336 Chapter 11 Additional TopicsHomework 11.1 1.77or 7xx x== = 3.3 x= Sincexis always nonnegative, the solution set for3 x= is the empty set,. 5.5 3 155 18185xxx === 18 18or5 5x x = =7.4 11 24 13134xxx = == Sincexis always nonnegative, the solution set for 134x= is the empty set,. 9.2 5 x + =2 5or 2 57or 3x xx x+ = + == = 11.5 05 05xxx = == 13.3 1 11 x =3 1 11or3 1 113 10or3 1210or 43x xx xx x = == == = 15.5 2 9 x =5 2 9or5 2 92 14or 2 47or 2x xx xx x = = = == = 17.2 9 6 x + = Since2 9 x +is always nonnegative, the solution set for2 9 6 x + = is the empty set,. 19.2 5 85 4xx+ =+ = 5 4or 5 49or 1x xx x+ = + == = 21.3 9 03 9 03 93xxxx = === 23.6 2 7 4 26 2 7 62 7 1xxx + = = = 2 7 1or2 7 12 6or2 83or 4x xx xx x = == == = 25. 1 5 72 3 6x =1 5 7 1 5 7or2 3 6 2 3 61 7 5 1 7 5or2 6 3 2 6 31 1 1 17 or 2 2 2 6171 or3x xx xx xx x = == + = += == = SSM: Intermediate AlgebraHomework 11.1 337 27.4.7 3.9 8.84.7 12.72.70xxx == 2.70or 2.70 x x 29.3.7 2.1 5.8 9.7 10.23.7 2.1 5.8 19.92.1 5.8 5.38xxx+ =+ =+ 2.1 5.8 5.38 or2.1 5.8 5.382.1 0.42 or 2.1 11.180.2 or5.32x xx xx x+ = + = = = = = 31.44 4xx< < < Interval:( ) 4, 4 4 0433.33or 3xx x Interval:( ] [ ) , 3 3, 3 03 35.3 x< Sincexis nonnegative, the inequality3 x< has an empty set solution,. 37.2 5 32 84xxx >>> 4or 4 x x < >Interval:( ) ( ) , 4 4, 4 0 439.2 5 85 102xxx 2or 2 x x Interval:( ] [ ) , 2 2, 2 0241.6 7 x 6 7or 6 71or 13x xx x Interval:( ] [ ) , 1 13, 1 0 13 43.2 5 15 x + 15 2 5 1520 2 1010 5xxx + Interval:[ ] 10, 5 0 5 10 45.7 15 4 x + > Since7 15 x +is always nonnegative, the solution set for the inequality7 15 4 x + > is the set of all real numbers. Interval:( ) , 047.2 3 1 172 3 163 8xxx + 3 8or 3 85or 11x xx x Interval:( ] [ ) , 5 11, 0 5 11 49.3 5 2 1 83 5 2 95 2 3xxx + 5 2 3or5 2 32 8or 2 24or 1x xx xx x Interval:( ] [ ) ,1 4, 0 4 1 Homework 11.1SSM: Intermediate Algebra 338 51.5 2 3 7 25 2 3 52 3 1xxx+ + >> 6.72or 6.72 x x < >Interval:( ) ( ) , 6.72 6.72, 6.720 6.72 57.6.2 3.5 1.3 14.53.5 1.3 2.34xx 2.34 3.5 1.3 2.341.04 3.5 3.640.30 1.04xxx [ ] 0.30,1.04 1.040 0.30 59.False. Fore example, let5 a and7 b . ( ) 2 7 2 7 5 5 + 2 7 2 7 9 + + Since5 9 , the statement a b a b + +is not true for all real numbers a and b. 61.a.2 1 112 1 11or2 1 112 12or2 106or 5xx xx xx x+ + + b.2 1 1111 2 1 1112 2 106 5xxxx+ < < + < < < < < c.2 1 112 1 11or2 1 112 12or2 106or 5xx xx xx x+ >+ < + >< >< > d.Written responses will vary. 0 6 5 When graphed together, this becomes 0 6 5 which covers all real numbers. 63.a bx c d ka bx c k dk dbx ca+ + + + ( ) ( )( ) ( )( ) ororor ork d k dbx c bx ca ak d k dbx c bx ca aac k d ac k dbx bxa aac k d ac k dx xab abac k dxab _+ + , + + + t SSM: Intermediate AlgebraSection 11.1 Quiz 339 65.No. 3 1010 3 1013 7xxx+ < < + < < < Section 11.1 Quiz 1.3 4 113 155xxx === 5or 5 x x = = 2.2 3 7 x =2 3 7or2 3 72 4or2 102or 5x xx xx x = == == = 3.2 3 5 9 12 3 5 83 5 4xxx + + = + = + = 3 5 4or3 5 43 9or3 113or3x xx xx x+ = + == = = = 4.5 6 5 156 5 3xx = = 6 5 3or6 5 36 2or6 81 4or3 3x xx xx x = == == = 5.2 5 02 5 02 552xxxx = === 6.7 1 3 x + = Since7 1 x +is always nonnegative, the solution set for7 1 3 x + = is the empty set,. 7.66 6xx Interval:[ ] 6, 6 60 6 8.44or 4xx x Interval:( ] [ ) , 4 4, 4 4 0 9.4 8 12 x >4 8 12or4 8 124 4or4 201or 5x xx xx x < >< >< > Interval:( ) ( ) , 1 5, 10 5 10.3 1 5 x + > Since2 9 x is always nonnegative, the solution set for 12 93x > is the set of all real numbers. Interval:( ) , 0 15.False. For example, let4 a =and9 b = . 4 9 5 5 a b = = =4 9 4 9 5 a b = = = Since5 5 , the statementa b a b = is not true for all real numbers a and b. 16.3 6 00 3 6 06 3 62 22xxxxx = As an alternative, remember that3 6 x is always nonnegative, which means it is never less than 0. Thus, the inequality can be written as 3 6 03 6 03 62xxxx = === Homework 11.2 1.2 4 y x Graph the line2 4 y x = with a solid line and shade the region above it.444 4yx 3. 132y x < +Graph the line 132y x = +with a dashed line and shade the region below it. 444 4yx 5. 253y x Graph the line 253y x = with a solid line and shade the region below it. 888 8yx SSM: Intermediate AlgebraHomework 11.2 341 7.y x >Graph the liney x =with a dashed line and shade the region above it. 444 4yx 9.2 5 105 2 10225x yy xy x+ Graph the line5 y = with a dashed line and shade the region above it. 444 4yx 19. 