3.1 The Rectangular Coordinate System · 2014. 8. 19. · 3.1 The Rectangular Coordinate System 97 11. (a) If , then both and have the sameBC ! B C sign. ÐBßCÑ B C is in quadrant

3.1 The Rectangular Coordinate System 95

CHAPTER 3 GRAPHS, LINEAREQUATIONS, ANDFUNCTIONS

3.1 The Rectangular Coordinate System3.1 Classroom Examples

1. $B %C œ "#

To complete the ordered pairs, substitute the givenvalue of or in the equation.B CFor Ð!ß ****Ñ B œ !, let .

$B %C œ "#

$Ð!Ñ %C œ "#

%C œ "#

C œ $

The ordered pair is .Ð!ß$Ñ

For Ð****ß !Ñ C œ ! let .

$B %C œ "#

$B %Ð!Ñ œ "#

$B œ "#

B œ %

The ordered pair is .Ð%ß !Ñ

For Ð****ß#Ñ C œ #, let .

$B %C œ "#

$B %Ð#Ñ œ "#

$B ) œ "#

$B œ %

B œ %$

The ordered pair is .ˆ ‰%$ ß #

For Ð'ß ****Ñ B œ ', let .

$B %C œ "#

$Ð'Ñ %C œ "#

") %C œ "#

%C œ $!

C œ œ $! "&% #

The ordered pair is .ˆ ‰'ß"&#

The completed table follows.

B C

! $

% !

#

'

%$

"&#

2. #B C œ %To find the B C œ !-intercept, let .

#B C œ %

#B ! œ %

#B œ %

B œ #

The B Ð#ß !Ñ-intercept is .

To find the -intercept, let .C B œ !

#B C œ %

#Ð!Ñ C œ %

C œ %

C œ %

The C Ð!ß%Ñ-intercept is .

Plot the intercepts, and draw the line throughthem.

3. C œ $

In standard form, the equation is .!B C œ $Every value of leads to , so the B C œ $ C-interceptis . There is no -intercept. The graph is theÐ!ß $Ñ Bhorizontal line through .Ð!ß $Ñ

4. B # œ !

In standard form, the equation is .B !C œ #Every value of leads to , the C B œ # B-interceptis . There is no -intercept. The graph isÐ#ß !Ñ Cthe vertical line through .Ð#ß !Ñ

96 Chapter 3 Graphs, Linear Equations, and Functions

5. $B C œ !To find the B C œ !-intercept, let .

$B ! œ !

$B œ !

B œ !

Since the B Ð!ß !Ñ C-intercept is , the -intercept isalso .Ð!ß !Ñ

Find another point. Let .B œ "

$Ð"Ñ C œ !

$ C œ !

C œ $

This gives the ordered pair . Plot andÐ"ß $Ñ Ð"ß $ÑÐ!ß !Ñ and draw the line through them.

6. By the midpoint formula, the midpoint of thesegment with endpoints and isÐ&ß )Ñ Ð#ß %Ñ

Œ Œ & # ) % $ "#

# # # #ß œ ß œ Ð"Þ&ß 'ÑÞ

7. In the leftmost graph in Exercise 73, the last line,X and Y , indicates that the œ # œ ! B-interceptis . Similarly, from the rightmost graph, theÐ#ß !ÑC Ð!ß $Ñ-intercept is .

3.1 Section Exercises1. (a) represents the year; represents the revenueB C

in billions of dollars.

The dot above the year 2002 appears to be at(b)about , so the revenue in 2002 was $")&! ")&!billion.

The ordered pair is (c) Ð#!!#ß ")&!ÑÞ

In 2000, federal tax revenues were about(d)$ billion.#!$!

2. (a) represents the year; represents the percentB Cof women in mathematics or computer scienceprofessions.

The line goes down from 1990 to 2000.(b)Therefore, from 1990–2000, the percent of womenin mathematics or computer science professionsdecreased.

The dot over the year 1990 appears to be at(c)about , so the ordered pair $' ÐBß CÑ œ Ð"**!ß $'ÑÞ

In 2000, the percent of women in mathematics(d)or computer science professions was %.$!

3. The point with coordinates is called theÐ!ß !Ñ origin of a rectangular coordinate system.

4. For any value of , the point B ÐBß !Ñ lies on the B -axis.

5. The B-intercept is the point where a line crossesthe -axis. To find the -intercept of a line, we letB B C B equal and solve for .!

The C-intercept is the point where a line crossesthe -axis. To find the -intercept of a line, we letC C B C equal and solve for .!

6. The equation horizontal line as its C œ % has a graph. The equation B œ % has a line asverticalits graph.

7. To graph a straight line, we must find a minimumof two points. A third point is sometimes foundto check the accuracy of the first two points.

8. Substitute for in the equation to% C #B $C œ !get . Solving for gives . The#B "# œ ! B B œ 'point Ð ' ß %Ñ #B $C œ ! is on the graph of .

9. (a) The point is located in quadrant I, sinceÐ"ß 'Ñthe - and s are both B C-coordinate positive.

(b) The point is located in quadrant III,Ð%ß#Ñsince the - and s are both B C-coordinate negative.

(c) The point is located in quadrant II,Ð$ß 'Ñsince the is B-coordinate negative and theC-coordinate is positive.

(d) The point is located in quadrant IV,Ð(ß&Ñsince the is B-coordinate positive and theC-coordinate is negative.

(e) The point is located on the -axis, soÐ$ß !Ñ Bit does not belong to any quadrant.

The point is located on the -axis, so(f) Ð!ß!Þ&Ñ Cit does not belong to any quadrant.

10. (a) The point is located in quadrant III,Ð#ß"!Ñsince the - and s are both B C-coordinate negative.

(b) The point is located in quadrant I, sinceÐ%ß )Ñthe - and s are both B C-coordinate positive.

(c) The point is located in quadrant II,Ð*ß "#Ñsince the is B-coordinate negative and theC-coordinate is positive.

(d) The point is located in quadrant IV,Ð$ß*Ñsince the is B-coordinate positive and theC-coordinate is negative.

(e) The point is located on the -axis, so itÐ!ß)Ñ Cdoes not belong to any quadrant.

The point is located on the -axis, so it(f) Ð#Þ$ß !Ñ Bdoes not belong to any quadrant.


11. (a) If , then both and have the sameBC ! B Csign.ÐBß CÑ B C is in quadrant I if and are positive.ÐBß CÑ is in negative.quadrant III if and are B C

(b) If , then and have different signs.BC ! B CÐBß CÑ B ! C ! is in quadrant II if and .ÐBß CÑ B ! C ! is in quadrant IV if and .

If (c)B

C ! B C, then and have different signs.

ÐBß CÑ is in either quadrant II or IV. (See part (b).)

If (d)B

C ! B C, then and have the same sign.

ÐBß CÑ is in either quadrant I or III. (See part (a).)

12. Any point that lies on an axis must have onecoordinate that is .!

For Exercises 13–22, see the rectangular coordinatesystem after Exercise 22.

13. To plot , go two units from zero to the rightÐ#ß $Ñalong the B-axis, and then go three units upparallel to the -axis.C

14. To plot , go unit in the negativeÐ"ß #Ñ "direction, that is, left, on the B #-axis, then unitsup.

15. To plot , go three units from zero to theÐ$ß#Ñleft along the B-axis, and then go two units downparallel to the -axis.C

16. To plot , go unit right on the Ð"ß%Ñ " B-axis, then% units down.

17. To plot , do not move along the Ð!ß &Ñ B-axis at allsince the is . Move five units upB-coordinate !along the -axis.C

18. To plot , go units left on the Ð#ß%Ñ # B-axis,then units down.%

19. To plot , go two units from zero to the leftÐ#ß %Ñalong the B-axis, and then go four units up parallelto the -axis.C

20. To plot , go units right on the Ð$ß !Ñ $ B-axis, thenstop since the is .C-coordinate !

21. To plot , go two units to the left along theÐ#ß !ÑB-axis. Do not move up or down since theC !-coordinate is .

22. To plot , go units right on the Ð$ß$Ñ $ B-axis,then units down.$

23. C œ B %To complete the table, substitute the given valuesfor and in the equation.B CFor :B œ ! C œ B %

C œ ! %

C œ % Ð!ß%Ñ

For :B œ "

Ð"ß$Ñ

C œ B %

C œ " %

C œ $

In a similar manner, substitute , , and toB œ # $ %get , , and .C œ # " !

B C

! %

" $

# #

$ "

% !

Plot the ordered pairs and draw the line throughthem.

24. C œ B $To complete the table, substitute the given valuesfor and in the equation.B CFor :B œ ! C œ B $

C œ ! $

C œ $ Ð!ß $Ñ

For :B œ "

Ð"ß %Ñ

C œ B $

C œ " $

C œ %

In a similar manner, substitute , , and toB œ # $ %get , , and .C œ & ' (

B C

! $

" %

# &

$ '

% (


continued


25. B C œ $To complete the table, substitute the given valuesfor and in the equation.B CFor : B œ ! B C œ $

! C œ $

C œ $ Ð!ß$Ñ

For : C œ !

Ð$ß !Ñ

B C œ $

B ! œ $

B œ $

For :B œ &

Ð&ß #Ñ

B C œ $

& C œ $

C œ #

C œ #

For :B œ #

Ð#ß"Ñ

B C œ $

# C œ $

C œ "

C œ "


26. B C œ &For :B œ !

Ð!ß&Ñ

! C œ &

C œ &

C œ &

For :C œ ! B ! œ &

B œ & Ð&ß !Ñ

For :B œ "

Ð"ß%Ñ

" C œ &

C œ %

C œ %

For :B œ $

Ð$ß#Ñ

$ C œ &

C œ #

C œ #


27. B #C œ &To complete the table, substitute the given valuesfor or in the equation.B CFor :B œ !

!ß

B #C œ &

! #C œ &

#C œ &

C œ &#

ˆ ‰&#

For :C œ !

Ð&ß !Ñ

B #C œ &

B #Ð!Ñ œ &

B ! œ &

B œ &

For :B œ #

#ß

B #C œ &

# #C œ &

#C œ $

C œ $#

ˆ ‰$#

For :C œ #

"ß #

B #C œ &

B #Ð#Ñ œ &

B % œ &

B œ " a b Plot the ordered pairs and draw the line through

them.

28. B $C œ &For :B œ ! ! $C œ &

$C œ &

C œ &$

ˆ ‰!ß&$

For :C œ ! B $Ð!Ñ œ &

B œ & a b&ß !

For :B œ " " $C œ &

$C œ '

C œ # a b"ß#

For :C œ " B $Ð"Ñ œ &

B $ œ &

B œ # a b#ß"



29. %B &C œ #!For :B œ ! %B &C œ #!

%Ð!Ñ &C œ #!

&C œ #!

C œ % Ð!ß%Ñ

For :C œ ! %B &C œ #!

%B &Ð!Ñ œ #!

%B œ #!

B œ & Ð&ß !Ñ

For :B œ # %B &C œ #!

%Ð#Ñ &C œ #!

) &C œ #!

&C œ "#

C œ "#&

ˆ ‰#ß"#&

For :C œ $ %B &C œ #!

%B &Ð$Ñ œ #!

%B "& œ #!

%B œ &

B œ &%

ˆ ‰&% ß$


30. 'B &C œ $!For :B œ ! 'Ð!Ñ &C œ $!

&C œ $!

C œ ' Ð!ß'Ñ

For :C œ ! 'B &Ð!Ñ œ $!

'B œ $!

B œ & Ð&ß !Ñ

For :B œ $ 'Ð$Ñ &C œ $!

") &C œ $!

&C œ "#

C œ "#&

ˆ ‰$ß"#&

For :C œ #

'B &Ð#Ñ œ $!

'B "! œ $!

'B œ #!

B œ œ#! "!' $

ˆ ‰"!$ ß#


31. C œ #B $

B #B C œ #B $

! ! $

" # "

# % "

$ ' $

Notice that as the value of increases by , theB "value of decreases by .C #

32. C œ $B "

B $B C œ $B "

! ! "

" $ #

# ' &

$ * )

Notice that as the value of increases by , theB "value of decreases by .C $

33. (a) The -values are , , , , and .C % $ # " !They increase by unit."

(b) The -values are , , , and . TheyC $ " " $decrease by units.#


(c) It appears that the -value increases (orCdecreases) by the value of the coefficient of . SoBfor , a conjecture is "For every increaseC œ #B %in by unit, increases by units."B " C #

34. In orderedquadrant III, both coordinates of the pairs are negative. If and is positive,B C œ 5 5then either or must be positive, because theB Csum of two negative.negative numbers is

35. Choose a value other than for either or and! B Cthen solve for the other variable. For%B &C œ !example, if , then andB œ & #! &C œ !C œ %. The student should then plot the pointsÐ!ß !Ñ Ð&ß %Ñ and and draw a line through them.

36. All points on the B C !-axis have -coordinate , sothe equation of the -axis is . All points onB C œ !the -axis have -coordinate , so the equation ofC B !the -axis is .C B œ !

37. #B $C œ "#To find the B C œ !-intercept, let .

#B $C œ "#

#B $Ð!Ñ œ "#

#B œ "#

B œ '

The B Ð'ß !ÑÞ-intercept is To find the -intercept, let .C B œ !

#B $C œ "#

#Ð!Ñ $C œ "#

$C œ "#

C œ %

The C Ð!ß %Ñ-intercept is .Plot the intercepts and draw the line through them.

38. &B #C œ "!To find the B C œ !-intercept, let .

&B #Ð!Ñ œ "!

&B œ "!

B œ #

The B Ð#ß !Ñ-intercept is .To find the -intercept, let .C B œ !

&Ð!Ñ #C œ "!

#C œ "!

C œ &

The C Ð!ß &Ñ-intercept is .Plot the intercepts and draw the line through them.

39. B $C œ 'To find the B C œ !-intercept, let .

B $C œ '

B $Ð!Ñ œ '

B ! œ '

B œ '

The B Ð'ß !Ñ-intercept is .To find the -intercept, let .C B œ !

B $C œ '

! $C œ '

$C œ '

C œ #

The C Ð!ß#Ñ-intercept is .Plot the intercepts and draw the line through them.

40. B #C œ %To find the B C œ !-intercept, let .

B #Ð!Ñ œ %

B œ %

The B Ð%ß !Ñ-intercept is .To find the -intercept, let .C B œ !

! #C œ %

#C œ %

C œ #

The C Ð!ß #Ñ-intercept is .Plot the intercepts and draw the line through them.

41. #$B $C œ (

To find the B C œ !-intercept, let .#$

#$

$#

B $Ð!Ñ œ (

B œ (

B œ • ( œ #"#


The B ß !-intercept is .ˆ ‰#"#


#$

($

a b! $C œ (

$C œ (

C œ

The C !ß-intercept is .ˆ ‰($

Plot the intercepts and draw the line through them.

42. & '( (B C œ #

To find the B C œ !-intercept, let .& '( (

&(

(&

B Ð!Ñ œ #

B œ #

B œ Ð#Ñ œ "%&

The B ß !-intercept is .ˆ ‰"%&


& '( (

'(

( (' $

a b! C œ #

C œ #

C œ Ð#Ñ œ

The C !ß-intercept is .ˆ ‰($


43. C œ &This is a horizontal line. Every point hasC C !-coordinate , so no point has -coordinate .&There is no -intercept.BSince every point of the line has -coordinate ,C &the -intercept is . Draw the horizontal lineC Ð!ß Ñ&through .Ð!ß Ñ&

44. C œ $This is a horizontal line. Every point hasC $ C !-coordinate , so no point has -coordinate .There is no -intercept.BSince every point of the line has -coordinate ,C $the -intercept is . Draw the horizontal lineC Ð!ß$Ñthrough .Ð!ß$Ñ

45. B œ #This is a vertical line. Every point hasB B Ð ß !Ñ-coordinate , so the -intercept is .# #Since every point of the line has -coordinate ,B #no point has -coordinate . There is noB !C Ð ß !Ñ-intercept. Draw the vertical line through .#

46. B œ $This is a vertical line. Every point hasB-coordinate , so the -intercept is .$ B Ð$ß !ÑSince every point of the line has -coordinate ,B $no point has -coordinate . There is noB !C Ð$ß !Ñ-intercept. Draw the vertical line through .

47. B % œ ! ÐB œ %ÑThis is a vertical line. Every point hasB % %-coordinate , so the -intercept is .B Ð ß !ÑSince every point of the line has -coordinate ,B %no point has -coordinate . There is noB !C Ð ß !Ñ-intercept. Draw the vertical line through .%


48. C # œ ! ÐC œ #ÑThis is a horizontal line. Every point hasC C !-coordinate , so no point has -coordinate .#There is no -intercept.BSince every point of the line has -coordinate ,C #the -intercept is . Draw the horizontal lineC Ð!ß Ñ#through .Ð!ß Ñ#

49. B &C œ !To find the B C œ !-intercept, let .

B &C œ !

B &Ð!Ñ œ !

B œ !

The B Ð!ß !Ñ B œ !-intercept is , and since , this isalso the -intercept. Since the intercepts are theCsame, another point is needed to graph the line.Choose any number for , say , and solveC C œ "the equation for .B

B &C œ !

B &Ð"Ñ œ !

B œ &

This gives the ordered pair . Plot Ð&ß"Ñ Ð&ß"Ñand Ð!ß !Ñ, and draw the line through them.

50. B $C œ !To find the B C œ !-intercept, let .

B $Ð!Ñ œ !

B œ !

The B Ð!ß !Ñ B œ !-intercept is , and since , this isalso the -intercept. Since the intercepts are theCsame, another point is needed to graph the line.Choose any number for , say , and solveC C œ "the equation for .B

B $Ð"Ñ œ !

B œ $

This gives the ordered pair . PlotÐ$ß"ÑÐ$ß"Ñ and Ð!ß !Ñ, and draw the line throughthem.

51. #B œ $CIf , then , so the - and B œ ! C œ ! B C-intercepts areÐ!ß !Ñ B œ $. To get another point, let .

#Ð$Ñ œ $C

# œ C

Plot and Ð$ß #Ñ Ð!ß !Ñ, and draw the line throughthem.

52. %C œ $B

If , then , so the - and B œ ! C œ ! B C-intercepts areÐ!ß !Ñ B œ %. To get another point, let .

%C œ $Ð%Ñ

C œ $

Plot and Ð%ß $Ñ Ð!ß !Ñ, and draw the line throughthem.

53. C œ B#$

If , then , so the - and B œ ! C œ ! B C-intercepts areÐ!ß !Ñ C œ $. To get another point, let .

Ð$Ñ œ B

# œ B

#$

Plot and Ð#ß $Ñ Ð!ß !Ñ, and draw the line throughthem.


54. $C œ B%$If , then , so the - and B œ ! C œ ! B C-intercepts areÐ!ß !Ñ B œ *. To get another point, let .

$C œ Ð*Ñ

$C œ "#

C œ %

%$

Plot and Ð*ß%Ñ Ð!ß !Ñ, and draw the line throughthem.

55. By the Midpoint Formula, the midpoint of thesegment with endpoints and isÐ)ß %Ñ Ð#ß'Ñ

Œ Œ ) Ð#Ñ % Ð'Ñ "! #

# # # #ß œ ß œ Ð&ß"ÑÞ

56. By the Midpoint Formula, the midpoint of thesegment with endpoints and isÐ&ß #Ñ Ð"ß )Ñ

Œ Œ & Ð"Ñ # ) % "!

# # # #ß œ ß œ Ð#ß &ÑÞ

57. By the Midpoint Formula, the midpoint of thesegment with endpoints and isÐ$ß'Ñ Ð'ß $Ñ

Œ Œ Œ $ ' ' $ * $ * $

# # # # # #ß œ ß œ ß Þ

58. By the Midpoint Formula, the midpoint of thesegment with endpoints and isÐ"!ß %Ñ Ð(ß "Ñ

Œ Œ Œ "! ( % " $ & $ &

# # # # # #ß œ ß œ ß Þ

59. By the Midpoint Formula, the midpoint of thesegment with endpoints and isÐ*ß $Ñ Ð*ß )Ñ

Œ Œ Œ * * $ ) ! "" ""

# # # # #ß œ ß œ !ß Þ

60. By the Midpoint Formula, the midpoint of thesegment with endpoints and isÐ%ß$Ñ Ð"ß $Ñ

Œ Œ Œ % Ð"Ñ $ $ $ ! $

# # # # #ß œ ß œ ß ! Þ

61. By the Midpoint Formula, the midpoint of thesegment with endpoints and Ð#Þ&ß $Þ"Ñ Ð"Þ(ß"Þ$Ñis

Œ Œ a b#Þ& "Þ( $Þ" Ð"Þ$Ñ %Þ# "Þ)

# # # #ß œ ß œ #Þ"ß !Þ* Þ

62. By the Midpoint Formula, the midpoint of thesegment with endpoints and Ð'Þ#ß &Þ)Ñ Ð"Þ%ß!Þ'Ñis

Œ Œ a b'Þ# "Þ% &Þ) Ð!Þ'Ñ (Þ' &Þ#

# # # #ß œ ß œ $Þ)ß #Þ' Þ

63. By the Midpoint Formula, the midpoint of thesegment with endpoints and isˆ ‰ ˆ ‰" " $ &

# $ # $ß ß

Œ a b" $ %# # #

" & '$ $ $

# # # # # #ß œ ß œ ß œ "ß " Þ

# #

64. By the Midpoint Formula, the midpoint of thesegment with endpoints and isˆ ‰ ˆ ‰#" # ( $

% & % &ß ß

Œ #" ( #)% % %

# $ && & &

# # # # # #ß œ ß œ ß Þ

( "

65. By the Midpoint Formula, the midpoint of thesegment with endpoints and isˆ ‰ ˆ ‰ ß ß" # " "

$ ( # "%

ˆ ‰ Œ

# # # # "# #)ß œ ß œ ß Þ

& &" " # " &$ # ( "% '

&"%

66. By the Midpoint Formula, the midpoint of thesegment with endpoints and isˆ ‰ ˆ ‰$ " " (

& $ # #ß ß

ˆ ‰ Œ $ " " ( "" #$& # $ # "! '

# # # # #! "#ß œ ß œ ß Þ

"" #$

67. midpoint of and TÐ&ß )Ñ UÐBß CÑ œ QÐ)ß #Ñ

& B ) C

# #ß œ Ð)ß #ÑŒ

The - and -coordinates must be equal.B C

& B ) C

# #œ ) œ #

& B œ "' ) C œ %

B œ "" C œ %

Thus, the endpoint is .U Ð""ß%Ñ

68. midpoint of and TÐ(ß "!Ñ UÐBß CÑ œ QÐ&ß $Ñ

( B "! C

# #ß œ Ð&ß $ÑŒ


( B "! C

# #œ & œ $

( B œ "! "! C œ '

B œ $ C œ %

Thus, the endpoint is .U Ð$ß%Ñ


69. midpoint of and T ß UÐBß CÑ œ Q ß "

B C

# #ß œ ß "

ˆ ‰ ˆ ‰ ˆ ‰" " $$ & #

" "$ & $

#


" "$ & B C

# #œ œ "

B œ $ C œ #

B œ C œ

$

#

B œ C œ

" "$ &" * " "!$ $ & &

) *$ &

Thus, the endpoint is .U ßˆ ‰) *$ &

70. midpoint of and TÐ#Þ&ß "Þ(&Ñ UÐBß CÑ œ QÐ$ß #Ñ

#Þ& B "Þ(& C

# #ß œ Ð$ß #ÑŒ


#Þ& B "Þ(& C

# #œ $ œ #

#Þ& B œ ' "Þ(& C œ %

B œ $Þ& C œ #Þ#&

Thus, the endpoint is .U Ð$Þ&ß #Þ#&Ñ

71. For 2003, .B œ #!!$ #!!! œ $

C œ "#$(B '! *$'

œ "#$(Ð$Ñ '! *$'

œ $("" '! *$'

C œ &( ##&

,,

,,

Let x = 3.

