College Algebra 3rd Edition Ratti Solutions Manual by kathysu98

Chapter 2 Graphs and Functions

Weareaskedtofindthedistancebetweenthe points A(60,0)and B(0,60).

2.1 Basic Concepts and Skills

1. Apointwithanegativefirstcoordinateanda positivesecondcoordinateliesinthesecond quadrant.

2. Anypointonthe x-axishassecondcoordinate 0

3. Thedistancebetweenthepoints ( ) 11 , Pxy and ( ) 22 , Qxy isgivenbytheformula

()() 22 2121 ,.dPQxxyy =−+−

4. Thecoordinatesofthemidpoint M(x, y)ofthe linesegmentjoining ( ) 11 , Pxy and ( ) 22 , Qxy aregivenby

1212 ,,. 22 xxyy

5. True

6. False.Thepoint(7,−4)is4unitstotheright and6unitsbelowthepoint(3,2).

144

The Coordinate

Plane

Practice

Problems 1.

2. (2005,17.9),(2006,17.8),(2007,17.8), (2008,17.6),(2009,17.6),(2010,16.3), (2011,16.0),(2012,18.0)

()() () () () () 22

=−+− =−−−+− =+−=≈

3. ( ) ( ) ( ) ( ) 1122 ,5,2;,4,1 xyxy=−=−

2121 22 22 4512 1121.4 dxxyy

() ()

= ()() ()() ()() 22 12121 22 22 2602 8268 dxxyy =−+− =−−+− =−+−= ()() ()() ()() 22 23131 22 22 1652 5334 dxxyy =−+− =−+− =−+= ()() () () () ()()

=−+− =−−+− =+=

4. ( ) ( ) ( ) ( )

1122

,6,2;,2,0 ,1,5 xyxy

==−

22 33232 22 22 1250 3534 dxxyy

Yes,thetriangleisanisoscelestriangle.

()()

(,)600060

=−+− =+−= =≈

()()

() 22 222

6060260 60284.85 dAB

⎛−+−⎞ + ==−⎛⎞ ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠

( )21 56113 ,, 2222 M

()

xy ++ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠

()

College Algebra 3rd Edition Ratti Solutions Manual Full Download: http://testbanktip.com/download/college-algebra-3rd-edition-ratti-solutions-manual/ Download all pages and all chapters at: TestBankTip.com

(2,2):Q1;(3,−1):Q4;(−1,0): x-axis (−2,−5):Q3;(0,0):origin;(−7,4):Q2 (0,3): y-axis;(−4,2):Q2

8. a. Answerswillvary.Sampleanswer: (2,0),(1,0),(0,0),(1,0),(2,0) The y-coordinateis0.

Thesetofallpointsoftheform(x,1)is ahorizontallinethatintersectsthe y-axis at1.

9. a. Ifthe x -coordinateofapointis0,thepoint liesonthe y-axis.

Thesetofallpointsoftheform(–1, y)isa verticallinethatintersectsthe x-axisat–1.

10. a. Averticallinethatintersectsthe x-axis at–3.

b. Ahorizontallinethatintersectsthe y -axis at4.

11. a. 0 y > b. 0 y <

c. 0 x < d. 0 x >

12. a. QuadrantIII b. QuadrantI

c. QuadrantIV d. QuadrantII

InExercises13−22,usethedistanceformula, ()() 22 2121 dxxyy =−+− andthemidpoint

formula,1212 (,),22 xxyy xy ++ ⎛ ⎞ = ⎜

13. a. 222 (22)(51)44 d =−+−==

b. () 2215 ,2,3 22 M ++ ⎛⎞ ==

14. a. 222 (23)(55)(5)5 d =−−+−=−=

b. () 3(2)55,0.5,5 22 M +−+ ⎛⎞ == ⎜⎟

15. a. 22 22 (2(1))(3(5)) 3213 d =−−+−−− =+=

b. () 125(3),0.5,4 22 M −+−+− ⎛⎞ ==−

16. a. 22 22 (7(4))(91) (3)(10)109 d =−−−+−− =−+−=

b. () 4(7)1(9),5.5,4 22 M −+−+− ⎛⎞ ==−−

17. a. 22 22 (3(1))(6.51.5) 4(8)8045 d =−−+−− =+−==

b. () 131.5(6.5),1,2.5 22 M −++− ⎛⎞ ==−

18. a. 22 22

d =−+−− =+−===

(10.5)(10.5) 510 (0.5)(1.5)2.522

b. () 0.510.5(1),0.75,0.25 22 M ++− ⎛⎞ ==− ⎜⎟ ⎝⎠

19. a. () 222 22(54)11 d =−+−==

b. () 2245 ,2,4.5 22 M ⎛⎞ ++ == ⎜⎟

=+−−+− ==

20. a. 22 2 (()())() (2)2 dvwvwtt ww

Section 2.1 The Coordinate Plane 145

⎟ ⎝ ⎠

⎜⎟ ⎝⎠

⎝⎠

⎜⎟ ⎝⎠

⎝⎠

⎜⎟

⎜⎟ ⎝⎠

⎝⎠

32. First,findthemidpoint M of PQ ( ) () 816412 ,4,4 22 M ⎛+−⎞ −+ ==− ⎜⎟

Nowfindthemidpoint R of PM. ( ) () 8444 ,2,0 22 R ⎛+−⎞ −+ ==− ⎜⎟

Finally,findthemidpoint S of MQ. ( ) () 416412 ,10,8 22 S ⎛−+−⎞ + ==−

Thus,thethreepointsare(−2,0),(4,−4),and (10,−8).