1233y xy x > + Graph 123y x = with a solid line and the line 3 y x = +with a dashed line. The solution region of the system is the intersection of the solution regions of 123y x and3 y x > + . 424yx2 Homework 11.2SSM: Intermediate Algebra 342 21.43y xy x Graph the line4 y x = with a solid line and the line3 y x = with a solid line. The solution region of the system is the intersection of the solution regions of4 y x and3 y x . 444yx4 23.3 92 300y xy xxy + Graph the lines3 9 y x = + ,2 3 y x = ,0 x = , and0 y =with solid lines. The solution region of the system is the intersection of the solution regions of3 9 y x + ,2 3 y x ,0 x , and 0 y . 888 8yx 25.55112y xy xy x< + +> + Graph the lines5 y x = +and 112y x = +with dashed lines, and the line5 y x = +with a solid line. The solution region of the system is the intersection of the solution regions of 5 y x < + ,5 y x + , and 112y x > + . 4248yx 27.35yy Graph the lines3 y = and5 y = with solid lines. The solution region of the system is the intersection of the solution regions of3 y and5 y . 444yx4 29.2 4 84 2 8122x yy xy x + 3 5 105 3 10325x yy xy x+ + + Graph the lines 122y x = and 325y x = +with solid lines. The solution region of the system is the intersection of the solution regions of 122y x and 325y x + . 444yx4 SSM: Intermediate AlgebraSection 11.2 Quiz 343 31.( )14 1212 12132y xy xy x ( ) 2 1 32 2 32 5y xy xy x< +< + +< + Graph the line 132y x = with a solid line and the line2 5 y x = +with a dashed line. The solution region of the system is the intersection of the solution regions of( )14 12y x and ( ) 2 1 3 y x < + . 444yx2 33.2 3 63 2 6223x yy xy x < < +> The graph of2 3 6 x y > +< Graph the line2 4 y x = with a dashed line and shade the region below it. 844yx4 Section 11.2 QuizSSM: Intermediate Algebra 344 4.3 5 153 5 15553y xy xy x Graph the line4 y x = +with a solid line and the line2 3 y x = with a dashed line. The solution region of the system is the intersection of the solution regions of4 y x +and 2 3 y x > . 444yx4 8. 215124y xy x +< + Graph the line 215y x = +with a solid line and the line 124y x = +with a dashed line. The solution region of the system is the intersection of the solution regions for 215y x +and 124y x < + . 444yx4 SSM: Intermediate AlgebraHomework 11.3 345 9.3 4 124 3 12334x yy xy x + 6 2 126 2 12123y xy xy x + + Graph the lines 334y x = and 123y x = +with solid lines. The solution region of the system is the intersection of the solution regions for 334y x and 123y x + . 168yx8 16 10.3 35 500x y y xx y y xxy > < + < < +>> Graph the lines3 y x = ,5 y x = + ,0 x = , and0 y =with dashed lines. The solution region of the system is the intersection of the solution regions for3 y x < ,5 y x < + ,0 x > , and 0 y > . 244yx4 11.2 5 105 2 10225x yy xy x > > +< Answers may vary. Three possible solutions: ( ) ( ) ( ) 5, 5 , 0, 4 ,and5, 2 ( )25 5 255 2 25 4true < < < ( )24 0 254 0 24 2true < < < ( )22 5 252 2 22 0true < < < Three possible non-solutions: ( ) ( ) ( ) 5, 0 , 0,1 ,and5, 2 ( )20 5 250 2 20 4false< < < ( )21 0 251 0 21 2false< < < ( )22 5 252 2 22 0false< < < 12.Answers may vary. One possible inequality5 y x +( ) ( ) ( ) 2, 5 , 2, 5 ,and2, 5 are solutions but ( ) 2, 5is not a solution. Homework 11.3 1. ( )( )( )( )3 3 32 2227 33 3 33 3 9x xx x xx x x+ = += + += + + 3. ( )( )( )( )3 3 32 2264 44 4 44 4 16x xx x xx x x = = + += + + 5. ( )( )( )( )3 3 32 221000 1010 10 1010 10 100x xx x xx x x = = + += + + Homework 11.3SSM: Intermediate Algebra 346 7. ( )( )( )( )3 3 32 22125 55 5 55 5 25x xx x xx x x+ = += + += + + 9. ( )( )( )( )3 3 32 221 11 11 1x xx x xx x x+ = += + += + + 11.( )( ) ( ) ( )( )( )( )( )33 322264 125 4 54 5 4 4 5 54 5 16 20 25x xx x xx x x = = + += + + 13.( )( ) ( ) ( )( )( )( )( )33 322227 64 3 43 4 3 3 4 43 4 9 12 16x xx x xx x x+ = += + += + + 15.( )( ) ( ) ( )( )( )( )( )33 32221000 1 10 110 1 10 10 1 110 1 100 10 1x xx x xx x x = = + += + + 17.( )( ) ( ) ( )( )( )( )( )33 32228 27 2 32 3 2 2 3 32 3 4 6 9x xx x xx x x = = + += + + 19.