.69 6 b. It is the same as part a). The average rate of change is $.4 billion dollars per year. y y m x x 40. 00 996 7..4 b. c. No. Note that change in education spending from one year to the next is not constant. It varies. d. No. Since the change in the y-values is not constant for constant changes in the x-values, the data can not be modeled exactly by a linear function. y y 8. a. m x x 6704 664 00 000 040 08 c. Each year between 960 and 996, the percentage of out-of-wedlock teenage births increased by approximately.69%. y y m( x x ) 6 y ( x 0) 6 6 60 y x 6 6 6 60 y x + 6 6 6 60 40 y x + 6 6 6 6 70 y x 6 6 y.69x.94 d. y y 60. a. m x x. 6. 99 960 8 0. b. It is the same as part a). The percentage of eligible people voting in presidential Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.4 4 elections is decreasing at a rate of 0.% per year. y y m( x x ) y 6. 0.( x 960) y 6. 0.x+ 490 y 0.x+. c. 6. a. f ( b) f( a) b a f () f () 49 08 4 409 4 0. 6. a. f( b) f( a) b a f(997) f(960) 997 960,97,90,9 997 960 984,67 7 6,6.8 b. The slope of the line connecting the two points is the same as the average rate of change between the two points. Based on part a), m 6,6.8. c. The equation of the secant line is given by: y y m( x x) 984,67 y,9 ( x 960) 7 y,9 6,6.8x,9,49 y 6,6.8x,946,96 d. No. The points on the scatter plot do not approximate a linear pattern. e. Points corresponding to 990 and 997. The points between those two years do approximate a linear pattern. b. Each year from year to year, the worth of the investment increases on average by $0.. c. The slope is the same as the average rate of change, 0.. y y m( x x ) y 08 0.( x ) y 08 0.x 0. y 0. + 980.7 d. 6. a. No. b. Yes. The points seem to follow a straight line pattern for years between 00 and 00. c. f( b) f( a) b a f(00) f(00) 00 00..9 00 00.7 0 0.08 y y m( x x ) y.9 0.08( x 00) y.9 0.08x+ 70.8 y 0.08x+ 74.7 d. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

4 CHAPTER Functions, Graphs, and Models; Linear Functions 64. a. No. The points in the scatter plot do not lie approximately in a line. b. c. f( b) f( a) b a f(90) f(890) 90 890 6.44.704 60.79 60 0. f( b) f( a) b a f(990) f(90) 990 90 9. 6.44 40 4.088 40.077 d. Yes. Since the graph curves, the average rate of change is not constant. The points do not lie exactly along a line. 6. a. Let x the number of years since 90, and let y the U.S. population in thousands. Then, the average rate of change in U.S. population, in thousands, between 90 and 99 is given by: f( b) f( a) b a f(4) f(0) 4 0 6,044,7 4 0 0,77 4 46.6 66. a. Changing the units from thousands to millions yields,46.8.466, 000 million per year. b. Remember to change the units into millions: y y m( x x) y.7.466( x 0) y.7.466x y.466x.7 c. 97 corresponds to x. y.466() +.7 y.8 or,8,000 No. The values are different. d. The table can not be represented exactly by a linear function. Years past 96 Percent 0 9.8 9.. 0 6 7 6.6 8. 4 9 8.9 8 4.8 0 4 44.9 48.4 Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.4 4 b. Smoking Cessation y, Percentage 60 0 40 0 0 0 y 0.7x + 9.46 0 0 0 0 0 x, Years past 96 c. Yes. A linear model seems to fit the data reasonably well. 67. a. Yes. The x-values have a constant change of $0, while the y-values have a constant change of $4. b. Since the table represents a linear function, the rate of change is the slope. y y m x x 7 0 0,00 0,000 4 0.8 0 Yes. The results from the equation match with the table. 68. a. Group Expense + Group Expense Total Expense 00x+ 00y 00,000 b. 00x+ 00y 00,000 00y 00x+ 00,000 00x + 00,000 y 00 00 00,000 y x+ 00 00 y.x+ 00 The y-intercept is 00. If no clients from the first group are served, then 00 clients from the second group can be served. The slope is.. For each one person increase in the number of clients served from the first group there is a corresponding decrease of. clients served from the second group. c. 0i. Fifteen fewer clients can be served from the second group. For every $.00 in income, taxes increase by $0.8. c. Equation: y y m( x x) y,0 0.8( x 0,000) y,0 0.8x 8400 y 0.8x,97 d. When x 0,00, y 0.8(0,00) 97. When x 0,00, y 0.8(0,00) 97 87. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

44 CHAPTER Functions, Graphs, and Models; Linear Functions Section. Skills Check. x 4 + 7x x 7x 4 + 7x 7x x 4 x 4+ 4 + 4 x 7 x 7 7 x x 8. Applying the intersections of graphs method, graph y x 4 and y + 7x. Determine the intersection point from the graph: [, ] by [ 00, 00]. x 7x 4 x 7x 7x 7x 4 4x 4 4x 4x 4 4 x 4 x. x 7x 4 x 7x + 4 0 4x + 0 Graph y 4x+ and determine the x- intercept. The x-intercept is the solution to the equation. [ 0, 0] by [ 0, 0]. ( x 7) 9 x x 9 x x + x 9 x+ x 4x 9 4x + 9+ 4x 40 4x 40 4 4 x 0 Applying the intersections of graphs method yields: [, ] by [ 0, 0] Applying the x-intercept method, rewrite the equation so that 0 appears on one side of the equal sign. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section. 4 6 8 y y 0 8 y 7y 48 4. ( y ) y 48 7 Applying the intersections of graphs method yields: [ 0, 0] by [ 0, 0]. [ 0, 0] by [ 0, 0] x x+ 6 4 LCM : x x+ 6 4 x 0 6x+ x 6x 0 6x 6x+ 4x 0 4x 0 + 0 + 0 4x 4x 4 4 x 4 6. x x+ 4 LCM : x x+ 4 6x 4 60x+ 9 4x 4 9 4x x 4 Applying the intersections of graphs method yields: [ 0, 0] by [ 0, 0] Applying the intersections of graphs method yields: Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

46 CHAPTER Functions, Graphs, and Models; Linear Functions 7. ( x ) x x 6 9 LCM :8 ( x ) x 8 x 8 6 9 ( x ) 8x 8 x x 4 8x 8 x x 4 8 x x 4 8 x 6 x 6 Applying the intersections of graphs method yields: [0, 70] by [ 0, ] 9..9t.78t 4.4.9t.78t 4.4 4.4t 4.4 4.4t 4.4 4.4 4.4 t Applying the intersections of graphs method yields: [ 00, 0] by [ 0, 0] 8. ( y ) 4 y y 6 LCM : 4( y ) y y 6 4( y ) y 90 y ( y ) y 90 y y 4 y 90 y y 4 90 y y 4 y 7 Applying the intersections of graphs method yields: [ 0, 0] by [ 0, 0] 0. 0.0x+ 0.8 0.6x.66 0.7x 6.066 6.066 x 0.7 x 8 Applying the intersections of graphs method yields: Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section. 47 The solution to f( x ) 0, the x-intercept of the function, and the zero of the function are all 0. 4. Answers a), b), and c) are the same. Let f( x ) 0 and solve for x. [0, 0] by [, ]. + x 4 x 4 LCM 60 4 60 + x 60 x 4 4 + x 0 48x 6x x 6 6. x 6 + x 6 LCM 0 6 0 x 0 + x 6 0x 6 + x x x x 60 0 x 60 x 4 The solution to f( x ) 0, the x-intercept of the function, and the zero of the function are all 4.. Answers a), b), and c) are the same. Let 0 f x and solve for x. x 6 0 LCM : x 6 0 x 0 x x 4 The solution to f( x ) 0, the x-intercept of the function, and the zero of the function are all 4.. Answers a), b), and c) are the same. Let f( x ) 0 and solve for x. +.6x 0.6x x.6 x 0 6. Answers a), b), and c) are the same. Let 0 f x and solve for x. x 0 4 LCM :4 x 4 4 0 4 x 0 x Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