The approximate number of U.S. travelers to othercountries in 2003 was , thousand (or&( ##&&( ##& !!!, , ).

72. For 2002, .B œ #!!# "**! œ "#

C œ $Þ$#B #)Þ(

œ $Þ$#Ð"#Ñ #)Þ(

œ $*Þ)% #)Þ(

C œ ')Þ&%

Let x = 12.

The approximate total amount spent on dentalservices in the United States in 2002 was $')Þ&%billion.

73. The graph goes through the point whichÐ#ß !Ñsatisfies only equations and . The graph alsoB Cgoes through the point which satisfies onlyÐ!ß $Ñequations and . Therefore, the correct equationA Bis .B

74. (a) From the table, the -intercept is .B Ð"Þ&ß !Ñ

The -intercept is .(b) C Ð!ß $Ñ

The point satisfies equations and .(c) C DÐ!ß $ÑThe point satisfies only equation .Ð!Þ&ß #Ñ DTherefore, the correct equation is .D

75. The screen on the right is more useful because itshows the intercepts.

76. Usually we need to solve for .C

B #C œ !

#C œ B

C œ B C œ !Þ&B"# or

77. We need to solve the given equation for .C

&B #C œ "!

#C œ &B "!

C œ B &

Subtract 5x.Divide by 2.&

#

Graph Y X in a standard viewing" œ #Þ& &window.

78. %B C œ "!

C œ %B "!

C œ %B "!

Subtract 4x.Multiply by 1.

Graph Y X in a standard viewing" œ % "!window.

79. $Þ'B C œ &Þ)

C œ $Þ'B &Þ)

C œ $Þ'B &Þ)

Subtract 3.6x.Multiply by 1.

Graph Y X in a standard viewing" œ $Þ' &Þ)window.

3.2 The Slope of a Line 105

80. !Þ$B !Þ%C œ !Þ'

$B %C œ '

%C œ $B '

C œ B

Multiply by 10.Subtract 3x.Divide by 4.$ $

% #

Graph Y X in a standard viewing" œ !Þ(& "Þ&window.

81. ' # %

& $ #œ œ #

82. & ( # # "

% # ' ' $œ œ œ

83. % Ð"Ñ % " &

$ Ð&Ñ $ & #œ œ

84. ' ! ' '

! Ð$Ñ ! $ $œ œ œ #

85. & Ð&Ñ & & !

$ # " "œ œ œ !

86. ( Ð#Ñ ( # *

$ Ð$Ñ $ $ !œ œ is sinceundefined

you cannot divide by zero.

3.2 The Slope of a Line3.2 Classroom Examples

1. If and ,a b a bB ß C œ Ð'ß *Ñ B ß C œ Ð$ß&Ñ" " # #

then

7 œ œ œ œ C C & * "% "%

B B $ Ð'Ñ * *# "

# ".

The slope is ."%*

2. To find the slope of the line with equation

$B %C œ "#,

first find the intercepts. The -intercept is ,B Ð%ß !Ñand the -intercept is . The slope is thenC Ð!ß$Ñ

7 œ œ œ$ ! $ $

! % % %.

The slope is .$%

3. (a) To find the slope of the line with equation

C $ œ !,

select two different points on the line, such asÐ!ß$Ñ Ð#ß$Ñ and , and use the slope formula.

7 œ œ œ !$ Ð$Ñ !

# ! #

The slope is .!

To find the slope of the line with equation(b)

B œ ',

select two different points on the line, such asÐ'ß !Ñ Ð'ß $Ñ and , and use the slope formula.

7 œ œ$ ! $

' Ð'Ñ !

Since division by zero is undefined, the slope isundefined.

4. Solve the equation for .C

$B %C œ *

%C œ $B *

C œ B

Subtract 3x.Divide by 4.$ *

% %

The slope is given by the coefficient of , so theBslope is .$

%

5. Through ; Ð$ß#Ñ 7 œ "#

Locate the point on the graph. Use theÐ$ß#Ñslope formula to find a second point on the line.

7 œ œC "

B #

change in change in

From , move unit and then units toÐ$ß#Ñ " #upthe to . Draw the line through theright Ð"ß"Ñtwo points.

6. Find the slope of each line.The line through and has slopeÐ"ß #Ñ Ð$ß &Ñ

7 œ œ& # $

$ Ð"Ñ %" .

The line through and has slopeÐ%ß (Ñ Ð)ß "!Ñ

7 œ œ Þ"! ( $

) % %#

The slopes are the same, so the lines are parallel.


7. Solve each equation for .C

$B &C œ ' &B $C œ #

&C œ $B ' $C œ &B #

C œ B C œ B

7 œ 7 œ

so so

$ ' & #& & $ $

" #$ && $

ˆ ‰ ˆ ‰ Since , the lines are7 7 œ œ "" #

$ && $

ˆ ‰ˆ ‰perpendicular.

8. anda bB ß C œ Ð#!!#ß )")Ñ" "a bB ß C œ Ð#!!$ß )%$Ñ# # .

average rate of change œC C

B B

œ œ œ #&)%$ )") #&

#!!$ #!!# "

# "

# "

The average rate of change is , which is#&approximately the same as in Example 8.

9. anda bB ß C œ Ð"**(ß $'Þ%Ñ" "a bB ß C œ Ð#!!$ß #"Þ*Ñ# # .

average rate of change

œC C

B B

œ œ#"Þ* $'Þ% "%Þ&

#!!$ "**( '¸ #Þ

# "

# "

%"(

Thus, the average rate of change from to"**(#!!$ #Þ%"( was % per year.

3.2 Section Exercises

1. slopechange in vertical position

change in horizontal position feet feet

œ

œ$!

"!!

Choices , , , , and , are all correct.A B D!Þ$ $ $!"! "!!

2. slopechange in vertical position

change in horizontal positionchange in vertical position

feet

œ

!Þ!& œ&!

So the change in vertical position is!Þ!&Ð&! Ñ œ #Þ& feet feet.

3. (a) C Graph indicates that sales leveled off duringthe second quarter.

Graph indicates that sales leveled off during(b) Athe fourth quarter.

(c) D Graph indicates that sales rose sharplyduring the first quarter, and then fell to the originallevel during the second quarter.

Graph is the only graph that indicates that(d) Bsales fell during the first two quarters.

4. To get to from , we must go up units andF E #move right unit. Thus,"

slope of .riserun

EF œ œ œ ##

"

5. slope of riserun

FG œ œ œ !!

%

6. slope of , which is undefined.riserun

GH œ œ(

!

7. slope of riserun

HI œ œ œ " "

$ $

8. slope of riserun

IJ œ œ œ "$

$

9. slope of riserun

JK œ œ œ %%

"

10. slope of riserun

FH œ œ(

%

11. 7 œ œ œ #' # %

& $ #

12. 7 œ œ œ& ( # "

% # ' $

13. 7 œ œ œ% Ð"Ñ % " &

$ Ð&Ñ $ & #

14. 7 œ œ œ #' ! '

! Ð$Ñ $

15. 7 œ œ œ œ !& Ð&Ñ & & !

$ # " "

16. 7 œ œ œ œ !# Ð#Ñ # # !

% Ð$Ñ % $ (

17. , which is7 œ œ œ$ ) & &

# Ð#Ñ # # !

undefined.

18. , which is7 œ œ œ& ' " "

) Ð)Ñ ) ) !

undefined.

19. , which is .7 œ œ

!

% " ""$ # '" "' '

undefined

20. B D A and are correct. Choice is wrong becausethe order of subtraction must be the same in thenumerator and denominator. Choice is wrongCbecause slope is defined as the change in dividedCby the change in .B

21. Let a b a bB ß C œ Ð#ß$Ñ B ß C œ Ð"ß &Ñ" " # # and .Then

7 œ œ œ œ )C C & Ð$Ñ )

B B " Ð#Ñ "# "

# ".

The slope is .)


22. Let a ba bB ß C œ Ð%ß $Ñ

B ß C œ Ð$ß %Ñ" "

# #

and. Then

7 œ œ œ œ "C C % $ "

B B $ Ð%Ñ "# "

# ".

The slope is ."

23. Let a ba bB ß C œ Ð%ß "Ñ

B ß C œ Ð#ß 'Ñ" "

# #

and. Then

7 œ œ œC C ' " &

B B # Ð%Ñ '# "

# ".

The slope is .&'24. Let a ba bB ß C œ Ð$ß$Ñ

B ß C œ Ð&ß 'Ñ" "

# #

and. Then

7 œ œ œC C ' Ð$Ñ *

B B & Ð$Ñ )# "

# ".

The slope is .*)25. Let a ba bB ß C œ Ð#ß %Ñ

B ß C œ Ð%ß %Ñ" "

# #

and. Then

7 œ œ œ œ !C C % % !

B B % # '# "

# ".

The slope is .!

26. Let a ba bB ß C œ Ð'ß $Ñ

B ß C œ Ð#ß $Ñ" "

# #

and. Then

7 œ œ œ œ !C C $ $ !

B B # Ð'Ñ )# "

# ".

The slope is .!

27. Let a ba bB ß C œ Ð"Þ&ß #Þ'Ñ

B ß C œ Ð!Þ&ß $Þ'Ñ" "

# #

and. Then

7 œ œ œ œ "C C $Þ' #Þ' "

B B !Þ& "Þ& "# "

# ".

The slope is ."

28. Let a ba bB ß C œ Ð$Þ%ß %Þ#Ñ

B ß C œ Ð"Þ%ß "!Þ#Ñ" "

# #

and. Then

7 œ œ œ œ $C C "!Þ# %Þ# '

B B "Þ% $Þ% ## "

# ".

The slope is .$

29. Let a b ˆ ‰a b ˆ ‰B ß C œ

B ß C œ

" "

# #

" "' #& *' #

ß

ß

and. Then

7 œ œ œ œ %C C

B B

# "

# "

* " )# # #& "' '

%'

• $# œ '.

The slope is .'

30. Let a b ˆ ‰a b ˆ ‰B ß C œ ß

B ß C œ

" "$ "% $

# #

and. Then& "!

% $ß

7 œ œ œ œ $C C

B B

# "

# "

"! " *$ $ $& $% %

#%

• # œ '.

The slope is .'

31. Let a b ˆ ‰a b ˆ ‰B ß C œ

B ß C œ

" "

# #

# &* ")

" &") *

ß

ß

and. Then

7 œ œ œC C

B B

œ

# "

# "

& & "&* ") ")

" #") *

&")

"&")

ˆ ‰• ")

& œ $.

The slope is .$

32. Let a b ˆ ‰a bB ß C œ

B ß C œ

" "

# #

ß

ß

% *& "!$ ""! &

and. Thenˆ ‰

7 œ œ œC C

B B

œ

# "

# "

" * (& "! "!$ %"! &

&"!

("!

ˆ ‰• "! (

& &œ .

The slope is .(&

33. The points shown on the line are andÐ$ß $Ña b"ß# . The slope is

7 œ œ œ # $ & &

" Ð$Ñ # #.

34. The points shown on the line are andÐ"ß"Ña b%ß & . The slope is

7 œ œ œ #& Ð"Ñ '

% " $.

35. The points shown on the line are andÐ$ß $Ña b$ß$ . The slope is

7 œ œ$ $ '

$ $ !, which is .undefined

36. Suppose the last name is Nguyen. Then and5 œ 'we graph the horizontal line .C œ '

The slope is .!

37. "The line has positive slope" means that the linegoes up from left to right. This is line B.

38. "The line has negative slope" means that the linegoes down from left to right. This is line C.


39. "The line has slope " means that there is no!vertical change; that is, the line is horizontal. Thisis line A.

40. "The line has undefined slope" means that there isno the line is .horizontal change; that is, verticalThis is line D.

41. To find the slope of

B #C œ %,

first find the intercepts. Replace with to findC !that the -intercept is ; replace with toB Ð%ß !Ñ B !find that the -intercept is . The slope is thenC Ð!ß #Ñ

7 œ œ œ # ! # "

! % % #.

To sketch the graph, plot the intercepts and drawthe line through them.


B $C œ ',

first find the intercepts. Replace with to findC !that the -intercept is ; replace with toB Ð'ß !Ñ B !find that the -intercept is . The slope isC Ð!ß#Ñthen

7 œ œ œ # ! # "

! Ð'Ñ ' $.



&B #C œ "!,

first find the intercepts. Replace with to findC !that the -intercept is ; replace with toB Ð#ß !Ñ B !find that the -intercept is . The slope isC Ð!ß&Ñthen

7 œ œ œ& ! & &

! # # #.



%B C œ %,

first find the intercepts. Replace with to findC !that the -intercept is ; replace with toB Ð"ß !Ñ B !find that the -intercept is . The slope isC Ð!ß%Ñthen

7 œ œ œ %% ! %

! " ".


45. In the equation

C œ %B,

replace with and then with to get theB ! B "ordered pairs respectively.a b a b!ß ! "ß % and ,(There are other possibilities for ordered pairs.)The slope is then

7 œ œ œ %% ! %

" ! ".

To sketch the graph, plot the two points and drawthe line through them.

46. In the equation

C œ $B,

replace with and then with to get theB ! B "ordered pairs respectively.a b a b!ß ! "ß$ and ,(There are other possibilities for ordered pairs.)The slope is then

7 œ œ œ $$ ! $

" ! ".


To sketch the graph, plot the two points and drawthe line through them.

47. B $ œ ! ÐB œ $ÑThe graph of is the B œ $ vertical line withB Ð$ß !Ñ-intercept . The slope of a vertical line isundefined.

48. B # œ ! ÐB œ #ÑThe graph of is the B œ # vertical line withB Ð#ß !Ñ-intercept . The slope of a vertical line isundefined.

49. C œ &The graph of is the C œ & horizontal line withC Ð!ß&Ñ-intercept . The slope of a horizontal lineis !.

50. C œ %The graph of is the C œ % horizontal line withC Ð!ß%Ñ-intercept . The slope of a horizontal lineis !.

51. #C œ $ C œˆ ‰$#

The graph of is the C œ $# horizontal line with

C !ß-intercept . The slope of a horizontal line isˆ ‰$#

!.

52. $B œ % B œˆ ‰%$

The graph of is the B œ %$ vertical line with

B ß !-intercept . The slope of a vertical line isˆ ‰%$

undefined.

53. To graph the line through with slopea b%ß #7 œ Ð%ß #Ñ"

# , locate on the graph. To find asecond point, use the definition of slope.

7 œ œC "

B #

change in change in

From go up unit. Then go units to thea b%ß # , " #right to get to . Draw the line throughÐ#ß $Ña b a b%ß # #ß $ and .

54. To graph the line through with slopea b#ß$7 œ Ð#ß$Ñ&

% , locate on the graph. To find asecond point, use the definition of slope.

7 œ œC &

B %

change in change in

From go up units. Then go units toa b#ß$ , & %the right to get to . Draw the line throughÐ#ß #Ña b a b#ß$ #ß # and .


55. To graph the line through with slopea b!ß#7 œ Ð!ß#Ñ#

$ , locate the point on the graph. Tofind a second point on the line, use the definitionof slope, writing as .# #

$ $

7 œ œC #

B $

change in change in

From move units down and then unitsa b!ß# , # $to the right. Draw a line through this second pointand (Note that the slope could also bea b!ß# .written as . In this case, move units up and #

$ # $

units to the left to get another point on the sameline.)

56. To graph the line through with slopea b!ß%7 œ Ð!ß%Ñ$

# , locate the point on the graph. Tofind a second point on the line, use the definitionof slope, writing as .$ $

# #

7 œ œC $

B #

change in change in

From move units down and then unitsa b!ß% , $ #to the right. Draw a line through this second pointand The slope could also be written asa b!ß% .$# . In this case, move units up and units to the$ #

left to get another point on the same line, as shownin the figure.

57. Locate . Then use to go Ð"ß#Ñ 7 œ $ œ $$"

units up and unit right to ." Ð!ß "Ñ

58. Locate . Then use to go Ð#ß%Ñ 7 œ % œ %%"

units up and unit right to ." Ð"ß !Ñ

59. Locate . A slope of means that the line isÐ#ß&Ñ !horizontal, so at every point. Draw theC œ &horizontal line through .Ð#ß&Ñ

60. Locate . A slope of means that the line isÐ&ß $Ñ !horizontal, so at every point. Draw theC œ $horizontal line through .Ð&ß $Ñ

61. Locate . Since the slope is undefined, theÐ$ß "Ñline is vertical. The -value of every point is .B $Draw the vertical line through .Ð$ß "Ñ

62. Locate . Since the slope is undefined, theÐ%ß "Ñline is vertical. The -value of every point is .B %Draw the vertical line through .Ð%ß "Ñ

63. If a line has slope , then any line parallel to it%*

has slope (the slope must be the same), and%*

any line perpendicular to it has slope (the slope*%

must be the negative reciprocal).


64. If a line has slope , then any line parallel to it!Þ#has slope (the slope must be the same), and!Þ#any line perpendicular to it has slope "

!Þ# œ &

(the slope must be the negative reciprocal).

65. The slope of the line through a b a b"&ß * "#ß( and is

7 œ œ œ( * "' "'

"# "& $ $.

The slope of the line througha b a b)ß% &ß#! and is

7 œ œ œ#! Ð%Ñ "' "'

& ) $ $.

Since the slopes are equal, the two lines areparallel.

66. The slope of the line through isa b a b%ß ' )ß ( and

7 œ œ œ ( ' " "

) % "# "#.

The slope of the line through isa b a b&ß & (ß % and

7 œ œ œ % & " "

( Ð&Ñ "# "#.

Since the slopes are equal, the two lines areparallel.

67. B %C œ ( %B C œ $

C

%C œ B ( C œ %B $

C œ B C œ %B $

andSolve the equations for .

" (% %

The slopes, %"% and , are negative reciprocals of

one another, so the lines are perpendicular.

68. #B &C œ ( &B #C œ "

C

&C œ #B ( #C œ &B "

C œ B C œ B


# ( & "& & # #

The slopes, # && # and , are negative reciprocals of


69. %B $C œ ' $B %C œ #

C

$C œ %B ' %C œ $B #

C œ B # C œ B

and Solve the equations for .

% $ "$ % #

The slopes are % $$ % and . The lines are neither

parallel nor perpendicular.

70. #B C œ ' B C œ %

C

C œ #B ' C œ B %

C œ B %


The slopes are and .# " The lines are neitherparallel nor perpendicular.

71. and B œ ' ' B œ )

The second equation can be simplified as .B œ #Both lines are vertical lines, so they are parallel.

72. and $B œ C #C 'B œ &The slope of the first line is the coefficient of ,Bnamely . Solve the second equation for .$ C

#C œ 'B &

C œ $B &#

The slope of the second line is also , so the lines$are parallel.

73. %B C œ ! &B ) œ #C

C

C œ %B B % œ C


&#

The slopes are % and . The lines are &# neither


74. #B &C œ ) ' #B œ &C

C

&C œ #B ) &C œ #B '

C œ B C œ B


# ) # '& & & &

The slopes are # #& & and . The lines are neither


75. #B œ C $ #C B œ $

C

#B $ œ C #C œ B $

C œ B


" $# #

The slopes, reciprocals of# and , are negative "#


76. %B $C œ ) %C $B œ "#

C

$C œ %B ) %C œ $B "#

C œ B C œ B $


% ) $$ $ %

The slopes, reciprocals of% $$ % and , are negative


77. Use the points and .Ð!ß #!Ñ Ð%ß %Ñ


œ œ œ œ %C % #! "'

B % ! %

change in change in

The average rate of change is $ per year, %!!!that is, the value of the machine is decreasing$ each year during these years.%!!!