( )( ) ( ) ( )( )( )( )( )33 3222125 8 5 25 2 5 5 2 25 2 25 10 4x xx x xx x x+ = += + += + + 21. ( )( )( )( )( )( )3 33 32 224 32 4 84 24 2 2 24 2 2 4x xxx x xx x x = = = + += + + 23. ( )( )( )( )( )( )3 33 32 227 7000 7 10007 107 10 10 107 10 10 100x xxx x xx x x = = = + += + + 25. ( )( )( )( )( )( )3 33 32 2210 640 10 6410 410 4 4 410 4 4 16x xxx x xx x x+ = += += + += + + 27. ( )( )( )( )( )( )4 33 32 2227 2733 3 33 3 9x x x xx xx x x xx x x x+ = += += + += + + 29. ( )( )( )( )( )( )4 33 32 223 24 3 83 23 2 2 23 2 2 4x x x xx xx x x xx x x x = = = + += + + 31. ( )( )( )( ) ( ) ( )( )( )( )( )4 33322232 4 4 8 14 2 14 2 1 2 2 1 14 2 1 4 2 1x x x xx xx x x xx x x x = = = + += + + 33. ( )( )( )3 22 50 2 252 5 5x x x xx x x = = + 35. ( )( )( )( )( )( )3 33 32 222 16 2 82 22 2 2 22 2 2 4x xxx x xx x x = = = + += + + SSM: Intermediate AlgebraSection 11.3 Quiz 347 37.( )2212 36 6 x x x + = 39.( ) ( )( )( )( )( )( )( )( )( )( )( )( ) ( )4 3 333 32 22221 1 1 11 11 11 1 11 1 11 1x x x x x xx xx xx x x xx x x xx x x+ + + = + + += + += + += + + += + + += + + 41.( )( )26 11 10 3 2 2 5 x x x x + = +43. ( )( )( )3 2 22 20 32 2 10 162 8 2x x x x x xx x x + = += 45. ( ) ( )( )( )( ) ( )( )( )( ) ( ) ( )( )( )( )( )( )4 333332228 32 48 4 1 44 8 14 2 14 2 1 2 2 1 14 2 1 4 2 1x x xx x xx xx xx x x xx x x x+ = + += + = + = + + += + + + 47. ( )( )( )3 2 230 306 5x x x x x xx x x = = + 49.( )( ) ( )( ) ( )( )( )( )( )( )33 32222125 64 5 45 4 5 5 4 45 4 25 20 164 5 16 20 25x xx x xx x xx x x = = + += + += + + 51.No. When factoring 3 3a b , the middle term of the trinomial is onlyab . ( )( )3 28 2 2 4 x x x x = + +53. No. The middle term of the trinomial is incorrect. ( )( )3 3 2 2x a x a x ax a + = + +55.Written response. Answers may vary. ( )( )( )( )3 3 2 23 3 2 2a b a b a ab ba b a b a ab b = + ++ = + + Section 11.3 Quiz 1. ( )( )( )( )3 3 32 22216 66 6 66 6 36x xx x xx x x = = + += + + 2. ( )( )( )( )3 3 32 22216 66 6 66 6 36x xx x xx x x+ = += + += + + 3.( )( ) ( ) ( )( )( )( )( )33 322264 1 4 14 1 4 4 1 14 1 16 4 1x xx x xx x x+ = += + += + + 4.( )( ) ( ) ( )( )( )( )( )33 322227 125 3 53 5 3 3 5 53 5 9 15 25x xx x xx x x = = + += + + 5. ( )( )( )( ) ( ) ( )( )( )( )( )4 33322216 54 2 8 272 2 32 2 3 2 2 3 32 2 3 4 6 9x x x xx xx x x xx x x x+ = += += + += + + 6. ( )( )( )( ) ( ) ( )( )( )( )( )4 333222375 24 3 125 83 5 23 5 2 5 5 2 23 5 2 25 10 4x x x xx xx x x xx x x x = = = + += + + 7. ( )( )( )( )( )4 3225 5 5 15 1 15 1 1x x x xx x x xx x x x = = + += + + Homework 11.4SSM: Intermediate Algebra 348 8. ( )( )( )( )( )( )( )( )3 33 32 22256 7 7 87 27 2 2 27 2 4 27 2 2 4x xxx x xx x xx x x = = = + += + += + + 9.( ) ( )( )( )( )( )( )( )( )( )( )( )4 3 333 32 222 8 16 2 8 22 82 22 2 2 22 2 2 4x x x x x xx xx xx x x xx x x x+ = + += + = + = + + += + + + 10.( ) ( )( )( )( )( )( )4 3 3325 5 5 1 55 15 1 1x x x x x xx xx x x x + = + = += + + Homework 11.4 1.36 36 6 i i = =3.25 25 5 i i = = 5.13 13 i =7.20 20 4 5 2 5 i i i = = =9. 4 24 4 248 84 4 684 2 684 2 68 81 62 4iiiii = === = 11. 20 50 20 5010 1020 25 21020 5 21020 5 210 10222iiiii+ +=+ =+== += + 13. ( )9 9 2 92 92 36iii + = === 15. ( ) ( )4 2 25 1 8 44 2 25 1 8 44 2 5 1 8 24 10 1 164 1 10 163 6i ii ii ii ii+ = + = + = + = + = 17. ( )24 25 4 252 51010 110i ii ii = = == = 19. 23 5 3 51515i ii = == 21.( ) ( ) 4 7 3 10 4 7 3 104 3 7 107 3i i i ii ii + + = + += + += + 23.( ) ( ) 6 5 2 13 6 5 2 136 2 5 134 18i i i ii ii + = = = SSM: Intermediate AlgebraHomework 11.4 349 25.( ) ( ) 2 3 5 4 2 6 6 10 8 246 8 10 242 14i i i ii ii = += += + 27. ( )22 9 1818 118i i i == = 29.( )( )210 5 5050 150i i i == = 31.( )( )25 3 2 5 3 5 215 1015 10 115 1010 15i i i i ii iiii = = = = += + 33.( )( )220 3 2 7 20 3 2 3 720 6 2120 6 21 120 6 211 6i i i i ii iiii = + = += + = = 35.( )( )( )22 5 3 4 2 3 2 4 5 3 5 46 8 15 206 23 20 16 23 2014 23i i i i i ii i iiii+ + = + + + = + + += + + = + = + 37.( )( )( )23 6 5 2 3 5 3 2 6 5 6 215 6 30 1215 24 12 115 24 1227 24i i i i i ii i iiii + = + = + = = += 39.( )( ) ( )( )2225 4 5 4 5 425 1625 16 125 1641i i ii+ = = = = += 41.( )( )( )2 21 1 11 11 12i i i + = = = += 43.( ) ( )( ) ( )( )2 2222 3 2 2 2 3 34 12 94 12 9 14 12 95 12i i ii iiii+ = + += + += + + = + = + 45.( ) ( )( )( )22 24 4 2 416 8 116 8 115 8i i iiii = += += = 47. ( )23 3 2 52 5 2 5 2 56 154 256 154 25 1ii i iiii= + + ==

6 154 256 15296 1529 29iii=+== 49. 27 7 6 36 3 6 3 6 342 2136 9ii i iii= + + =

( )42 2136 9 142 2136 942 214514 715 15iiii= =+== Homework 11.4SSM: Intermediate Algebra 350 51. ( )( )222 3 2 3 77 7 714 2 21 34914 23 3 149 111 235011 2350 50i i ii i ii i iiiii = + + += + === 53. 223 4 3 4 3 43 4 3 4 3 49 12 12 169 16i i ii i ii i ii+ + += ++ + +=

