(Solution Manual) Rogawski's Calculus Early Transcendentals for AP

Rogawski,Cannon, 2

Math & Statistics

Solution Manual

2nd Edition

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Solution Manual for Rogawski's Calculus Early Transcendentals for AP, The downloadable file consists of 760 pages, and the total size of the file is 52.97 MB. Purchase now only for highly discounted price.

Author(s): Rogawski, Cannon

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Rogawski's Calculus Early Transcendentals ED.2 Jonathan D. Rogawski , Ray Cannon Test Bank 2021 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. CHAPTER 1 Functions, Graphs, and Limits Section 1.1 The Cartesian Plane and the Distance Formula .................................... 2 Section 1.2 Graphs of Equations ............................................................................... 8 Section 1.3 Lines in the Plane and Slope ................................................................ 17 Quiz Yourself ............................................................................................................... 27 Section 1.4 Functions............................................................................................... 30 Section 1.5 Limits .................................................................................................... 38 Section 1.6 Continuity ............................................................................................. 45 Review Exercises .......................................................................................................... 51 Test Yourself ............................................................................................................... 62 2 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. CHAPTER 1 Functions, Graphs, and Limits Section 1.1 The Cartesian Pl ane and the Distance Formula 1. 2. Skills Warm Up 1. () ( ) ()2 22 236 1 5 3 6 93 6 45 35−+ ⎡ − − ⎤ = − +⎣⎦ =+ = = 2. () ( ) () ()2 22 220 7 3 2 4 41 6 20 25−− + ⎡ −−−⎤= − +−⎣⎦ =+ = = 3. () 54 1 22+−= 4. () 31 4222−+− −== − 5. 27 12 3 3 2 3 5 3+=+= 6. 81 8 2 2 3 2 2−=−= − 7. () () () () () () () () () ()22 2222 22 2 2 2 237 4 4 5 37 4 4 5 37 4 4 5 33 4 5 39 4 5 33 6 36 36 36 3, 9x x x x x x x x x x−+ −= ⎛⎞−+ − =⎜⎟⎝⎠ −+ −= −+ = −+ = −= −= ± −= −± = =−∓ 8. () ( ) () ( ) () () ( ) () () ()22 2222 22 2 2 2 262 2 5 2 62 2 5 2 62 2 5 2 42 5 2 16 2 52 23 6 26 62 62 8, 4y y y y y y y y y y−+ − −= ⎛⎞−+ − − =⎜⎟⎝⎠ −+ − −= +−− = +−− = −− = −− = ± −= ±+ =− =−∓ 9. () ()572 51 4 19x x x+−= +− = = 10. 