22 22

22 22 22

Section 2.1 The Coordinate Plane 147

⎝⎠

⎜⎟

33. ()() ()() ()() 22 22

(,)1(5)4517 (,)4(1)1432 (,)4(5)1517 dPQ dQR dPR =−−−+−= =−−−+−= =−−−+−= Thetriangleisisosceles. 34. ()() ()() ()() 22

(,)63625 (,)165652 (,)13525 dPQ dQR dPR =−+−= =−−+−= =−−+−= Thetriangleisanisoscelestriangle. 35.

(,)0(4)7817 (,)305713 (,)3(4)5810 dPQ dQR dPR =−−+−= =−−+−= =−−−+−= Thetriangleisscalene.

(,)161672 (,)5(1)3(1)42 (,)5636130 dPQ dQR dPR =−−+−−= =−−−+−−= =−−+−= Thetriangleisscalene.

dPQ

dPQ dQR dPR

Thetriangleisisosceles. 39. ()() () () () () 22 22 (,)111(1)22 (,)3(1)31 32313231 822 dPQ dQR =−−+−−= =−−−+−− =−++++ == () () () () 22 (,)313(1) 32313231 822 dPR =−−+−−− =+++−+ == Thetriangleisequilateral. 40. ()() () () 22 2 2 (,)1.5(0.5)1(1) 5 3 (,)31(1.5)1 2 dPQ dQR =−−−+−− = =−−−+−⎛⎞ ⎜⎟ ⎝⎠ () () () () () () () () () () 2 2 2 2 313312.25 3 31 4 32313332.25 1.753 5 31(0.5) (,)3 2(1) 31310.25 3 31 4 3231310.25 1.753 5 dPR ⎛ ⎞ −+−+ ⎝ ⎠ = +−+⎛⎞ ⎜⎟ ⎝⎠ −++−+ = +− = = +−−⎛⎞ ⎜⎟ ⎝⎠ ⎛⎞ −+−+ ⎝⎠ = +++⎛⎞ ⎜⎟ ⎝⎠ =−++−+ ++ = Thetriangleisequilateral. 41. Firstfindthelengthsofthesides: ()() ()() ()() ()() 22 22 22 22 (,)173(12)17 (,)14(1)11317 (,)221441117 (,)2274(12)17 dPQ dQR dRS dSP =−−+−−= =−−+−= =−+−−= =−+−−−= (continuedonnextpage)

22 22

()() ()() ()() 22 22 22

36. ()() ()() ()() 22 22 22

37. ()() ()() ()() 22

(,)909(1)145 (,)591(9)229 (,)501(1)29

dQR dPR =−+−−−= =−+−−= =−+−−= Thetriangleisscalene. 38. ()() ()() ()()

(,)4(4)5465 (,)042565 (,)0(4)24213

=−−+−= =−+−−= =−−+−−=

Allthesidesareequal,sothequadrilateralis eitherasquareorarhombus.Nowfindthe lengthofthediagonals:

dPR dQS

()()2222

(5)02(2)(03)

()()

(,)14711(12)172

(,)22(1)43172

()() 22 22

=−+−−= =−−+−−=

Thediagonalsareequal,sothequadrilateralis asquare.

42. Firstfindthelengthsofthesides:

Thecoordinatesof R are 8 ,0 7

dRS dSP

()()

(,)9811(10)2

(,)8912(11)2 dPQ dQR

=−+−−−= =−+−−−=

()() 22 22

46. P =(7,–4), Q =(8,3), R =(0, y)(R isonthe y-axis,sothe x -coordinateis0).

(,)7811(12)2

(,)8710(11)2

dPR dQS

=−+−−−= =−+−−−=

()() ()() 22 22

Allthesidesareequal,sothequadrilateralis eitherasquareorarhombus.Nowfindthe lengthofthediagonals.

()() 22 22 (,)07(4) (,)(08)(3) dPRy dQRy =−+−− =−+−

()() 22 22 07(4) (08)(3) y y

−+−− =−+−

(,)8812(10)2

(,)7911(11)2

()() ()() 22 22

=−+−−−= =−+−−−=

Thediagonalsareequal,sothequadrilateralis asquare. 43. ()() 22

yy yyyy yy +−−=+− +++=+−+ +=−+ = =

Thecoordinatesof R are 4 0, 7

⎛⎞ ⎜⎟ ⎝⎠ .

47.

48.

45. P =(–5,2), Q =(2,3), R =(x,0)(R isonthe x-axis,sothe y -coordinateis0).

()() 22 22 (,)(5)02 (,)(2)(03) dPRx dQRx =−−+− =−+−

()

20002012282314 , 22 2006,298

M ++ ⎛ ⎞ = ⎝ ⎠ = Themidpointofthesegmentgivesagood approximationtotheactualvalue,about300 million.

148 Chapter 2 Graphs and Functions

(continued)

2 22 2

449 541325413

2or6 x xx xxxx xxxx xx =−+−− =−++⇒ =−+⇒=−+⇒ =−−⇒=−+⇒ =−= 44. ()() 22 2 2 2 2 132(10)(3) 14469 6153 1696153 06160(8)(2) 8or2 y yy yy yy yyyy yy =−−+−− =+++ =++⇒ =++ =+−⇒=+−⇒ =−=

522(1)

04120(6)(2)

xx x x −−+−=−+− ++−=−+− +++=−++ +=−+ =− =−

⎛⎞ ⎜⎟ ⎝⎠

2222 22

(5)(02)(2)(03) 10254449 1029413 1416 8 7 xx xx xxxx

() 22 22 49(4)64(3) 498166469 865673 148 4 7 y y

2.1 Applying the Concepts

53. 16,92914,61215,770.5 2 M + ==

Therewereabout15,771murdersin2009.

54. 2009isthemidpointoftheinitialrange,so 2009 228320 274 2 M + == .

2008isthemidpointoftherange [2007,2009],so2008 228274 251 2 M + ==

2010isthemidpointoftherange [2009,2011],so2010 274320 297 2 M + == .