( )( )9 24 16 19 16 17 24257 2425 25iii+ + = +== + 55. 223 5 3 5 2 92 9 2 9 2 96 27 10 454 81i i ii i ii i ii+ + = + + + = ( )( )6 17 45 14 81 151 178551 1785 853 15 5iiii = == = 57. 22 3 0 x x + + =( ) ( )( )( )22 2 4 1 32 12 822 822 2 221 2xiii = = = == 59. 22 5 0 x x + =( ) ( ) ( )( )( )22 2 4 1 52 12 1622 1622 421 2xiii = ==== 61. 22 1 0 x x + + =( ) ( )( )( )21 1 4 2 12 21 741 741 74 4xii = === SSM: Intermediate AlgebraHomework 11.4 351 63. 225 4 15 4 1 0x xx x = + = ( ) ( ) ( )( )( )24 4 4 5 12 54 4104 4104 2102 15 5xiii = ==== 65.( )222 2 1 34 2 34 2 3 0x xx xx x = = + = ( ) ( ) ( )( )( )22 2 4 4 32 42 4482 4482 2 1181 114 4xiii = ==== 67.( )( )221 22 22 2 0x x xx x x xx x+ + =+ + + =+ + = ( ) ( )( )( )22 2 4 1 22 12 422 422 221xiii = = = == 69.( ) ( )223 2 2 2 33 2 2 2 63 4 4 0x x xx x xx x = + = + + = ( ) ( ) ( )( )( )24 4 4 3 42 34 3264 3264 4 262 2 23 3xiii = ==== 71. 29993xxx ix i= = = = 73. 223 553xx ==

53535 33 3153153xx ix ix ix i= = = = = 75.( )25 3 75 3 75 3 75 3 73 75 5xxx ix ix i+ = + = + = = = Section 11.4 QuizSSM: Intermediate Algebra 352 77.( )22 3 1 4 12 x + = ( )( )222 3 1 83 1 43 1 43 1 43 1 23 1 21 23 3xxxx ix ix ix i = = = = = = = 79.Student 2 did the work correctly. When working with even roots of negative numbers, it is necessary to first write the problem using complex numbers of the forma bi + . Not all rules that apply to the real numbers will apply to complex numbers. 81.Answers may vary. Possible answers: a.( ) ( ) 1 1 2 1 1 2 2 i i i i i + + = + + = +b.( ) ( ) 1 1 1 1 2 i i i i + + = + + =c.( ) ( ) 1 1 1 1 2 i i i i i + ++ = + + =83.If the discriminant is positive, there are two distinct real solutions. If the discriminant is zero, there is one real solution (a repeated root). If the discriminant is negative, there are two distinct imaginary solutions. Section 11.4 Quiz 1. ( ) ( )3 4 36 2 8 36 3 4 36 2 8 363 4 6 2 8 63 24 2 483 2 24 481 24i ii ii ii ii+ = + = + = + = + = 2. 22 7 2 71414i ii = == 3.( ) ( ) 6 2 3 4 6 2 3 46 3 2 49 6i i i ii ii + = + = + = 4. ( )24 3 1212 112i i i = = = 5.( )( )( )25 3 7 5 7 5 3 7 335 5 21 335 16 3 135 16 338 16i i i i i ii i iiii + = + = + = = += 6.( ) ( ) ( )( ) ( )( )2 2 224 3 4 2 4 3 316 24 916 24 9 116 24 97 24i i ii iiii = += += + = = 7.( )( ) ( ) ( )( )2 228 5 8 5 8 564 2564 25 164 2589i i ii+ = = = = += 8. ( )22 2 44 4 48 2168 216 18 2178 217 17ii i iiiiii+= ++=+=+== + 9. ( )( )223 2 3 2 5 45 4 5 4 5 415 12 10 825 1615 22 8 125 16 17 22417 2241 41i i ii i ii i iiiii+ + += ++ + +=+ + = +== +SSM: Intermediate AlgebraSection 11.4 Quiz 353 10. ( )( )2 22 21 1 11 1 1111 2 11 122i i ii i ii i iiiii = + + += +=== 11. 22 3 0 x x + =( ) ( ) ( )( )( )22 2 4 1 32 12 822 822 2 221 2xiii = ==== 12. 23 4 0 x x + =( ) ( )( )( )21 1 4 3 42 31 4761 4761 476 6xii = === 13.( )( )224 54 5 208 20 0x x xx x x xx x+ + =+ + + =+ + = ( ) ( )( )( )28 8 4 1 202 18 1628 1628 424 2xiii = = = == 14.( )( )( )2223 2 5 1 133 2 5 122 5 4xxx + = = = 2 5 42 5 42 5 22 5 252xx ix ix ix i = = = = = 15.The discriminant of the quadratic equation 20 ax bx a + =is( )( )2 2 24 4 b a a b a = + . Since0 a for quadratic equations, this discriminant will always equal a positive number and each solution will be a real number. 16.False. The number3i(or0 3i + ) is a complex number andiis a pure imaginary number. Then ( )( )23 3 3 i i i = = . Since3 is not an imaginary number, the statement is false.Homework 11.5SSM: Intermediate Algebra 354 Homework 11.5 1. x=dollars invested in CD y=dollars invested in mutual fund a.10000or10000.0287 .081( ) .0287 .081(10000 ).0287 .081 10000 .081( ) .0523 810 0 10000x y y xI x yf x x xx xf x x x+ = = = += + = + = + b.fis decreasing.The more money invested in the CD, the less interest will be earned. c.We want ( ) 400.0523 810 400.0523 410410$7839.39.0523f xxxx= + = = = = You should invest $7839.39 in the CD and $2160.61 in the mutual fund in order to earn $400 interest. 3. x=dollars invested in CD y=dollars invested in mutual fund a.9000 or 9000.025 .0945( ) .025 .0945(9000 ).025 .0945 9000 .0945( ) .0695 850.5 0 9000x y y xI x yf x x xx xf x x x+ = = = += + = + = + b.(500) .0695(500) 850.5(500) $815.75ff= +=