732 76 1y y y−+=− −+= − = x y −1 −2 −3 −4 −5 2 3 −2 −3 −4 −5 −6 1 2 3 (1, −6) (−2, −4) (1, −1) (2, 0) (−5, 3) x y −2 −4 2 4 6 −2 −4 −6 2 4 6 (−3, 5) (−6, −1) (5, 1) (2, −2) (0, −4) Section 1.1 The Cartesian Plane and the Distance Formula 3 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 3. (a) (b) () ()2253 51 41 6 2 5 d=− + − = + = (c) Midpoint ()35 15,4 , 322++⎛⎞==⎜⎟⎝⎠ 4. (a) (b) () ( )223 3 2 2 36 16 2 13 d=− −++ = += (c) Midpoint ()()22 33,0 , 022⎛+ − ⎞−+==⎜⎟ ⎝⎠ 5. (a) (b) ()() ( )2 232 12 5 1 43 6 21 0d=⎡ − ⎤ +−⎣⎦ =+ = (c) Midpoint ()()() 12 32 1 5 1,, 222 2⎛+ − + − ⎞ ⎛⎞== − −⎜⎟ ⎜⎟⎝⎠ ⎝⎠ 6. (a) (b) 225 2 1 1 16 65163 3 3 6 9 6d⎛⎞ ⎛ ⎞=− + + =+ =⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ (c) Midpoint ()()() 56 23 1 13 31,,22 4 3⎛+ −⎞ ⎛⎞==⎜⎟ ⎜⎟⎝⎠ ⎝⎠ 7. (a) (b) () ( )2242 1 42 4 144 23 7d=− + − =+ = (c) Midpoint ()24 21 4,3 , 822++⎛⎞==⎜⎟⎝⎠ 8. (a) (b) () ( )223 1 7 1 16 64 4 5 d=− −++= += (c) Midpoint ()31 71,1 , 322−+ −⎛⎞== −⎜⎟⎝⎠ 9. (a) (b) ()() ()()2 2 2275 32 12 5 144 25 169 13=− − + − − =+ =+ ==d (c) ()() 75 32Midpoint , 22 21, 22 11, 2⎛+ − + −⎞=⎜⎟ ⎝⎠ ⎛⎞=⎜⎟⎝⎠ ⎛⎞=⎜⎟⎝⎠ 5 4 3 2 1 5 4 3 2 1x(5, 5) (4, 3) (3, 1)y x ( 3, 2) − (0, 0) (3, 2) − − 3 − 2 − 12 3 3 2 1 − 1 − 2 − 3 y ( )( )2 −6−44 2x1 , 12 y () −2 −4 − , − 2 1 2 − , − 53 2 x 11y ( ) , 1 5 6 ( ) , 3 41 3 ( ) , −2 31 314 10 6 2 8 6 4 2x(4, 14) (3, 8) (2, 2)y x ( 3, 7) − (1, 1) − ( 1, 3) − − 6 − 4 − 22 4 − 2 8 6 4 y −4 −6 −8 2468−1 −2 −3 −41234 x (−5, −2)(7, 3) 1 2(1, )y4 Chapter 1 Functions, Graphs, and Limits © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 10. (a) (b) () ( )()22 2282 4 1 2 68 36 64 100 10=− + − − − =+ =+ ==d (c) ()() ()12 4 28Midpoint , 22 10 16, 22 5, 8⎛− + − ⎞+=⎜⎟ ⎝⎠ −⎛⎞=⎜⎟⎝⎠ =− 11. (a) (b) () ( ) ()2 20.5 0 6 4.8 0.25 116.64 116.89d=− + − − =+ = (c) Midpoint ()00 . 54 . 86,0 . 2 5 , 0 . 622+− +⎛⎞==⎜⎟⎝⎠ 12. (a) (b) () ( )222.7 5.2 1.8 6.4 62.41 21.16 83.57d=−− + − =+ = (c) Midpoint () ()5.2 2.7 6.4 1.8,22 1.25, 4.1⎛+ − ⎞ +=⎜⎟ ⎝⎠ = 13. (a) 4 a= 3b= () ()2240 30 1 69 5 c=−+ −= + = (b) 22 216 9 25 ab c+=+ == 14. (a) () ( )2213 1 1 1 144 0 12 a=− + − =+ = () ( )2213 13 6 1 0 25 5 b=−+ − =+ = () ( )2213 1 6 1 144 25 13 c=− + − =+ = (b) 22 2144 25 169 ab c+= += = 15. (a) 10 a= 3b= () ()227 3 4 1 100 9 109 c=+ + − = + = (b) 22 2100 9 109 ab c+= + = = 16. (a) () ( )2262 22 1 60 4 a=−+ − += + = () ()2222 52 04 9 7 b=−+ +=+= () ()222 6 5 2 16 49 65 c=− + + =+ = (b) 22 216 49 65 ab c+=+== 17. () ()22 1 30 71 93 6 45 35d=− + − =+ = = () ( )22 2 40 1 1 16 4 20 25d=− + − − =+ = = () ( )2 2 3 34 7 1 16 4 65d=−+ ⎡ − − ⎤⎣⎦ =+ = Because 22 2 12 3 , dd d+= the figure is a right triangle. −2 2468 1 0 1 2−2 −4 −6 −8 −10 −122 x (2, −12)(5, −8)(8, −4)y