So,in2008,$251billionwasspent;in2009, $274billionwasspent,and$297wasspentin 2010.

55. 2008isthemidpointoftheinitialrange,so

2008 548925 736.5 2 M + ==

2006isthemidpointoftherange [2004,2008],so

2006 548736.5 642.25 2 M + ==

2005isthemidpointoftherange [2004,2006],so

2005 548642.25 595.125 2 M + == .

2007isthemidpointoftherange [2006,2008],so

2007 642.25736.5 689.375 2 M + ==

Usesimilarreasoningtofindtheamountsfor 2009,2010,and2011.Defensespendingwas asfollows:

YearAmountspent

2004$548billion

2005$595.125billion

2006$642.25billion

2007$689.375billion

2008$736.5billion

2009$783.625billion

2010$830.75billion

2011$877.875billion

2012$925billion

56. Denotethediagonalconnectingtheendpoints oftheedges a and b by d.Then a, b,and d formarighttriangle.BythePythagorean theorem,222 abd += .Theedge c andthe diagonals d and h alsoformarighttriangle,so 222 cdh += .Substituting2 d fromthefirst equation,weobtain2222 abch ++=

Section 2.1 The Coordinate Plane 149

49.

50. 51. 52.

60. a. If AB isoneofthediagonals,then DC isthe otherdiagonal,andbothdiagonalshavethe samemidpoint.Themidpointof AB is

⎜⎟ ⎝⎠

.Themidpointof

⎛⎞

⎜⎟ ⎝⎠

dDM dMP

(,)(800200)(1200400) 1000

=−+− = =−+−

b. 22 22

(,)(2000800)(3001200) 1500

Thedistancetraveledbythepilot =1000+1500=2500miles.

Sowehave 3 3.54 2 x x + =⇒= and 8 3.5 2 y + =⇒ 1 y =−

Thecoordinatesof D are(4,–1).

b. If AC isoneofthediagonals,then DB isthe otherdiagonal,andbothdiagonalshavethe samemidpoint.Themidpointof AC is

c. 22 (,)(2000200)(300400)

dDP =−+−

58. First,findtheinitiallengthoftheropeusing thePythagoreantheorem:

22 241026 c =+=

⎛⎞

.Themidpointof

⎜⎟ ⎝⎠

Sowehave 5 2.50 2 x x + =⇒= and 4 5.5 2 y + =⇒ 7 y =

Thecoordinatesof D are(0,7).

tx ttx ttx ttx −=+ −+=+ −+= −+=

22 22 2 (263)10 6761569100 5761569 5761569

59. Themidpointofthediagonalconnecting(0,0) and(a + b, c)is,. 22 abc + ⎛⎞ ⎜⎟ ⎝⎠ Themidpointof thediagonalconnecting(a,0)and(b, c)is also,. 22 abc + ⎛⎞ ⎜⎟ ⎝⎠ Becausethemidpointsof thetwodiagonalsarethesame,thediagonals bisecteachother.

c. If BC isoneofthediagonals,then DA isthe otherdiagonal,andbothdiagonalshavethe samemidpoint.Themidpointof BC is

⎜⎟ ⎝⎠

.Themidpointof DA is 23 (4,6),. 22 xy++ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ Sowehave 2 46 2 x x + =⇒= and 3 69 2 y y + =⇒= Thecoordinatesof D are(6,9).

61. Themidpointofthediagonalconnecting(0,0) and(x, y)is,22 xy ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ .Themidpointofthe diagonalconnecting(a,0)and(b, c)is ,.22 abc + ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ Becausethediagonalsbisecteach other,themidpointscoincide.So

22 xab + =⇒ xab =+ ,and 22 yc yc=⇒= Therefore,thequadrilateralisaparallelogram.

150 Chapter 2 Graphs and Functions

57. a.

3,250,00050013 1802.78miles == ≈

After t seconds,thelengthoftheropeis 26–3t. Nowfindthedistancefromtheboat tothedock, x,usingthePythagoreantheorem againandsolvingfor x: 222

2.1 Beyond the Basics

() 2534 ,3.5,3.5 22 ++ ⎛⎞ =

DC = 38 (3.5,3.5),. 22 xy++

⎜⎟ ⎝⎠

() 2338 ,2.5,5.5 22 ++

DB = 54 (2.5,5.5),. 22 xy++ ⎛⎞ =

() 5348 ,4,6 22 ++ ⎛⎞ =

62. a. Themidpointofthediagonalconnecting (1,2)and(5,8)is () 1528 ,3,5. 22 ++ ⎛⎞ = ⎜⎟ ⎝⎠

Themidpointofthediagonalconnecting (–2,6)and(8,4)is

2864 ,(3,5).22 −++ ⎛⎞ = ⎜⎟ ⎝⎠ Becausethe midpointsarethesame,thefigureisa parallelogram.

b. Themidpointofthediagonalconnecting (3,2)and(x, y)is 32 ,.22 xy++ ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ The midpointofthediagonalconnecting(6,3) and(6,5)is(6,4).So 3 69 2 x x + =⇒= and 2 46 2 y y + =⇒=