78. Use the points and .Ð!ß !Ñ Ð%ß #!!Ñ


œ œ œ œ &!C #!! ! #!!

B % ! %

change in change in

The average rate of change is $ per month, that&!is, the amount saved is increasing $ each month&!during these months.


79. We can see that there is no change in the percentof pay raise. Thus, the average rate of change is!% per year, that is, the percent of pay raise is notchanging—it is % each year during these years.$

80. If the graph of a linear equation rises from left toright, then the average rate of change is positive .If the graph of a linear equation from left tofallsright, then the average rate of change is . negative

81. Let be the C vertical rise.Since the slope is the vertical rise divided by thehorizontal run,

!Þ"$ œC

"&!.

Solving for givesC

C œ !Þ"$Ð"&!Ñ œ "*Þ&.

The vertical rise could be a maximum of ft."*Þ&

82. The vertical change is ft, and the horizontal'$change is ft.#&! "'! œ *!The slope is '$ (

*! "!œ .

83. For 1999–2000:(a)

7 œ œ #$ %$""!* %() )' !%(

#!!! "***

, ,,

For 2000–2001:

7 œ œ ") )*("#) $(& "!* %()

#!!" #!!!

, ,,

For 2001–2002:

7 œ œ "# $*#"%! ('( "#) $(&

#!!# #!!"

, ,,

For 2002–2003:

7 œ œ "( *&&"&) (## "%! ('(

#!!$ #!!#

, ,,

For 2003–2004:

7 œ œ #$ %")")# "%! "&) (##

#!!% #!!$

, ,,

The average rates of change, , are measured in7thousands.

(b) The average rate of change in successive yearsis approximately the same. Therefore, annotapproximately straight line could not be drawnthrough the plotted ordered pairs.

84. (a) For 1995–1996:

7 œ œ œ "!!!#! !!! "* !!! "!!!

"**' "**& "

, ,

For 1995–1999:

7 œ œ œ "!!!#$ !!! "* !!! %!!!

"*** "**& %

, ,

For 1998–2000:

7 œ œ œ "!!!#% !!! ## !!! #!!!

#!!! "**) #

, ,

The answers have units of million per year.

(b) The average rate of change is the same. Whengraphed, the data points lie on a straight line.

85. (a) Use and Ð"**)ß ##*Þ$Ñ Ð#!!%ß $$)Þ)ÑÞ

7 œ œ œ ")Þ#&$$)Þ) ##*Þ$ "!*Þ&

#!!% "**) '

The average rate of change is $ billion per")Þ#&year.

(b) The positive slope means that personalspending on recreation in the United Statesincreased by an average of $ billion each")Þ#&year.

86. (a) Use and .Ð#!!!ß &"#Ñ Ð#!!%ß $*#Ñ

7 œ œ œ $!$*# &"# "#!

#!!% #!!! %

The average rate of change is theaters per$!year.

(b) The negative slope means that the number ofdrive-in theaters by an average of decreased $!each year.

87. Use and Ð"**(ß &!!Ñ Ð#!!#ß "&&ÑÞ

7 œ œ œ '*"&& &!! $%&

#!!# "**( &

The average rate of change in price is $ per '*year, that is, the price decreased an average of $'*each year from 1997 to 2002.

88. Use and Ð"**(ß !Þ$%*Ñ Ð#!!#ß "&Þ&ÑÞ

7 œ œ œ $Þ!$!#"&Þ& !Þ$%* "&Þ"&"

#!!# "**( &

The average rate of change in sales is about $Þ!$million per year, that is, the sales of DVD playersincreased an average of million each year$Þ!$from 1997 to 2002.


89. Label the points as shown in the figure.

In order to determine whether is aABCDparallelogram, we need to show that the slope ofAB CD equals the slope of and that the slope ofAD BC equals the slope of .

Slope of AB œ œ œ %* Ð"Ñ )

"$ Ð""Ñ #

Slope of CD œ œ œ %' Ð#Ñ )

% # #

Slope of AD œ œ Ð"Ñ (

% Ð""Ñ "&

'

Slope of BC œ œ# Ð*Ñ (

# Ð"$Ñ "&

Thus, the figure is a parallelogram.

90. Label the points as shown in the figure.

In order to determine whether is aABCDparallelogram, we need to show that the slope ofAB CD equals the slope of and that the slope ofAD BC equals the slope of .

Slope of AB œ œ œ "* Ð&Ñ "% "%

# Ð""Ñ * *

Slope of CD œ œ œ % Ð"!Ñ "% "%

$ "# * *

Slope of AD œ œ% Ð&Ñ *

$ Ð""Ñ "%

Slope of BC œ œ"! Ð"*Ñ *

"# Ð#Ñ "%

Thus, the figure is a parallelogram. If two adjacentsides form a right angle, the parallelogram is arectangle. A right angle is formed byperpendicular lines. Notice perpendicular toAB is BC since œ ""% *

* "%ˆ ‰ .

Therefore, the figure is a rectangle.

91. Line has negative slope and line has positiveE Fslope, so line must be and line E C œ #B $ F"

must be .C œ $B %#

92. Both lines have a positive slope, but line isEsteeper than line . Thus, its slope (the F coefficientof ) must be larger, so line is andB E C œ %B &#

line is .F C œ #B &"

93. For and , the slope of EÐ$ß "Ñ FÐ'ß #Ñ AB is

7 œ œ# " "

' $ $.

94. For and , the slope of FÐ'ß #Ñ GÐ*ß $Ñ BC is

7 œ œ$ # "

* ' $.

95. For and , the slope of EÐ$ß "Ñ GÐ*ß $Ñ AC is

7 œ œ œ$ " # "

* $ ' $.

96. The slope of slope of slope of

.

AB BCAC

œ

œ

œ "$

97. For and , the slope of EÐ"ß#Ñ FÐ$ß"Ñ AB is

7 œ œ" Ð#Ñ "

$ " #.

For and , the slope of FÐ$ß"Ñ GÐ&ß !Ñ BC is

7 œ œ! Ð"Ñ "

& $ #.

For and , the slope of EÐ"ß#Ñ GÐ&ß !Ñ AC is

7 œ œ œ! Ð#Ñ # "

& " % #.

Since the three slopes are the same, the threepoints are collinear.

98. For and , the slope of EÐ!ß 'Ñ FÐ%ß&Ñ AB is

7 œ œ œ & ' "" ""

% ! % %.

For and , the slope of FÐ%ß&Ñ GÐ#ß "#Ñ BC is

7 œ œ œ "# Ð&Ñ "( "(

# % ' '.

Since these two slopes are not the same, the threepoints are not collinear.


99. $B #C œ )

#C œ $B )

C œ B %$#

100. %B $C œ !

$C œ %B

C œ B%$

101. C # œ %ÐB $Ñ

C # œ %B "#

C œ %B "%

102. C Ð#Ñ œ ÐB &Ñ

#ÐC #Ñ œ $ÐB &Ñ

#C % œ $B "&

"* œ $B #C

$#

103. C Ð"Ñ œ ÒB Ð%ÑÓ

$ÐC "Ñ œ &ÐB %Ñ

$C $ œ &B #!

"( œ &B $C

&$

104. C ( œ ÒB Ð$ÑÓ

%ÐC (Ñ œ "ÐB $Ñ

%C #) œ B $

B %C œ #&

"%

105. C Ð"Ñ œ ÒB Ð#ÑÓ

C " œ ÐB #Ñ

#ÐC "Ñ œ "ÐB #Ñ

#C # œ B #

B #C œ %

"#"#

3.3 Linear Equations in Two Variables3.3 Classroom Examples

1. Slope ; -intercept # C Ð!ß$ÑHere and . Substitute these values7 œ # , œ $into the slope-intercept form.

C œ 7B ,

C œ #B Ð$Ñ

C œ #B $

2. Solve the equation for .B #C œ %

C

#C œ B %

C œ B #"#

Plot the C Ð!ß#Ñ-intercept . The slope can beinterpreted as either or . Using , move" " "

# # #

from unit and to the unitsÐ!ß#Ñ " #down rightto locate the point . Draw a line throughÐ#ß$Ñthe two points.

3. Through ; slopeÐ$ß%Ñ œ 7 œ #&

Use the point-slope form with a bB ß C œ Ð$ß%Ñ" "

and .7 œ #&

C C œ 7 B B

C Ð%Ñ œ ÐB $Ñ

C % œ ÐB $Ñ

&ÐC %Ñ œ #ÐB $Ñ

&C #! œ #B '

#B &C œ #'

#B &C œ #'

" "#&#&

a b

Multiply by 5.

Multiply by 1.

The last step was performed to get the equation instandard form, , with positive.EB FC œ G E

4. Through and Ð#ß 'Ñ Ð"ß %Ñ

7 œ œ œ % ' # #

" Ð#Ñ $ $

Let .a bB ß C œ Ð"ß %Ñ" "

C C œ 7ÐB B Ñ

C % œ ÐB "Ñ

$C "# œ #B #

#B $C œ "%

" "#$

Multiply by 3.Standard form

5. (a) Through ; parallel to the lineÐ)ß $Ñ#B $C œ "!

Find the slope of the given line.

#B $C œ "!

$C œ #B "!

C œ B # "!$ $

The slope is , so a line parallel to it also has slope#$

# #$ $. Use and 7 œ a bB ß C œ Ð)ß $Ñ" " in the point-slope form.

C C œ 7ÐB B Ñ

C $ œ ÒB Ð)ÑÓ

C $ œ ÐB )Ñ

C $ œ B

C œ B

C œ B

" "#$#$# "'$ $# "' *$ $ $# #&$ $

3.3 Linear Equations in Two Variables 115

(b) Through ; perpendicular toÐ)ß $Ñ#B $C œ "!

The slope of is . The negative#B $C œ "! #$

reciprocal of is , so the slope of the line# $$ #

through is .Ð)ß $Ñ $#

C C œ 7ÐB B Ñ

C $ œ ÒB Ð)ÑÓ

C $ œ ÐB )Ñ

C $ œ B "#

C œ B *

" "$#$#$#$#

6. Since the price you pay is $ per minute plus a!Þ"!flat rate of $ , an equation for minutes is!Þ#! B

C œ !Þ"B !Þ#.

7. We will use the years 1940 and 2000, whichcorrespond to the ordered pairs andÐ!ß #%Þ&ÑÐ'!ß )!Þ%Ñ.

7 œ œ œ !Þ*$"')!Þ% #%Þ& &&Þ*

'! ! '!

Since we have the -intercept, , we canC Ð!ß #%Þ&Ñwrite the slope-intercept form of the line.

C œ 7B ,

C œ !Þ*$"'B #%Þ&

Rounded off, we have . A choiceC œ !Þ*$B #%Þ&of different ordered pairs will lead to a differentequation.

8. Use and .Ð""ß "'%Ñ Ð"$ß #!$Ñ

7 œ#!$ "'% $*

"$ "" #œ œ "*Þ&

Use the point-slope form withÐB ß C Ñ œ Ð""ß "'%Ñ" " .

C C œ 7 B B

C "'% œ "*Þ&ÐB ""Ñ

C "'% œ "*Þ&B #"%Þ&

C œ "*Þ&B &!Þ&

" "a b

9. %ÐB $Ñ B œ B '

%ÐB $Ñ B B ' œ !

Let Y X X X . Graph Y in the" "œ % $ 'a bwindow by and find theÒ"!ß "!Ó Ò"!ß "!ÓB-intercept. In the figure, we see that theB Ð$ß !Ñ-intercept is , so the solution of the equationis and the solution set is .$ Ö$×

3.3 Section Exercises1. A Choice , , is in the form$B #C œ &

EB FC œ G E ! E F with and integers , , andG " having no common factor (except ).

2. C Choice , , is in the formC $ œ #ÐB "ÑC C œ 7 B B" "a b.

3. A Choice , , is in the form .C œ 'B # C œ 7B ,

4. C # œ $ÐB %Ñ

C # œ $B "#

C œ $B "!

5. C # œ $ÐB %Ñ

C # œ $B "#

$B C œ "! Standard form

6. Solve for ."!B (C œ (! C

(C œ "!B (!

C œ B "!"!(

7. C œ #B $This line is in slope-intercept form with slope7 œ # C Ð!ß ,Ñ œ Ð!ß $Ñ and -intercept . The onlygraph with positive slope and with a positiveC C-coordinate of its -intercept is .A

8. C œ #B $This line is in slope-intercept form with slope7 œ # C Ð!ß ,Ñ œ Ð!ß $Ñ and -intercept . The onlygraph with negative slope and with a positiveC C-coordinate of its -intercept is .D

9. C œ #B $This line is in slope-intercept form with slope7 œ # C Ð!ß ,Ñ œ Ð!ß$Ñ and -intercept . Theonly graph with negative slope and with a negativeC C-coordinate of its -intercept is .C

10. C œ #B $This line has slope and -intercept7 œ # CÐ!ß ,Ñ œ Ð!ß$Ñ. The only graph with positiveslope and with a negative C-coordinate of itsC-intercept is .F

11. C œ #BThis line has slope and -intercept7 œ # CÐ!ß ,Ñ œ Ð!ß !Ñ. The only graph with positive slopeand with C Ð!ß !Ñ-intercept is .H

12. C œ #BThis line has slope and -intercept7 œ # CÐ!ß ,Ñ œ Ð!ß !Ñ. The only graph with negative slopeand with C Ð!ß !Ñ-intercept is .G

13. C œ $This line is a horizontal line with -interceptCÐ!ß $Ñ C. Its -coordinate is positive. The only graphthat has these characteristics is .B


14. C œ $This line is a horizontal line with -interceptCÐ!ß$Ñ C. Its -coordinate is negative. The onlygraph that has these characteristics is .E

15. ; 7 œ & , œ "&Substitute these values in the slope-intercept form.

C œ 7B ,

C œ &B "&

16. ; 7 œ # , œ "#Substitute these values in the slope-intercept form.

C œ 7B ,

C œ #B "#

17. ; 7 œ , œ# %$ &

Substitute these values in the slope-intercept form.

C œ 7B ,

C œ B # %$ &

18. ; 7 œ , œ & ") $

The equation is

C œ B & ") $ .

19. Slope ; #& C Ð!ß &Ñ-intercept

Here, and . 7 œ , œ &#& Substitute these values in

the slope-intercept form.

C œ 7B ,

C œ B &#&

20. Slope ; $% C Ð!ß (Ñ-intercept

7 œ , œ ($% ;

The equation is

C œ B ($% .

21. To get to the point from the Ð$ß $Ñ C-interceptÐ!ß "Ñ # $, we must go up units and to the right units, so the slope is slope-intercept form is#

$ . The

C œ B "#$ .

22. To get to the point from the Ð$ß "Ñ C-interceptÐ!ß#Ñ $ $, we must go up units and to the left units, so the slope is slope-intercept$

$ œ ". The form is

C œ "B #.

23. B C œ %(a) Solve for to get the equation in slope-Cintercept form.

B C œ %

C œ B %

(b) The slope is the coefficient of , .B "

The -intercept is the point , or .(c) C Ð!ß ,Ñ Ð!ß %Ñ

(d)

24. B C œ '(a) Solve for to get the equation in slope-Cintercept form.

B C œ '

C œ B '

(b) The slope is the coefficient of , .B "

The -intercept is the point , or .(c) C Ð!ß ,Ñ Ð!ß 'Ñ

(d)

25. 'B &C œ $!(a) Solve for to get the equation in slope-Cintercept form.

'B &C œ $!

&C œ 'B $!

C œ B ''&

(b) The slope is the coefficient of , B Þ'&

The -intercept is the point , or .(c) C Ð!ß ,Ñ Ð!ß 'Ñ

(d)

26. $B %C œ "#(a) Solve for to get the equation in slope-Cintercept form.

$B %C œ "#

%C œ $B "#

C œ B $$%

(b) The slope is the coefficient of , B Þ$%

The -intercept is the point , or .(c) C Ð!ß ,Ñ Ð!ß $Ñ

(d)


27. %B &C œ #!(a) Solve for to get the equation in slope-Cintercept form.

%B &C œ #!

&C œ %B #!

C œ B %%&

(b) The slope is the coefficient of , B %& Þ

The -intercept is the point , or .(c) C Ð!ß ,Ñ Ð!ß%Ñ

(d)

28. (B $C œ $(a) Solve for to get the equation in slope-Cintercept form.

(B $C œ $

$C œ (B $

C œ B "($

(b) The slope is the coefficient of , B ($ Þ

The -intercept is the point , or .(c) C Ð!ß ,Ñ Ð!ß"Ñ

(d)

29. B #C œ %(a) Solve for to get the equation in slope-Cintercept form.

B #C œ %

#C œ B %

C œ B #"#

(b) The slope is the coefficient of , B Þ"#

The -intercept is the point , or .(c) C Ð!ß ,Ñ Ð!ß#Ñ

(d)

30. B $C œ *(a) Solve for to get the equation in slope-Cintercept form.

B $C œ *

$C œ B *

C œ B $"$

(b) The slope is the coefficient of , B Þ"$

The -intercept is the point , or .(c) C Ð!ß ,Ñ Ð!ß$Ñ

(d)

31. %B $C œ "#(a) Solve for to get the equation in slope-Cintercept form.

%B $C œ "#

$C œ %B "#

C œ B %%$

(b) The slope is the coefficient of , B %$ Þ

The -intercept is the point , or .(c) C Ð!ß ,Ñ Ð!ß %Ñ

(d)

32. Solve the equation for .C

EB FC œ G

FC œ EB G

C œ B E G

F F

The slope is the coefficient of , B E

F.

33. Through ; slope (a) Ð&ß )Ñ #Use the point-slope form with a bB ß C œ Ð&ß )Ñ" "

and 7 œ #.

C C œ 7ÐB B Ñ

C ) œ # B &

C ) œ #B "!

#B C œ ")

" "a b

(b) Solve the last equation from part (a) for .C

#B C œ ")

C œ #B ")


34. Through ; slope (a) Ð"#ß "!Ñ "

C "! œ " B "#

C "! œ B "#

B C œ #

B C œ #

a b


B C œ #

C œ B #

C œ B #

35. Through ; slope (a) Ð#ß %Ñ $%

Use the point-slope form with a bB ß C œ Ð#ß %Ñ" "

and 7 œ $% .

C C œ 7ÐB B Ñ

C % œ B Ð#Ñ

% C % œ $ÐB #Ñ

%C "' œ $B '

$B %C œ "!

" "$% c da b


$B %C œ "!

%C œ $B "!

C œ B

C œ B

$ "!% %$ &% #

36. Through ; slope (a) Ð"ß 'Ñ &'

Use the point-slope form with a bB ß C œ Ð"ß 'Ñ" "

and 7 œ &' .

C C œ 7ÐB B Ñ

C œ B Ð"Ñ

' C œ &ÐB "Ñ

'C $ œ &B &

&B 'C œ $"

" "

'

'

'

&' c da b


&B 'C œ $"

'C œ &B $"

C œ B & $"' '

37. Through ; slope (a) Ð&ß %Ñ "#

Use the point-slope form with a bB ß C œ Ð&ß %Ñ" "

and 7 œ "# .

C C œ 7ÐB B Ñ

C œ B Ð&Ñ

# C œ "ÐB &Ñ

#C ) œ B &

B #C œ "$

B #C œ "$

" "

%

%

"# c da b


B #C œ "$

#C œ B "$

C œ B " "$# #

38. Through ; slope (a) Ð(ß#Ñ "%

C Ð#Ñ œ B (

%ÐC #Ñ œ "ÐB (Ñ

%C ) œ B (

B %C œ "&

B %C œ "&

"% a b


B %C œ "&

%C œ B "&

C œ B " "&% %

39. Through ; slope (a) Ð$ß !Ñ %Use the point-slope form with a bB ß C œ Ð$ß !Ñ" "

and 7 œ %.

C C œ 7ÐB B Ñ

C ! œ %ÐB $Ñ

C œ %B "#

%B C œ "#

%B C œ "#

" "


%B C œ "#

C œ %B "#

C œ %B "#

40. (a) -intercept ; slope B Ð#ß !Ñ &

C ! œ & B Ð#Ñ

C œ &ÐB #Ñ

C œ &B "!

&B C œ "!

c d


&B C œ "!

C œ &B "!

41. Through ; slope (a) Ð#ß 'Þ)Ñ "Þ%Use the point-slope form with a bB ß C œ Ð#ß 'Þ)Ñ" "

and 7 œ "Þ%.

C C œ 7ÐB B Ñ

C 'Þ) œ "Þ%ÐB #Ñ

C 'Þ) œ ÐB #Ñ

&ÐC 'Þ)Ñ œ (ÐB #Ñ

&C $% œ (B "%

(B &C œ #!

(B &C œ #!

" "

(&



(B &C œ #!

&C œ (B #!

C œ B % C œ "Þ%B %(& or

42. Through ; slope (a) Ð'ß"Þ#Ñ !Þ)

C C œ 7ÐB B Ñ

C Ð"Þ#Ñ œ !Þ)ÐB 'Ñ

C "Þ# œ ÐB 'Ñ

&ÐC "Þ#Ñ œ %ÐB 'Ñ

&C ' œ %B #%

%B &C œ $!

%B &C œ $!

" "

%&


%B &C œ $!

&C œ %B $!

C œ B ' C œ !Þ)B '%& or

43. Through ; slope Ð*ß &Ñ !A line with slope is a ! horizontal line. Ahorizontal line through the point hasÐBß 5Ñequation . Here , so an equation isC œ 5 5 œ &C œ &.

44. Through ; slope Ð%ß#Ñ !An equation of this line is .C œ #

45. Through ; undefined slopeÐ*ß "!ÑA vertical line has undefined slope and equationB œ - B Ð*ß "!Ñ *. Since the -value in is , theequation is .B œ *

46. Through ; undefined slopeÐ#ß )ÑA line with undefined slope is a vertical line in theform . The equation of this line is .B œ - B œ #

47. Through ; verticalÐ!Þ&ß !Þ#ÑA vertical line through the point hasÐ5ß CÑequation . Here , so the equation isB œ 5 5 œ !Þ&B œ !Þ&.