If$500 is invested in the CD and $8500 in the mutual fund, $815.75 in interest will be earned. c.We want( ) 500.0695 850.5 500.0695 350.5350.5$5043.17.0695f xxxx= + = = = = You should invest $5043.17 in the CD and $3956.83 in the mutual fund in order to earn $500 interest. d.(10000) .0695(10000) 850.5$155.50f = += Although the formula for( ) f x yields a plausible result at10000 x = , this value is outside the domain of f, since only $9000 is available for investment.Thus model breakdown has occurred. 5. x=dollars invested in CD y=dollars invested in mutual fund a. 8000or 8000-.015 .116( ) .015 .116(8000 ).015 .116 8000 .116( ) .101 9282500 8000 ( .)x y y xI x yf x x xx xf x xx see part b+ = == += + = + = + b.Since f is a decreasing function and 2500 8000 x the minimum interest possible will be (8000) .101(8000) 928 $120.00 f = + = and the maximum interest possible will be (2500) .101(2500) 928 $675.50 f = + = Thus the investment can earn from $120 to $675.50 in interest. c.We want( ) 400.101 928 400.101 528528$5227.72.101f xxxx= + = = = = You should invest $5227.72 in the CD and $2772.28 in the mutual fund in order to earn $400 interest. SSM: Intermediate AlgebraHomework 11.5 355 7. x=dollars invested in CD y=dollars invested in mutual fund a. 6000or 6000-.0285 .09( ) .0285 .09(6000 ).0285 .09 6000 .09( ) .0615 540 0 6000x y y xI x yf x x xx xf x x x+ = == += + = + = + b.We evaluate (0) $540. f =Thus the I-intercept is(0, 540).This means that if the entire $6000 is invested in the mutual fund, $540 in interest will be earned. Since f is a decreasing function, $540 is also the maximum interest that can be earned. c.The x-intercept occurs when f (x)=0. So.0615 540 0.0615 540540$8780.49.0615xxx + = = = = Thus the x-intercept is(8780.49, 0). Although the formula for( ) f x yields a plausible result forf (x) = 0 , this value is outside the domain off , since only $6000 is available for investment. It is also illogical to invest a large sum of money that returns no interest.Thus model breakdown has occurred. d.The slope offis 0.0615 .It means that for every additional dollar invested in the CD, 6.15 cents less interest is received. 9. x=number of $50 tickets sold y=number of $75 tickets sold a.20000 or 2000050 75( ) 50 75(20000 )50 75 20000 75( ) 25 1500000 0 20000x y y xR x yf x x xx xf x x x+ = = = += + = + = + b.fis decreasing.The more $50 tickets sold, the less revenue will be earned. c. (16000) 25(16000) 15000001,100, 000.f = += If 16000 of the $50 tickets are sold and 4000 of the $75 tickets are sold, the total revenue will be$1,100,000. d.We want R= cost + profit=$1,075,000.25 1500000 1075000.25 425000.4250001700025xxx + = = = = Thus we should sell 17000 of the $50 tickets and 3000 of the $75 ticketsto make $600,000 profit. 11. x=number of $45 tickets sold y=number of $70 tickets sold a.12000 or 1200045 70( ) 45 70(12000 )45 70 12000 70( ) 25 840000 0 12000x y y xR x yf x x xx xf x x x+ = = = += + = + = + b. c.Since f is a decreasing function and 0 12000 x , the minimum revenue possible will be (12000) 25(12000) 840000 540, 000 f = + = and the maximum revenue possible will be (0) 25(0) 840000 840, 000. f = + = Thus the earned revenue will be between $540,000 and$840,000. Homework 11.5SSM: Intermediate Algebra 356 d.We want R=$602,500. 25 840000 602500.25 237500237500950025xxx + = = = = Thus we should sell 9500 of the $45 tickets and 2500 of the $70 tickets to earn $602,500 in revenue. 13. x=average price of a coach ticket y=average price of a first-class ticket a. 242126 8( ) 126 8( 242)126 8 8 242( ) 134 1936 0y xR x yf x x xx xf x x x= += += + += + + = + b.The slope offis 134.This means that for each dollar increase in the average price of a coach ticket (and therefore a dollar increase in the average price of a first-class ticket also), the revenue from the flight will increase by$134. c.We want R=$14,130. 134 1936 14130134 1219412194$91.134xxx+ === = The average selling price for a coach ticket should be $91 and the average price for a first-class ticket should be $333 in order to earn revenue of $14,130. 15. x=number of part-time students per semester y=number of full-time students per semester 313(3 14 )( ) 13 3 13 14(3 )39 546( ) 585 0y xR x yf x x xx xf x x x== += + = += For revenue of$900,000 per semester, we need( ) 900000585 9000009000001538.5585f xxx=== = That is, we need 1539 part-time students and 4617 full-time students. 