48 CHAPTER Functions, Graphs, and Models; Linear Functions The solution to f( x ) 0, the x-intercept of the function, and the zero of the function are all. 7. a. The x-intercept is, since an input of creates an output of 0 in the function. The solution is the x-coordinate of the intersection point or x.. Applying the intersections of graphs method yields: b. The y-intercept is 4, since the output of 4 corresponds with an input of 0. c. The solution to f( x ) 0 is equal to the x-intercept position for the function. Therefore, the solution to f( x ) 0 is. 8. a. The x-intercept is 0., since an input of 0. creates an output of 0 in the function. b. The y-intercept is 7, since the output of 7 corresponds with an input of 0. [ 0, 0] by [ 0. 0] The solution is the x-coordinate of the intersection point or x.. Applying the intersections of graphs method yields: c. The solution to f( x ) 0 is equal to the x-intercept position for the function. Therefore, the solution to f( x ) 0 is. 9. The answers to a) and b) are the same. The graph crosses the x-axis at x 40. 0. The answers to a) and b) are the same. The graph crosses the x-axis at x 0.8. [ 0, ] by [ 70, 0] The solution is the x-coordinate of the intersection point or s.. Applying the intersections of graphs method yields: [ 0, 0] by [ 0. 0] Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section. 49 4. Applying the intersections of graphs method yields: The solution is the x-coordinate of the intersection point or x 6. 7. Applying the intersections of graphs method yields: [ 0, ] by [ 70, 0] The solution is the x-coordinate of the intersection point or x 4.. Applying the intersections of graphs method yields: [ 0, 0] by [, ] The solution is the x-coordinate of the 7 intersection point, which is x 4.. 4 8. Applying the intersections of graphs method yields: [ 0, ] by [ 0, 0] The solution is the x-coordinate of the intersection point. t 4. 6. Applying the intersections of graphs method yields: [ 0, 0] by [, ] The solution is the x-coordinate of the 9 intersection point, which is x.. 9 [ 0, 0] by [ 0, 0] Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

0 CHAPTER Functions, Graphs, and Models; Linear Functions 9. a. A P( + rt) A P+ Prt A P P P+ Prt A P Prt A P Prt Pr Pr A P A P t or t Pr Pr b. A P( + rt) A P( + rt) + rt + rt A A P or P + rt + rt 0. V π r h LCM : ( V) π r h V π r h V π r h πr πr V V h or h π r π r. F 9C 60 F 9C+ 9C 60+ 9C F 60+ 9C F 60+ 9C 9 60 F C+ 9 F C+ c. 4( a x) x+ LCM : c 4 ( ( a x) ) x+ ( a x) x+ c a 4x x+ c a 4x x x x+ c a 9x c a a 9x c a 9x c a c a x or 9 c a a c x 9 9 P. + A m n LCM : P + A m n P+ A 0m 4n P+ A 0m 0m 4n 0m P+ A 0m 4n 4 4 P+ A 0m n 4 P A 0m n + 4 4 4 m P A n 4 4. y y m( x x ) y y mx mx y y + mx mx mx + mx y y+ mx mx m m y y + mx x m Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.. x y y x+ x + y y x [ 0, 0] by [ 0, 0] [ 0, 0] by [ 0, 0] 8. 4 8 x + y y x + 4 8 x + y y x + 4 8 4 6. x+ y 6 y x+ 6 x + 6 y y x+ [ 0, 0] by [ 0, 0] [ 0, 0] by [ 0, 0] 7. x + y 6 y 6 x 6 x y y x y x or + Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Functions, Graphs, and Models; Linear Functions Section. Exercises 9. Let y 690,000 and solve for x. 690,000 88,000 00x 8,000 00x 8, 000 x 00 x 60 After 60 months or years the value of the building will be $690,000. 40. Let C 0 and solve for F. F 9C 60 F 9( 0) 60 F 80 60 F 40 40 F 68 68 Fahrenheit equals 0 Celsius. 4. S P( + rt) 9000 P( + (0.0)()) 9000 P( + 0.0) 9000.P 9000 P 6000. $6000 must be invested as the principal. 4. I t 0.( h)( t 8) 79 80 0.( h)( 80 8) 79 80 0.( h)( ) 79 80.( h) 79 80.+.h 79 67.9 +.h.h.. h. h 0.9779 0.9 A relative humidity of 9% gives an index of 79. 4. M 0.99W.6 0,60 0.99W.6 0.99W 0,6.6 0,6.6 W,7 0.99 The median annual salary for whites is approximately $7,7. 44. Let B(t) 4.44, and calculate t. 4.44.0t 69.6 4.44 + 69.6.0t 6606.0t 6606 t.0 t 000 The model predicts that in 000 there will be 4.44 million accounts. 4. Recall that F 9C 60. Let F C, and solve for C. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section. C 9C 60 4C 60 60 C 4 C 40 Therefore, F C when the temperature is 40...96x +.7.96x.8.8 x 4.96 Based on the model, the average monthly mobile phone bill is $. in 999. 46. Let y 80, and solve for x. 80.78x.998.78x 8.998 8.998 x 47.9.78 Based on the model, the level will reach 80% in approximately 997. 47. Let y 9.4, and solve for x. 9.4 0.x +.7 9.4.7 0.x 4.0 0.x 4.0 x 0. x 6 An x-value of 6 corresponds to the year 996. The average reading score is 9.4 in 996. 48. Let B(t) 47.88, and calculate t. 47.88 0.7 +.84t.84t 7. 7. t.84 In 99 the per capita tax burden is $4788. 49. Let B(x)., and calculate x. 0. y 0.076x+ 8.84 6.09 0.076x + 8.84 0.076x.484.484 x 0.076 When x is, the year is 98. The model predicts the marriage rate to be 6.09% in 98.. Note that p is in thousands. A population of 8,4,000 corresponds to a p-value of 8,4. Let p 8,4 and solve for x. 8,4 x + 0,87 8,4 0,87 x 8,44 x 6,44 x x 4 An x-value of 4 corresponds to the year 994. Based on the model, in 994 the population is estimated to be 8,4,000.. Let P(x) 44%, and solve for x. 44 6.x 6 44 + 6 6.x 06 6.x 06 x 4 6. An x-value of 4 corresponds to the year 999. Based on the model, forty-four Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

4 CHAPTER Functions, Graphs, and Models; Linear Functions percent of firms recruited on the Internet in 999.. When the number of prisoners is 797,0, then y 797.0. Let y 797.0, and calculate x. 797.0 68.476x + 78.64 797.0 78.64 68.476x 68.476 68.476x 68.476 x 68.476 An x-value of one corresponds to the year 99. The number of inmates was 797,0 in 99. 4. Let p.%, and solve for x. 0. 9x 96 9x 9x 9 9 x 9 When x is, the year is 99. During 99 the percentage using marijuana daily was.%. a. Let p 49%, and solve for x. 49 6.404 0.x 49 6.404 0.x 6.404 0.x 6.404 x 0. x 46. An x-value of 46 corresponds to the year 996. The model predicts that in 996 the percent voting in a presidential election is 49% b. Year 000 corresponds with an x-value of 0. Let x 0, and solve for p. p 6.404 0.(0) p 6.404 7.76 p 47.644 Based on the model the percentage of people voting in the 000 election was approximately 47.6%. The prediction is different from reality. Models do not always yield accurate predictions. 6. Let I(x) 790, and solve for x. 790 4.8x + 449.78 790 449.78 4.8x 70. 4.8x 70. x 4.8 x 7.99 8 An x-value of 8 corresponds to the year 998. Based on the model, U.S. personal income reached $790 billion in 998. 7. Let y 6000, and solve for x. 6000 77.8x 44.766 77.8x 744.766 744.766 x 6.77 77.8 An x-value of 7 corresponds to the year 997. Cigarette advertising exceeds $6 billion in 997. 8. If the number of customers is 68,00,000, then the value of S(x) is 68.. Let S(x) 68., and solve for x. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section. 68..7x +.9 68..9.7x..7x. x.7 An x-value of corresponds to the year 998. There were 68. million subscribers in 998. 9. Let x represent the score on the fifth exam. 9 + 86 + 79 + 96 + x 90 LCM: 9 + 86 + 79 + 96 + x 90 40 + x x 97 The student must score 97 on the fifth exam to earn a 90 in the course. 60. Since the final exam score must be higher than 79 to earn a 90 average, the 79 will be dropped from computation. Therefore, if the final exam scores is x, the student s average x + 86+ 96 is. 4 To determine the final exam score that produces a 90 average, let x + 86+ 96 90. 4 LCM :4 x + 86+ 96 4 4 90 4 x + 86+ 96 60 x + 8 60 x 78 78 x 89 The student must score at least an 89 on the final exam. 6. Let x the company s 999 revenue in billions of dollars. 0.94x 74 74 x 0.94 x 78.7 The company s 999 revenue was approximately $78.7 billion. 6. Let x the company s 999 revenue in billions of dollars. 6. 4.79x 6 6 x 4.79 x 7. The company s 999 revenue was approximately $7. billion. Commission Reduction ( 0% )( 0,000) 0,000 New Commission 0,000 0,000 40,000 To return to a $0,000 commission, the commission must be increased $0,000. The percentage increase is now based on the $40,000 commission. L Let x represent the percent increase from the second year. 40, 000x 0, 000 x 0. % Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