63. Let P(0,0), Q(a,0), R(a + b, c),and S(b, c)be theverticesoftheparallelogram. PQ = RS = 22 (0)(00). aa −+−= QR = PS =

() 2222 ()(0). abacbc +−+−=+

Thesumofthesquaresofthelengthsofthe sides=2222(). abc ++

(,)(). dPRabc =++

() 22 (,)(0). dQSabc =−+−

Thesumofthesquaresofthelengthsofthe diagonalsis ( ) ( ) 2222()() abcabc +++−+=

222222 22 aabbcaabbc ++++−++= 222222 2222(). abcabc ++=++

ab dRMb abab ab dPM abab

22 2222 22 2222

65. Let P(0,0), Q(a,0),and R(0, a)bethe verticesofthetriangle.

=−+−⎛⎞⎛⎞

(,)022 222

ab dQMa abab =−+−⎛⎞⎛⎞ ⎜⎟⎜⎟ ⎝⎠⎝⎠ + ⎛⎞⎛⎞ =−+= ⎜⎟⎜⎟ ⎝⎠⎝⎠ =−+−⎛⎞⎛⎞ ⎜⎟⎜⎟ ⎝⎠⎝⎠ + ⎛⎞⎛⎞ =−+−= ⎜⎟⎜⎟ ⎝⎠⎝⎠

caacaca acc

=+⇒=⇒=⇒ ==

UsingthePythagoreantheorem,wehave 2222222 12 22

66. Since ABC isanequilateraltriangleand O is themidpointof AB,thenthecoordinatesof A are(−a,0).

AB = AC = AB =2a.Usingtriangle BOC and thePythagoreantheorem,wehave ()2 22222 22222 2 433 BCOBOCaaOC aaOCaOCOCa =+⇒=+⇒ =+⇒=⇒= Thus,thecoordinatesof C are ( ) 0,3a and thecoordinatesof D are ( ) 0,3. a

Section 2.1 The Coordinate Plane 151

⎛⎞ ⎜⎟ ⎝⎠

64. Let P(0,0), Q(a,0),and R(0,b)bethevertices oftherighttriangle.Themidpoint M ofthe hypotenuseis,. 22 ab ⎜⎟⎜⎟ ⎝⎠⎝⎠ + ⎛⎞⎛⎞ =+−= ⎜⎟⎜⎟ ⎝⎠⎝⎠

22 2222

(,)022 222 (,)0022 222

Toshowthat M isthemidpointoftheline segment PQ,weneedtoshowthatthe distancebetween M and Q isthesameasthe distancebetween M and P andthatthis distanceishalfthedistancefrom P to Q (continuedonnextpage)

152 Chapter 2 Graphs and Functions

67. a. 2222 12121212

(,)3333 xxyyxxyydACxy ++−− ⎛⎞⎛⎞⎛⎞⎛⎞ =−+−=+ ⎜⎟⎜⎟⎜⎟⎜⎟ ⎝⎠⎝⎠⎝⎠⎝⎠ 2222 12121212 22 222222 (,)3333 xxyyxxyydCBxy ++−− ⎛⎞⎛⎞⎛⎞⎛⎞ =−+−=+ ⎜⎟⎜⎟⎜⎟⎜⎟ ⎝⎠⎝⎠⎝⎠⎝⎠ ()() ()() ()() ()() ()() ()() 2222 12121212 2222 1212121222 1212 22 1212 2222 (,)(,)33 2 33 (,) dACdCBxxyyxxyy xxyyxxyy xxyy dABxxyy −+−−+− +=+ −+−−+− =+=−+− =−+− So A, B,and C arecollinear. ()() 2222 121212121 (,)(,). 3333 xxyyxxyy dACdAB −+− ⎛⎞⎛⎞ =+== ⎜⎟⎜⎟ ⎝⎠⎝⎠ b. 2222 12121212 11 222222 (,)3333 xxyyxxyydADxy ++++ ⎛⎞⎛⎞⎛⎞⎛⎞ =−+−=+ ⎜⎟⎜⎟⎜⎟⎜⎟ ⎝⎠⎝⎠⎝⎠⎝⎠ 2222 12121212 22 22 (,)3333 xxyyxxyydDBxy ++−− ⎛⎞⎛⎞⎛⎞⎛⎞ =−+−=+ ⎜⎟⎜⎟⎜⎟⎜⎟ ⎝⎠⎝⎠⎝⎠⎝⎠ ()() ()() ()() ()() ()() () 2222 12121212 2222 12121212 22 1212 2222 (,)(,)33 2 33 , dADdDBxxyyxxyy xxyyxxyy xxyydAB +++−+− +=+ −+−−+− =+ =−+−= So A, B,and C arecollinear. ()() 2222 12121212 2 22222 (,)(,). 3333 xxyyxxyy dADdAB +++ +− ⎛⎞⎛⎞ =+== ⎜⎟⎜⎟ ⎝⎠⎝⎠ c. 12 12 12 12 22(1)42 333 22(2)15 333 212(4)7 333 222(1)4 333 xx yy xx yy +−+ == + + == + −+ == + + == Thepointsoftrisectionare 25 , 33 ⎛⎞ ⎜⎟ ⎝⎠ and 74 ,.33 ⎛⎞ ⎜⎟ ⎝⎠ 68.

11 22

2.1 Critical Thinking/Discussion/Writing

69. a. y-axis

b. x-axis

70. a. Theunionofthe x-and y-axes

b. Theplanewithoutthe x-and y-axes

71. a. QuadrantsIandIII

b. QuadrantsIIandIV

72. a. Theorigin

b. Theplanewithouttheorigin

73. a. Righthalf-plane

b. Upperhalf-plane

74. Let(x, y)bethepoint. Thepointliesinif

QuadrantI x >0and y >0

QuadrantII x <0and y >0

QuadrantIII x <0and y <0

QuadrantIV x >0and y <0

2.1 Maintaining Skills

75. a. 22 22111121 224442 xy ⎛⎞⎛⎞ +=+=+==

b. 22 2222221 2244 xy

Section 2.1 The Coordinate Plane 153

(continued) ()() 22 2121 PQxxyy =−+− () () ()() 22 1212 11 22 2121 22 2121 22 2121 22 22 44 1 2 xxyy MPxy xxyy xxyy xxyy ++ =−+−⎛⎞⎛⎞ ⎜⎟⎜⎟ ⎝⎠⎝⎠ =+⎛⎞⎛⎞ ⎜⎟⎜⎟ ⎝⎠⎝⎠ =+ =−+− Thus,wehave 1

MPPQ = Similarly,wecanshowthat

= Thus, M isthemidpointof

() 1212 ,,22 xxyy xy ++ ⎛⎞ = ⎜⎟ ⎝⎠ .