17. Demand:2.25 3162.5 n p = + a. Revenue=demand price 2( ) ( 2.25 3162.5)( ) 2.25 3162.5R n pf p p pf p p p= = += + b.Set ( ) 0 f p = ( )22.25 3162.5 02.25 ( 1405.56) 01405.56 00 or 1405.56Thus -intercepts are (0, 0) and (1405.56, 0)p pp pp pp pp + = = == = If p = $0., we will give the guitars away; if p = $1405.56, there will be no demand.In either case, no revenue will be generated. c.The revenue function is part of the parabola 22.25 3162.5 y p p = + which has standard form 222.25( 1405.56 493896.60)1111267.362.25( 702.78) 1111267.36y p py p= ++= + The maximum R value will occur at the vertex of the parabola, which is(702.78, 1111267.36).Thus the maximum monthly revenue is $1,111,267.36, which occurs when the price per guitar is $702.78 19.a. Start by plotting the data. A linear model seems appropriate.Using linear regression we get 136.2 5861.64 n p = + SSM: Intermediate AlgebraSection 11.5 Quiz 357 b.Revenue=demand price 2( ) ( 136.2 5861.64)( ) 136.2 5861.64R n pf p p pf p p p= = += + c.Set( ) 0 f p = ( )2136.2 5861.64 0136.2 ( 43.04) 043.04 00 or 43.04Thus the-intercepts are (0, 0) and(43.04, 0).p pppp pp pp + = = == =If p = $0., we will give the CDs away; if p = $43.04, there will be no demand.In either case, no revenue will be generated. d.The revenue function is part of the parabola 2136.2 5861.64 y p p = + which has standard form 22136.2( 43.04 463.11) 63075.64136.2( 21.52) 63075.64y p py p= + += + The maximum R value will occur at the vertex of the parabola, which is(21.52, 63075.64).Thus the maximum monthly revenue is $63,075.64, which occurs when the price per CD is $21.52. 21.a.Start by plotting the data. A linear model seems appropriate.Using linear regression we get 2.164 2493.6 n p = +b.The variable n is the dependent variable because the number of demands depends on the price. c. Revenue=demand price 2( ) ( 2.164 2493.6)( ) 2.164 2493.6R n pf p p pf p p p= = += + d.Set( ) $240, 000. f p =2222.164 2493.6 2400002.164 2493.6 240000 0Using the quadratic formula:2493.6 (2493.6) 4( 2.164)( 240000)2( 2.164)2493.6 2034.85or so4.328106 or 1046.31p pp pppp p + = + = = == = Thus if we sell the printers for either $106 apiece or $1046.31 apiece, our revenue will be $240,000. e.The revenue function is part of the parabola 22.164 2493.6 y p p = + which has standard form 222.164( 1152.31 331954.89) 718350.392.164( 576.16) 718350.39y p py p= + += + The maximum R value will occur at the vertex of the parabola, which is(576.16, 718350.39).Thus the maximum monthly revenue is $718,350.39, which occurs when the price per printer is $576.16. Section 11.5 Quiz 1. x=dollars invested in CD y=dollars invested in mutual fund a. 8000or 8000.029 .127( ) .029 .127(8000 ).029 .127 8000 .127( ) .098 1016 0 8000x y y xI x yf x x xx xf x x x+ = = = += + = + = + b. (500) .098(500) 1016(500) $967.ff= += If$500 is invested in the CD and $7500 in the mutual fund, $967 in interest will be earned. Homework 11.6SSM: Intermediate Algebra 358 c.We want( ) 500.098 1016 500.098 516516$5265.31.098f xxxx= + = = = = You should invest $5265.31 in the CD and $2734.69 in the mutual fund in order to earn $500 interest. 2.x=number of $30 tickets sold y=number of $50 tickets sold a. 16000or 1600030 50( ) 30 50(16000 )30 50 16000 50( ) 20 800000 0 16000x y y xR x yf x x xx xf x x x+ = = = += + = + = + b.The slope off is -20; this means that for every (additional) $30 ticket sold, the revenue will decrease by $20. c.We wantR=$500,000. 20 800000 500000.20 300000.3000001500020xxx + = = = = Thus we should sell15,000$30 tickets and 1000$50 ticketsto earn $500,000 in revenue. 3. Demand:120 65000 n p = +a.Revenue=demand price 2( ) ( 120 65000)( ) 120 65000R n pf p p pf p p p= = += + b.Set ( ) 0 f p = ( )2120 65000 0120 ( 541.67) 0541.67 00 or 541.67Thus the -intercepts are(0, 0)and(541.67, 0).p pppp pp pp + = = == = If p = $0., we will give the DVD players away; if p = $541.67, there will be no demand.In either case, no revenue will be generated. c.The revenue function is part of the parabola 2120 65000 y p p = + which has standard form 22120( 541.67 73351.6) 8802191.67120( 270.84) 8802191.67y p py p= + += + The maximum R value will occur at the vertex of the parabola, which is(270.84, 8802191.67).Thus the maximum monthly revenue is $8,802,191.67, which occurs when the price per DVD player is $270.84. Homework 11.6 1. 