x y 6 6 4 2 −2 −4 −6 −2 −4 −6 4 2 (0, −4.8) (0.25, 0.6) (0.5, 6) x y −2 2 4 6 2 4 6 8 (−2.7, 1.8) (1.25, 4.1) (5.2, 6.4) −2 6 4 2 6 2 x d 2 d 1 3 d (4, −1) (3, 7) (0, 1) y Section 1.1 The Cartesian Plane and the Distance Formula 5 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 18. () ( ) () () () ( )22 22 2223 42 2 54 2 9 31 2 3 4 2 5 2 9 21 43 9 4 9 5 8a b c=− −+− = + = =− + + =+ = =− −++ = += Because ab=the figure is an isosceles triangle. [ Note: It is also a right triangle since 22 2. ab c+= ] 19. () () () () () () () ()22 1 22 2 22 3 22 410 20 14 5 31 32 41 5 23 13 14 5 02 01 41 5d d d d=− + −= + = =− + − =+ = =− + − = + = =− + − = + = Because 1234 , dddd=== the figure is a parallelogram. 20. () () () () () () () ( )22 22 22 2230 71 93 6 3 5 34 74 19 1 0 41 4 2 93 6 3 5 10 21 19 1 0a b c d=− + − = + = =− + − = + = =− + + = + = =− + − − = + = Because ac=and ,bd= the figure is a parallelogram. 21. d=() ( ) () ()22 2 2 214 0 5 21 7 5 21 7 2 5 28 0 42 0 4, 2x xx xx xx xx x−+ − − = −+= −+= −− = −+ = =− 22. d=() () () ()22 2 2 222 1 5 41 3 5 41 3 2 5 41 2 0 26 0 2, 6x xx xx xx xx x−+ += −+= −+= −−= +− = =− 23. d=() ()22 2 2 230 0 8 98 96 4 55 55y y y y y−+ −= += += = =± 24. d=() () () ()22 2 255 1 8 18 16 4 18 18 7, 9y y y y y y−+ −= −= −= −= ± =± =− 25. () ()22 2250 12 42 18 38 24 2020 2 505 44.9 ydd=− + − =+ = =≈ 26. () ()22 2233 12 37 18 21 19 441 361 802 28.3 ydd=− + − =+ =+ =≈ x ( 2, 4)− − 3 − 2 − 13 4 14 3 2 − 2 − 3 (3, 2)d 1 d 3 d 2 (1, 3)−y 3 2 1 3 2 1 x d 4 d 2 d 3 d 1 (3, 3) (2, 1) (1, 2) (0, 0) y x (3, 7) d 2 d 1 d 3 d 4 (0, 1) (4, 4) (1, 2) − 8 6 4 − 2 − 24 68 y 6 Chapter 1 Functions, Graphs, and Limits © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 27. The number of cable high- speed Internet customers increases each year. 28. The number of cellular telephone subscribers increases each year. 29. (a) March 2009: 7600 July 2009: 9200 July 2010: 10,500 (b) April 2010: 11,000 May 2010: 10,100 Decrease: 10,100 11,000 900−= Percent decrease: 9000.082 8.2%11,000≈= 30. (a) 1996: $120,000 2003: $180,000 2008: $200,000 (b) 2001: $155,000 2002: $165,000 Increase: 165,000 155,000 $10,000− = Percent increase: 10,0000.065155,000 6.5%≈ = 31. (a) Revenue ()2007 2009 329.7 538.9, 22 2008, 434.3++⎛⎞=⎜⎟⎝⎠ = Revenue estimate fo r 2008: $434.3 million Profit ()2007 2009 19.7 30.7, 22 2008, 25.2++⎛⎞=⎜⎟⎝⎠ = Profit estimate for 2008: $25.2 million (b) Actual 2008 revenue: $422.4 million Actual 2008 profit: $24.4 million (c) Yes, the revenue and profit increased in a linear pattern from 2007 to 2009. (d) 2007 expense: 329.7 19.7 $310 million −= 2008 expense: 422.4 