48. Through ; ˆ ‰& #) *ß vertical

The equation of this line is B œ &) .

49. Through ; Ð(ß )Ñ horizontalA horizontal line through the point hasÐBß 5Ñequation , so the equation is .C œ 5 C œ )

50. Through ; Ð#ß (Ñ horizontalThe equation of this line is .C œ (

51. (a) and a b a b$ß % &ß )

Find the slope.

7 œ œ œ #) % %

& $ #

Use the point-slope form with a bB ß C œ Ð$ß %Ñ" "

and .7 œ #

C C œ 7ÐB B Ñ

C % œ #ÐB $Ñ

C % œ #B '

#B C œ #

#B C œ #

" "


#B C œ #

C œ #B #

C œ #B #

Note: You could use any of the equations in part(a) to solve for . In this exercise, choosingC

C % œ #B ' or ,#B C œ #

easily leads to the equation .C œ #B #

52. (a) and a b a b&ß# $ß "%

7 œ œ œ #"% Ð#Ñ "'

$ & )

Use the point-slope form with .a bB ß C œ Ð&ß#Ñ" "

C C œ 7ÐB B Ñ

C Ð#Ñ œ #ÐB &Ñ

C # œ #B "!

#B C œ )

" "


#B C œ )

C œ #B )

53. (a) and a b a b'ß " #ß &

Find the slope.

7 œ œ œ & " % "

# ' ) #

Use the point-slope form with a bB ß C œ Ð'ß "Ñ" "

and 7 œ "# .

C C œ 7ÐB B Ñ

C " œ ÐB 'Ñ

# C " œ "ÐB 'Ñ

#C # œ B '

B #C œ )

" ""#a b


B #C œ )

#C œ B )

C œ B %"#


54. (a) and a b a b#ß & )ß "

7 œ œ œ" & % #

) Ð#Ñ ' $

Let a bB ß C œ Ð#ß &Ñ" " .

C & œ B Ð#Ñ

$ C & œ #ÐB #Ñ

$C "& œ #B %

#B $C œ "*

#B $C œ "*

#$ c da b


#B $C œ "*

$C œ #B "*

C œ B # "*$ $

55. (a) and ˆ ‰ ˆ ‰ ß ß# # % #& & $ $

Find the slope.

7 œ

œ

# # "!'$ & "&

% #$ &

#!'"&

%"&#'"&

ˆ ‰ œ

œ œ% #

#' "$

Use the point-slope form witha b ˆ ‰B ß C œ ß 7 œ" "# # #& & "$ and .

C œ B

"$ C œ # B

"$C œ #B

#B "$C œ

#B "$C œ '

# # #& "$ &

# #& &#' %& &

$!&

‘ˆ ‰ˆ ‰ ˆ ‰


#B "$C œ '

"$C œ #B '

C œ B # '"$ "$

56. (a) and ˆ ‰ ˆ ‰$ ) # #% $ & $ß ß

7 œ

œ

# ) #)$ $ $# $ )"&& % #!

'$(#!

œ

œ Ð#Ñ œ#! %!

( (Œ

Let a b ˆ ‰B ß C œ ß" "$ )% $ .

C œ B

( C œ %! B

(C œ %!B $!

%!B (C œ

"#!B #"C œ $%

) %! $$ ( %

) $$ %&'$

$%$

ˆ ‰ˆ ‰ ˆ ‰


"#!B #"C œ $%

#"C œ "#!B $%

C œ B C œ B "#! $% %! $%#" #" ( #" or

57. (a) and a b a b#ß & "ß &Find the slope.

7 œ œ œ !& & !

" # "

A line with slope is ! horizontal. A horizontal linethrough the point has equation , so theÐBß 5Ñ C œ 5equation is .C œ &

is already in the slope-intercept form.(b) C œ &

58. (a) and a b a b#ß # %ß #

7 œ œ œ !# # !

% Ð#Ñ '

A line with slope is ! horizontal. A horizontal linethrough the point has equation , so theÐBß 5Ñ C œ 5equation is .C œ #

is already in the slope-intercept form.(b) C œ #

59. (a) and a b a b(ß ' (ß)Find the slope.

7 œ œ) ' "%

( ( ! Undefined

A line with undefined slope is a vertical line. Theequation of a vertical line is , where is theB œ 5 5common -value. So the equation is .B B œ (

It is not possible to write in slope-(b) B œ (intercept form.

60. (a) and a b a b"$ß & "$ß"

7 œ œ" & '

"$ "$ ! Undefined

A line with undefined slope is a vertical line. Theequation of a vertical line is , where is theB œ 5 5common -value. So the equation is .B B œ "$

It is not possible to write in slope-(b) B œ "$intercept form.

61. (a) and ˆ ‰ ˆ ‰" ## $ß $ ß$

Find the slope.

7 œ œ œ !$ Ð$Ñ !

# "$ #

('

A line with slope is ! horizontal. A horizontal linethrough the point has equation , so theÐBß 5Ñ C œ 5equation is .C œ $

is already in the slope-intercept form.(b) C œ $


62. (a) and ˆ ‰ ˆ ‰ ß' ß'% "#* (

7 œ œ œ !' Ð'Ñ !

"# %( *

"$''$

ˆ ‰ A line with slope is ! horizontal. A horizontal line

through the point has equation , so theÐBß 5Ñ C œ 5equation is .C œ '

is already in the slope-intercept form.(b) C œ '

63. Through ; parallel to the graph of the(a) Ð(ß #Ñline having equation $B C œ )Find the slope of .$B C œ )

C œ $B )

C œ $B )

The slope is , so a line parallel to it also has slope$$ 7 œ $. Use and a bB ß C œ Ð(ß #Ñ" " in the point-slope form.

C C œ 7ÐB B Ñ

C # œ $ÐB (Ñ

C # œ $B #"

C œ $B "*

" "

(b) C œ $B "*

$B C œ "*

$B C œ "*

64. Through ; parallel to (a) Ð%ß "Ñ #B &C œ "!Find the slope of .#B &C œ "!

&C œ #B "!

C œ B ##&

The slope is . We are to find the #& equation of a

line parallel to this line so its slope is also #& .

Using the point-slope form, we have

C " œ ÐB %Ñ

C " œ B

C œ B

#&# )& &# "$& &

(b) C œ B

&C œ #B "$

#B &C œ "$

# "$& &

Multiply by 5.

65. Through ; parallel to (a) Ð#ß#Ñ B #C œ "!Find the slope of .B #C œ "!

#C œ B "!

C œ B &"#

The slope is , so a "# line parallel to it also has slope

" "# #. Use and 7 œ a bB ß C œ Ð#ß#Ñ" " in thepoint-slope form.

C C œ 7ÐB B Ñ

C Ð#Ñ œ B Ð#Ñ

C # œ B #

C # œ B "

C œ B "

" ""#"#"#"#

c da b

(b) C œ B "

#C œ B #

B #C œ #

B #C œ #

"#

Multiply by 2.

66. Through ; parallel to (a) Ð"ß $Ñ B $C œ "#

$C œ B "#

C œ B %"$

The slope of the required line is the same as theslope of this line: ."

$

C $ œ B Ð"Ñ

C $ œ B "

C $ œ B

C œ B

"$"$" "$ $" "!$ $

c da b

(b) C œ B

$C œ B "!

B $C œ "!

B $C œ "!

" "!$ $

Multiply by 3.

67. Through ; perpendicular to (a) Ð)ß &Ñ #B C œ (Find the slope of .#B C œ (

C œ #B (

C œ #B (

The slope of the line is . Therefore, the slope of#the line perpendicular to it is since"

#

# œ " 7 œ ˆ ‰" "# #. Use and a bB ß C œ Ð)ß &Ñ" "

in the point-slope form.

C C œ 7ÐB B Ñ

C & œ ÐB )Ñ

C & œ B %

C œ B *

" ""#"#"#

(b) C œ B *

#C œ B ")

B #C œ ")

"#

Multiply by 2.

68. Through ; perpendicular to(a) Ð#ß(Ñ&B #C œ ")

#C œ &B ")

C œ B *

7 œ

&#

"&#

We wish to find such that , or7 7 7 œ "# " #

7 œ"

7#

".

continued


7 œ"

# &

#

œ Ð"Ñ œ# #

& &Œ

Use the point-slope form.

C Ð(Ñ œ ÐB #Ñ

C ( œ B

C œ B

#&# %& &# $*& &

(b) C œ B

&C œ #B $*

#B &C œ $*

#B &C œ $*

# $*& &

Multiply by 5.

69. Through ; perpendicular to (a) Ð#ß (Ñ B œ *B œ * is a vertical line so a line perpendicular to itwill be a horizontal line. It goes through Ð#ß (Ñso its equation is

C œ (Þ

(b) is already in standard form.C œ (

70. Through ; perpendicular to (a) Ð)ß %Ñ B œ $B œ $ is a vertical line so a line perpendicular toit will be a horizontal line. It goes through ,Ð)ß %Ñso its equation is

C œ %.

(b) is already in standard form.C œ %

71. Distance (rate)(time), soœ

C œ %&BÞ

B C œ %&B! %&Ð!Ñ œ !& %&Ð&Ñ œ ##& Ð&ß ##&Ñ"! %&Ð"!Ñ œ %&! Ð"!ß %&!Ñ

Ordered PairÐ!ß !Ñ

72. Total cost (cost/disc)(number of discs), soœ

C œ "'BÞ

B C œ "'B! "'Ð!Ñ œ !& "'Ð&Ñ œ )! Ð&ß )!Ñ"! "'Ð"!Ñ œ "'! Ð"!ß "'!Ñ


73. Total cost (cost/gal)(number of gallons), soœ

C œ $Þ!"BÞ

B C œ $Þ!"B! $Þ!"Ð!Ñ œ !& $Þ!"Ð&Ñ œ "&Þ!& Ð&ß "&Þ!&Ñ"! $Þ!"Ð"!Ñ œ $!Þ"! Ð"!ß $!Þ"!Ñ


74. Total cost (cost/day)(number of days), soœ

C œ $Þ&!BÞ

B C œ $Þ&!B! $Þ&!Ð!Ñ œ !& $Þ&!Ð&Ñ œ "(Þ&! Ð&ß "(Þ&!Ñ"! $Þ&!Ð"!Ñ œ $&Þ!! Ð"!ß $&Þ!!Ñ


75. (a) The fixed cost is $ , so that is the value of .** ,The variable cost is $ , so$*

C œ 7B , œ $*B **Þ

If , The ordered(b) B œ & C œ $*Ð&Ñ ** œ #*%Þpair is . The cost of a 5-monthÐ&ß #*%Ñmembership is $ .#*%

If , The cost(c) B œ "# C œ $*Ð"#Ñ ** œ &'(Þof the first year's membership is $&'(Þ

76. (a) The fixed cost is $ , so that is the value"&*of . The variable cost is $ , so, '!

C œ 7B , œ '!B "&*Þ

If , The(b) B œ & C œ '!Ð&Ñ "&* œ %&*Þordered pair is The cost of a familyÐ&ß %&*ÑÞmembership for a family of six (with fiveadditional members) is $ .%&*

For a four-person family, , so(c) B œ $C œ '!Ð$Ñ "&* œ $$*. The cost of a familymembership for a four-person family is $ .$$*

77. (a) The fixed cost is $ $ $ , so"*Þ*& #& œ %%Þ*&that is the value of . The variable cost is $ , so, $&

C œ 7B , œ $&B %%Þ*&Þ

If , The(b) B œ & C œ $&Ð&Ñ %%Þ*& œ #"*Þ*&Þordered pair is . The cost of the plan forÐ&ß #"*Þ*&Ñ& #"*Þ*& months is $ .

For a 1-year contract, , so(c) B œ "#C œ $&Ð"#Ñ %%Þ*& œ %'%Þ*&. The cost of theplan for year is $ ." %'%Þ*&

78. (a) The fixed cost is $ , so that is the value of .#& ,The variable cost is $ , so&!

C œ 7B , œ &!B #&Þ

If , The ordered(b) B œ & C œ &!Ð&Ñ #& œ #(&Þpair is . The cost of the plan for monthsÐ&ß #(&Ñ &is $#(&Þ

For a 2-year contract, , so(c) B œ #%C œ &!Ð#%Ñ #& œ "##&Þ The cost of the plan for# "##&Þ years is $

79. (a) The fixed cost is $ , so that is the value of .$! ,The variable cost is $ , so'

C œ 7B , œ 'B $!Þ


If , The ordered(b) B œ & C œ 'Ð&Ñ $! œ '!Þpair is . It costs $ to rent the saw for Ð&ß '!Ñ '! &days.

(c) "$) œ 'B $!

"!) œ 'B

B œ œ ")"!)

'

Let y 138.œ

The saw is rented for days.")

80. (a) The fixed cost is $ , so that is the value of .&! ,The variable cost is $ , so!Þ#!

C œ 7B , œ !Þ#!B &!Þ

If , . The ordered(b) B œ & C œ !Þ#!Ð&Ñ &! œ &"pair is . The charge for driving miles isÐ&ß &"Ñ &$&"Þ

(c) )%Þ'! œ !Þ#!B &!

$%Þ'! œ !Þ#!B

B œ œ "($$%Þ'!

!Þ#!

Let y 84.60.œ

The car was driven miles."($

81. (a) Use and .Ð!ß *"Ñ Ð&ß '$Ñ

7 œ œ œ &Þ''$ *" #)

& ! &

The equation is . The slope tellsC œ &Þ'B *"us that the percent of households accessing theInternet by dial-up is decreasing % per year.&Þ'

(b) The year 2006 corresponds to , soB œ 'C œ &Þ'Ð'Ñ *" œ &(Þ% ¸ &(%.

82. (a) Use and .Ð!ß *Ñ Ð&ß $(Ñ

7 œ œ œ &Þ'$( * #)

& ! &

The equation is . The slope tells usC œ &Þ'B *that the percent of households accessing theInternet by broadband is increasing % per year.&Þ'

(b) The year 2006 corresponds to , soB œ 'C œ &Þ'Ð'Ñ * œ %#Þ' ¸ %$%.

83. (a) Use , and , .Ð&ß ## $*$Ñ Ð"$ß #* '%&Ñ

7 œ œ œ *!'Þ&"$ & )

(#&##* '%& ## $*$, ,

Now use the point-slope form.

C

C

C œ B "( )'!Þ&

## $*$ œ *!'Þ&ÐB &Ñ

## $*$ œ *!'Þ&B %&$#Þ&

*!'Þ&

,,

,

(b) The year 1999 corresponds to , soB œ *C œ *!'Þ&Ð*Ñ "( )'!Þ& œ #' !"*, , . This value isslightly lower than the actual value of $ , .#( *"!

84. (a) Use , and , .Ð&ß #) $*#Ñ Ð"!ß #( )('Ñ

7 œ œ œ "!$Þ##( )(' #) $*# &"'

"! & &

, ,

Now use the point-slope form.

C #( )(' œ "!$Þ#ÐB "!Ñ

C #( )(' œ "!$Þ#B "!$#

C œ "!$Þ#B #) *!)

,,

,

(b) The year 1998 corresponds to , soB œ )C œ "!$Þ#Ð)Ñ #) *!) œ #) !)#Þ% ¸ #) !)#Þ, , ,This value is slightly higher than the actual valueof , .#( *&#

85. (a) #B ( B œ %B #

B ( œ %B #

$B * œ !

$B * œ C

(b) From the screen, we see that is theB œ $solution.

(c) #B ( B œ %B #

B ( œ %B #

* œ $B

$ œ BThe solution set is .e f$

86. (a) (B #B % & œ $B "

&B " œ $B "

#B # œ !

#B # œ C

(b) From the screen, we see that is theB œ "solution.

(c) (B #B % & œ $B "

&B " œ $B "

#B œ #

B œ "The solution set is .e f"

87. (a) $Ð#B "Ñ #ÐB #Ñ

'B $ #B % & œ !

%B # œ !

%B # œ C

œ &

(b) From the screen, we see that is theB œ !Þ&solution.

(c) $Ð#B "Ñ #ÐB #Ñ

'B $ #B % œ &

%B ( œ &

%B œ #

B œ !Þ&

œ &

"# or

The solution set is .e f!Þ&


88. (a) %B $Ð% #BÑ

%B "# 'B œ #B ' 'B #

"!B "# œ )B %

#B ) œ !

#B ) œ C

œ #ÐB $Ñ 'B #

(b) From the screen, we see that is theB œ %solution.

(c) %B $Ð% #BÑ

%B "# 'B œ #B ' 'B #

"!B "# œ )B %

#B œ )

B œ %

œ #ÐB $Ñ 'B #

The solution set is .e f%89. The solution to the equation is theC œ !"

B-coordinate of the -intercept. In this case, theBB "!-intercept is greater than , so the correct choicemust be .D

90. (a) #ÐB &Ñ œ B #

#B "! œ B #

"# œ BThe solution set is .e f"#

(b) Because the solution is , it will not appear"#as the -intercept in the standard window. ManyBchoices are possible, but every choice mustinclude . One such choice is from to"# B œ &B œ "&.

91. When C F Cœ ! œ "!!°, °,œ $#° , and when F œ #"#° . These are the freezing and boilingtemperatures for water.

92. The two points of the form would beÐ ÑC FßÐ!ß $#Ñ Ð"!!ß #"#Ñ and .

93. 7 œ œ œ#"# $# ")! *

"!! ! "!! &

94. Let and 7 œ *& a bB ß C œ Ð!ß $#Ñ" " .

C C œ 7ÐB B Ñ

$# œ Ð !Ñ

$# œ

œ $#

" "*&*&*&

F CF C

F C

95. F CF C

F C

œ $#

$# œ

Ð $#Ñ œ

*&*&

&*

96. A temperature of &!° corresponds to aCtemperature of ° ."## F

97. #B & *

#B %

B #

The solution set is .Ð_ß #Ñ

98. B % $

B "

B "

The solution set is .Ð_ß "Ñ

99. & $B *

$B %

B Ÿ %$

The solution set is .ˆ ‘_ß%$

100. B Ÿ !

B !

The solution set is .Ò!ß_Ñ

Summary Exercises on Slopes andEquations of Lines

1. For , slope$B &C œ * œ œ E $

F &.

2. For , slope%B (C œ $ œ œ E %

F (.

3. For , slope#B C œ & œ œ œ #E #

F ".

4. For , slope&B #C œ % œ œ œE & &

F # #.

5. For ,!Þ#B !Þ)C œ !

slope œ œ œ !Þ#&E !Þ#

F !Þ).

6. For , slope" E

# F œ œ #

B C œ ! œ

"

%

"#"%

.

7. Through and Ð#ß 'Ñ Ð%ß "Ñ

(a) The slope is

7 œ œ œ " ' & &

% Ð#Ñ ' '.

Use the point-slope form.

C C œ 7ÐB B Ñ

C ' œ ÒB Ð#ÑÓ

C ' œ B

C œ B

C œ B

" "&'& &' $& & ")' $ $& "$' $

(b) C œ B

'C œ &B #'

&B 'C œ #'

& "$' $

Multiply by 6.

8. Through ; parallel to Ð#ß &Ñ $B C œ %

Find the slope of .(a) $B C œ %

C œ $B %

C œ $B %

Summary Exercises on Slopes and Equations of Lines 125

The slope is , so a line parallel to it also has slope$$ 7 œ $. Use and a bB ß C œ Ð#ß &Ñ" " in the point-slope form.

C C œ 7ÐB B Ñ

C & œ $ÒB Ð#ÑÓ

C & œ $ÐB #Ñ

C & œ $B '

C œ $B ""

" "

(b) C œ $B ""

$B C œ ""

$B C œ ""

9. Through ; perpendicular to Ð!ß !Ñ #B &C œ '

Find the slope of .(a) #B &C œ '

&C œ #B '

C œ B # '& &

The slope of the line is . Therefore, the slope of#&

the line perpendicular to it is since&#

# & && # #ˆ ‰ œ " 7 œ . Use and a bB ß C œ Ð!ß !Ñ" "

in the point-slope form.

C C œ 7ÐB B Ñ

C ! œ ÐB !Ñ

C œ B

" "&#&#

(b) C œ B

#C œ &B

&B #C œ !

&#

10. Through ; parallel to Ð&ß)Ñ C œ %

(a) is a horizontal line, so a line parallel toC œ %it must be horizontal and have an equation of theform . It goes through , so itsC œ 5 Ð&ß)Ñequation is

C œ ).

is already in standard form.(b) C œ )

11. Through ; perpendicular to ˆ ‰$ ( #% * $ß B œ

(a) B œ #$ is a vertical line so a line perpendicular

to it will be a horizontal line. It goes throughˆ ‰$ (% *ß so its equation is

C œ Þ(*

(b) C œ

*C œ (

(*

Multiply by 9.

12. The perpendicular bisector of the segment withendpoints and Ð%ß 'Ñ Ð"#ß %Ñ

(a) The midpoint of the endpoints is

Œ Œ % "# ' % ) "!

# # # #ß œ ß œ Ð%ß &Ñ.

The slope of the line through the endpoints is

7 œ œ œ œ % ' # # "

"# Ð%Ñ "# % "' ).

The slope of the perpendicular bisector is since) Ð)Ñ œ ""

) . Use the point-slope form withÐB ß C Ñ œ Ð%ß &Ñ 7 œ )" " and .

C C œ 7ÐB B Ñ

C & œ )ÐB %Ñ

C & œ )B $#

C œ )B #(

" "

(b) C œ )B #(

)B C œ #(

)B C œ #(

13. Through ; parallel to the line through Ð%ß #Ñ Ð$ß *Ñand Ð'ß ""Ñ

(a) The slope of the line through and Ð$ß *Ñ Ð'ß ""Ñis

7 œ œ"" * #

' $ $.