2 2 22 2 2225 1225 14416913c a bcccc= += += +== 3. 2 2 22 2 2224 516 254141c a bcccc= += += +== 5. 2 2 22 2 2223 89 645555a b cbbbb+ =+ =+ === 7. 2 2 22 2 2225 725 4924242 6a b caaaaa+ =+ =+ ==== SSM: Intermediate AlgebraHomework 11.6 359 9. ( ) ( )2 2 22 22222 52 577c a bcccc= += += +== 11. 2 2 22 2 22221 7441 494904907 10c a bccccc= += += +=== 13.( ) ( ) 2, 5and6, 7( ) ( )2 22 26 2 7 54 216 4204.47d = + = += += 15.( ) ( ) 3, 5and4,2 ( ) ( ) ( )( )2 2224 3 2 57 349 9587.62d = + = += += 17.( ) ( ) 4, 1and1, 5 ( ) ( ) ( ) ( )( )2 2221 4 5 13 49 16255d = + = += +== 19.( ) ( ) 2.1, 8.9and5.6,1.7( ) ( )( )2 2225.6 2.1 1.7 8.93.5 7.212.25 51.8464.098.01d = + = += += 21.( ) ( ) 2.18, 5.74and3.44, 6.29 ( ) ( ) ( ) ( )2 22 23.44 2.18 6.29 5.745.62 12.0331.5844 144.7209176.305313.28d = + = += += 23.( ) 0, 0and5 C r =( ) ( )( ) ( )2 222 222 20 0 525x h y k rx yx y + = + =+ = 25.( ) 0, 0and6.7 C r =( ) ( )( ) ( )2 222 222 20 0 6.744.89x h y k rx yx y + = + =+ = 27.( ) 5, 3and2 C r =( ) ( )( ) ( )( ) ( )2 222 222 25 3 25 3 4x h y k rx yx y + = + = + = 29.( ) 2,1and4 C r =( ) ( )( ) ( ) ( )( ) ( )2 222 222 22 1 42 1 16x h y k rx yx y + = + =+ + = Homework 11.6SSM: Intermediate Algebra 360 31.( ) 7, 3and3 C r =( ) ( )( ) ( ) ( ) ( ) ( )( ) ( )2 2222 22 27 3 37 3 3x h y k rx yx y + = + =+ + + = 33. 2 225 x y + =The equation has the form 2 2 2x y r + = . Therefore,( ) 0, 0 C =and 225255rrr=== 444yx4 35. 2 28 x y + =The equation has the form 2 2 2x y r + = . Therefore,( ) 0, 0 C =and2882 2rrr=== 444yx4 37.( ) ( )2 23 5 16 x y + =The equation is in the form ( ) ( )2 22x h y k r + = .The center is( ) , h kor( ) 3, 5 Cand216164rrr=== 84yx4 8 39.( ) ( )( ) ( ) ( )( )2 222 26 1 76 1 7x yx y+ + = + = The equation is in the form ( ) ( )2 22x h y k r + = . The center is( ) 6,1 Cand the radius is7 r = . 888yx8 41.( ) ( )( ) ( ) ( ) ( )2 22 223 2 13 2 1x yx y+ + + = + = The equation is in the form ( ) ( )2 22x h y k r + = . The center is( ) 3, 2 C and the radius is1 r = . 444yx4 SSM: Intermediate AlgebraHomework 11.6 361 43.5, 20, length of ladder a b c = = =2 2 22 2 2225 2025 40042542520.62c a bccccc= += += +== The ladder must be approximately 20.62 feet long. 45.465, 964,distance a c b = = =2 2 22 2 22465 964713071713071844.44a b cbbbb+ =+ === 465 844.44 9642273.44a b c + + = + += The total distance of the road trip would be approximately 2273.44 miles. 47.( ) 1, 2 C and3 r = . ( ) ( )( ) ( ) ( )( ) ( )2 222 222 21 2 31 2 9x h y k rx yx y + = + = + + = 49.The radius is the distance from the center to any point on the circle. The distance between ( ) 3, 2 Cand( ) 5, 6is given by: ( ) ( )2 22 25 3 6 22 44 1620d = + = += += The radius is20 r = . The equation of the circle is ( ) ( )( ) ( )( )( ) ( )2 2222 22 23 2 203 2 20x h y k rx yx y + = + = + = 51.Answers may vary. One possible answer: ( ) ( )( ) ( )2 22 22 3 95 6 9x yx y + = + = 84yx4 8(5, 3) 53.Find the equation of the circle that has center ( ) 3, 2 Cand radius4 r = . ( ) ( )( ) ( )( ) ( )2 222 222 23 2 43 2 16x h y k rx yx y + = + = + = Find the coordinates of five points,( ) , xy , that satisfy this equation. Two possible answers are: ( ) 3, 6and( ) 7, 2 . 55.No. The graph of the relation is a circle with radius 7 and centered at the origin. The graph fails the vertical line test. 57.a.The square root of a nonnegative number is a nonnegative real number and the square root of a negative number is an imaginary number. Therefore,0 y for real number values of y. b. 444yx4 59.a.Sketches may vary. One example: 88 Section 11.6 QuizSSM: Intermediate Algebra 362 b., a kb k = =2 2 22 2 22 2222222c a bc k kc kc kc kc k= += +==== c.23 2c k == d.25 2525 22 25 22c kkkkk==== = Section 11.6 Quiz 1.4, 8 a c = =2 2 22 2 2224 816 6448484 3a b cbbbbb+ =+ =+ ==== The other leg is4 3inches. 2.16, 19 b c = =2 2 22 2 22216 19256 36110510510.25a b caaaaa+ =+ =+ === The height of the screen is about 10.25 inches. 3.( ) 2, 5 and( ) 3, 1 ( ) ( ) ( ) ( )2 22 23 2 1 55 425 1641d = + = += += 4.( ) ( ) 3, 2and4, 2 ( ) ( ) ( )2 22 24 3 2 27 0497d = + = +== 5.( ) 3, 2and6 C r =( ) ( )( ) ( ) ( )( ) ( )2 222 222 23 2 63 2 36x h y k rx yx y + = + =+ + = 6.( ) 0, 0and2.8 C r =( ) ( )( ) ( )2 222 222 20 0 2.87.84x h y k rx yx y + = + =+ = 7. 