1 2 MQPQ

PQ,and

⎝⎠⎝⎠

⎜⎟⎜⎟

⎛⎞⎛⎞ +=+=+= ⎜⎟⎜⎟ ⎝⎠⎝⎠

() 2

234913

−++=⎡−−⎤++ ⎣⎦ =−+=+=

76. a. ()()()()

222

121112

3491625

=+=+=

a. 3 223 110 2323

+=+=+=−=

b. ()()()() 2222 22 124122

xy−++=−++

77.

xy xy

=−+−=−

b. 3 443 110 4343 xy xy +=+=+=−+= 78. a. () 1212 1212 112 xy xy +=+=+

+=+=+=+=

2 69

xx ⎛⎞ −+=−+ ⎜⎟ ⎝⎠ =−+ 80. 2

2 8

2 816

xx ⎛⎞ −+=−+ ⎜⎟ ⎝⎠ =−+ 81. 2 22239

24yyyyyy ⎛⎞ +=++=++ ⎜⎟ ⎝⎠ 82. 2

525

24yyyy ⎛⎞ ++=++ ⎜⎟ ⎝⎠ 83. 22 22 24 aa xaxxax ⎛⎞ −+=−+ ⎜⎟ ⎝⎠ 84. 22 22 24 yy

⎛⎞ ++=++ ⎜⎟ ⎝⎠

3232 112 3232 xy xy

79.

222

6 663

xxxx

222

884

xxxx

333

xxyxxy

Section 2.2 Graphs of Equations

2.2 Practice Problems

1. 21 yx=−+

x y = −x 2 + 1 (x, y)

−2 y =(−2)2+1(−2,−3)

−1 y =(−1)2+1(−1,0)

0 y =(0)2+1(0,1)

1 y =(1)2+1(1,0)

2 y =(2)2+1(2,−3)

2. Tofindthe x-intercept,let y =0,andsolvethe equationfor x: 2 0232 xx =+−⇒

()() 1 0212or2 2 xxxx =−+⇒==− .To findthe y-intercept,let x =0,andsolvethe equationfor y:

()() 2 203022.yy =+−⇒=−

The x-interceptsare 1 2 and−2;the y-intercept is−2.

3. Totestforsymmetryaboutthe y-axis,replace x with–x todetermineif(–x, y)satisfiesthe equation.

() 222211xyxy −−=⇒−= ,whichisthe sameastheoriginalequation.Sothegraphis symmetricaboutthe y-axis.

4. x-axis: () 3 223 , xyxy =−⇒=− whichisnot thesameastheoriginalequation,sothe equationisnotsymmetricwithrespecttothe x-axis.

y-axis: () 2323 , xyxy −=⇒= whichisthe sameastheoriginalequation,sotheequation issymmetricwithrespecttothe y-axis.

origin: ()()2323 , xyxy −=−⇒=− whichis notthesameastheoriginalequation,sothe equationisnotsymmetricwithrespecttothe origin.

5. 4277324ytt=−++

a. First,findtheintercepts.If t =0,then y =324,sothe y-interceptis(0,324). If y =0,thenwehave

So,the

Next,checkforsymmetry.

t-axis:4277324ytt −=−++ isnotthe sameastheoriginalequation,sothe equationisnotsymmetricwithrespectto the t-axis.

y-axis: ()() 4277324ytt =−−+−+⇒ 4277324,ytt=−++ whichisthesameas theoriginalequation.Sothegraphis symmetricwithrespecttothe y-axis.

origin: ()() 4277324ytt −=−−+−+⇒ 4277324,ytt −=−++ whichisnotthe sameastheoriginalequation.Sothegraph isnotsymmetricwithrespecttotheorigin. Now,makeatableofvalues.Sincethe graphissymmetricwithrespecttothe y-axis,if(t, y)isonthegraph,thensois (−t, y).However,thegraphpertainingto thephysicalaspectsoftheproblemconsists onlyofthosevaluesfor t ≥ 0.

t y = −t4 + 77t2 + 324 (t, y)

0324(0,324)

1400(1,400)

2616(2,616)

3936(3,936)

41300(4,1300)

51624(5,1624)

61800(6,1800)

71696(7,1696)

81156(8,1156)

90(9,0)

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154 Chapter 2 Graphs and Functions

()() ()() () 42 42 22 2 077324 773240 8140 99409,9,2 tt tt tt tttti =−++ −−= −+= +−+=⇒=−±

t-interceptsare(−9,0)and(9,0).

c. Thepopulationbecomesextinctafter9 years.

6. Thestandardformoftheequationofacircle is222 ()() xhykr −+−= (h, k)=(3,−6)and r =10 Theequationofthecircleis 22 (3)(6)100. xy−++=

7. ()()()() 22 2136,2,1,6 xyhkr −++=⇒=−= Thisistheequationofacirclewithcenter (2,−1)andradius6.