6 CHAPTER Functions, Graphs, and Models; Linear Functions 64. Salary Reduction ( % )( 00,000) 000 New Salary 00,000 000 9,000 To return increase to a $04,00 salary, the new $9,000 must be increased $9,00. The percentage increase is now based on the $9,000 salary. Let x represent the percent raise from the reduced salary. 9,000x 900 900 x 0.0 0% 9,000 6. Total cost Original price + Sales tax Let x original price. 9,998 x+ 6% x 9,998 x+ 0.06x 9,998.06x 9,998 x 8,00.06 Sales tax 9,998 8,00 $698 66. Let x total in population. x 0 0 0 x 0 000 000 0 0 0x 00 00 x 0 The estimated population is sharks. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.6 7 Section.6 Skills Check. No. The data points do not lie approximately in a straight line.. Yes. The data points lie approximately in a straight line. y 0 y.x +. 0 0 0 4 6 8 0 x. 4. y y 0 0 0 0 4 6 8 0 0 8 6 4 x 8. Using a spreadsheet program yields y 0 9 8 7 6 4 0 y 0.8x +.484 0 0 x 0 0 0 x 9. 0 40 0. Exactly. The first differences are constantly three. y 0 0 0 0 0 0 0 6. No. The first differences are not constant. Also, a line will not connect perfectly the points on the scatter plot. x 7. Using a spreadsheet program yields 0. Yes. The points appear to lie approximately along a line. Using a spreadsheet program yields Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

8 CHAPTER Functions, Graphs, and Models; Linear Functions y 0 y.49x -.74 40 0 0 0 0 0 0 0 0 x y 0 0 0 0 y.77x +.89 0 0 0 x. See problem 0 above. y.49x.7. f ().49().74 7.7.74.686.7 f ().49().74.09.74 6.6 6. 6. f ().77()+.89 4.7+.89 6.6 6.6 f ().77()+.89 7.88 +.89 9.777 9.8 7.. y 0 0 0 0 0 0 0 x 4. Yes. The points appear to lie approximately along a line. The second equation, y.x+ 8, is a better fit to the data points.. Using a spreadsheet program yields Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.6 9 8. Section.6 Exercises. a. Discrete. The ratio is calculated each year, and the years are one unit apart. b. No. A line would not fit the points on the scatter plot. c. Yes. Beginning in 00, a line would fit the points on the scatter plot well.. a. Discrete. There are gaps between the years. The first equation, y.x+ 4, is a better fit to the data points. 9. a. Exactly linear. The first differences are constant b. Nonlinear. The first differences increase continuously. c. Approximately linear. The first differences vary, but don t grow continuously. 0. The difference between inputs is not constant. The inputs are not equally spaced. b. Continuous. Gaps between the years no longer exist. c. No. A line would not fit the points on the scatter plot. A non-linear function is best.. a. Yes. There is a one unit gap between the years and a constant 60 unit gap in future values. b. Yes. Since the first differences are constant, the future value can be modeled by a linear function c. Using the graphing calculator yields [, 0] by [ 00, 00] 4. p (.7) does not make sense, since.7 does not correspond exactly to a specific month. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

60 CHAPTER Functions, Graphs, and Models; Linear Functions. a. Let x. Then, y.90x+ 4.78.90() + 4.78 6.9 + 4.78 608.688 Based on the model, 608,688 people were employed in dentist s offices in 99. Since 99 is within the range of the data used to generate the model (970-998), this calculation is an interpolation. b. Let x 0. Then, y.90x+ 4.78.90(0) + 4.78 477. + 4.78 70.08 Based on the model, 70,08 people were employed in dentist s offices in 000. Since 000 is not within the range of the data used to generate the model (970-998), this calculation is an extrapolation. 7. a. Using a spreadsheet program yields Percent of Adults who Quit Smoking 60 0 40 0 0 0 0 Smoking Cessation y 0.7x +.6 0 0 0 0 Years past 960 b. Solve 0.7x +.6 9. Using the intersections of graphs method to determine x when y 9 yields 6. a. Since x represents years past 990, the model is discrete. b. Yes, since 9 corresponds exactly to a specific year. P (9) represents the percentage of Fortune Global 00 firms that actively recruit workers in 999. c. No. P (9.4) is not valid, since 9.4 does not correspond exactly to a specific year. Based on the model, 9% of adults had quit smoking 8.46 years past 960. Therefore, the year was approximately 978. c. Since the data is not exactly linear (see the scatter plot in part a) above), the model will yield only approximate solutions. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.6 6 8. a. Using a spreadsheet program yields y, Percentage 60 0 40 0 0 0 Smoking Cessation y 0.7x + 9.46 0 0 0 0 0 b. See part a) above. x, Years since 96 c. The slope is the same, but y-intercept is different. 9. a. Using a spreadsheet program yields Postal Rate (dollars).8.6.4. Bound Printed Matter Rates y 0.09x +.08 0 6 9 8 Weight (pounds) b. The model fits the data very well. Notice the small residuals in the following table. Weight Actual postal rate Predicted Residual postal rate (difference based on the between the regression actual and equation predicted values) (pounds) (dollars) (dollars) (dollars).6.967 0.000.0.9874 0.0066 4.4.7794 0.0006.8.768 0.0047 6..9 0.009 7.6.497 0.0008 8.9.940 0.0040 9.4.409 0.0009 0.47.470 0.000..0 0.000..069 0.00069.9.898 0.00067 4.6.6888 0.006.67.667447 0.00 0. Using a spreadsheet program yields Average Math SAT Score 474 47 470 468 466 464 46 SAT Scores y 0.4x + 466.67 0 4 6 8 0 Years past 990 Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