24.4 $398 million −= 2009 expense: 538.9 30.7 $508.2 million −= (e) Answers will vary. 32. (a) Revenue ()2007 2009 6484.5 7773.3, 22 2008, 7128.9++⎛⎞=⎜⎟⎝⎠ = Revenue estimate fo r 2008: $7128.9 million Profit ()2007 2009 23.7 285.6, 22 2008, 154.65++⎛⎞=⎜⎟⎝⎠ = Profit estimate for 2008: $154.65 million (b) Actual 2008 revenue: $7230.1 million Actual 2008 profit: $149.7 million (c) Yes, the revenue and profit increased in a linear pattern from 2007 to 2009. (d) 2007 expense: 6484.5 23.7 $6460.8 million −= 2008 expense: 7230.1 149.7 $7080.4 million −= 2009 expense: 7773.3 285.6 $7487.7 million −= (e) Answers will vary. 2000 2010 050 2000 2010 050 2000 2010 050 2000 2010 0 300 2000 2010 0300 2000 2010 0300 Section 1.1 The Cartesian Plane and the Distance Formula 7 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 33. (a) (b) The larger the clinic, the more patients a doctor can treat. 34. (a) 500 pickups were sold in the year 2007. (b) About 500 pickups were sold in the year 2007. (c) The number of pickups sold each year is decreasing. 35. The vertex () 3, 1−− is translated to () 6, 6 .−− The vertex ()0, 0 is translated to () 3, 5 .−− The vertex ()1, 2−− is translated to ()4, 7 .−− 36. The vertex ()0, 2 is translated to ()2, 6 . The vertex ()1, 3 is translated to ()3, 7 . The vertex ()3, 1 is translated to ()5, 5 . The vertex ()2, 0 is translated to ()4, 4 . 37. Midpoint 12 12,22x xy y++⎛⎞=⎜⎟⎝⎠ The point one-fourth of the way between ()11,xy and ()22,xyis the midpoint of the line segment from ()11,xyto 12 12,,22x xy y++⎛⎞ ⎜⎟⎝⎠ which is 12 12 1112 1233 22,, .22 4 4xx yyxyx xy y++⎛⎞++⎜⎟ ++⎛⎞= ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ The point three-fourths of the way between () 11,xy and ()22,xy is the midpoint of the line segment from 12 12,22x xy y++⎛⎞ ⎜⎟⎝⎠to () 22,,xywhich is 12 12 2212 1233 22,, .22 4 4xx yyxyx xy y++⎛⎞++⎜⎟ ++⎛⎞= ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ Thus, 1212 12 1233,, , ,44 2 2x xyy xx yy++ + +⎛⎞ ⎛ ⎞ ⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠and 12 1233,44x xy y++⎛⎞ ⎜⎟⎝⎠ are the three points that divide the line segment joining () 11,xy and () 22,xy into four equal parts. 38. (a) ()() () ()31 4 3 2 1 77,,44 4 4 1421 5 3,,22 2 2 13 4 23 1 13 5,,44 4 4⎛+ − − ⎞ ⎛⎞=− ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ +− −⎛⎞ ⎛ ⎞=− ⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ ⎛+ −+ −⎞ ⎛⎞=− ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ (b) ()() () ()32 033 0 39,,44 2 4 2030 3,1 ,22 2 23 0 33 0 13,,44 2 4⎛−+ −+⎞ ⎛⎞=− − ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ −+ −+⎛⎞ ⎛ ⎞=−− ⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ ⎛− + − + ⎞ ⎛⎞=− − ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ 39. To show 121222,33x xyy++⎛⎞ ⎜⎟⎝⎠is a point of trisection of the line segment joining () 11,xyand () 22,,xywe must show that 121 2dd= and 12 3 . dd d+= () ()22 12 12 111 22 21 21 22 21 2122 33 33 1 3xx yydxy xx yy xx yy++⎛⎞ ⎛⎞=− + −⎜⎟ ⎜⎟⎝⎠ ⎝⎠ −−⎛⎞ ⎛⎞=+⎜⎟ ⎜⎟⎝⎠ ⎝⎠ =− + − () ()22 12 12 22 2 22 21 21 22 21 2122 33 22 22 33 2 3xx yydx y xx yy xx yy++ ⎛⎞ ⎛⎞=− + −⎜⎟ ⎜⎟⎝⎠ ⎝⎠ −−⎛⎞ ⎛⎞=+⎜⎟ ⎜⎟⎝⎠ ⎝⎠ =− + − () ()22 32 1 2 1dx x y y=− + − Therefore, 121 2dd= and 12 3 . dd d+= The midpoint of the line segment joining 121222,33x xyy++⎛⎞ ⎜⎟⎝⎠ and () 22,xy is Midpoint 12 12 22 12 1222 33,22 22,.33xx yyx y xx yy++⎛⎞++⎜⎟ =⎜⎟⎝⎠ ++⎛⎞=⎜⎟⎝⎠ Small clinicLarge clinic Number of ear infections20 30 40 50 60 70 10 Number of doctors1 2 3 4 Medium clinic x d 3 d 1 d 2 (, ) xy 11 (, ) xy 22 2+ xx 12 2+ yy 12 33 , ( ( y 8 Chapter 1 Functions, Graphs, and Limits © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 40. (a) ()()() () ()()21 4 2 2 1,2 , 133 12 4 22 1,3 , 033⎛+ − + ⎞=− ⎜⎟ ⎝⎠ ⎛+ −+ ⎞= ⎜⎟ ⎝⎠ (b) ()() () ()22 0 23 0 4,, 233 3 22 0 32 0 2,, 133 3⎛−+ −+⎞ ⎛⎞=− − ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎛− + − + ⎞ ⎛⎞=− − ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ Section 1.2 Graphs of Equations 1. The graph of 2 yx=− is a straight line with -intercepty at ()0, 2 .− So, it matches (e). 2. The graph of 1 22 yx=− + is a straight line with -intercepty at()0, 2 . So, it matches (b). 3. The graph of 22 yx x=+ is a parabola opening up with vertex at () 1, 1 .−− So, it matches (c). 4. The graph of 29 yx=− is a semicircle with intercepts ()() 0, 3 , 3, 0 , and ()3, 0 .− So, it matches (f). 5. The graph of 2 yx=−has a y-intercept at () 0, 2−and has x-intercepts at ()2, 0− and ()2, 0 . So, it matches (a). Skills Warm Up 1. 51 2 51 2 12 5yx yx xy−= =+ += 2. 15 15yx yx−= − =− 3. ()3 3 321 21 1 2xy y yx yx+= += =+ 4. 22 22 22 260 6 6 6xx y yx x yx x yx x+− −= −=−− =+ − =+ − 5. () () () () () () ()22 22 2 2 2 221 9 19 2 19 2 92 1 94 4 1 54 1xy yx yx yx xx xx−+ += += − − += − − ⎛⎞=− −−⎜⎟⎝⎠ =−− + − = +−− 6. () () () () () () ()22 22 2 2 2 265 8 1 58 1 6 58 1 6 58 1 6 58 1 1 2 3 6 54 5 1 2xy yx yx yx xx xx++ −= −= − + −= − + =+ − + =+ − + + =+ − − 7. () 52 1 0=−= − y 8. ()33 4 5=−= y 9. ()222 1 9=+= y 10. () ()242 4 7 1=−+ − − = y 11. () ()232 12xx xx−+ −− 12. () ()256 23xx xx++ ++ 13. ()2 29 4 3 23 yy y−+ − 14. ()2 249 4 7 27 yy y−+ − Section 1.2 Graphs of Equations 9 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 6. The graph of 3yx x=− has intercepts at () ()0 ,0, 1 ,0, a n d ()1, 0 .