2.2 Basic Concepts and Skills

1. Thegraphofanequationintwovariables, suchas x and y,isthesetofallorderedpairs (a, b)thatsatisfytheequation

2. If(–2,4)isapointonagraphthatis symmetricwithrespecttothe y-axis,thenthe point(2,4)isalsoonthegraph.

3. If(0,−5)isapointofagraph,then−5isa y- interceptofthegraph.

4. Anequationinstandardformofacirclewith center(1,0)andradius2is () 22 14xy−+= .

5. False.Theequationofacirclehasbothan 2 x -termanda2 y -term.Thegivenequation doesnothavea2 y -term.

6. False.Thegraphbelowisanexampleofa graphthatissymmetricaboutthe x-axis,but doesnothavean x-intercept.

8. 2246120 xyxy++−−=⇒

224612xxyy++−= Nowcompletethesquare:

2244691249xxyy+++−+=++⇒

()() 22 2325xy++−= Thisisacirclewithcenter(−2,3)and radius5.

Inexercises7−14,todetermineifapointliesonthe graphoftheequation,substitutethepoint’s coordinatesintotheequationtoseeiftheresulting statementistrue.

7. onthegraph:(–3,–4),(1,0),(4,3);notonthe graph:(2,3)

8. onthegraph:(–1,1),(1,4), 5 ,0 3 ⎛⎞

;noton thegraph:(0,2)

9. onthegraph:(3,2),(0,1),(8,3);notonthe graph:(8,–3)

10. onthegraph:(1,1), 1 2, 2 ⎛⎞ ⎜⎟ ⎝⎠ ;notonthegraph: (0,0), 1 3, 3

11. onthegraph:(1,0), ( ) ( ) 2,3,2,3;noton thegraph:(0,–1)

12. Eachpointisonthegraph.

13. a. x-intercepts:–3,3;no y-intercepts

Section 2.2 Graphs of Equations 155

(continued)

⎜⎟ ⎝⎠

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

b. Symmetricaboutthe x -axis, y -axis,and origin.

14. a. No x-intercepts; y-intercepts:–2,2

b. Symmetricaboutthe x -axis, y -axis,and origin.

15. a. x-intercepts:−3,3; y-intercepts:–2,2

b. Symmetricaboutthe x -axis, y -axis,and origin.

16. a. x-intercepts:−3,3; y-intercepts:–3,3

b. Symmetricaboutthe x -axis, y -axis,and origin.

17. a. x-intercept:0; y-intercept:0

b. Symmetricaboutthe x -axis.

18. a. x-intercepts:,0,; ππ y-intercept:0

b. Symmetricabouttheorigin.

19. a. x-intercepts:−3,3; y-interceps:2

b. Symmetricaboutthe y -axis.

20. a. x-intercepts: 33 ,,,;2222 ππππ y-intercept:2

b. Symmetricaboutthe y -axis.

21. a. No x-intercept; y -intercept:1

b. Nosymmetries

22. a. x-intercept:1;no y-intercepts

b. Nosymmetries

23. a. x-intercepts:0,3; y-intercept:0

b. Symmetricaboutthe x -axis.

24. a. x-intercept:0; y-intercept:0

b. Symmetricaboutthe x -axis, y -axis,and origin.

156 Chapter 2 Graphs and Functions

26. 27. 28. 29. 30.

25.

Section 2.2 Graphs of Equations 157

31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41.

42.

47. Tofindthe x -intercept,let y =0,andsolvethe equationfor x:34(0)124. xx +=⇒= To findthe y-intercept,let x =0,andsolvethe equationfor y:3(0)4123. yy +=⇒= The x-interceptis4;the y -interceptis3.

48. Tofindthe x -intercept,let y =0,andsolvethe equationfor x: 0 15. 53 x x +=⇒= Tofindthe y-intercept,let x =0,andsolvetheequation for y: 0 13. 53 y y +=⇒=

The x-interceptis5;the y -interceptis3.

49. Tofindthe x -intercept,let y =0,andsolvethe equationfor x: 2 0684or xxx =−+⇒=

2. x = Tofindthe y -intercept,let x =0,and solvetheequationfor y: 2 06(0)8 y =−+⇒

8. y = The x-interceptsare2and4;the y-interceptis8.

50. Tofindthe x -intercept,let y =0,andsolvethe equationfor x: 2 05(0)66.xx =−+⇒=

Tofindthe y -intercept,let x =0,andsolvethe equationfor y: 2 0562or yyy =−+⇒=

3. y = The x-interceptis6;the y -intercepts are2and3.

51. Tofindthe x -intercept,let y =0,andsolvethe equationfor x: 22042.xx+=⇒=±

Tofindthe y -intercept,let x =0,andsolvethe equationfor y: 22 042. yy +=⇒=±

The x-interceptsare–2and2;the y-intercepts are–2and2.

52. Tofindthe x -intercept,let y =0,andsolvethe equationfor x: 2 093. xx =−⇒=±

Tofindthe y -intercept,let x =0,andsolvethe equationfor y: 2 903.yy=−⇒= The x-interceptsare–3and3;the y-intercept is3.

53. Tofindthe x -intercept,let y =0,andsolvethe equationfor x: 2 011. xx =−⇒=±

Tofindthe y -intercept,let x =0,andsolvethe equationfor y: 2 01 y =−⇒ nosolution. The x-interceptsare–1and1;thereisno y-intercept.

54. Tofindthe x -intercept,let y =0,andsolvethe equationfor x:(0)1 x =⇒ nosolution.

Tofindthe y -intercept,let x =0,andsolvethe equationfor y:(0)1 y =⇒ nosolution. Thereisno x-intercept;thereisno y-intercept.

55. Tofindthe x -intercept,let y =0,andsolvethe equationfor x: 2 01 xx =++⇒

()() () 2 1141113 212, x −±− −±− == which arenotrealsolutions.Therefore,thereareno x-intercepts.Tofindthe y-intercept,let x =0, andsolvetheequationfor y:

2 0011. y =++=

Thereareno x-intercepts;the y-interceptis1.

158 Chapter 2 Graphs and Functions

43. 44. 45. 46.