6 CHAPTER Functions, Graphs, and Models; Linear Functions Year Actual data Model prediction Difference 994 47 468.77.88 99 464 469.. 996 470 469.748 0.4 997 47 470.60 0.799 998 47 470.7744.6 999 470 47.887.887 The actual and predicted values are closest in 996.. a. Using a spreadsheet program yields Average Daily Number of Inmates 00 0 00 0 Jail Population y 9.09x + 70.4 0 0 4 6 8 0 Years past 990. a. Let x 6, and solve for y. y 0.09(6) +.08 y 0.64 +.08 y.706.7 Using the unrounded model yields b. Let x 00 990, and solve for y. y 9.09( ) + 70.4 y 0.778 06 In 00 the model predicts an average daily prison population of 06. Using the unrounded model yields [, 0] by [0, ] b. Let y., and solve for x.. 0.09x +.08..08 0.09x 0.09x 0.468 0.468 x pounds 0.09 c. The slope of the linear model is 0.09. Therefore, for when the weight changes by one pound, the postal rate changes by approximately.9 cents. d. Considering the data, the change in postal rate between 9 pounds and 4 pounds is 4 cents per pound. [0, ] by [0, 00] c. Let y 6, and solve for x. 6 9.09x + 70.4 6 70.4 9.086x 9.09x 4.67 4.67 x.07 9.09 The model predicts that in 99 the average daily prison population is 6. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.6 6. a. Using a spreadsheet program yields Marriage Rate (per 000 unmarried women) 90 80 70 60 0 40 0 0 0 0 Marriage Rate y -0.76x + 8.84 0 0 0 0 40 4 0 Years since 90 b. Let x 98 90, and solve for y. y 0.76() + 8.84 y 8.607 8.6 The model predicts 8.6 marriages per 000 unmarried women. Using the unrounded model yields the marriage rate is 0 between 99 and 996. 4. a. Using a spreadsheet program yields Patients Treated 000 4000 000 000 000 Volunteers in Medicine y 4.700x + 468.00 0 0 4 6 8 0 Years past 990 b. Based on the slope of model, each year the number of patients treated increases by approximately 4.. a. Using a spreadsheet program yields 6000 Education 00 [ 0, 60] by [0, 00] c. Let y 0, and solve for x. 0 0.76 x + 8.84 0 8.84 0.76x 0.76x.84.84 x 46.04 46. 0.76 0 marriages per 000 unmarried women occurs in approximately 996. d. The answer in part c) is an approximation based on the model. Considering the data, Disabled Children Served (in thousands) 000 400 4000 00 y 4.000x + 0.667 000 0 4 8 6 0 Years past 980 b. Based on the model, when the time increases by one year, the number of disabled children served increases by 4,000. c. In 00, x. Using the unrounded model yields Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

64 CHAPTER Functions, Graphs, and Models; Linear Functions 7. a. Using a spreadsheet program yields Gross Domestic Product [, 0] by [4000, 6000] Approximately 6,86,000 children will be served in 00 based on the model. This is an extrapolation, since 00 is beyond the scope of the available data. 6. a. Using a spreadsheet program yields y, Percentage with Internet Access U.S. Households with Internet Access 60 0 40 0 0 0 0 y 8.400x -. 0 0 x, Years since 99 b. See 7 a) above. The equation is y 8.400x.. c. See 7 a) above. The model fits the data set reasonably well. Billions of Dollars y 49.9x - 68.0 000 0000 8000 6000 4000 000 0-000 0 0 40 60 80 Years beyond 940 b. See 7 a) above. The equation is y 49.9x 68.00. c. Considering the scatter plot, the model does not fit the data very well. 8. a. Using a spreadsheet program yields Donations in millions 0 40 0 0 0 0 Giving at IBM y.990x + 6.0 0 4 6 7 8 9 Years past 990 b. Let y 0, and solve for x. 0.990x + 6.0 0 6.0.990x.990x.680.680 x.89949749.9.990 Donations reach $0 million in 00. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.6 6 9. a. Using a spreadsheet program yields Salary (dollars) 0000 000 0000 000 0000 000 Salaries y 80.89x + 096. 0 0 6 9 8 Number of Years since 970 b. See 9 a) above. The equation is y 80.89x+ 096.. c. See 9 a) above. The linear function fits the data points very well. The line is very close to the data points on the scatter plot. d. i. Discrete ii. Discrete iii. Continuous. The model is not limited to data from the table. 40. a. Using a spreadsheet program yields c. On a per person basis, cigarette production increased until 990. Then it began to decrease. d. No. If the decreasing trend that began in 990 continues, the linear model calculated in part a) will not be valid between 99 and 00. 4. a. Using a spreadsheet program yields Population, Millions 600 00 400 00 00 00 0 U.S. Population Projections y.90x + 6.864 0 0 40 60 80 00 0 Years past 000 b. Using the unrounded model, f 6 4.64. World Cigarette Production World Production (cigarettes per person) 00 000 900 800 700 600 00 y 9.4x + 649.8 400 0 0 0 0 40 4 Number of Years since 90 The equation is y 9.4x+ 649.8. [0, 00] by [0, 800] In 06, the projected U.S. population is 4.64 million or 4,640,000. c. In 080, x 80. b. Continuous. The model is not restricted to values of x from the table of given values. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

66 CHAPTER Functions, Graphs, and Models; Linear Functions When median household income for whites is $0,000, median household income for blacks is $4,86.4. Using the rounded model yields: [0, 00] by [0, 800] Based on the model, the U.S. population in 080 will be 499.4 million. The prediction in the table is slightly smaller, 497.8 million. 4. a. Using a spreadsheet program yields Household Incomes for Blacks 8000 000 000 9000 6000 000 0000 Median Household Income y.7x - 78.89 000 9000 4000 47000 000 y.7 0,000,78.89 y 4,84. 4. a. Using a spreadsheet program yields Brokerage Acccounts (millions) 0 0 0 Internet Brokerage Accounts y.0x - 8.874 0 0 Number of Years past 990 b. Considering the slope of the linear model, a one year increase in time corresponds to a. million unit increase in Internet brokerage accounts. Household Incomes for Whites b. See 4 a) above. The model is y.7x,78.89 c. Let x 0,000. Using the unrounded model yields y 4,86.4. [,000, 0,000] by [0,000] by [40,000] 9. c. Let y 0, and solve for x. 0.0x 8.874 0 + 8.874.0x 8.874.0x 8.874 x.769.8.0 Based on the model, the number of Internet brokerage accounts will be 0 million near the end of 00. d. The model may no longer be valid due to economic and social instability resulting from the terrorist attacks on September, 00. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.6 67 44. a. Using a spreadsheet program yields Dollars per Person 0 00 0 00 0 0 Consumer Spending y.447x + 7.004 0 0 Years past 990 b. Let x, and solve for y. y 0.46( ) +.487 y 0.40 Based on the model the anticipated prison time in 97 is approximately 0.4 days. b. Let y 0, and solve for x. 0.447x + 7.004.447x 76.004 76.004 x.46..447 Per capita spending for television services first exceeds $0 during 00. c. Since the calculation in part b) yields an answer beyond the scope of the original data, an assumption must be made that the model remains valid for years beyond 00. The calculation in part b) is an extrapolation. 4. a. Using a spreadsheet program yields Prison Sentences Anticipated prison time (days) 0 0 0 y 0.46x +.487 0 0 0 0 40 Years past 960 Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

68 CHAPTER Functions, Graphs, and Models; Linear Functions Section.7 Skills Check. Applying the substitution method y x and y x x x x x x x x x x Substituting to find y y (), is the intersection point.. Applying the elimination method x+ y ( Eq) x y ( Eq) 9x+ 6y ( Eq) 0x 6y 4 ( Eq) 9x 7 7 x 9 Substituting to find y + y 9+ y y 4 y, is the intersection point. The solution is ( 4, 4). 4. Solving the equations for y x 4y 6 4y x+ 6 x + 6 y 4 y x and x+ y 0 y x+ 0 x + 0 y y x+ 4 Applying the intersection of graphs method [ 0, 0] by [ 0, 0] The solution is (, ).. Applying the intersection of graphs method [ 0, 0] by [ 00, 0] Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.7 69. Solving the equations for y 4x y 4 y 4x 4 4x 4 y 4 4 y x+ and x y 4 y x 4 x 4 y 4 y x+ x 6y 6y x+ x + y 6 y x 6 and 4x 4y 6 4y 4x+ 6 4x + 6 y 4 y x 4 Applying the intersection of graphs method Applying the intersection of graphs method [ 0, 0] by [ 0, 0] The solution is ( 0.74,0.74) or 4 4, 7 7. 6. Solving the equations for y 7. [ 0, 0] by [ 0, 0] The solution is (, ). x+ y 6 Eq x +.y Eq x+ y 6 Eq x y 6 Eq 0 0 There are infinitely many solutions to the system. The graphs of both equations represent the same line. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