Use the point-slope form with ÐB ß C Ñ œ Ð%ß #Ñ" "

and (since the slope of the desired line7 œ #$

must equal the slope of the given line).

C C œ 7ÐB B Ñ

C # œ ÒB Ð%ÑÓ

C # œ ÐB %Ñ

C # œ B

C œ B

C œ B

" "#$#$# )$ $# ) '$ $ $# "%$ $

(b) C œ B

$C œ #B "%

#B $C œ "%

#B $C œ "%

# "%$ $

14. Through ; perpendicular to the line throughÐ%ß#ÑÐ$ß (Ñ Ð&ß 'Ñ and

and (a) The slope of the line through Ð$ß (Ñ Ð&ß 'Ñis

7 œ' ( " "

& $ # #œ œ .

The slope of a line perpendicular to the given lineis (the negative reciprocal of ). Use the point-# "

#

slope form with and .ÐB ß C Ñ œ Ð%ß#Ñ 7 œ #" "

C C œ 7ÐB B Ñ

C Ð#Ñ œ #ÐB %Ñ

C # œ #B )

C œ #B "!

" "

(b) C œ #B "!

#B C œ "!

#B C œ "!


15. Through and the midpoint of the segmentÐ%ß "#Ñwith endpoints and Ð&ß )Ñ Ð$ß #Ñ

(a) The midpoint of the segment with endpointsÐ&ß )Ñ Ð$ß #Ñ and is

Œ Œ a b& Ð$Ñ ) # # "!

# # # #ß œ ß œ "ß & .

The slope of the line through and Ð%ß "#Ñ Ð"ß &Ñis

7 œ œ œ & "# ( (

" Ð%Ñ & &.

Use the point-slope form with ÐB ß C Ñ œ Ð%ß "#Ñ" "

and .7 œ (&

C C œ 7ÐB B Ñ

C "# œ ÒB Ð%ÑÓ

C "# œ ÐB %Ñ

C "# œ B

C œ B

C œ B

" "

(&(&( #)& &( #) '!& & &( $#& &

(b) C œ B

&C œ (B $#

(B &C œ $#

( $#& &

16. Through and ˆ ‰ ˆ ‰" " # &$ # $ #ß ß

(a) 7 œ œ œ œ #

#

"

& " %# # ## " $$ $ $

Use the point-slope form with ÐB ß C Ñ œ ß" "" "$ #

ˆ ‰and .7 œ #

C C œ 7ÐB B Ñ

C œ # B

C œ #B

C œ #B

C œ #B

" "" "# $" ## $

% $' '('

ˆ ‰

(b) C œ #B

'C œ "#B (

"#B 'C œ (

('

17. Through ; parallel to ˆ ‰$ " " (( ' & %ß C œ B

(a) The slope of the desired line is the same as theslope of the given line, so use the point-slope formwith and .ÐB ß C Ñ œ 7 œ" "

"&

ˆ ‰$ "( 'ß

C C œ 7ÐB B Ñ

C œ B

C œ B

C œ B

C œ B

" "" " $' & (" " $' & $&

" ") $&& #"! #"!" "(& #"!

ˆ ‰

(b) C œ B

#"!C œ %#B "(

%#B #"!C œ "(

%#B #"!C œ "(

" "(& #"!

Multiply by 210.

18. Through ; perpendicular to ˆ ‰% & % $( ' $ )ß C œ B

(a) The slope of a line perpendicular to the givenline is (the negative reciprocal of ). Use the$ %

% $

point-slope form with andÐB ß C Ñ œ" " ˆ ‰% &( 'ß

7 œ $% .

C C œ 7ÐB B Ñ

C œ B

C œ B

C œ B

C œ B

" "

& $ %' % (& $ $' % (

$ ") $&% %# %#$ &$% %#

ˆ ‰

(b) C œ B

)%C œ '$B "!'

'$B )%C œ "!'

$ &$% %#

Multiply by 84.

19. Through and Ð!Þ$ß "Þ&Ñ Ð!Þ%ß "Þ(Ñ

(a) 7 œ œ œ #"Þ( "Þ& !Þ#

!Þ% !Þ$ !Þ"

Use the point-slope form withÐB ß C Ñ œ Ð!Þ$ß "Þ&Ñ 7 œ #" " and .

C C œ 7ÐB B Ñ

C "Þ& œ #ÐB !Þ$Ñ

C "Þ& œ #B !Þ'

C œ #B !Þ*

" "

(b) C œ #B !Þ*

"!C œ #!B *

#!B "!C œ *

#!B "!C œ *

Multiply by 10.

20. Through ; parallel to a b#Þ&ß "Þ(& C œ !Þ&B $Þ#&

(a) The slope of the desired line is the same as theslope of the given line, so use the point-slope formwith and .ÐB ß C Ñ œ 7 œ !Þ&" " a b#Þ&ß "Þ(&

C C œ 7ÐB B Ñ

C "Þ(& œ !Þ& B #Þ&

C "Þ(& œ !Þ&B "Þ#&

C œ !Þ&B !Þ&

" "a b

(b) C œ !Þ&B !Þ&

#C œ B "

B #C œ "

B #C œ "

Multiply by 2.

21. Slope , !Þ& , œ #

The slope-intercept form of a line, ,C œ 7B ,becomes ,C œ !Þ&B # or ,C œ B #"

#

which is choice .B

3.4 Linear Inequalities in Two Variables 127

22. B Ð%ß !Ñ C Ð!ß #Ñ-intercept , -intercept

7 œ œ œ # ! # "

! % % #

Using and a -intercept of , we get7 œ C Ð!ß #Ñ"#

C œ B #"# . Changing this equation to the

standard form gives us , or#C œ B %B #C œ %, which is choice .D

23. Passes through and Ð%ß#Ñ Ð!ß !Ñ

7 œ! Ð#Ñ # "

! % % #œ œ

Using and a -intercept of , we get7 œ C Ð!ß !Ñ"#

C œ B !"# , which is choice .A

24. , passes through 7 œ Ð#ß#Ñ"#

Use the point-slope form withÐB ß C Ñ œ Ð" " #ß#Ñ 7 œ and ."

#

C C œ 7ÐB B Ñ

C Ð#Ñ œ ÒB Ð#ÑÓ

C # œ ÐB #Ñ

#ÐC #Ñ œ B #

#C % œ B #

# œ B #C

" ""#"#

This is choice .C

25. , passes through the origin7 œ "#

Use the point-slope form with ÐB ß C Ñ œ Ð!ß !" " Ñand .7 œ "

#

C C œ 7ÐB B Ñ

C ! œ ÐB !Ñ

C œ B #C œ B

" ""#"# or

This is choice .E

3.4 Linear Inequalities in Two Variables3.4 Classroom Examples

1. B C Ÿ %

Step 1Graph the line, , which has interceptsB C œ %Ð%ß !Ñ Ð!ß %Ñ and , as a solid line since the inequalityinvolves " ".Ÿ

Step 2Test .Ð!ß !Ñ

B C Ÿ %

! ! Ÿ %

! Ÿ %

?True

Step 3Since the result is true, shade the region thatcontains .Ð!ß !Ñ

2. $B %C "#

Solve the inequality for .C

%C $B "#

C B $

Subtract 3x.Divide by 4.$

%

Graph the boundary line, [which hasC œ B $$%

slope and $% C Ð!ß $Ñ-intercept ], as a dashed line

because the inequality symbol is . Since theinequality is solved for and the inequalityCsymbol is , we shade the half-plane below theboundary line.

3. and B C Ÿ % B #Graph , which has intercepts andB C œ % Ð%ß !ÑÐ!ß%Ñ, as a solid line since the inequalityinvolves " ". Test Ÿ Ð!ß !Ñ ! Ÿ %, which yields , atrue statement. Shade the region that includesÐ!ß !Ñ.

Graph as a solid vertical line throughB œ #Ð#ß !Ñ B œ #. Shade the region to the right of .

The graph of the intersection is the regioncommon to both graphs.

continued


4. or (B $C #" B #Graph as a dashed line through its(B $C œ #"intercepts and . Test Ð$ß !Ñ Ð!ß(Ñ Ð!ß !Ñ, whichyields , a true statement. Shade the region! #"that includes .Ð!ß !Ñ

Graph as a dashed vertical line throughB œ #Ð#ß !Ñ B œ #. Shade the region to the right of .

The graph of the union is the region that includesall the points in both graphs.

3.4 Section Exercises1. The boundary of the graph of will beC Ÿ B #

a solid line (since the inequality involves ),Ÿand the shading will be below the line (since theinequality sign is or ).Ÿ

2. The boundary of the graph of will beC B #a dashed line (since the inequality involves ),and the shading will be below the line (since theinequality sign is or ).Ÿ

3. The boundary of the graph of will beC B #a dashed line (since the inequality involves ),and the shading will be above the line (since theinequality sign is or ).

4. The boundary of the graph of will beC B #a solid line (since the inequality involves ), and the shading will be above the line (since theinequality sign is or ).

5. The graph of divides the plane intoEB FC œ Gtwo regions. In one of these regions, the orderedpairs satisfy ; in the other, theyEB FC Gsatisfy .EB FC G

6. One method is choosing any test point not on theboundary line and substituting the coordinates intothe inequality. If the test point satisfies theinequality, the solution set is the region where it islocated; if not, the solution set is the other region.The second method is writing the inequality in theform or . The formC 7B , C 7B ,C 7B , indicates the solution set is the regionbelow the boundary line; indicates theC 7B ,solution set is the boundary line.above

7. B C Ÿ #Graph the line by drawing a solid lineB C œ #(since the inequality involves ) through theŸintercepts and .a b a b#ß ! !ß #Test a point not on this line, such as Ð!ß !Ñ.

B C Ÿ #

! ! Ÿ #

! Ÿ #

? True

Shade that side of the line containing the test pointÐ!ß !Ñ.

8. B C Ÿ $Graph the line by drawing a solid lineB C œ $(since the inequality involves ) through theŸintercepts and .a b a b$ß ! !ß$Test a point not on this line, such as Ð!ß !Ñ.

B C Ÿ $

! ! Ÿ $

! Ÿ $

? False

Shade that side of the line containing the testnot point .Ð!ß !Ñ


9. %B C %Graph the line by drawing a dashed%B C œ %line (since the inequality involves ) through theintercepts and .a b a b"ß ! !ß% Instead of using atest point, we will solve the inequality for .C

C %B %

C %B %

Since we have " " in the last inequality, shadeC the region above the boundary line.

10. $ $B C Graph the line by drawing a dashed$ $B C œline (since the inequality involves ) through theintercepts and .a b a b"ß ! !ß$ Instead of using atest point, we will solve the inequality for .C

C B

C B

$ $

$ $

Since we have " " in the last inequality, shadeC the region above the boundary line.

11. B $C #Graph the solid line (since theB $C œ #inequality involves ) through the interceptsa b ˆ ‰#ß ! !ß and .#

$

Test a point not on this line such as Ð!ß !Ñ.

! $Ð!Ñ #

! #

? True


12. B %C $Graph the solid line (since theB %C œ $inequality involves ) through the interceptsa b ˆ ‰$ß ! !ß and .$

%

Test a point not on this line such as Ð!ß !Ñ.

! %Ð!Ñ $

! $

? True


13. B C !Graph the line , which includes theB C œ !points Ð!ß !Ñ Ð#ß#Ñ and , as a dashed line (sincethe inequality involves ). Solving the inequalityfor gives usC

C B,

So shade the region above the boundary line.

14. B #C !Graph the line , which includes theB #C œ !points Ð!ß !Ñ Ð%ß #Ñ and , as a dashed line (sincethe inequality involves ). Solving the inequalityfor gives usC

C B"# ,

So shade the region above the boundary line.


15. B $C Ÿ !Graph the solid line through the pointsB $C œ !a b a b!ß ! $ß " and .Solve the inequality for .C

$C Ÿ B

C B"$

Shade the region above the boundary line.

16. B C Ÿ !&Graph the solid line through the pointsB C œ !&a b a b!ß ! ß " and .&Solve the inequality for .C

C Ÿ B

C B

&"&

Shade the region above the boundary line.

17. C BGraph the dashed line through C œ B Ð!ß !Ñ andÐ#ß #Ñ C . Since we have " " in the inequality,shade the region below the boundary line.

18. C Ÿ %BGraph the solid line through C œ %B Ð!ß !Ñ andÐ"ß %Ñ C Ÿ. Since we have " " in the inequality,shade the region below the boundary line.

19. and B C Ÿ " B "Graph the solid line through B C œ " a b!ß " anda b"ß ! . The inequality can be written asB C Ÿ "C Ÿ B ", so shade the region below theboundary line.Graph the solid vertical line through B œ " Ð"ß !Ñand shade the region to the right. The requiredgraph is the common shaded area as well as theportions of the lines that bound it.

20. and B C # B $Graph as a solid line through B C œ # a b#ß ! anda b!ß# Ð!ß !Ñ. . Test

! ! #

! #

? False

The graph is the region that does not contain Ð!ß !Ñ B œ $. Graph as a solid vertical line throughÐ$ß !Ñ. The graph of the inequality is the region tothe right of the line. Shade the region that includesthe overlap of the two graphs.

21. and #B C # C %Graph the solid line through the#B C œ #intercepts and . to geta b a b"ß ! !ß# Ð!ß !Ñ Test ! #, a false statement. Shade that side of thegraph not containing . To graph on theÐ!ß !Ñ C %same axes, graph the dashed horizontal linethrough . Test to get , a trueÐ!ß %Ñ Ð!ß !Ñ ! %statement. Shade that side of the dashed linecontaining .Ð!ß !ÑThe word "and" indicates the intersection of thetwo graphs. The final solution set consists of theregion where the two shaded regions overlap.


22. and $B C $ C $Graph as a solid line through $B C œ $ a b"ß !and . .a b!ß$ Ð!ß !Ñ Test

$Ð!Ñ ! $

! $

? False

The graph is the region that does not containÐ!ß !Ñ.Graph as a dashed horizontal line throughC œ $Ð!ß $Ñ. The graph of the inequality is the regionbelow the dashed line.Shade the region that includes the overlap of thetwo graphs.

23. and B C & C #Graph , which has B C œ & intercepts Ð&ß !Ñand , as a dashed line. Test , whichÐ!ß&Ñ Ð!ß !Ñyields , a true statement. Shade the region! &that includes .Ð!ß !ÑGraph as a dashed horizontal line. ShadeC œ #the region below . The required graph ofC œ #the intersection is the region common to bothgraphs.

24. and 'B %C "! C #Graph the dashed line through'B %C œ "!ˆ ‰ ˆ ‰& &$ #ß ! !ß Ð!ß !Ñ and . . Test

'Ð!Ñ %Ð!Ñ "!

! "!

? True

The graph includes the region that includes Ð!ß !Ñ.Graph the dashed horizontal line throughC œ #Ð!ß #Ñ. The graph includes the region above theline. Shade the region that includes the overlap ofthe two graphs.

25. k kB $ $ B $ can be rewritten as . Theboundaries are the dashed vertical lines B œ $and . Since is between and , the graphB œ $ B $ $includes all points between the lines.

26. k kC & & C & can be rewritten as . Theboundaries are the dashed horizontal lines C œ &and . Since is between and , the graphC œ & C & &includes all points between the lines.

27. k kB " # can be rewritten as

# B " #

$ B ".

The boundaries are the dashed vertical linesB œ $ B œ " B $ " and . Since is between and ,the graph includes all points between the lines.

28. k kC $ # can be rewritten as

# #C $

C" &.

The boundaries are the dashed horizontal linesC C Cœ " œ & " & and . Since is between and , thegraph includes all points between the lines.


29. or B C " C #Graph the solid line , which crosses theB C œ "C " B " Ð!ß !Ñ-axis at and the -axis at . Use as atest point, which yields , a false statement.! "Shade the region that does not include .Ð!ß !ÑNow graph the solid line . Since theC œ #inequality is , shade above this line.C #The required graph of the union includes all theshaded regions, that is, all the points that satisfyeither inequality.

30. or B C Ÿ # C $Graph the solid line through B C œ # a b#ß ! anda b!ß # Ð!ß !Ñ. Use as a test point, which yields! Ÿ #, a true statement. Shade the region thatincludes .Ð!ß !ÑGraph the solid horizontal line throughC œ $Ð!ß $Ñ. Shade the region above the line.The required graph of the union includes all theshaded regions, that is, all the points that satisfyeither inequality.

31. or B # C B "Graph , which has B # œ C intercepts anda b#ß !a b!ß# Ð!ß !Ñ, , which yields as a dashed line. Test # !, a false statement. Shade the region thatdoes not include .Ð!ß !ÑGraph as a dashed B œ " vertical line. Shade theregion to the left of .B œ "The required graph of the union includes all theshaded regions, that is, all the points that satisfyeither inequality.

32. or B $ C B $Graph the dashed line through B $ œ C a b$ß !and . as a test point, which yieldsa b!ß $ Ð!ß !Ñ Use $ !, a false statement. Shade the region thatdoes not include Ð!ß !Ñ.Graph the dashed vertical line throughB œ $Ð$ß !Ñ. Shade the region to the right of the line.The required graph of the union includes all theshaded regions, that is, all the points that satisfyeither inequality.

33. or $B #C ' B #C #Graph , which has $B #C œ ' intercepts a b#ß !and , , whicha b!ß $ Ð!ß !Ñ as a dashed line. Test yields , a true statement. ! ' Shade the regionthat includes Ð!ß !Ñ.Graph , which has intercepts andB #C œ # #ß !a ba b!ß" Ð!ß !Ñ, , which yields as a dashed line. Test ! #, a false statement. Shade the region thatdoes not include .Ð!ß !ÑThe required graph of the union includes all theshaded regions, that is, all the points that satisfyeither inequality.

34. or B C " B C Ÿ %Graph the solid line through B C œ " a b"ß ! anda b!ß" Ð!ß !Ñ ! ". , which yields , a false Test statement. Shade the region that does not includeÐ!ß !Ñ.Graph the solid line through andB C œ % %ß !a ba b!ß % Ð!ß !Ñ ! Ÿ %. , which yields , a true Test statement. .Shade the region that includes Ð!ß !ÑThe required graph of the union includes all theshaded regions, that is, all the points that satisfyeither inequality.


35. C Ÿ $B 'The boundary line, , has slope andC œ $B ' $C '-intercept . This would be graph or graph .B CSince we want the region less than or equal to$B ', we want the region on or below theboundary line. The answer is graph .C

36. C $B 'The boundary line, , has slope andC œ $B ' $C '-intercept . This would be graph or graph .B CSince we want the region greater than or equal to$B ', we want the region on or above theboundary line. The answer is graph .B

37. C Ÿ $B 'The slope of the boundary line isC œ $B '$, and the C '-intercept is . This would begraph or graph . The A D inequality sign is , soŸwe want the region on or below the boundary line.The answer is graph .A

38. C $B 'The slope of the boundary line isC œ $B '$, and the C '-intercept is . This would begraph or graph . The A D inequality sign is , so we want the region on or above the boundary line.The answer is graph .D

39. (a) The -intercept is , so the solution setB Ð%ß !Ñfor is .C œ ! %e f

The solution set for is , since(b) C ! Ð_ß%Ñthe graph is below the -axis for these values of .B B

The solution set for is , since(c) C ! Ð%ß_Ñthe graph is above the -axis for these values of .B B

40. (a) The -intercept is , so the solution set forB Ð'ß !ÑC œ ! ' is .e f

The solution set for is , since(b) C ! Ð_ß 'Ñthe graph is below the -axis for these values of .B B

The solution set for is , since the(c) C ! Ð'ß_Ñgraph is above the -axis for these values of .B B

41. (a) The -intercept is , so the solution setB Ð$Þ&ß !Ñfor is .C œ ! $Þ&e f

The solution set for is , since(b) C ! Ð$Þ&ß_Ñthe graph is below the -axis for these values of .B B

The solution set for is , since(c) C ! Ð_ß $Þ&Ñthe graph is above the -axis for these values of .B B

42. (a) The -intercept is , so the solutionB Ð"Þ#&ß !Ñset for is .C œ ! "Þ#&e f

The solution set for is ,(b) C ! Ð"Þ#&ß_Ñsince the graph is below the -axis for theseBvalues of .B

The solution set for is ,(c) C ! Ð_ß"Þ#&Ñsince the graph is above the -axis for these valuesBof .B

43. (a) &B $ œ !

&B œ $

B œ œ !Þ'$&

The solution set is .e f!Þ'

or

(b) &B $ !

&B $

B !Þ'$&

The solution set is .a b!Þ'ß_

or

(c) &B $ !

&B $

B !Þ'$&

The solution set is .a b ß!Þ'_

The -intercept is , as in part (a). TheB Ð!Þ'ß !Ñgraph is above the -axis for , as in partB B !Þ'(b), and below the -axis for , as in partB B !Þ'(c).

44. (a) 'B'B

$ œ !

œ $

B œ œ !Þ&$'

The solution set is .e f!Þ&

or

(b) 'B'B

$ !

$

B !Þ&"#

The solution set is .a b!Þ&ß_

or

(c) 'B'B

$ !

$

B !Þ&"#

The solution set is .a b ß!Þ&_

continued


The -intercept is , as in part (a). TheB Ð!Þ&ß !Ñgraph is above the -axis for , as in partB B !Þ&(b), and below the -axis for , as in partB B !Þ&(c).

45. (a) )B Ð#B "#Ñ œ !

)B #B "# œ !

"!B "# œ !

"!B œ "#

B œ "Þ#The solution set is .e f"Þ#

(b) )B Ð#B "#Ñ !

)B #B "# !