56. Tofindthe x -intercept,let y =0,andsolvethe equationfor x:

() 333300111.xxxx ++=⇒=⇒= Tofind the y-intercept,let x =0,andsolvethe equationfor y:

() 333 030111. yyyy ++=⇒=⇒= The x-interceptis1;the y -interceptis1. Inexercises57–66,totestforsymmetrywithrespect tothe x-axis,replace y with–y todetermineif(x,–y) satisfiestheequation.Totestforsymmetrywith respecttothe y-axis,replace x with–x todetermineif (–x, y)satisfiestheequation.Totestforsymmetry withrespecttotheorigin,replace x with–x and y with–y todetermineif(–x,–y)satisfiestheequation.

57. 21 yx −=+ isnotthesameastheoriginal equation,sotheequationisnotsymmetric withrespecttothe x -axis.

2 ()1yx=−+⇒ 21 yx=+ ,sotheequationis symmetricwithrespecttothe y-axis.

2 ()1yx −=−+⇒ 21 yx −=+ ,isnotthe sameastheoriginalequation,sotheequation isnotsymmetricwithrespecttotheorigin.

58. 22 ()11xyxy =−+⇒=+ , sotheequationis symmetricwithrespecttothe x-axis.

21 xy −=+ isnotthesameastheoriginal equation,sotheequationisnotsymmetric withrespecttothe y -axis.

2 ()1xy −=−+⇒ 21 xy −=+ isnotthesame astheoriginalequation,sotheequationisnot symmetricwithrespecttotheorigin.

59. 3 yxx−=+ isnotthesameastheoriginal equation,sotheequationisnotsymmetric withrespecttothe x -axis.

()3 yxx=−−⇒ 3 yxx=−−⇒

3 ()yxx =−+ isnotthesameastheoriginal equation,sotheequationisnotsymmetric withrespecttothe y -axis.

()33 yxxyxx −=−−⇒−=−−⇒

33 (), yxxyxx −=−+⇒=+ sotheequation issymmetricwithrespecttotheorigin.

60. 23 yxx−=− isnotthesameastheoriginal equation,sotheequationisnotsymmetric withrespecttothe x -axis.

3 2()() yxx=−−−⇒ 23 yxx=−+⇒

3 2() yxx =−− isnotthesameastheoriginal equation,sotheequationisnotsymmetric withrespecttothe y -axis.

332()()2 yxxyxx −=−−−⇒−=−+⇒

332()2 yxxyxx −=−−⇒=− ,sothe equationissymmetricwithrespecttothe origin.

61. 4252 yxx−=+ isnotthesameasthe originalequation,sotheequationisnot symmetricwithrespecttothe x-axis.

425()2() yxx =−+−⇒ 4252 yxx =+ ,so theequationissymmetricwithrespecttothe y-axis.

442 5()2()52 yxxyxx −=−+−⇒−=+ is notthesameastheoriginalequation,sothe equationisnotsymmetricwithrespecttothe origin.

62. 64232 yxxx −=−++ isnotthesameasthe originalequation,sotheequationisnot symmetricwithrespecttothe x-axis.

642 3()2()() yxxx =−−+−+−⇒

64232 yxxx =−++ ,sotheequationis symmetricwithrespecttothe y-axis.

642 3()2()() yxxx −=−−+−+−⇒

64232 yxxx −=−++ isnotthesameasthe originalequation,sotheequationisnot symmetricwithrespecttotheorigin.

63. 5332 yxx−=−+ isnotthesameasthe originalequation,sotheequationisnot symmetricwithrespecttothe x-axis.

53 3()2() yxx =−−+−⇒ 5332 yxx =− is notthesameastheoriginalequation,sothe equationisnotsymmetricwithrespecttothe y-axis.

5353 3()2()32 yxxyxx −=−−+−⇒−=−⇒

5353 (32)32 yxxyxx −=−−+⇒=−+ ,so theequationissymmetricwithrespecttothe origin.

Section 2.2 Graphs of Equations 159

64. 22 yxx−=− isnotthesameastheoriginal equation,sotheequationisnotsymmetric withrespecttothe x -axis.

2()2 yxx=−−−⇒ 22 yxx =− ,sothe equationissymmetricwithrespecttothe y-axis.

2()2 yxx −=−−−⇒ 22 yxx−=− isnot thesameastheoriginalequation,sothe equationisnotsymmetricwithrespecttothe origin.

65. 2222 ()2()121 xyxyxyxy −+−=⇒−= is notthesameastheoriginalequation,sothe equationisnotsymmetricwithrespecttothe x-axis.

2222 ()2()121 xyxyxyxy −+−=⇒−= is notthesameastheoriginalequation,sothe equationisnotsymmetricwithrespecttothe y-axis.

22 ()()2()()1 xyxy −−+−−=⇒

2221xyxy+= ,sotheequationissymmetric withrespecttotheorigin.

66. 2222()1616xyxy+−=⇒+= ,sothe equationissymmetricwithrespecttothe x-axis.

2222 ()1616 xyxy −+=⇒+= ,sothe equationissymmetricwithrespecttothe y-axis.

2222 ()()1616 xyxy −+−=⇒+= ,sothe equationissymmetricwithrespecttothe origin.

Forexercises67–80,usethestandardformofthe equationofacircle,222 ()() xhykr −+−= .