70 CHAPTER Functions, Graphs, and Models; Linear Functions 8. 9. 6x+ 4y Eq x+ y Eq 6x+ 4y Eq 6x 4y 6 Eq 0 There is no solution to the system. The graphs of the equations represent parallel lines. x y x+ 4y Solving the first equation for x x y x y+ Substituting into the second equation ( y+ ) + 4y y+ 6 + 4y 9y + 6 9y 8 y Substituting to find x x ( ) x + 0 x The solution is (, ). 0. x y x y 8 Solving the second equation for y x y 8 y x+ 8 y x 8 Substituting into the first equation ( x ) x 8 x x+ 4 x + 4 x x 4 Substituting to find y 4 y 8 y y 6 y The solution is 4,. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.7 7. x y x+ 4y Solving the first equation for x x y x y+ y + x Substituting into the second equation y + + 4y y + + 4y [] y+ + 8y y + y y Substituting to find x x ( ) x + x x The solution is,.. 4x y 7 x+ y 7 Solving the first equation for x 4x y 7 4x y 7 y 7 x 4 Substituting into the second equation y 7 + y 7 4 y 7 4 + y 4[ 7] 4 y + 8y 8 y 8 y y Substituting to find x 4x () 7 4x 7 4x x The solution is,.. x+ y Eq x+ 4y 8 Eq () x 6y 0 Eq x+ 4y 8 Eq y y Substituting to find x x + x + x The solution is,. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

7 CHAPTER Functions, Graphs, and Models; Linear Functions 4. 4x y ( Eq) x+ 6y ( Eq) 8x 6y 6 ( Eq) x+ 6y ( Eq). x x Substituting to find x 4( ) y 4 y y 9 y The solution is,. ( ) x+ y 8 Eq x+ 4y 8 Eq 0x 6y 6 Eq 0x + 0y 40 Eq 4y 4 4 y 4 7 Substituting to find x x + 4 8 7 48 7 x + 7[ 8] 7 4x + 48 6 4x 8 8 4 x 4 7 4 The solution is,. 7 7 6. x+ y ( Eq) x+ 4y 8 ( Eq) 7. x y 0 4 Eq 6x + y 4 Eq 6x 4 4 y 6 Substituting to find x + y + y y 7 7 y 7 The solution is,. 0.x+ 0.4y.4 Eq x y Eq 9x + y 7 0 Eq 0x y 44 4 Eq 9x 6 6 x 4 9 Substituting to find x 4 y 0 y y 9 y The solution is 4,. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.7 7 8. 8x 4y 0 ( Eq) 0.x+ 0.y. ( Eq) 4x y 0 ( Eq) 0x y 88 40 ( Eq) 9. 0.. 44x 88 x Substituting to find x 8 4y 0 6 4y 0 4y 6 y 4 The solution is,4. x+ 6y Eq x+ 4y 8 Eq 6x y 4 Eq 6x+ y 4 Eq 0 0 Infinitely many solutions. The lines are the same. This is a dependent system. 4x+ 6y Eq 0x y 0 Eq 0x+ 0y 60 Eq 0x 0y 60 Eq 0 0 Infinitely many solutions. The lines are the same. This is a dependent system. 6x 9y Eq x 4.y 6 Eq 6x 9y Eq 6x + 9y Eq 0 4 No solution. Lines are parallel... 4. 4x 8y Eq 6x y 0 Eq x 4y Eq x + 4y 0 Eq 0 No solution. Lines are parallel. y x y x 6 Substituting the first equation into the second equation x x 6 x 6 x 4 x Substituting to find y y 6 4 The solution is,4. y 8x 6 y 4x Substituting the first equation into the second equation 8x 6 4x 6x 6 x Substituting to find y () y 4 y The solution is,. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

74 CHAPTER Functions, Graphs, and Models; Linear Functions. 6. 4x+ 6y 4 x 4y+ 8 Substituting the second equationinto the first equation 44 ( y+ 8) + 6y 4 6y+ + 6y 4 y + 4 y 8 8 4 y Substituting to find x 4 x 4 + 8 6 88 x + 4 The solution is,. y 4x x 4y 7 Substituting the first equation into the second equation x 4( 4x ) 7 x 6x+ 0 7 x x Substituting to find y y 4 () y The solution is,. ( ) 7. 8. ( ) x y 6 Eq 6x 8y 4 Eq 6x + y 48 Eq 6x 8y 4 Eq 7y 4 y Substituting to find x x ( ) 6 x + 0 6 x 6 x The solution is,. 4x y 4 Eq 6x+ y Eq x y Eq 6x+ y Eq 8x 7 7 x 8 Substituting to find y 4 y 4 6 y 4 y y The solution is,. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.7 7 9. x 7y Eq 4x+ y Eq x + 8y 4 4 Eq x + 9y Eq 7y 7 7 y 7 Substituting to find x x 7() x 7 x 6 x The solution is,... 4x y 9 Eq 8x 6y 6 Eq 8x + 6y 8 Eq 8x 6y 6 Eq 0 No solution. Lines are parallel. x 4y 8 Eq x + y Eq x y 4 Eq x + y Eq 0 No solution. Lines are parallel. 0. x y Eq x y 8 Eq x + 9y 6 Eq x y 40 Eq 6y 4 4 y 6 4 Substituting to find x x 4 x + 4 4 x + 4 4 0x + 48 0x 4 4 9 x 0 4 9 The solution is,. 4 4 Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

76 CHAPTER Functions, Graphs, and Models; Linear Functions Section.7 Exercises. R C 76.0x 970 + 7x 49.0x 970 970 x 49.0 x 60 Applying the intersections of graphs method yields x 60.. a. Let p 60 and solve for q. Supply function 60 q + 0 q 40 q 8 Demand function 60 8 4q 4q 68 68 q 7 4 When the price is $60, the quantity supplied is 8, while the quantity demanded is 7. 4. [ 0, 00] by [ 0, 0,000] Break-even occurs when the number of units produced and sold is 60. R C 89.7x.0x+ 9.0 66.x 9.0 9.0 x 66. x 8 b. Equilibrium occurs when the demand equals the supply, q+ 0 8 4q 9q + 0 8 9q 08 08 q 9 Substituting to calculate p p + 0 80 When the price is $80, units are produced and sold. This level of production and price represents equilibrium. Applying the intersections of graphs methods yields x 8. [ 0, 00] by [ 000, 000] Break-even occurs when the number of units produced and sold is 8. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.7 77 6. p+ q 0 ( Eq) p 8q 0 ( Eq) 4 p + 8q 80 4 ( Eq) p 8q 0 ( Eq) p 00 00 p 60 Substituting to find q 60 8q 0 8q 40 40 q 0 8 The solution is 60,0. Equilibrium occurs when 0 units are demanded and supplied at a price of $60 per unit. 7. a. Applying the intersection of graphs method [0, 0] by [ 00, 000] The solution is ( 7.,.787 ). The number of active duty Navy personnel equals the number of active duty Air Force personnel in 987. b. Considering the solution in part a, approximately,787 people will be on active duty in each service branch. 8. Applying the intersection of graphs method 9. a. [0, 0] by [ 0, 00] The solution is ( 8.7,.6 ). The percentage of male students enrolled in college within months of high school graduation equals the percentage of female students enrolled in college within months of high school graduation in 979. y 4.x+ 9. y 0.x + 007 Substituting the first equation into the second equation 4.x+ 9. 0.x+ 007 0 4. 9. 0 0. 007 4x+ 9 x+ 0,070 47x + 9 0,070 47x 9 9 x 6.988 7 47 ( x+ ) ( x+ ) b. In 990 + 7 07, mint sales and gum sales are equal. c. The graphs are misleading. Notice that the scales are different. Mint sales are measured between $0 and $00 million, while gum sales are measured between $0 and $000 million. Also note that the first tick mark on the y-axis for each graph represents inconsistent units when compared with the remainder of the graph. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