− So, it matches (d). 7. 23 yx=+ 8. 32yx=− + 9. 23 yx=− 10. 26 yx=+ 11. ()21 yx=− 12. ()25 yx=− 13. 32 yx=+ 14. 31 yx=− x −2 3 2− −1 0 1 2 y −1 0 1 3 5 7 x −1 0 2 3 1 2 y 5 2 0 −1 −4 x −2 −1 0 1 2 3 y 1 −2 −3 −2 1 6 x −2 −1 0 1 2 y 10 7 6 7 10 x −2 −1 0 1 2 y 9 4 1 0 1 x 3 4 5 6 7 y 4 1 0 1 4 x −2 −1 0 1 2 y −6 1 2 3 10 x 0 1 −1 2 y 1 0 2 −7 ( ) − 4 − 3 − 1 2 1 3 4 4 3 1 − 2 − 3 − 4 y x (0, 3) , 0 3 2 − x − 2 3 1 2 1 − 1 − 2 (0, 2) y 1 ( ) , 02 3 y x 3, 0 − ) ( 3, 0 ) ( (0, −3) −4 −3 −1 1 2 3 4 −4 1 3 4 x (0, 6) − 6 − 3 6 3 12 9 3 y(0, 1) (1, 0) 1 2 3 4 5 −1 −1 4 5 6 −2 −2 −3 y x x (5, 0) 10 8 6 4 2 10 8 2 4 y 6 3 2 1 −3−25 4 3 1 −1x(0, 2) ),, 0 ( 2−3y x (0, 1) (1, 0) − 1 2 y10 Chapter 1 Functions, Graphs, and Limits © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 15. 1 yx=− − 16. 1 yx=+ 17. 1 yx=+ 18. 2 yx=− − 19. 1 3yx=− 20. 1 1yx=+ 21. 24 xy=− 22. 24 x y=− x 1 2 3 4 5 y 0 −1 −1.41 −1.73 −2 x −1 0 1 2 3 y 0 1 2 3 4 x −3 −2 −1 0 1 y 2 1 0 1 2 x 2 0 1 3 4 y 0 −2 −1 −1 −2 x −1 0 1 2 2.5 3.5 4 5 6 y 1 4− 1 3− 1 2− −1 −2 2 1 1 2 1 3 x −4 −3 −2 −1.5 −0.5 0 1 2 y 1 3− 1 2− −1 −2 2 1 1 2 1 3 x 5 0 −3 −4 y ±3 ±2 ±1 0 x 0 3 4 y ±2 ±1 0 x y 5 3 2 1 −1 −2 −3 −1 4 3 2 (1, 0) x − 2 3 1 1 2 3 4 (0, 1) ( 1, 0) − y −4−3−2−1−1123456 −21234y x (0, 1)(−1, 0) x (2, 0) (0, 2) − 12 34 1 − 1 − 2 − 3 y−3−2−13 2 1 6 5 4xy (0, − )1 3 x y −2 1 2 −1 −2 −3 1 3 (0, 1) 1 −1 −2 −3 −3−13 1 x(0, 2) (−4, 0)y (0, −2) x (0, 2) (0, 2) − 12 3 1 − 1 (4, 0) y Section 1.2 Graphs of Equations 11 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 23. Let 0. y= Then, () 3 220 3 0 .x x−− = = Let 0. x= Then, ()20 3 0 3.y y−−= =− -intercept:x()3 2,0 -intercept: y ()0, 3− 24. Let 0. y= Then, () 5 442 0 5 0 .x y−− = = Let 0. x= Then, () 5 240 2 5 0 y y−− = =− x-intercept: ()5 4,0 y-intercept: ()5 20,− 25. Let 0. y= Then, 202 xx=+ − () () 02 1 xx=+ − 2, 1. x=− Let 0. x= Then, () ()200 2 2y y=+ − =− x-intercepts: () ( )2, 0 , 1, 0− y-intercept: ()0, 2− 26. Let 0. y= Then, () ()204 3 03 1 1, 3.xx xx x=−+ =− − = Let 0. x= Then, () ()204 0 3 3.y y=−+ = x-intercepts: () ()1, 0 , 3, 0 y-intercept: ()0, 3 27. Let 0. y= Then, 2 204 4 2.x x x=− = =± Let 0. x= Then, ()240 y=− 2. y= x-intercepts: ()() 2, 0 , 2, 0− y-intercept: ()0, 2 28. Let 0.=y Then, ()()2209 09 09 9x x x x=+ =+ =+ −= Let 0. =x Then, 09 9 3=+ = =y y y x-intercept: ()9, 0− y-intercept: ()0, 3 29. Let 0. y= Then, () ()2402 02 2 2.x x xx x−=− =− + =± Let 0. x= Then, () ()204 02 2.y y−=− = x-intercept: Because the equation is undefined when 2, x= the only x-intercept is ()2, 0 .