67. Center(2,3);radius=6

68. Center(–1,3);radius=4

69. Center(–2,–3);radius=11

70. Center 13 , 22 ⎛⎞ ⎜⎟ ⎝⎠ ;radius= 3 2

71. 22(1)4xy+−=

72. 22 (1)1 xy−+=

73. 22 (1)(2)2 xy++−=

74. 22 (2)(3)7 xy+++=

75. Findtheradiusbyusingthedistanceformula: 22 (13)(5(4))97 d =−−+−−= .

Theequationofthecircleis 22 (3)(4)97. xy−++=

160 Chapter 2 Graphs and Functions

76. Findtheradiusbyusingthedistanceformula:

22 (12)(15)255 d =−−+−== .The equationofthecircleis

22 (1)(1)25. xy++−=

77. Thecircletouchesthe x-axis,sotheradius is2.Theequationofthecircleis 22 (1)(2)4. xy−+−=

80. Findthecenterbyfindingthemidpointofthe diameter: () 2835,5,122 C +−+

Findthelengthoftheradiusbyfindingthe lengthofthediameteranddividingthatby2. 22 (28)(35)10010 d =−+−−==

Thus,thelengthoftheradiusis5,andthe equationofthecircleis 22 (5)(1)25. xy−+−=

78. Thecircletouchesthe y-axis,sotheradius is1.Theequationofthecircleis 22 (1)(2)1 xy−+−= .

81. a. 222240 xyxy+−−−=⇒

22224xxyy−+−=

Nowcompletethesquare:

222121411xxyy−++−+=++⇒

22 (1)(1)6 xy−+−= .Thisisacirclewith center(1,1)andradius6.

b. Tofindthe x-intercepts,let y =0andsolve for x:

79. Findthediameterbyusingthedistance formula:

22 (37)(64)104226 d =−−+−== Sotheradiusis26.Usethemidpoint formulatofindthecenter:

() 7(3)46,2,5 22 M +−+ ⎛⎞ == ⎜⎟

.Theequation ofthecircleis22(2)(5)26 xy−+−=

()

() 222 2 (1)(01)6116 151515 xx xxx −+−=⇒−+=⇒ −=⇒−=±⇒=±

Thus,the x-interceptsare ( ) 15,0 + and ( ) 15,0.

Tofindthe y-intercepts,let x =0andsolve for y:

()

() 222 2 (01)(1)6116 151515 yy yyy −+−=⇒+−=⇒ −=⇒−=±⇒=±

Thus,the y-interceptsare ( ) 0,15 + and ( ) 0,15.

Section 2.2 Graphs of Equations 161

⎝⎠

⎛⎞ == ⎝⎠

82. a. 2242150 xyxy+−−−=⇒

224215xxyy−+−=

Nowcompletethesquare:

2244211541xxyy−++−+=++⇒

22 (2)(1)20. xy−+−= Thisisacircle withcenter(2,1)andradius25.

b. Tofindthe x-intercepts,let y =0andsolve

for x: 22 (2)(01)20 x −+−=⇒

()() 22 2120219 219219 xx xx −+=⇒−=⇒ −=±⇒=±

Thus,the x-interceptsare ( ) 219,0 + and ( ) 219,0.

Tofindthe y-intercepts,let x =0andsolve

for y: 22 (02)(1)20 y −+−=⇒

() 22 4(1)20116 143,5 yy yy +−=⇒−=⇒ −=±⇒=−

Thus,the y-interceptsare(0,−3)and(0,5).

83. a. 22 2240 xyy++=⇒

22 2(2)0 xyy++=⇒ 2220.xyy++=

Nowcompletethesquare: 22222101(1)1.xyyxy +++=+⇒++=

Thisisacirclewithcenter(0,–1)and radius1.

b. Tofindthe x-intercepts,let y =0andsolve

for x: 222(01)100xxx ++=⇒=⇒=

Thus,the x-interceptis(0,0).

Tofindthe y-intercepts,let x =0andsolve

for y:

22 0(1)1110,2 yyy ++=⇒+=±⇒=−

Thus,the y-interceptsare(0,0)and(0,−2).

84. a. 22 3360 xyx++=⇒

22 3(2)0 xyx++=⇒ 2220.xxy++=

Nowcompletethesquare:

2222 2101(1)1.xxyxy +++=+⇒++=

Thisisacirclewithcenter(–1,0)and radius1.

b. Tofindthe x-intercepts,let y =0andsolve for x: 22 (1)01110,2 xxx ++=⇒+=±⇒=− Thus,the x-interceptsare(0,0)and(−2,0). Tofindthe y-intercepts,let x =0andsolve for y: 222 (01)100 yyy ++=⇒=⇒= Thus,the y-interceptis(0,0).

85. a. 222200.xyxxxy+−=⇒−+= Nowcompletethesquare:

2211 0 44xxy−++=+⇒ 2 2 11 . 24xy

Thisisacirclewith center 1 ,0 2 ⎛ ⎞

andradius 1 2

b. Tofindthe x-intercepts,let y =0andsolve for x: 2 2 1111 0 2422 xx ⎛⎞−+=⇒−=±⇒

0,1. x = Thus,the x-interceptsare(0,0) and(1,0).Tofindthe y-intercepts,let x =0 andsolvefor y: 2 22 1111 0 2444 yy

200.yy=⇒= Thus,the y-interceptis(0,0).

86. a. 2222101.xyxy++=⇒+=− The radiuscannotbenegative,sothereisno graph.

b. Therearenointercepts.

2.2 Applying the Concepts

87. Thedistancefrom P(x, y)tothe x-axisis x whilethedistancefrom P tothe y-axisis. y

Sotheequationofthegraphis xy =

88. Thedistancefrom P(x, y)to(1,2)is 22 (1)(2) xy−+− whilethedistancefrom P to(3,–4)is22 (3)(4) xy−++

Sotheequationofthegraphis

2222 (1)(2)(3)(4) xyxy −+−=−++⇒ 2222 (1)(2)(3)(4) xyxy −+−=−++⇒

162 Chapter 2 Graphs and Functions

⎛⎞−+= ⎜⎟ ⎝⎠

⎟

⎠

⎜

⎝

⎜⎟ ⎝⎠

⎛⎞−+=⇒+=⇒ ⎜⎟ ⎝⎠

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