78 CHAPTER Functions, Graphs, and Models; Linear Functions 40. Applying the intersection of graphs method 4. [90, 000] by [ 00, 000] The solution is ( 960.9,96.07 ). The number of nursing homes in Massachusetts equals the number of nursing homes in Illinois at the end of 960 or approximately 96. The number of nursing homes in each state is approximately 96. Let l the low stock price, and let h the high stock price. h+ l 8. Eq h l.88 Eq h 0.8 0.8 h.69 Substituting to calculate l.69 + l 8. l 0.8 The high stock price is $.69, while the low stock price is $0.8. 4. Let l the 998 revenue, and let h the 999 revenue. h+ l 44.9 ( Eq) h l. ( Eq) h+ l 44.9 ( Eq) h l 7 ( Eq) h 4.9 4.9 h 80. Substituting to calculate l 80. l. l 669.8 l 669.8 The 998 revenue is $669.8 million, while the 999 revenue is $80. million. 4. a. x+ y 400 b. 0x c. 4y d. 0x+ 4y 84,000 e. x+ y 400 Eq 0x+ 4y 84,000 Eq 0x 0y 7,000 0 Eq 0x+ 4y 84,000 Eq y,000,000 y 800 Substituting to calculate x x + 800 400 x 600 The promoter needs to sell 600 tickets at $0 per ticket and 800 tickets at $4 per ticket. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.7 79 44. a. x+ y 0,000 b. 0% x or 0.0x c. % y or 0.y d. 0.0x+ 0.y 6,00 e. x+ y 0, 000 Eq 0.0x+ 0.y 6,00 Eq 0.0x 0.0y, 000 0.0 Eq 0.0x+ 0.y 6,00 Eq 0.0y 00 00 y 7, 000 0.0 Substituting to calculate x x + 7, 000 0, 000 x 7, 000 $7,000 is invested in the 0% property, and $7,000 is invested in the % property. 4. a. Let x the amount in the safer account, and let y the amount in the riskier account. x+ y 00, 000 Eq 0.08x+ 0.y 9000 Eq 0.08x 0.08y 8000 0.08 Eq 0.08x+ 0.y 9000 Eq 0.04y 000 000 y,000 0.04 Substituting to calculate x x +,000 00,000 x 7, 000 $7,000 is invested in the 8% account, and $,000 is invested in the % account. b. Using two accounts minimizes investment risk. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

80 CHAPTER Functions, Graphs, and Models; Linear Functions 46. Let x the amount in the safer fund, and let y the amount in the riskier fund. x+ y,000 Eq 0.0x+ 0.4y 70 Eq 0.0x 0.0y 00 0.0 Eq 0.0x+ 0.4y 70 Eq 0.04y 0 0 y,000 0.04 Substituting to calculate x x +,000,000 x 9,000 $9,000 is invested in the 0% fund, and $,000 is invested in the 4% fund. 47. Let x the number of glasses of milk, and let y the number of quarter pound servings of meat. Protein equation: 8.x+ y 69. Iron equation: 0.x+.4y 7. + ( q) 8.x+ y 69. Eq 0.x+.4y 7. Eq 8.x y 69. E 8.x 89y 60. 8 Eq 67 y 4 4 y 67 Substituting to calculate x 8.x + 69. 8.x + 44 69. 8.x.. x 8. The person on the special diet needs to consume glasses of milk and quarter pound portions of meat to reach the required iron and protein content in the diet. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.7 8 48. Let x the amount of substance A, and let y the amount of substance B. Nutrient equation: 6% x+ 0% y 00% x+ y 4 Eq 0.06x+ 0.0y Eq 0.06x 0.06y 0.84 0.06 Eq 0.06x+ 0.0y Eq 0.04y 0.6 0.6 y 4 0.04 Substituting to calculate x x + 4 4 x 0 The patient needs to consume 0 ounces of substance A and 4 ounces of substance B to reach the required nutrient level of 00%. 49. Let x the amount of 0% solution, and let y the amount of % solution. x+ y 0 Medicine concentration: 0% x+ % y 8% 0 x+ y 0 Eq 0.0x+ 0.0y.6 Eq 0.0x 0.0y 0.0 Eq 0.0x+ 0.0y.6 Eq 0.0y 0.4 0.4 y 8 0.0 Substituting to calculate x x + 8 0 x The nurse needs to mix cc of the 0% solution with 8 cc of the % solution to obtain 0 cc of an 8% solution. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

8 CHAPTER Functions, Graphs, and Models; Linear Functions 0. Let x the amount of the 0% solution, and let y the amount of the % solution. x+ y 4 Medicine concentration: 0% x+ % y 0% 4 x+ y 4 Eq 0.0x+ 0.y 9 Eq 0.0x 0.0y. 0.0 Eq 0.0x+ 0.y 9 Eq 0.y 4. 4. y 0 0. Substituting to calculate x x + 0 4 x The nurse needs to mix cc of the 0% solution with 0 cc of the % solution to obtain 4 cc of a 0% solution.. c. Applying the intersection of graphs method a. Demand function: Finding a linear model using L as input and L as output yields p q+. [0, 00] by [ 0, 00] b. Supply function: Finding a linear model using L as input and L as output yields p q+ 0. 4 When the price is $8, 40 units are both supplied and demanded. Therefore, equilibrium occurs when the price is $8 per unit. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.7 8.. Applying the intersection of graphs method Demand function: Finding a linear model using L as input and L as output yields p q+ 600. [0, 0] by [, 7] Note that the lines do not intersect. The slopes are the same, but the y-intercepts are different. There is no solution to the system. Based on the two models, the percentages are never equal. Supply function: Finding a linear model using L as input and L as output yields p q+ 0 or p q. 4. Let x the amount of federal tax, and let y the amount of Alabama tax. a. The federal taxable income is,000,000 y. (,000,000 y)( 4% ) x x 40,000 0.4y Applying the intersection of graphs method b. c. Alabama taxable income is, 000, 000 x. (,000,000 x)( % ) y y 0, 000 0.0x x 40,000 0.4y y 0,000 0.0x [0, 800] by [ 00, 800] When the price is $00, 600 units are both supplied and demanded. Therefore equilibrium occurs when the price is $00 per unit Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

84 CHAPTER Functions, Graphs, and Models; Linear Functions d. x 40, 000 0.4y y 0,000 0.0x Substituting the second equation into the first equation yields: ( x) x 40, 000 0.4 0, 000 0.0 x 40, 000 7, 000 + 0.07x x, 000 + 0.07x 0.98x, 000, 000 x 8,8.96 0.98 Substituting to find y y 0, 000 0.0( 8,8.96) y,70.70 Federal income tax is $8,8.96, while Alabama income tax is $,70.70.. a. 00x+ 00y 00,000 b. x y 00( y) + 00y 00,000 800y 00,000 00,000 y 800 Substituting to calculate x x 0 There are 0 clients in the first group and clients in the second group. 6. 0% x+ % y.% ( x+ y) 0.0x + 0.0y 0.x+ 0.y 0.04x 0.0y 0.04x 0.0y 0.0 0.0 y x 7 Therefore, the amount of y must equal of the amount of x. 7 If x 7, y ( 7). 7 The % concentration must be increased by cc. 7. The slope of the demand function is y y 0 60 0 m. x x 900 400 00 0 Calculating the equation: y y m( x x) y 0 ( x 900) 0 y 0 x+ 90 0 y x+ 00 or 0 p q+ 00 0 Likewise, the slope of the supply function is y y 0 0 0 m. x x 700 400 700 70 Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.7 8 Calculating the equation y y m x x y 0 700 70 y 0 x 0 70 y x+ 0 or 70 p q+ 0 70 ( x ) The quantity, q, that produces market equilibrium is 700. q+ 00 q+ 0 0 70 70 q+ 00 70 q+ 0 0 70 7q+ 7000 q+ 700 9q 600 600 q 700 9 The price, p, at market equilibrium is $0. p ( 700 ) + 0 70 p 0 + 0 p 0 700 units priced at $0 represents the market equilibrium. 8. The slope of the demand function is y y 0 00 0 m. x x 800 00 400 8 Calculating the equation y y m x x y 0 800 8 y 0 x+ 00 8 y x+ 40 or 8 p q+ 40 8 ( x ) Likewise, the slope of the supply function is y y 80 8 0 m. x x 700 400 700 0 Calculating the equation y y m( x x) y 80 ( x 700) 0 y 80 x 0 0 y x+ 7 or 0 p q+ 7 0 The quantity, q, that produces market equilibrium is 000. q+ 40 q+ 7 8 0 LCM : 40 40 q+ 40 40 q+ 7 8 0 q+ 8,000 6q+ 7000 q,000,000 q 000 The price, p, at market equilibrium is $. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