"!B "# !

"!B "#

B Ÿ "Þ#

The solution set is Ð_ß"Þ#Ó.

(c) )B Ð#B "#Ñ !

)B #B "# !

"!B "# !

"!B "#

B "Þ#

Ÿ

Ÿ

Ÿ

Ÿ

The solution set is Ò"Þ#ß_Ñ.

The -intercept is , as in part (a). TheB Ð"Þ#ß !Ñgraph is on or above the -axis for , asB Ð_ß"Þ#Óin part (b), and on or below the -axis forBÒ"Þ#ß_Ñ, as in part (c).

46. (a) Ð#B Ñ œ !

#B œ !

B œ !

B œ

B œ

%B ")

%B ")

' ")

' ")

$The solution set is .e f$

(b) Ð#B Ñ !

#B !

B !

B

B Ÿ

%B ")

%B ")

' ")

' ")

$The solution set is Ð_ß Ó$ .

(c) Ð#B Ñ !

#B !

B !

B

B

%B ") Ÿ

%B ") Ÿ

' ") Ÿ

' Ÿ ")

$The solution set is Ò ß$ _Ñ.

The -intercept is , as in part (a). TheB Ð ß !Ñ$graph is on or above the -axis for , asB Ð_ß Ó$in part (b), and on or below the -axis forBÒ ß_Ñ$ , as in part (c).

47. "A factory can have no more than #!! workers ona shift, but must have " can beat least "!!translated as and . "MustB Ÿ #!! B "!!manufacture units" can be translatedat least $!!!as .C $!!!

48.

49. The total daily cost consists of $ per workerG &!and $ to manufacture unit, so"!! "G œ &!B "!!C.

3.5 Introduction to Functions 135

50. Some examples of points in the shaded region areÐ"&!ß %!!!Ñ Ð"#!ß $&!!Ñ Ð")!ß '!!!Ñ, , and .Some examples of points on the boundary areÐ"!!ß &!!!Ñ Ð"&!ß $!!!Ñ Ð#!!ß %!!!Ñ, , and .The corner points are andÐ"!!ß $!!!ÑÐ#!!ß $!!!Ñ.

51. ,,,

ÐBß CÑ &!B "!!C œ G

Ð"&!ß %!!!Ñ &!Ð"&!Ñ "!!Ð%!!!Ñ œ %!( &!!Ð"#!ß $&!!Ñ &!Ð"#!Ñ "!!Ð$&!!Ñ œ $&' !!!Ð")!ß '!!!Ñ &!Ð")!Ñ "!!Ð'!!!Ñ œ '!* !!!Ð"!!ß &!!!Ñ &!Ð"!!Ñ "!!Ð&!!!Ñ œ &!& !!!Ð"&!ß $!!!Ñ &!Ð"&!Ñ "!!Ð$!!!Ñ œ $!( &!!Ð#!!ß %!!!Ñ &!Ð#!!Ñ "!!Ð%!!!Ñ œ %"! !!!Ð"!!ß $!!!Ñ &!Ð"!!Ñ "!!Ð$!!

,,,

!Ñ œ $!& !!!

Ð#!!ß $!!!Ñ &!Ð#!!Ñ "!!Ð$!!!Ñ œ $"! !!!

,(least value)

,

52. The company should use workers and"!!manufacture units to achieve the least$!!!possible cost.

53. C œ (B "#

C œ (Ð$Ñ "#

C œ *

Let x = 3.

54. C œ &B %

C œ &Ð$Ñ %

C œ "*

Let x = 3.

55. C œ $B )

C œ $Ð$Ñ )

C œ "

Let x = 3.

56. C œ #B *

C œ #Ð$Ñ *

C œ $

Let x = 3.

57. $B (C œ )

(C œ $B )

C œ B $ )( (

58. #B %C œ (

%C œ #B (

C œ B " (# %

59. "#

"#

" &) %

B %C œ &

%C œ B &

C œ B

60. $%

$%$ *) #

B #C œ *

#C œ B *

C œ B

3.5 Introduction to Functions3.5 Classroom Examples

1. (a) e fÐ!ß $Ñß Ð"ß #Ñß Ð"ß $Ñ

The last two ordered pairs have the -valuesame Bpaired with -values ( is pairedtwo different C "with both and ), so this relation # $ is not afunction.

(b) ÖÐ&ß %Ñß Ð'ß %Ñß Ð(ß %Ñ×The relation because for eachis a functiondifferent -value there is exactly one -value. It isB Cacceptable to have different -values paired withBthe same -value.C

2. The domain of this relation is the set of all firstcomponents, that is, Ö"***ß #!!!ß #!!"ß #!!#ß#!!$ß #!!%×. The range of this relation is the set ofall second components, that is,Ö)' !%(ß "!* %()ß "#) $(&ß "%! ('(ß "&) (##ß, , , , ,")# "%!×, . This relation because foris a functioneach different first component, there is exactly onesecond component.

3. The arrowheads indicate that the graph extendsindefinitely left and right, as well as downward.The domain includes all real numbers, writtenÐ_ß_Ñ C %. Because there is a greatest -value, ,the range includes all numbers less than or equalto , .% Ð_ß %Ó

4. Any vertical line would intersect the graph inClassroom Example 3 at most once, so the relationis a function.

5. (a) is a function because eachC œ #B (value of corresponds to exactly one value of .B CIts domain is the set of all real numbers,Ð_ß_Ñ.

is a function because each(b) C œ &B 'Èvalue of corresponds to exactly one value of .B CSince the quantity under the radical must benonnegative, the domain is the set of real numbersthat satisfy the condition

&B ' !

&B '

B '& .

Therefore, the domain is . ‰'& ß_

(c) is not a function. If , forC œ B B œ "%

example, and or . Since C œ " C œ " C œ " C% %

must be nonnegative, the domain is the set ofnonnegative real numbers, .Ò!ß_Ñ

is not a function because if ,(d) C %B # B œ !then . Thus, the -value corresponds toC # B !many -values. Its domain is the set of all realCnumbers, .Ð_ß_Ñ


(e) C œ'

& $BGiven any value of in the domain, we find byB Cmultiplying by , adding , and then dividing the$ &result into .' This process produces exactly onevalue of for each value in the domain, so theCgiven equation defines a function. The domainincludes all real numbers except those that makethe denominator . We find those numbers by!setting the denominator equal to and solving for!B.

& $B œ !

$B œ &

B œ &$

The domain includes all real numbers except ,&$

written as .ˆ ‰ ˆ ‰_ß ß_& &$ $

6. 0ÐBÑ œ$B &

#

(a) 0Ð$Ñ œ œ œ ($Ð$Ñ & "%

# #

(b) 0Ð>Ñ œ$> &

#

7. 1ÐBÑ œ &B "

1Ð7 #Ñ œ &Ð7 #Ñ "

œ &7 "! "

œ &7 *

8. (a) When , , so .B œ # 0ÐBÑ œ % 0Ð#Ñ œ %

When , , so .(b) B œ # C œ ' 0Ð#Ñ œ '

(c) 0ÐBÑ œ B

0Ð#Ñ œ Ð#Ñ œ Ð%Ñ œ %

#

#

(d) When , , so .B œ # C œ $ 0Ð#Ñ œ $

9. B %C œ $#

Solve for .C

%C œ B $

C œ B C œ" $ B $

% % %

#

##

or

0ÐBÑ œB $

%

0Ð"Ñ œ œ œ œ Ð"Ñ $ " $ # "

% % % #

0Ð+Ñ œ+ $

%

#

#

#

Replace y with f(x).

3.5 Section Exercises1. We give one of many possible answers here. A

function is a set of ordered pairs in which eachfirst component corresponds to exactly one secondcomponent. For example, ÖÐ!ß "Ñß Ð"ß #Ñß Ð#ß $ÑßÐ$ß %Ñá× is a function.

2. Here is one of many possible answers. The domainof a function is the set of all first elements in theordered pairs. The domain of the function in theanswer to Exercise 1 is .e f!ß "ß #ß $ßá

3. In an ordered pair of a relation, the first element isthe independent variable.

4. A relation that is not a function having domainÖ$ß #ß '× Ö%ß '× and range must be a relation inwhich, for some value in the domain, there is morethan one value in the range. An example of such arelation is

e fÐ$ß %Ñß Ð#ß %Ñß Ð#ß 'Ñß Ð'ß %Ñ Þ

Notice that , from the domain, is paired with both#% ' and , from the range. (There are other correctanswers possible.)

5. e fÐ&ß "Ñß Ð$ß #Ñß Ð%ß *Ñß Ð(ß 'ÑThe relation is a function since for each B-value,there is only one -value.C

6. e fÐ)ß !Ñß Ð&ß %Ñß Ð*ß $Ñß Ð$ß )ÑThe relation is a function since for each B-value,there is only one -value.C

7. ÖÐ#ß %Ñß Ð!ß #Ñß Ð#ß &Ñ×The relation is not a function since the B #-value has two different -values associated with it, andC %&.

8. ÖÐ*ß#Ñß Ð$ß &Ñß Ð*ß #Ñ×The relation is not a function since the B *-value has two different -values associated with it, C #and .#

9. e fÐ$ß "Ñß Ð%ß "Ñß Ð#ß (ÑThe relation is a function since for each B-value,there is only one -value.C

10. e fÐ"#ß &Ñß Ð"!ß $Ñß Ð)ß $ÑThe relation is a function since for each B-value,there is only one -value.C

11. ÖÐ"ß "Ñß Ð"ß"Ñß Ð#ß %Ñß Ð#ß%Ñ×Ð!ß !ÑßThe relation is not a function since the B "-value has two different -values associated with it, andC "" B œ #Þ. (A similar statement can be made for )

The domain is the set of -values: .B !ß "ß #e fThe range is the set of -values: .C %ß"ß !ß "ß %e f

12. e fÐ#ß &Ñß Ð$ß (Ñß Ð%ß *Ñß Ð&ß ""ÑThe relation is a function since for each B-value,there is only one -value.C

The domain is the set of -values: .B #ß $ß %ß &e fThe range is the set of -values: .C &ß (ß *ß ""e f


13. The relation can be described by the set of orderedpairs

e fÐ#ß "Ñß Ð&ß "Ñß Ð""ß (Ñß Ð"(ß #!Ñß Ð$ß #!Ñ Þ

The relation is a function since for each B-value,there is only one -value.C

The domain is the set of B #ß $ß &ß ""ß "(-values: .e fThe range is the set of -values: C "ß (ß #! Þe f


e fÐ"ß "!Ñß Ð#ß "&Ñß Ð#ß "*Ñß Ð$ß "*Ñß Ð&ß #(Ñ Þ

The relation is not a function since the B #-value has two different -values associated with it, C "&and ."*

The domain is the set of B "ß #ß $ß &-values: .e fThe range is the set of -values: C "!ß "&ß "*ß #( Þe f


e fÐ"ß &Ñß Ð"ß #Ñß Ð"ß"Ñß Ð"ß%Ñ Þ

The relation is not a function since the B "-value has four different -values associated with it, , ,C & #" %, and .

The domain is the set of B "-values: .e fThe range is the set of -values: C &ß #ß"ß% Þe f


e fÐ%ß$Ñß Ð#ß$Ñß Ð!ß$Ñß Ð#ß$Ñ Þ


The domain is the set of B %ß #ß !ß#-values: .e fThe range is the set of -values: C $ Þe f

17. Using the vertical line test, we find any verticalline will intersect the graph at most once. Thisindicates that the graph represents a function.This graph extends indefinitely to the left Ð_Ñand indefinitely to the right . Therefore, theÐ_Ñdomain is . This graph extendsÐ_ß_Ñindefinitely downward , and indefinitelyÐ_Ñupward . Thus, the range is .Ð_Ñ _ß_ÑÐ

18. The relation is not a function since a vertical linemay intersect the graph in more than one point.The domain is the set of B Ò#ß #Ó-values, . Therange is the set of -values, .C Ò#ß #Ó

19. Using the vertical line test, we find any verticalline will intersect the graph at most once. Thisindicates that the graph represents a function.This graph extends indefinitely to the left Ð_Ñand indefinitely to the right . Therefore, theÐ_Ñdomain is . This graph extendsÐ_ß_Ñindefinitely downward , and reaches a highÐ_Ñpoint at . Therefore, the range is .C œ % Ð_ß %Ó

20. Since any vertical line that intersects the graphintersects it in no more than one point, the relationrepresented by the graph is a function.The domain is , and the range is .Ò#ß #Ó Ò!ß %Ó

21. Since a vertical line can intersect the graph of therelation in more than one point, the relation is nota function. The domain, the B-values of the pointson the graph, is . The range, the -values ofÒ%ß %Ó Cthe points on the graph, is .Ò$ß $Ó

22. Since a vertical line, such as , intersects theB œ %graph in two points, the relation is not a function.The domain is Ò$ß_Ñ _ß_Ñ, and the range is .Ð

23. C œ B#

Each value of B corresponds to one -value. ForCexample, if , then . Therefore,B œ $ C œ $ œ *#

C œ B C B# defines as a function of .Since any -value, B positive, negative, or zero, canbe squared, the domain is Ð_ß_Ñ.

24. defines as a function of because eachC œ B C B$

value, , in the domain has exactly one value, , inB Cthe range.Since any -value, B positive, negative, or zero, canbe cubed, the domain is Ð_ß_Ñ.

25. B œ C'

The ordered pairs both satisfya b a b'%ß # '%ß# and the equation. Since one value of , ,B '%corresponds to two values of , and , theC # #relation does not define a function. Because isBequal to the sixth power of , the values of mustC Balways be nonnegative. The domain is .Ò!ß_Ñ

26. B œ C%

If , may be either or . doesB œ "' C # # B œ C%

not define as a function of .C BAll the B-values must be nonnegative, so thedomain is .Ò!ß_Ñ

27. C œ #B '

For any value of , there is exactly one value of ,B Cso this equation defines a function. The domain isthe set of all real numbers, .Ð_ß_Ñ

28. C œ 'B )

For any value of , there is exactly one value of ,B Cso this equation defines a function. The domain isthe set of all real numbers, .Ð_ß_Ñ


29. B C %For a particular B C-value, more than one -valuecan be selected to satisfy . For example,B C %if and , thenB œ # C œ !

# ! %. True

Now, if and , thenB œ # C œ "

# " %. Also true

Therefore, does not define as aB C % Cfunction of .BThe graph of consists of the shadedB C %region below the dashed line , whichB C œ %extends indefinitely from left to right. Therefore,the domain is .Ð_ß_Ñ

30. B C $Let .B œ "

" C $

C #

C #

So for , may be any number greater thanB œ " C#.B C $ C B does not define as a function of .The B-values may be any number. The domain isÐ_ß_Ñ.

31. C œ BÈFor any value of , there is exactly oneBcorresponding value for , so this relation definesCa function. Since the radicand must be anonnegative number, must always beBnonnegative. The domain is .Ò!ß_Ñ

32. defines as a function of becauseC œ B C BÈeach value, , in the domain has exactly one value,BC.Since must be a BÈ real number, , so theB !

domain is .Ò!ß_Ñ

33. BC œ "

Rewrite as . Note that can neverBC œ " C œ B"

Bequal , otherwise the denominator would equal .! !The domain is Ð_ß !Ñ Ð!ß_Ñ.Each nonzero -value gives exactly one -value.B CTherefore, defines as a function of .BC œ " C B

34. may be rewritten as .BC œ $ C œ $

BThis relation defines as a function of becauseC Beach value of in the domain has exactly oneB

C-value. Since$

BB Á ! must be defined, . The

domain is Ð_ß !Ñ Ð!ß_Ñ.

35. C œ %B #ÈTo determine the domain of , recallC œ %B #Èthat the radicand must be nonnegative. Solve theinequality , which gives us .%B # ! B "

#

Therefore, the domain is .‰ ß_"#

Each B-value from the domain produces exactlyone -value. Therefore, C C œ %B #È defines afunction.

36. defines as a function of becauseC œ * #B C BÈeach value of in the domain has exactly oneBC-value.È* #B must be a real number, so

* #B !

#B *

B Ÿ *#

The domain is ˆ ‘_ß *# .

37. C œ#

B %Given any value of , is found by subtracting ,B C %then dividing the result into . This process#produces exactly one value of for each -value,C Bso the relation represents a function. The domainincludes all real numbers except those that makethe denominator , namely . The domain is! %Ð_ß %Ñ Ð%ß_Ñ.

38. defines as a function of . C œ C B(

B #The

domain includes all real numbers except those thatmake the denominator , namely . The domain! #is Ð_ß#Ñ Ð#ß_Ñ.

39. is the value of the 0Ð$Ñ dependent variable whenthe independent variable is —choice .$ B

40. The amount of income tax you pay depends onyour taxable income, so income tax is a functionof taxable income.

41. 0ÐBÑ œ $B %

0Ð!Ñ œ $Ð!Ñ %

œ ! %

œ %

42. 0ÐBÑ œ $B %

0Ð$Ñ œ $Ð$Ñ %

œ * %

œ "$

43. 1ÐBÑ œ B %B "

1Ð#Ñ œ Ð#Ñ %Ð#Ñ "

œ Ð%Ñ ) "

œ ""

#

#

44. 1ÐBÑ œ B %B "

1Ð"!Ñ œ Ð"!Ñ %Ð"!Ñ "

œ "!! %! "

œ &*

#

#


45. 0ÐBÑ œ $B %

0 œ $ %

œ " %

œ $

ˆ ‰ ˆ ‰" "$ $

46. 0ÐBÑ œ $B %

0 œ $ %

œ ( %

œ $

ˆ ‰ ˆ ‰( ($ $

47. 1ÐBÑ œ B %B "

1Ð!Þ&Ñ œ Ð!Þ&Ñ %Ð!Þ&Ñ "

œ !Þ#& # "

œ #Þ(&

#

#

48. 1ÐBÑ œ B %B "

1Ð"Þ&Ñ œ Ð"Þ&Ñ %Ð"Þ&Ñ "

œ #Þ#& ' "

œ %Þ(&

#

#

49. 0ÐBÑ œ $B %

0Ð:Ñ œ $Ð:Ñ %

œ $: %

50. 1ÐBÑ œ B %B "

1Ð5Ñ œ 5 %5 "

#

#

51. 0ÐBÑ œ $B %

0ÐBÑ œ $ÐBÑ %

œ $B %

52. 1ÐBÑ œ B %B "

1ÐBÑ œ ÐBÑ %ÐBÑ "

œ B %B "

#

#

#

53. 0ÐBÑ œ $B %

0ÐB #Ñ œ $ÐB #Ñ %

œ $B ' %

œ $B #

54. 0ÐBÑ œ $B %

0ÐB #Ñ œ $ÐB #Ñ %

œ $B ' %

œ $B "!

55. 1ÐBÑ œ B %B "

1Ð Ñ œ % "

#

#1 1 1

56. 1ÐBÑ œ B %B "

1Ð/Ñ œ / %/ "

#

#

57. 0ÐBÑ œ $B %

0ÐB 2Ñ œ $ÐB 2Ñ %

œ $B $2 %

58. Use the result from the previous exercise.

0ÐB 2Ñ 0ÐBÑ

œ $B $2 % Ð$B %Ñ

œ $B $2 % $B %

œ $2

a b

59. 0Ð%Ñ 1Ð%Ñ

œ Ò$Ð%Ñ %Ó Ð%Ñ %Ð%Ñ "

œ Ò)Ó Ò"Ó

œ *

‘#

60. 0Ð"!Ñ 1Ð"!Ñ

œ Ò$Ð"!Ñ %Ó Ð"!Ñ %Ð"!Ñ "

œ Ò#'Ó Ò&*Ó

œ $$

‘#

61. (a) When , , so B œ # C œ # 0Ð#Ñ œ #Þ

When , , so (b) B œ " C œ $ 0Ð"Ñ œ $Þ

62. (a) When , , so B œ # C œ & 0Ð#Ñ œ &Þ

When , , so (b) B œ " C œ "" 0Ð"Ñ œ ""Þ

63. (a) When , , so B œ # C œ "& 0Ð#Ñ œ "&Þ

When , , so (b) B œ " C œ "! 0Ð"Ñ œ "!Þ

64. (a) When , , so B œ # C œ " 0Ð#Ñ œ "Þ

When , , so (b) B œ " C œ ( 0Ð"Ñ œ (Þ

65. (a) The point is on the graph of , soÐ#ß $Ñ 00Ð#Ñ œ $Þ

(b) The point is on the graph of , soÐ"ß$Ñ 00Ð"Ñ œ $Þ

66. (a) The point is on the graph of , soÐ#ß$Ñ 00Ð#Ñ œ $Þ

(b) The point is on the graph of , soÐ"ß #Ñ 00Ð"Ñ œ #Þ

67. (a) Solve the equation for .C

B $C œ "#

$C œ "# B

C œ"# B

$

Since ,C œ 0ÐBÑ

0ÐBÑ œ œ B %"# B "

$ $.

(b) 0Ð$Ñ œ œ œ $"# $ *

$ $

68. (a) B %C œ )

%C œ ) B

C œ) B

%

0ÐBÑ œ œ B #) B "

% %


(b) 0Ð$Ñ œ œ œ ) $ & &

% % %


C #B œ $

C œ $ #B

#

#

Since ,C œ 0ÐBÑ

0ÐBÑ œ $ #B#.

(b) 0Ð$Ñ œ $ #Ð$Ñ

œ $ #Ð*Ñ

œ "&

#

70. (a) C $B œ #

C œ # $B

0ÐBÑ œ # $B

#

#

#

(b) 0Ð$Ñ œ # $Ð$Ñ

œ # $Ð*Ñ

œ #*

#


%B $C œ )

$C œ ) %B

C œ) %B

$

Since ,C œ 0ÐBÑ

0ÐBÑ œ œ B ) %B % )

$ $ $.