− y-intercept: ()0, 2 12 Chapter 1 Functions, Graphs, and Limits © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 30. Let 0. y= Then, ()2302 03 3, 0.x x x xx x+= =+ =− Let 0. x= Then, () () ()203 0 20 undefined.y y+= = x-intercept: Because the equation is undefined when 0, x= the only x-intercept is ()3, 0 .− y -intercept: Because the equation is undefined when 0, y= there is no -intercept. y 31. Let 0. y= Then, ()()22 204 0 0 0 0.xx x x−+ = = = Let 0. x= Then, () ()2200 4 0 0.yy y−+ = = x-intercept: ()0, 0 y-intercept: ()0, 0 32. Let 0. y= Then, ()()22 220 8 0 1 1 1.xx x x+− = =− =± − Let 0. x= Then, () ()22 1 820 8 0 1 .yy y+− = = -intercept: x Because the equation has no real roots when 0, y= there is no -intercept.x y-intercept: ()1 80, 33. () ()22 2 2200 4 16−+ −= +=xy xy 34. () ()22 2 2200 5 25−+ −= +=xy xy 35. () ( )() () ()2 2 2 2221 3 21 9−+ − −= −+ +=xy xy 36. ()() () () ()2 2 2 2243 2 43 4−− + − = ++ −=xy xy 37. The radius is the distance between ()1, 5− and ()1, 1 .− ()() ()2 2 2211 5 1 04 16 4=− − −+− =+ ==r Using the center ()1, 1− and the radius 4:=r ()() () () ()2 2 2 2211 4 11 1 6−− + − = ++−=xy xy −1 −2 −3 123−1 −2 −3123 x(0, 0)yxy −2 −4 −62 4 6 −2 −4 −6246 (0, 0) 1234 −1 −2 −42 x (2, −1)y xy (−4, 3) −1 −2 −3 −4 −5 −6 −7 −8112345678 −1 −2 −3 −4 −51 2 3−1 −21234 x(−1, 5) (−1, 1)y Section 1.2 Graphs of Equations 13 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 38. The radius is the distance between ()1, 1− and () 3, 2 .− ()() () ()2 2 2 231 2 1 43 16 9 25 5r=− − + − − =+ − =+ = = Using the center () 3, 2−and the radius 5:=r () ( )() () ()2 2 2 2232 5 32 2 5−+ − −= −+ +=xy xy 39. The diameter is the distance between ()6, 8−− and ()6, 8 . ()() ()()22 2266 88 12 16 144 256 400 20d=− − + − − =+ =+ = = The radius is one-half the diameter: 2010.2==r The center is the midpoint of the diameter: ()6688, 0, 0 .22−+ −+⎛⎞= ⎜⎟⎝⎠ () ()22 2 2200 1 0 100−+ −= +=xy xy 40. The diameter is the distance between () 0, 4−and ()6, 4 . () ( )()2 2 2260 4 4 68 36 64 100 10d=− + − − =+ =+ = = The radius is one-half the diameter: 105.2==r The center is the midpoint of the diameter: ()06 44, 3, 022++⎛⎞= ⎜⎟⎝⎠ () () ()22 2 2 230 5 32 5−+ −= −+=xy xy 41. Set the two equations equal to each other. 22 1 33 1−+= − −= − =xx x x Substitute 1 =x into one of the equations. ()12 1=−+= y The point of intersection is ()1, 1 . 42. Set the two equations equal to each other. 3 278 21 4 31 6 53 0 6−+= − −+= − −= − =xx xx x x Substitute 6 =x into one of the equations. ()67 1=−+= y The point of intersection is ()6, 1 . xy (3, −2)−4 246 1 0 −2 −6 −8 −1024 −2 −4 −6 −8 2468 −4 −6 −82468 x(0, 0) (−6, −8)(6, 8)yxy (3, 0) −3 1234567 9 −1 −4 −5 −612456