86 CHAPTER Functions, Graphs, and Models; Linear Functions p ( 000 ) + 7 0 p 0 + 7 p 000 units priced at $ each represents market equilibrium. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.8 87 Section.8 Skills Check. Algebraically: x 7 x 4x 7 4x x x 4 Graphically: y x y x 7 Graphically: y 4x+ y x+ 6 [, ] by [, ] x+ 6 < 4x+ implies that the solution region is x >. The interval notation is,. The graph of the solution is [ 0, 0] by [ 0, 0] x 7 x implies that the solution region is x. The interval notation is (,]. The graph of the solution is 4 0 4 / / / 0 / / /. Algebraically: ( x ) 4 x 9 x 8 x 9 7x 7x 7 7 x 7. Algebraically: x+ 6< 4x+ x + 6< x < x > (Note the inequality sign switch) x > Graphically: y 4(x ) [, ] by [, ] y 4x+ 6 Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

88 CHAPTER Functions, Graphs, and Models; Linear Functions ( x ) 4 x 9 implies that the solution region is x. 7 The interval notation is,. 7 The graph of the solution is / 0 / 7 4. Algebraically:. Algebraically: 4x+ < x+ 4 ( x+ ) < x+ 0x+ < x+ x + < x < 0 0 x < ( x ) > 4x+ 6 0x > 4x+ 6 6x > 6 6x > x > x > 6 7 Graphically: y x+ y 4x+ Graphically: y 4x+ 6 y (x ) [ 0, 0] by [ 0, 0] 4x+ < x+ implies that the solution 0 region is x <. 0 The interval notation is,. [ 0, 0] by [, ] ( x ) >4x+ 6 implies that the solution 7 region is x >. 7 The interval notation is,. The graph of the solution is 0 0 The graph of the solution is 4 0 4 7 Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.8 89 6. Algebraically: x 4x + x 6 4x 6 + 4x + x x x 9 9 x 7. Algebraically: x 8 < x 8 0 < 0 ( x ) < ( 8) x < 6 x < 6 6 x < Graphically: y 4x Graphically: y 8 x y + y x [ 0, 0] by [ 0, 0] x 4x + implies that the solution 9 region is x. 9 The interval notation is,. The graph of the solution is 4 0 4 9 [ 0, 0] by [ 0, 0] x 8 < implies that the solution 6 region is x <. 6 The interval notation is,. The graph of the solution is 6 0 4 6 8 0 4 Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

90 CHAPTER Functions, Graphs, and Models; Linear Functions 8. Algebraically: y 6 < 4 y 6 < 4 < 64 ( y ) y 9< 64 y < 7 y < Graphically: 7 y [ 0, 0] by [ 0, 0] y 6 < implies that the solution 4 7 region is x <. 7 The interval notation is,. The graph of the solution is y 6 y 4 4 0 4 8 6 0 4 8 7 9. Algebraically: ( x ) 6 x ( x 6) x 0 0 6 0 ( ( x )) ( x) ( x ) x 6 4 0 x 90 4x 0 x 90 0 x 0 0 x Graphically: [ 0, 0] by [, ] ( x ) 6 x implies that the solution 0 region is x. 0 The interval notation is,. The graph of the solution is 0 ( x ) 6 y x y 4 0 4 Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.8 9 0. Algebraically: ( y ) 4 y 8 ( y 4) y 8 0y 40 9y 0 y 80 Graphically: y y 8 4 y ( y ) Graphically: y 6 0.8x [ 0, 0] by [ 0, 0] y.x.6.x.6 6 0.8 x implies that the solution region is x.86. The interval notation is.86,. The graph of the solution is.86 ) - -4 - - - 0 4 [ 90, 70] by [ 60, 0] ( y ) 4 y 8 implies that the solution region is y 80. The interval notation is [ 80, ). The graph of the solution is. Algebraically:.x 6. 8 0.x 4x 4. 4. x 4 x. Graphically: 00 80 40 0 40 80. Algebraically:.x.6 6 0.8x.0x.6 6.0x 8.6 8.6 x.0 x.86 00 y 8 0.x [ 0, 0] by [ 0, 0] y.x 6..x 6. 8 0. x implies that the solution region is x.. The interval notation is (-,.]. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

9 CHAPTER Functions, Graphs, and Models; Linear Functions The graph of the solution is. Let f( x) 4x+, and determine graphically where f( x) 0. - -4 - - - 0 4. Applying the intersection of graphs method yields: y 7x+ f( x) 4x + [ 0, 0] by [ 0, 0] y x 7 [ 0, 0] by [ 0, 0] f( x) 0 implies that the solution region is x 8. The interval notation is [ 8, ). 7x+ < x 7 implies that the solution region is x <. The interval notation is (, ). 4. Applying the intersection of graphs method yields: y x+ 4 y 6x 6. To apply the x-intercept method, first rewrite the inequality so that zero is on one side of the inequality. ( x 4) ( x ) x+ 6x 9x + 4 0 Let f( x) 9x+ 4, and determine graphically where f( x) 0. f( x) 9x+ 4 [ 0, 0] by [ 0, 0] x+ 4 6x implies that the solution region is x <. The interval notation is [, ).. To apply the x-intercept method, first rewrite the inequality so that zero is on one side of the inequality. ( x+ 4) 6( x ) 0x+ 0 6x 4x + 0 [ 0, 0] by [ 0, 0] f( x) 0 implies that the solution region is x.. The interval notation is (,.. 7. a. The x-coordinate of the intersection point is the solution.. x b. (, ) Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.8 9 8. a. x 0 b. (,0] c. No solution. f ( x ) is never less than h( x ). 9. 7 x < 7 + x + < + x < 6 x 6 < x < The interval notation is,. 0. 0 < 0x 40 60 0 + 40 < 0x 40 + 40 60 + 40 60 < 0x 00 8< x 0 The interval notation is 8,0. ( ]. x+ 6 and x+ 6 x + x 0 x 0 x and x 0 The interval notation is,0.. 6x 8 > and 6x 8 < < 6x 8 < 0 < 6x < 40 0 6x 40 < < 6 6 6 < x < 4 x> and x< 4 The interval notation is,. 4. x+ < 7 and x > 6 Inequality x + < 7 x < 8 8 x < Inequality x > 6 x > x > 8 x< and x> 4. 6x or x+ 4> 9 Inequality 6x 6x x Inequality x + 4> 9 x > x > x or x> Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

94 CHAPTER Functions, Graphs, and Models; Linear Functions. x 6 x or x x 4 Inequality 4 x 4( 6 x) 4 x 8 4 8x x x Inequality x x 6x 6 4x x 4 x or x 4 ( x ) 6. x < x or x > 6x Inequality x < ( x) x 6< 0x 9x < 6 x > Inequality x > ( 6x) x > 0x 8x > x < 8 x> or x< 8 7. 7.00 0.4x.886 77.998 7.00 +.886 0.4x.886 +.886 77.998 +.886 9.888 0.4x 80.884 9.888 0.4x 80.884 0.4 0.4 0.4 7 x 46 [ ] The interval notation is 7,46. 8. 60 + 88 + 7 + 6 + x 70 < 80 86 + x 70 < 80 86 + x 70 80 0 86 + x 400 0 86 86 86 + x 400 86 64 x < 4 The interval notation is 64,4. [ ) Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.8 9 Section.8 Exercises 9. a. p 0. b. Considering x as a discrete variable representing the number of drinks, then if x 6, the 0-lb male is intoxicated. 0. a. V < 8000 b. t. F 9 C + 9 C 0 C 0. A Celsius temperature at or below zero degrees is freezing. C 00 ( F ) 00 9 9 9 00 9 F 60 900 F 060 F ( F ) [ ] A Fahrenheit temperature at or above degrees is boiling.. Position income 00 Position income 000 + 0.0 x,where x represents the sales within a given month When does the income from the second position exceed the income from the first position? Consider the inequality 000 + 0.0x > 00 0.0x > 00 00 x > 0.0 x >,000 When monthly sales exceed $,000, the second position is more profitable than the first position. 4. Original value ( 000)( ) $,000 Adjusted value,000 (,000)( 0% ),000 4400 7,600 Let x percentage increase 7,600 + 7,600x >,000 7,600x > 4400 4400 x > 7,600 x > 0. x > % The percentage increase must be greater than % in order to ensure a profit. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