(b) 0Ð$Ñ œ œ) %Ð$Ñ ) "#

$ $

œ œ% %

$ $

72. (a) #B &C œ *

&C œ * #B

C œ* #B

&

0ÐBÑ œ œ B * #B # *

& & &

(b) 0Ð$Ñ œ* #Ð$Ñ

&

œ œ œ $* ' "&

& &

73. The equation has a straight line as its#B C œ %graph. To find in , let in theC Ð$ß C Ñ B œ $

equation.

#B C œ %

#Ð$Ñ C œ %

' C œ %

C œ #

To use functional notation for , solve#B C œ %for to getC

C œ #B %.

Replace with to getC 0ÐBÑ

0ÐBÑ œ #B %

0Ð$Ñ œ #Ð$Ñ % œ #

.

Because when , the point C œ # B œ $ Ð$ß#Ñ lies on the graph of the function.

74. A Choice , , is the only choice that0ÐBÑ œ B " &% %

defines as a linear function of .C B

75. 0ÐBÑ œ #B &The graph will be a line. The intercepts are a b!ß &and .ˆ ‰&

# ß !

The domain is Ð Ð_ß_Ñ _ß_Ñ. The range is .

76. 1ÐBÑ œ %B "Using a C Ð!ß "Ñ-intercept of and a slope of7 œ % œ %

" , graph the line. From the graph we seethat the domain is Ð_ß_ÑÞ The range isÐ_ß_Ñ.

77. 2ÐBÑ œ B #"#

The graph will be a line. The intercepts are a b!ß #and .a b%ß !The domain is Ð Ð_ß_Ñ _ß_Ñ. The range is .

78. JÐBÑ œ B ""%

Using a C Ð!ß "Ñ-intercept of and a slope of7 œ "

% , we graph the line. From the graph wesee that the domain is Ð_ß_Ñ. The range isÐ_ß_Ñ.


79. KÐBÑ œ #BThis line includes the points , , andÐ!ß !Ñ Ð"ß #ÑÐ#ß %Ñ. The domain is Ð_ß_Ñ. The range isÐ_ß_Ñ.

80. LÐBÑ œ $BUsing a C Ð!ß !Ñ-intercept of and a slope of7 œ $ œ $

" , we graph the line. From the graphwe see that the domain is Ð_ß_Ñ. The range isÐ_ß_Ñ.

81. 1ÐBÑ œ %Using a C Ð!ß%Ñ-intercept of and a slope of7 œ !, we graph the horizontal line. From thegraph we see that the domain is Ð_ß_Ñ. Therange is e f% Þ

82. 0ÐBÑ œ &Draw the horizontal line through the point .Ð!ß &ÑOn the horizontal line the value of can be anyBreal number, so the domain is . TheÐ_ß_Ñrange is e f& Þ

83. 0ÐBÑ œ !Draw the horizontal line through the point .Ð!ß !ÑOn the horizontal line the value of can be anyBreal number, so the domain is . TheÐ_ß_Ñrange is e f! Þ

84. , or , is the0ÐBÑ œ ! C œ ! B-axis.

85.

(dollars)

(a) 0ÐBÑ œ #Þ(&B

0Ð$Ñ œ #Þ(&Ð$Ñ

œ )Þ#&

(b) is the value of the independent variable,$which represents a package weight of pounds;$0Ð$Ñ is the value of the dependent variablerepresenting the cost to mail a 3-pound package.

$ . , the cost to mail a 5-lb(c) #Þ(&Ð&Ñ œ "$ (&package. Using function notation, we have0Ð&Ñ œ "$Þ(&Þ

86. (a) B 0ÐBÑ! !" #Þ&!# &Þ!!$ (Þ&!

$$$

(b) Since the charge equals the cost per mile,$ , times the number of miles, ,#Þ&! B0ÐBÑ œ #Þ&!B $ .

To graph for , plot(c) C œ 0ÐBÑ B − !ß "ß #ß $e fthe points , , , and Ð!ß !Ñ Ð"ß #Þ&!Ñ Ð#ß &Þ!!Ñ Ð$ß (Þ&!Ñfrom the chart.

87. Since the length of a man's femur is given, use(a) the formula .2Ð<Ñ œ '*Þ!* #Þ#%<

2Ð&'Ñ œ '*Þ!* #Þ#%Ð&'Ñ

œ "*%Þ&$

Let r 56.œ

The man is cm tall."*%Þ&$

(b) Use the formula .2Ð>Ñ œ )"Þ'* #Þ$*>

2Ð%!Ñ œ )"Þ'* #Þ$*Ð%!Ñ

œ "((Þ#*

Let t 40.œ

The man is cm tall."((Þ#*


(c) Since the length of a woman's femur is given,use the formula .2Ð<Ñ œ '"Þ%" #Þ$#<

2Ð&!Ñ œ '"Þ%" #Þ$#Ð&!Ñ

œ "((Þ%"

Let r 50.œ

The woman is cm tall."((Þ%"

(d) Use the formula .2Ð>Ñ œ (#Þ&( #Þ&$>

2Ð$'Ñ œ (#Þ&( #Þ&$Ð$'Ñ

œ "'$Þ'&

Let t 36.œ

The woman is cm tall."'$Þ'&

88. (a) 0ÐBÑ œ !Þ*"Ð$Þ"%ÑB

0Ð!Þ)Ñ œ !Þ*"Ð$Þ"%ÑÐ!Þ)Ñ

œ "Þ)#)($'

#

#

To the nearest hundredth, the volume of the poolis m ."Þ)$ $

(b) 0ÐBÑ œ !Þ*"Ð$Þ"%ÑB

0Ð"Þ!Ñ œ !Þ*"Ð$Þ"%ÑÐ"Þ!Ñ

œ #Þ)&(%

#

#

To the nearest hundredth, the volume of the poolis m .#Þ)' $

(c) 0ÐBÑ œ !Þ*"Ð$Þ"%ÑB

0Ð"Þ#Ñ œ !Þ*"Ð$Þ"%ÑÐ"Þ#Ñ

œ %Þ""%'&'

#

#

To the nearest hundredth, the volume of the poolis m .%Þ"" $

(d) 0ÐBÑ œ !Þ*"Ð$Þ"%ÑB

0Ð"Þ&Ñ œ !Þ*"Ð$Þ"%ÑÐ"Þ&Ñ

œ 'Þ%#*"&

#

#

To the nearest hundredth, the volume of the poolis m .'Þ%$ $

89. (a) The independent variable is , the number of>hours, and the possible values are in the setÒ!ß "!!Ó 1. The dependent variable is , the numberof gallons, and the possible values are in the setÒ!ß $!!!Ó.

The graph rises for the first hours, so the(b) #&water level increases for hours. The graph falls#&for to , so the water level decreases> œ &! > œ (&for hours.#&

There are gallons in the pool when(c) #!!!> œ *!.

is the number of gallons in the pool at(d) 0Ð!Ñtime . Here, , which means the pool> œ ! 0Ð!Ñ œ !is empty at time .!

; After hours, there are (e) 0Ð#&Ñ œ $!!! #& $!!!gallons of water in the pool.

90. (a) For every hour, there is one and only onemegawatt reading. Thus, the graph passes thevertical line test, so it is the graph of a function.

(b) We start the day at midnight and end the dayat midnight. The domain is [ ].!ß #%

At in the morning, it appears that the number(c) )of megawatts used is halfway between and""!!"$!! "#!!. So is a good estimate.

The most electricity was used at hours(d) "((5 ) and the least electricity was used at P.M. %hours (4 )A.M. Þ

; At 12 noon, electricity use is(e) 0Ð"#Ñ œ #"!!#"!! megawatts.

91. The graph shows and . In functionB œ $ C œ (notation, this is

0Ð$Ñ œ (.

92. (a) 0Ð#Ñ œ "Þ"

Y is when X is .(b) " $Þ( 'So, if X , then X .0Ð Ñ œ $Þ( œ '

Let and. Then

(c) a ba bB ß C œ Ð!ß $Þ&Ñ

B ß C œ Ð"ß #Þ$Ñ" "

# #

7 œ œ œ "Þ#$Þ& #Þ$ "Þ#

! " ".

The slope is ."Þ#

(d) When X , Y is , so the -intercept isœ ! $Þ& C"

$Þ&.

Use the slope-intercept form of the equation of(e)a line and the information found in parts (c) and(d).

0ÐBÑ œ 7B ,

0Ð Ñ œ "Þ# $Þ&X X

93. 'B &C œ #

'Ð$Ñ &C œ #

") &C œ #

&C œ #!

C œ %

Let x = 3.

94. %$

%$

B C œ *

Ð$Ñ C œ *

% C œ *

C œ "$

Let x = 3.

95. "Þ&B #Þ&C œ &Þ&

"Þ&Ð$Ñ #Þ&C œ &Þ&

%Þ& #Þ&C œ &Þ&

#Þ&C œ "!

C œ %

Let x = 3.

Chapter 3 Review Exercises 143

96. # #B œ %$B "

#

$B " #B œ %

B " œ %

B œ $

Œ

The solution set is .Ö$×

97. % #B œ )B $

$

%ÐB $Ñ 'B œ #%

%B "# 'B œ #%

#B "# œ #%

#B œ "#

B œ '

Œ Multiply by 3.

The solution set is .Ö'×

98. # B œ %$B "

&

#Ð$B "Ñ &B œ #!

'B # &B œ #!

B # œ #!

B œ ")

B œ ")

Œ Multiply by 5.

The solution set is .Ö")×

Chapter 3 Review Exercises1. $B #C œ "!

For :B œ !

$Ð!Ñ #C œ "!

#C œ "!

C œ & Ð!ß &Ñ

For :C œ !

$B #Ð!Ñ œ "!

$B œ "!

B œ "!$ ˆ ‰"!

$ ß !

For :B œ #

$Ð#Ñ #C œ "!

' #C œ "!

#C œ %

C œ # Ð#ß #Ñ

For :C œ #

$B #Ð#Ñ œ "!

$B % œ "!

$B œ "%

B œ "%$ ˆ ‰"%

$ ß#

Plot the ordered pairs, and draw the line throughthem.

2. B C œ )For :B œ #

# C œ )

C œ '

C œ ' Ð#ß'Ñ

For :C œ $

B Ð$Ñ œ )

B $ œ )

B œ & Ð&ß$Ñ

For :B œ $

$ C œ )

C œ &

C œ & Ð$ß&Ñ

For :C œ #

B Ð#Ñ œ )

B # œ )

B œ ' Ð'ß#Ñ

Plot the ordered pairs, and draw the line throughthem.

3. %B $C œ "#To find the B C œ !-intercept, let .

%B $C œ "#

%B $Ð!Ñ œ "#

%B œ "#

B œ $

The B Ð$ß !Ñ-intercept is .To find the -intercept, let .C B œ !

%B $C œ "#

%Ð!Ñ $C œ "#

$C œ "#

C œ %

The C Ð!ß%Ñ-intercept is .Plot the intercepts and draw the line through them.

continued


4. &B (C œ #)To find the B C œ !-intercept, let .

&B (C œ #)

&B (Ð!Ñ œ #)

&B œ #)

B œ #)&

The B ß !-intercept is .ˆ ‰#)&

To find the C B œ !-intercept, let .

&B (C œ #)

&Ð!Ñ (C œ #)

(C œ #)

C œ %


5. #B &C œ #!To find the B C œ !-intercept, let .

#B &C œ

#B &Ð!Ñ œ

#B œ

B œ "!

#!

#!

#!

The B "!ß !-intercept is .a bTo find the C B œ !-intercept, let .

#B &C œ

#Ð!Ñ &C œ

&C œ

C œ %

#!

#!

#!


6. B %C )œTo find the B C œ !-intercept, let .

B %C )

B % )

B )

œ

Ð!Ñ œ

œ

The B )ß !-intercept is .a bTo find the C B œ !-intercept, let .

! œ

%C œ

C œ #

%C )

)

The C Ð!ß#Ñ-intercept is .Plot the intercepts and draw the line through them.

7. If both positive, the point lies incoordinates are quadrant I. If the first coordinate is negative andthe second is positive, the point lies in quadrant II.To lie in quadrant III, the point must have bothcoordinates quadrant IV, thenegative. To lie in first coordinate must be positive and the secondmust be negative.

8. Through a b a b"ß # %ß& and

7 œ œ œ œ C & # ( (

B % Ð"Ñ & &

change in change in

9. Through a b a b!ß $ #ß % and Let a b a bB ß C œ Ð!ß $Ñ B ß C œ Ð#ß %Ñ" " # # and .

7 œ œ œ œ C C % $ " "

B B # ! # ## "

# "

10. The slope of is , the C œ #B $ # coefficient of .B

11. $B %C œ &Write the equation in slope-intercept form.

%C œ $B &

C œ B $ &% %

The slope is .$%12. is a vertical line and has slope.B œ & undefined

13. Parallel to $C œ #B &Write the equation in slope-intercept form.

$C œ #B &

C œ B # &$ $

The slope of is ; all lines parallel to$C œ #B & #$

it will also have a slope of .#$


14. Perpendicular to $B C œ %Solve for .C

C œ $B %

The slope is ; the slope of a line perpendicular to$it is since"

$

$ œ "ˆ ‰"$ .

15. Through a b a b"ß & "ß% and

7 œ œ œC % & *

B " Ð"Ñ !

?

? Undefined

This is a vertical line; it has undefined slope.

16. Through a b a b$ß" $ß " and

7 œ œ œ œ C " Ð"Ñ # "

B $ $ ' $

?

?.

17. The B Ð#ß !Ñ C-intercept is and the -intercept isÐ!ß #ÑÞ The slope is

7 œ œ œ œ "ÞC # ! #

B ! # #

change in change in

18. The line goes up from left to right, so it haspositive slope.

19. The line goes down from left to right, so it hasnegative slope.

20. The line is slope.vertical, so it has undefined

21. The line is horizontal, so it has slope.!

22. The slope is which can be written as %,#"! !Þ#ß #!

#! ""!! &, or .

The correct responses are , , , , and .A B C D F

23. To rise foot, we must move feet in the" %horizontal direction. To rise feet, we must move$$Ð%Ñ œ "# feet in the horizontal direction.

24. Let a bB ß C œ Ð"*)!ß #" !!!Ñ" " , anda bB ß C œ Ð#!!$ß &# (!!Ñ# # , . Then

7 œ œ œC &# (!! #" !!! $" (!!

B #!!$ "*)! #$¸ "$()

?

?

, , ,

.

The average rate of change is $ per year."$()

25. Slope , (a) -intercept "$ C Ð!ß"Ñ

Use the slope-intercept form with 7 œ "$ and

, œ "Þ

C œ 7B ,

C œ B ""$

(b) C œ B "

$C œ B $

B $C œ $

"$

26. Slope , (a) -intercept ! C Ð!ß#ÑUse the slope-intercept form with 7 œ ! and, œ #Þ

C œ 7B ,

C œ Ð!ÑB #

C œ #

(b) is already in standard form.C œ #

27. Slope , through (a) Ð#ß (Ñ%$

Use the andpoint-slope form with 7 œ %$a bB ß C œ Ð#ß (Ñ" " .

C C œ 7ÐB B Ñ

C ( œ ÐB #Ñ

C ( œ B

C œ B

" "

%$% )$ $% #*$ $

(b) C œ B

$C œ %B #*

%B $C œ #*

% #*$ $

28. Slope , through (a) $ Ð"ß %ÑUse the andpoint-slope form with 7 œ $a bB ß C œ Ð"ß %Ñ" " .

C C œ 7ÐB B Ñ

C % œ $ B Ð"Ñ

C % œ $ÐB "Ñ

C % œ $B $

C œ $B (

" "c d

(b) C œ $B (

$B C œ (

$B C œ (

29. Vertical, through (a) Ð#ß &ÑThe vertical line is in the formequation of any B œ 5 Ð#ß &Ñ. Since the line goes through , theequation is . (Slope-intercept form is notB œ #possible.)

is already in standard form.(b) B œ #

30. Through (a) and a b a b#ß& "ß %Find the slope.

7 œ œ œ œ *C % Ð&Ñ *

B " # "

?

?

Use the point-slope form with and7 œ *a bB ß C œ Ð#ß&Ñ" " .

C C œ 7ÐB B Ñ

C Ð&Ñ œ *ÐB #Ñ

C & œ *B ")

C œ *B "$

" "

(b) C œ *B "$

*B C œ "$


31. Through (a) and a b a b$ß" #ß 'Find the slope.

7 œ œ œC ' Ð"Ñ (

B # Ð$Ñ &

?

?

Use the andpoint-slope form with 7 œ (&a bB ß C œ Ð#ß 'Ñ" " .

C C œ 7ÐB B Ñ

C ' œ ÐB #Ñ

C ' œ B

C œ B

" "

(&( "%& &( "'& &

(b) C œ B

&C œ (B "'

(B &C œ "'

(B &C œ "'

( "'& &

32. From Exercise 17, we have and a(a) 7 œ "C Ð!ß #Ñ-intercept of . The slope-intercept form is

C œ "B # C œ B #Þ or

(b) C œ B #

B C œ #

33. Parallel to and through (a) %B C œ $ Ð(ß"ÑWriting in %B C œ $ slope-intercept form givesus C œ %B $ %, which has slope . Lines parallelto it will also have slope . The line with slope % %through is :Ð(ß"Ñ

C C œ 7ÐB B Ñ

C Ð"Ñ œ %ÐB (Ñ

C " œ %B #)

C œ %B #*

" "

(b) C œ %B #*

%B C œ #*

%B C œ #*

34. Perpendicular to and through(a) #B &C œ (Ð%ß $ÑWrite the equation in slope-intercept form.

#B &C œ (

&C œ #B (

C œ B # (& &

has slope and is perpendicular toC œ B # ( #& & &

lines with slope .&#

The line with slope through is Ð%ß $Ñ&#

C C œ 7ÐB B Ñ

C $ œ ÐB %Ñ

C $ œ B "!

C œ B "$

" "

&#&#&#

(b) C œ B "$

#C œ &B #'

&B #C œ #'

&#

35. (a) The fixed cost is $ , so that is the value of"&*, &(. The variable cost is $ , so

C œ 7B , œ &(B "&*Þ

The cost of a 1-year membership can be found bysubstituting for ."# B

C œ &(Ð"#Ñ "&*

œ ')% "&* œ )%$

The cost is $)%$Þ

As in part (a),(b)

C œ %(B "&*Þ

C œ %(Ð"#Ñ "&*

œ &'% "&* œ (#$

The cost is $(#$Þ

36. (a) Use and .Ð'ß "#Þ'Ñ Ð"#ß $&Þ!Ñ

7 œ œ œ ¸ $Þ($C $&Þ! "#Þ' ##Þ%

B "# ' '

?

?

Use the point-slope form of a line.

C C œ 7ÐB B Ñ

C "#Þ' œ $Þ($ÐB 'Ñ

C "#Þ' œ $Þ($B ##Þ$)

C œ $Þ($B *Þ()

C œ $Þ($B *Þ)

" "

or

The slope, , indicates that the number of$Þ($e-filing taxpayers increased by % each year$Þ($from 1996 to 2002.

(b) The year 2005 corresponds to B œ "&Þ

C œ $Þ($Ð"&Ñ *Þ)

œ &&Þ*& *Þ) œ %'Þ"&

According to the equation from part (a), %%'Þ"&of tax returns will be filed electronically in 2005.

37. $B #C Ÿ "#Graph as a solid line through$B #C œ "#a b a b!ß' %ß ! Ð!ß !Ñ and . as a test point. Since Use Ð!ß !Ñ satisfies the inequality, shade the region onthe side of the line containing Ð!ß !Ñ.


38. &B C 'Graph as a dashed line through&B C œ 'a b ˆ ‰!ß' ß ! Ð!ß !Ñ and . as a test point. Since'

& Use Ð!ß !Ñ does not satisfy the inequality, shade theregion on the side of the line that does not containÐ!ß !Ñ.

39. and #B C Ÿ " B #CGraph as a solid line through#B C œ "ˆ ‰"# ß ! Ð!ß "Ñand , and shade the region on the side

containing inequality.Ð!ß !Ñ since it satisfies the Next, graph as a solid line through B œ #C Ð!ß !Ñand , and shade the region on the sideÐ#ß "Ñcontaining since or is true.Ð#ß !Ñ # #Ð!Ñ # !The intersection is the region where the graphsoverlap.

40. or B # C #

Graph as a solid vertical line through .B œ # Ð#ß !ÑShade the region to the right of .B œ #

Graph as a solid horizontal line throughC œ #Ð!ß #Ñ C œ #. Shade the region above . The graph of

B # C # or

includes all the shaded regions.

41. In , the " " symbol indicates that theC %B $ graph has a dashed boundary line and that theshading is below the line, so the correct choice isD.

42. e fÐ%ß #Ñß Ð%ß#Ñß Ð"ß &Ñß Ð"ß&ÑThe domain, the set of B %ß "-values, is .e fThe range, the set of -values, is .C #ß#ß &ß&e fSince each -value has more than one -value, theB Crelation is not a function.


.e fÐ*ß $#Ñß Ð""ß %(Ñß Ð%ß %(Ñß Ð"(ß '*Ñß Ð#&ß "%Ñ


The domain is the set of B-values:e f*ß ""ß %ß "(ß #& .The range is the set of -values: C $#ß %(ß '*ß "% Þe f

44. The domain, the B-values of the points on thegraph, is . The range, the -values of thec d%ß % Cpoints on the graph, is . Since a c d!ß # vertical lineintersects the graph of the relation in at most onepoint, the relation is a function.

45. The B-values are negative or zero, so the domainis . The -values can be any real number,Ð_ß !Ó Cso the range is . A Ð_ß_Ñ vertical line, such asB œ $, will intersect the graph twice, so by thevertical line test, the relation is not a function.