96 CHAPTER Functions, Graphs, and Models; Linear Functions. 6. Let x Jill's final exam grade. 78 + 69 + 9 + 8+ x 80 89 6 680 78 + 69 + 9 + 8+ x 6 6 689 480 0 + x 4 480 0 0 0 + x 4 0 60 x 4 60 x 4 80 x 07 If the final exam does not contain any bonus points, Jill needs to score between 80 and 00 to earn a grade of B for the course. Let x John's final exam grade. 78 + 6 + 8 + x 70 79 70 78 + 6 + 8 + x 79 0 + x 9 8 x 7 8 x 7 64 x 86. 7. Let x 6, and solve for p. 0 p 9(6) 0 p 4 0 p p.8 0 Let x 0, and solve for p. 0 p 0(9) 0 p 90 0 p 9 9 p 6.6 0 Therefore, between 996 and 000, the percentage of marijuana use is between.8% and 6.7%. In symbols,.8 p 6.7. John needs to score between 64 and 86. to earn a grade of C for the course. 8. If the years are between 90 and 99, then 0 x 4. Therefore, cigarette production is given by: y 9.4 0 + 649.8 9.4 4 + 649.8 649.8 y 9.494 + 649.8 649.8 y 04.87 or rounding to zero decimal places 649 y 04 Between 90 and 99, cigarette production is between 649 and 04 cigarettes per person per year inclusive. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.8 97 9. y 000 0.97x + 8.89 000 0.97x 87.67 87.67 x 0.97 x 898.7 or approximately x 899 Old scores greater than or equal to 899 are equivalent to new scores. 4. Let x represent the actual life of the HID headlights. x 00 0% 00 00 + 0% 00 00 0 x 00 + 0 0 x 60 40. y < 000 9.4x + 649.8 < 000 9.4x < 0.66 0.66 x < 9.4 x < 7.68 Prior to 997 cigarette production is less than 000 cigarettes per person per year. 4. Remember to convert years into months. < y < 4 6 48 < 0.4x.886 < 7 48 +.886 < 0.4x.886 +.886 < 7 +.886 0.886 < 0.4x < 74.886 0.886 0.4x 74.886 < < 0.4 0.4 0.4 9.8986 < x <.78 or approximately, 9 < x < The prison sentence needs to be between 9 months and months. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

98 CHAPTER Functions, Graphs, and Models; Linear Functions 4. a. Let y > 0. 0.76x + 8.84 > 0 Applying the intersection of graphs method: y 0.76x+ 8.84 y 0 44. Note that W and M are in thousands. M 00 0.99W.6 00 0.99W 0.6 0.6 W 0.99 W 0.708 The median salary for whites needs to be approximately $0, or greater. [, 7] by [, 00] When x< 46.4, y> 0. The marriage rate per 000 women is greater than 0 prior to 996. b. Let y < 4. 0.76x + 8.84 > 4 Applying the intersection of graphs method: y 0.76x+ 8.84 y 4 [, 7] by [, 00] When x>.80, y< 4. The marriage rate per 000 women will be less than 4 beyond 00. 4. a. Since the rate of increase is constant, the equation modeling the value of the home is linear. y y m x x 70,000 90,000 4 0 80,000 4 0,000 Solving for the equation: y y m( x x) y 90, 000 0, 000 x 0 y 90, 000 0, 000x y 0,000x+ 90,000 b. y > 400,000 0,000x + 90,000 > 400,000 0,000x > 0,000 0,000 x > 0,000 x > 0. 00 corresponds to x 00 996 4. Therefore, x < 4. Or, y > 400,000 between 007 and 00. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.8 99 The value of the home will be greater than $400,000 between 007 and 00. 46. Cost of cars,00 90,000 Cost of cars,00 7,00 th Let x profit on the sale of the car. (.% )( 7,00) + x ( 6% )( 90,000) 9,66.0 + x,400 x 77.0 th The price of the car needs to be at least,00 + 77.0 6,7.0 or $6,8. 47. Px > 0,900 6.4x 000 > 0,900 6.4x >,900,900 x > 6.4 x > 000 A production level above 000 units will yield a profit greater than $0,900. 49. Px 0 6.4x 967 0 6.4x 967 967 x 6.4 x 00 Sales of 00 feet or more of PVC pipe will avoid a loss for the hardware store. 0. Generating a loss implies that Px< 0. Px < 0 40, + 9.80x < 0 9.80x < 40, 40, x < 9.80 x < 407.606 Rounding since the data is discrete: x < 408 Producing and selling fewer than 408 units results in a loss. 48. Px > 84, 40, + 9.80x > 84, 9.80x > 4,60 4,60 x > 9.80 x >,7.06 Rounding since the data is discrete: x >,7 The number of units sold needs to exceed,7.. Recall that Profit Revenue Cost. Let x the number of boards manufactured and sold. Px Rx Cx Rx 489x Cx x+ 4, 000 Px 489x x+ 4,000 Px 489x x 4,000 Px 64x 4, 000 To make a profit, Px > 0. 64x 4, 000 > 0 Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

00 CHAPTER Functions, Graphs, and Models; Linear Functions. T 8 0.4m + 76.8 8 0.4m 8. 8. m 0.4 m 9.06976744 or approximately, m 9 The temperature will be at most 8 F for the first 9 minutes.. 4 < y < 48 4 < 0.x + 44.7 < 48 4 44.7 < 0.x + 44.7 44.7 < 48 44.7 0.6 < 0.x <.6 0.6 0..6 < x < 0. 0. 0. 4.06 < x <.4 Considering x as a discrete variable yields 4 < x<. From 974 until 99 the reading scores were between 4 and 48. 4. a. b. 6.404 0.x < 0 0.x <.404.404 x > 0. x > 99.6798649 00 ( Note the inequality sign switch. ) The voting percentage is less than 0 after the year 00. 6.404 0.x > 7 0.x > 9.98 9.98 x < 0. x < 7.007 ( Note the inequality sign switch. ) x 7 The voting percentage is greater than 7 before the year 9. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

CHAPTER Section.8 0 c. 0 6.404 0.x 60 0 6.404 6.404 6.404 0.x 60 6.404.404 0.x.404.404 0.x.404 0. 0. 0. 4.676808 x.9696 4 x ( Note the inequality sign switch. ) x 4 The voting percentage is between 0and 60 between the years 96 and 99 inclusive.. a. t 998 97 p() 7.409 0.706948() 7.409 6.9804 9.9096 9.% In 998 the percent of high school seniors who have tried cigarettes is estimated to be 9.%. b. 0 p 00 p 00% p 7.409 0.706948 t p 0% 6. a. 4 t 06 c. Considering part b above, the model is valid between 97 4 94 and 97 + 06 08 inclusive. It is not valid before 94 or after 08. t 00 90 p() 6.404 0. 4.868 4.9% In 00 the percent of the voting population who vote in the presidential election is estimated to be 4.9%. b. 0 p 00 p 00% p 6.404 0.x p 0% Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.

0 CHAPTER Functions, Graphs, and Models; Linear Functions 97 x 84 c. Considering part b above, the model is valid between 90 97 8 and 90 + 84 4 inclusive. It is not valid before 8 or after 4. Copyright 007 Pearson Education, publishing as Pearson Addison-Wesley.