46. C œ $B $For any value of , there is exactly one value of ,B Cso the equation defines a function, actually a linearfunction. The domain is the set of all real numbers,Ð_ß_Ñ.

47. C B #For any value of , there are many values of . B C Forexample, and are both solutions of theÐ"ß !Ñ Ð"ß "Ñinequality that have the same B-value but differentC-values. The inequality does not define afunction. The domain is the set of all real numbers,Ð_ß_Ñ.

48. C œ k kBFor any value of , there is exactly one value of ,B Cso the equation defines a function. The domain isthe set of all real numbers, .Ð_ß_Ñ

49. C œ %B (ÈGiven any value of , is found by B C multiplying Bby , adding , and taking the square root of the% (result. This process produces exactly one value ofC B for each -value, so the equation defines afunction. Since the radicand must be nonnegative,

%B ( !

%B (

B (% .

The domain is ‰ ß_(% .

50. B œ C#

The ordered pairs both satisfya b a b%ß # %ß# and the equation. Since one value of , , correspondsB %to two values of , and , the equation doesC # #not define a function. Because is equal to theBsquare of , the values of must always beC Bnonnegative. The domain is .Ò!ß_Ñ


51. C œ(

B 'Given any value of , is found by B C subtracting ,'then dividing the result into . This process(produces exactly one value of for each -value,C Bso the equation defines a function. The domainincludes all real numbers except those that makethe denominator , namely . The domain is! 'Ð_ß 'Ñ Ð'ß_Ñ.

52. If no vertical line intersects the graph in more thanone point, then it is the graph of a function.

In Exercises 53–56, use

0ÐBÑ œ #B $B '# .

53. 0Ð!Ñ œ #Ð!Ñ $Ð!Ñ ' œ '#

54. 0Ð Ñ œ #Ð Ñ $Ð Ñ '

œ )Þ)# 'Þ$ ' œ )Þ&#

#Þ" #Þ" #Þ"#

55. 0 œ # $ '

œ ' œ )

ˆ ‰ ˆ ‰ ˆ ‰" " "# # #

#

" $# #

56. 0Ð5Ñ œ #5 $5 '#

57. (a) For each year, there is exactly one lifeexpectancy associated with the year, so the tabledefines a function.

The domain is the set of years, that is,(b)Ö"*%$ß "*&$ß "*'$ß "*($ß "*)$ß "**$ß #!!$×.The range is the set of life expectancies, that is,Ö'$Þ$ß ')Þ)ß '*Þ*ß ("Þ%ß (%Þ'ß (&Þ&ß ((Þ'×Þ

Answers will vary. Two possible answers are(c)Ð"*%$ß '$Þ$Ñ Ð"*&$ß ')Þ)ÑÞ and

. In 1973, life expectancy at(d) 0Ð"*($Ñ œ ("Þ%birth was yr.("Þ%

Since , (e) 0Ð"**$Ñ œ (&Þ& B œ "**$Þ

58.

Since ,,

and .

#B C œ !

C œ #B

C œ #B

C œ 0ÐBÑ

0ÐBÑ œ #B

0Ð$Ñ œ #Ð$Ñ œ #Ð*Ñ œ ")

#

#

#

#

#

59. Solve for in terms of .C B

#B &C œ (

#B ( œ &C

B œ C# (& &

Thus, choice is correct.C

60. No, because the equation of a line with anundefined slope is . The ordered pairs haveB œ +the form , where is a constant and is aÐ+ß CÑ + Cvariable. Thus, the number corresponds to an+infinite number of values of .C

61. The slope is negative since the line falls from leftto right.

62. Use the points a b a b"ß & $ß" and .

7 œ œ œ œ C " & ' $

B $ Ð"Ñ % #

?

?

63. Since , the slope of any line parallel to7 œ $#

this line is also , whereas the slope of any line$#

perpendicular to this line is since is the# #$ $

negative .reciprocal of $#

64. #C œ $B (To find the B C œ !-intercept, let .

#Ð!Ñ œ $B (

$B œ (

B œ ($

The .B-intercept is ˆ ‰($ ß !

65. #C œ $B (To find the C B œ !-intercept, let .

#C œ $Ð!Ñ (

#C œ (

C œ (#

The .C-intercept is ˆ ‰!ß (#

66. Solve for .#C œ $B ( C

C œ B $ (# #

Since ,C œ 0ÐBÑ

0ÐBÑ œ B $ (# # .

67. 0ÐBÑ œ B

0Ð)Ñ œ Ð)Ñ

œ œ

$ (# #$ (# ##% ( "(# # #

68. 0ÐBÑ œ B $ (# #

) œ B

"' œ $B (

#$ œ $B

B œ

$ (# #

#$$

Let f(x) 8.Multiply by 2.Subtract 7.Divide by 3.

œ

69. 0ÐBÑ !

B !

B

B Ÿ

B Ÿ

$ (# #

$ (# #

( ## $

($

ˆ ‰ˆ ‰

Chapter 3 Test 149

70. is 0ÐBÑ œ ! equivalent to , which is theC œ !equation we solved in Exercise 64 to find theB-intercept.The solution set is .˜ ™(

$

71. The graph is below the , so theB-axis for B ($

solution set of is 0ÐBÑ ! _ˆ ‰($ ß .

72. The graph is above the , so theB-axis for B ($

solution set of is 0ÐBÑ ! _ßˆ ‰ ($ .

Chapter 3 Test1. #B $C œ "#

For :B œ "

#Ð"Ñ $C œ "#

# $C œ "#

$C œ "!

C œ "!$ ˆ ‰"ß"!

$

For :B œ $

#Ð$Ñ $C œ "#

' $C œ "#

$C œ '

C œ # a b$ß#

For :C œ %

#B $Ð%Ñ œ "#

#B "# œ "#

#B œ !

B œ ! a b!ß%

2. $B #C œ #!To find the B C œ !-intercept, let .

$B #Ð!Ñ œ #!

$B œ #!

B œ #!$

The .B-intercept is ˆ ‰#!$ ß !

To find the C B œ !-intercept, let .

$Ð!Ñ #C œ #!

#C œ #!

C œ "!

The C Ð!ß"!Ñ-intercept is .Draw the line through these two points.

3. The graph of is the horizontal line withC œ &slope and ! C Ð!ß &Ñ-intercept . There is noB-intercept.

4. The graph of is the vertical line withB œ #B Ð#ß !Ñ C-intercept at . There is no -intercept.

5. Through a b a b'ß % %ß" and

7 œ œ œ œC " % & "

B % ' "! #

?

?

The slope of the line is ."#

6. The graph of a line with undefined slope is thegraph of a vertical line.

7. Find the slope of each line.

&B C œ )

C œ &B )

C œ &B )

The slope is .&

&C œ B $

C œ B " $& &

The slope is ."&

Since , the two slopes are negative& œ "ˆ ‰"&

reciprocals and the lines are perpendicular.

8. Find the slope of each line.

#C œ $B "#

C œ B '$#

The slope is .$#$C œ #B &

C œ B # &$ $

The slope is .#$The lines are neither parallel nor perpendicular.


9. Use the points , andÐ"*)!ß ""* !!!ÑÐ#!!&ß )* !!!Ñ, .


œ œC )* !!! ""* !!!

B #!!& "*)!

œ œ "#!!$! !!!

#&

change in , ,change in

,

The average rate of change is about farms"#!!per year, that is, the number of farms decreased byabout each year from 1980 to 2005."#!!

10. Through ; Ð%ß"Ñ 7 œ &

(a) in theLet and 7 œ & a bB ß C œ Ð%ß"Ñ" "

point-slope form.

C C œ 7ÐB B ÑC Ð"Ñ œ &ÐB %Ñ

C " œ &B #!C œ &B "*

" "

(b) C œ &B "*

&B C œ "*

From part (a)Standard form

11. Through ; Ð$ß "%Ñ horizontal

(a) A equation . Herehorizontal line has C œ 55 œ "% C œ "%, so the line has equation .

is already in standard form.(b) C œ "%

12. Through and parallel to Ð(ß #Ñ $B &C œ '

(a) To find the slope of , write the$B &C œ 'equation in slope-intercept form by solving for .C

$B &C œ '&C œ $B '

C œ B $ '& &

The slope is , so a line parallel to it also has$&

slope . Let and 7 œ $ $& & a bB ß C œ Ð(ß #Ñ" " in

the point-slope form.

C C œ 7ÐB B Ñ

C # œ B Ð(Ñ

C # œ B (

" "$&$&

c da b

C # œ B

C œ B

$ #"& &$ ""& &

(b) C œ B

B C œ

& B C œ &

$B &C œ ""

$ ""& &

$ ""& &

$ ""& &

From part (a)Variable termson one sideMultiply by 5.Standard form

ˆ ‰ ˆ ‰

13. Through and perpendicular to Ð(ß #Ñ C œ #B

(a) slope-intercept formSince is in C œ #BÐ, œ !Ñ 7 C œ #B #, the slope, , of is . A lineperpendicular to it has a slope that is the negativereciprocal of , that is, # 7 œ " "

# #. Let anda bB ß C œ Ð(ß #Ñ" " in the point-slope form.

C C œ 7ÐB B Ñ

C # œ ÐB (Ñ

C # œ B

C œ B

" ""#" (# #" $# #

(b) C œ B

B C œ

# B C œ #

B #C œ $

" $# #

" $# #

" $# #


ˆ ‰ ˆ ‰14. Through and a b a b#ß $ 'ß"

(a) First find the slope.

7 œ œ œ œ C " $ % "

B ' Ð#Ñ ) #

?

?

Use and 7 œ "# a bB ß C œ Ð#ß $Ñ" " in the point-

slope form.

C C œ 7ÐB B Ñ

C $ œ B Ð#Ñ

C $ œ ÐB #Ñ

C $ œ B "

C œ B #

" ""#"#"#

"#

c d

(b) C œ B #

B C œ #

# B C œ # #

B #C œ %

"#

"#

"#


ˆ ‰ a b15. Through ; verticalÐ&ß'Ñ

(a) equation of any The vertical line is in the formB œ 5 Ð&ß'Ñ. Since the line goes through , theequation is . Writing in slope-B œ & B œ &intercept form is since there is nonot possibleC-term.

(b) From part (a), the standard form is .B œ &

16. Positive slope means that the line goes up fromleft to right. The only line that has positive slopeand a negative C-coordinate for its -intercept isCchoice .B

17. (a) Use the points , and , .Ð&ß ## )'!Ñ Ð""ß $$ &'&Ñ

7 œ œ œC $$ &'& ## )'! "! (!&

B "" & '¸ "()%Þ"(

?

?

, , ,

Cumulative Review Exercises (Chapters 1–3) 151

Use the point-slope form with and7 œ "! (!&',

a bB ß C œ Ð&ß ## )'!Ñ" " , .

C ## )'! œ ÐB &Ñ

C ## )'! œ B

C ¸ "()%Þ"(B "$ *$*Þ"(

,,

,

"! (!&'

"! (!& &$ &#&' '

,

, ,

(b) ,, $ , ,

C œ "()%Þ"(Ð*Ñ "$ *$*Þ"(

œ #* **'Þ( ¸ #* **(

Let x 9.œ

which is slightly less than the actual value.

18. $B #C 'Graph the line , which has $B #C œ ' interceptsa b a b#ß ! !ß$ and , as a dashed line since theinequality involves . Test Ð!ß !Ñ, which yields! ', a false statement. Shade the region thatdoes not include Ð!ß !Ñ.

19. and C #B " B C $First graph as a dashed line throughC œ #B "a b a b#ß $ !ß" Ð!ß !Ñ and . , which yields Test ! ", a false statement. Shade the side of theline not containing .Ð!ß !ÑNext, graph as a dashed line throughB C œ $a b a b$ß ! !ß$ Ð!ß !Ñ ! $ and . , which yields , Test a true statement. Shade the side of the linecontaining . The intersection is the regionÐ!ß !Ñwhere the graphs overlap.

20. D Choice is the only graph that passes the verticalline test.

21. D Choice does not define a function, since itsdomain (input) element is paired with two!different range (output) elements, and ." #

22. greater than or equal to zero, soThe -values are Bthe domain is Since can be any value, theÒ!ß_ÑÞ Crange is Ð_ß_Ñ.

23. The domain is the set of -values: . TheB Ö!ß#ß %×range is the set of -values: .C Ö"ß $ß )×

24. 0ÐBÑ œ B #B "#

(a) 0Ð"Ñ œ Ð"Ñ #Ð"Ñ "

œ " # "

œ !

#

(b) 0Ð+Ñ œ + #+ "#

25. 0ÐBÑ œ B "#$

This function represents a line with -interceptCÐ!ß"Ñ B ß ! and -intercept .ˆ ‰$

#

Draw the line through these two points.The domain is Ð_ß_Ñ, and the range isÐ_ß_Ñ.

Cumulative Review Exercises(Chapters 1–3)

1. The absolute value of a negative number is apositive number and the additive inverse of thesame negative number is the same positivenumber. For example, suppose the negativenumber is :&

k k& œ Ð&Ñ œ &

and Ð&Ñ œ &

The statement is .always true

2. The statement is ; in fact, it is thealways truedefinition of a rational number.

3. The sum of two negative numbers is anothernegative number, so the statement is never true.

4. The statement is . For example,sometimes true

$ Ð$Ñ œ !

$ Ð"Ñ œ # Á !

,but .

5.

œ ' $ (

œ $ (

œ %

k k k k# % $ ( œ # % $ (

6. Ð!Þ)Ñ œ Ð!Þ)ÑÐ!Þ)Ñ œ !Þ'%#

7. È'% is not a real number.

8. The product of two numbers that have the samesign is positive.

œ œ œ# "# # # )

$ & $ & &Œ • •

•

"# %

&

9. Ð%7 $Ñ œ Ð%7Ñ $

œ %7 $


10. $B %B % *B B

œ $B B %B *B %

œ #B &B %

# #

# #

#

11. a b% % Ð"Ñ( Ð"' %Ñ Ð(Ñ

% Ð'Ñ #œ

œ œ "# ( "*

# #

#

12. $ B Ÿ &This is the set of numbers between and , not$ &including (use a parenthesis), but including $ &(use a bracket). In interval notation, the set isÐ$ß &ÓÞ

13. Ê Ê Ê# % # #

& & &œ œ

This is not a real number since the number underthe radical sign is negative.

For Exercises 14–16, let , , and .: œ % ; œ < œ "'"#

14. $Ð#; $:Ñ œ $ # $Ð%Ñ

œ $Ð" "#Ñ

œ $Ð"$Ñ

œ $*

‘ˆ ‰"#

15. k k k k¸ ¸ ˆ ‰¹ ¹¸ ¸ˆ ‰k k

: "'; œ % "'

"'

#

$ $$ "#

$

")œ %

œ '%

œ '% # œ '#

$

16.

, which is

È<

): #< )Ð%Ñ #Ð"'Ñœ

œ%

$# $#

œ%

!

È"'

undefined.

17. #D & $D œ # D

&D & œ # D

'D œ (

D œ ('

The solution set is .˜ ™('

18.

Multiply both sides by the LCD,

$+ " + # $

& # "! œ

"!Þ

"! œ "! $+ " + # $

& # "!

#Ð$+ "Ñ &Ð+ #Ñ œ $

'+ # &+ "! œ $

""+ ) œ $

""+ œ ""

+ œ "

Œ Œ

The solution set is .e f"

19. Solve for .Z œ < 2 2"

$$Z œ < 2$Z

<œ 2

1

1

1

#

#

#

Multiply by 3.

Divide by r .1 2

20. J œ G $#

J œ Ð&!Ñ $#

œ *! $# œ &)

*&*& Let C= –50.

&! &)°C is equivalent to °F.

21. Let denote the side of the original square and B %Bthe perimeter. Now is the side of the newB %square and is its perimeter.%ÐB %Ñ"The perimeter would be inches less than twice)the perimeter of the original square " translates as

%ÐB %Ñ œ #Ð%BÑ )

%B "' œ )B )

#% œ %B

' œ B

.

The length of a side of the original square is 'inches.

22. Let the time it takes for the planes to be B œ #"!!miles apart.Make a table. Use the formula .. œ <>

r t dEastbound PlaneWestbound Plane

&&! B &&!B&!! B &!!B

The total distance is miles.#"!!

&&!B &!!B œ #"!!

"!&!B œ #"!!

B œ #

It will take hr for the planes to be mi apart.# #"!!

23.

Divide by ; reverse the inequality symbols. or

% $ #5 *

( #5 '

#

5 $ $ 5 ( (# #

The solution set is .ˆ ‰$ß (#

Cumulative Review Exercises (Chapters 1–3) 153

24.

!Þ$B #Þ"ÐB %Ñ Ÿ 'Þ'

$B #"ÐB %Ñ Ÿ ''

$B #"B )% Ÿ ''

")B )% Ÿ ''

")B Ÿ ")

B Ÿ "

Multiply by 10.

The solution set is Ð_ß "Ó.

25. and and

" " )# $ $B $ B

B ' B )The graph of the solution set is all numbers bothgreater than less than . This is the' )and intersection. The elements common to both setsare the numbers between and , not including the' )endpoints. The solution set is .Ð'ß )Ñ

26. or

or

&B " "" $B & #'

&B "! $B #"

B Ÿ # B (The graph of the solution set is all numbers eitherless than or equal to greater than . This is# (or the union. The solution set is .Ð_ß#Ó Ð(ß_Ñ

27. k kk k#5 ( % œ ""

#5 ( œ (#5 ( œ ( #5 ( œ (

#5 œ "% #5 œ !

5 œ ( 5 œ !

or

or The solution set is .e f!ß (

28. k k$7 ' !The absolute value of an expression is alwaysnonnegative, so the realinequality is true for any number .7The s .olution set is Ð_ß_Ñ

29. The union of the three solution sets is ;Ð_ß_Ñthat is, the set of all real numbers.

30. To complete the table of ordered pairs, substitutethe given values for or in the equationB C$B %C œ "#.

For : B œ ! $B %C œ "#

$Ð!Ñ %C œ "#

%C œ "#

C œ $

The ordered pair is Ð!ß$ÑÞ

For : C œ ! $B %C œ "#

$B %Ð!Ñ œ "#

$B œ "#

B œ %

The ordered pair is Ð%ß !ÑÞ

For : B œ # $B %C œ "#

$Ð#Ñ %C œ "#

' %C œ "#

%C œ '

C œ œ ' $% #

The ordered pair is ˆ ‰#ß Þ$#

31. $B &C œ "#To find the B C œ !-intercept, let .

$B &Ð!Ñ œ "#

$B œ "#

B œ %

The B Ð%ß !Ñ-intercept is .To find the -intercept, let .C B œ !

$Ð!Ñ &C œ "#

&C œ "#

C œ "#&

The C !ß-intercept is .ˆ ‰"#&


32. and EÐ#ß "Ñ FÐ$ß&Ñ(a) The slope of line isAB

7 œ œ œ œ C & " ' '

B $ Ð#Ñ & &

?

?.

perpendicular to line (b) The slope of a line isABthe reciprocal of negative , which is .' &

& '


33. #B C 'Graph the line , which has#B C œ 'intercepts and ,a b a b$ß ! !ß' as a dashed line sincethe inequality involves . Test Ð!ß !Ñ, whichyields , a false statement. ! ' Shade the regionthat does not include Ð!ß !Ñ.

34. Slope ; (a) -intercept $% C Ð!ß"Ñ

To write an equation of this line, let and7 œ $%

, œ " in the slope-intercept form.

C œ 7B ,

C œ B "$%

(b) C œ B "

%C œ $B %

$B %C œ %

$%

35. Horizontal; through (a) Ð#ß#ÑA horizontal line through the point hasÐ-ß .Ñequation . Here , so the equation ofC œ . . œ #the line is C œ #Þ

is already in standard form.(b) C œ #

36. Through and (a) Ð%ß$Ñ Ð"ß "ÑFirst find the slope of the line.

7 œ œ œ œ C " Ð$Ñ % %

B " % $ $

?

?

Now substitute and ina bB ß C œ Ð%ß$Ñ 7 œ " "%$

the point-slope form. Then solve for .C

C C œ 7ÐB B Ñ

C Ð$Ñ œ ÐB %Ñ

C $ œ B

C œ B

" "

%$% "'$ $% ($ $

(b) C œ B

$C œ %B (

%B $C œ (

% ($ $

37. ; through (a) Vertical Ð%ß'ÑA vertical line through the point hasÐ-ß .Ñequation . Here , so the equation of theB œ - - œ %line is It is not possible to write thisB œ %Þequation in slope-intercept form.

is in standard form.(b) B œ %

38. The domain of the relation consists of the elementsin the leftmost figure; that is, Ö"%ß *"ß (&ß #$×Þ

The range of the relation consists of the elementsin the rightmost figure; that is, .Ö*ß (!ß &'ß &×

Since the element in the domain is paired with(&two different values, and , in the range, the(! &'relation is not a function.

39. 0ÐBÑ œ %B "!

(a) The variable can be any real number, so theBdomain is . The function is a non-Ð_ß_Ñconstant linear function, so its range is .Ð_ß_Ñ

(b) 0Ð$Ñ œ %Ð$Ñ "! œ "# "! œ ##

40. Use , and ,Ð"**(ß "# !!!Ñ Ð#!!%ß )' !!!ÑÞ

, , ,

,

7 œ œ œC )' !!! "# !!! (% !!!

B #!!% "**( (¸ "! &("

?

?

So the average rate of change is , per year;"! &("that is, the number of motor scooters sold in theUnited States increased by an average of ,"! &("per year from 1997 to 2004.

Documents

3.1 The Rectangular Coordinate System · 2014. 8. 19. · 3.1 The Rectangular Coordinate System 97 11. (a) If , then both and have the sameBC ! B C sign. ÐBßCÑ B C is in quadrant