2 212 x y + =The equation is in the form 2 2 2x y r + = . The center is( ) 0, 0 Cand212122 3rrr=== 444yx4 SSM: Intermediate AlgebraHomework 11.7 363 8.( ) ( )( ) ( ) ( )2 22 224 3 254 3 5x yx y+ + = + = The equation is in the form ( ) ( )2 22x h y k r + = . The center is( ) 4, 3 Cand the radius is5 r = . 12 44yx12 9.( ) 2, 1 C The radius is the distance from the center, ( ) 2, 1 , to the point( ) 4, 7that lies on the circle. ( ) ( ) ( )2 22 24 2 7 12 84 6468d = + = += += The equation of the circle is( ) ( )( ) ( ) ( ) ( )( ) ( )2 2222 22 22 1 682 1 68x h y k rx yx y + = + = + + = 10.Answers may vary. One possible answer: ( )( )22222 43 9x yx y+ + = + = 444yx4(0, 0) Homework 11.7 1. 2 2136 9x y+ =236, 6 a a = =x-intercepts:( ) ( ) 6, 0 , 6, 0 29, 3 b b = =y-intercepts:( ) ( ) 0, 3 , 0, 3 44yx4 4 3. 2 214 36x y+ =24, 2 a a = =x-intercepts:( ) ( ) 2, 0 , 2, 0 236, 6 b b = =y-intercepts:( ) ( ) 0, 6 , 0, 6 44yx4 4 5. 2 21100 16x y+ =2100, 10 a a = =x-intercepts:( ) ( ) 10, 0 , 10, 0 216, 4 b b = =y-intercepts:( ) ( ) 0, 4 , 0, 4 88yx4 4 Homework 11.7SSM: Intermediate Algebra 364 7. 2 22 22 225 4 10025 4 100100 100 10014 25x yx yx y+ =+ =+ = 24, 2 a a = =x-intercepts:( ) ( ) 2, 0 , 2, 0 225, 5 b b = =y-intercepts:( ) ( ) 0, 5 , 0, 5 22yx4 4 9. 2 22 22 29 100 9009 100 900900 900 9001100 9x yx yx y+ =+ =+ = 2100, 10 a a = =x-intercepts:( ) ( ) 10, 0 , 10, 0 29, 3 b b = =y-intercepts:( ) ( ) 0, 3 , 0, 3 4 488yx 11. 2 22 22 2363636 36 36136 36x yx yx y+ =+ =+ = 236, 6 a a = =x-intercepts:( ) ( ) 6, 0 , 6, 0 236, 6 b b = =y-intercepts:( ) ( ) 0, 6 , 0, 6 44yx4 4 13. 2 22 22 225 2525 2525 25 25125 1x yx yx y+ =+ =+ = 225, 5 a a = =x-intercepts:( ) ( ) 5, 0 , 5, 0 21, 1 b b = =y-intercepts:( ) ( ) 0, 1 , 0,1 44yx4 4 15. 2 22 22 25 16 805 16 8080 80 80116 5x yx yx y+ =+ =+ = 216, 4 a a = =x-intercepts:( ) ( ) 4, 0 , 4, 0 25, 5 b b = =y-intercepts: ( ) ( )0, 5 , 0, 5 442 2yx SSM: Intermediate AlgebraHomework 11.7 365 17. 2 22 216 25 111/16 1/ 25x yx y+ =+ = 21 1,16 4a a = =x-intercepts: 1 1, 0 , , 04 4 21 1,25 5b b = =y-intercepts: 1 10, , 0,5 5 111yx1 19. 2 2116 4x y =216, 4 a a = =x-intercepts:( ) ( ) 4, 0 , 4, 0 24, 2 b b = =Sketch a dashed rectangle that contains the points( ) ( ) ( ) ( ) 4, 0 , 4, 0 , 0, 2 ,and0, 2 , and then sketch the inclined asymptotes. 443 3yx 21. 2 2116 25y x =216, 4 b b = =y-intercepts:( ) ( ) 0, 4 , 0, 4 225, 5 a a = =Sketch a dashed rectangle that contains the points( ) ( ) ( ) ( ) 5, 0 , 5, 0 , 0, 4 ,and0, 4 , and then sketch the inclined asymptotes. 4 4yx22 23. 2 2125 81x y =225, 5 a a = =x-intercepts:( ) ( ) 5, 0 , 5, 0 281, 9 b b = =Sketch a dashed rectangle that contains the points( ) ( ) ( ) ( ) 5, 0 , 5, 0 , 0, 9 ,and0, 9 , and then sketch the inclined asymptotes. 88yx88 25. 2 22 22 216 4 6416 4 6464 64 6414 16x yx yx y = = = 24, 2 a a = =x-intercepts:( ) ( ) 2, 0 , 2, 0 216, 4 b b = =Sketch a dashed rectangle that contains the points( ) ( ) ( ) ( ) 2, 0 , 2, 0 , 0, 4 ,and0, 4 , and then sketch the inclined asymptotes. 88yx88 Homework 11.7SSM: Intermediate Algebra 366 27. 2 22 22 29 99 99 9 919 1x yx yx y = = = 29, 3 a a = =x-intercepts:( ) ( ) 3, 0 , 3, 0 21, 1 b b = =Sketch a dashed rectangle that contains the points( ) ( ) ( ) ( ) 3, 0 , 3, 0 , 0, 1 ,and0,1 , and then sketch the inclined asymptotes. 4yx45 5 29. 2 22 22 2444 4 414 4y xy xy x = = = 24, 2 b b = =y-intercepts:( ) ( ) 0, 2 , 0, 2 24, 2 a a = =Sketch a dashed rectangle that contains the points( ) ( ) ( ) ( ) 2, 0 , 2, 0 , 0, 2 ,and0, 2 , and then sketch the inclined asymptotes. 44yx44 31. 2 22 22 216 1616 1616 16 1611 16y xy xy x = = = 21, 1 b b = =y-intercepts:( ) ( ) 0, 1 , 0,1 216, 4 a a = = Sketch a dashed rectangle that contains the points( ) ( ) ( ) ( ) 4, 0 , 4, 0 , 0, 1 ,and0,1 , and then sketch the inclined asymptotes. 4yx45 5 33. 2 22 22 225 7 17525 7 175175 175 17517 25x yx yx y = = = 27, 7 a a = =x-intercepts: ( ) ( )7, 0 , 7, 0 225, 5 b b = =Sketch a dashed rectangle that contains the points ( ) ( ) ( ) ( ) 7, 0 , 7, 0 , 0, 5 ,and0, 5 , and then sketch the inclined asymptotes. 88yx88 SSM: Intermediate AlgebraHomework 11.7 367 35. 2 2164 4x y+ =Ellipse 264, 8 a a = =x-intercepts:( ) ( ) 8, 0 , 8, 0 24, 2 b b = =y-intercepts:( ) ( ) 0, 2 , 0, 2 88yx88 37. 2 22 2111 1x yx y = = Hyperbola 21, 1 a a = =x-intercepts:( ) ( ) 1, 0 , 1, 0 21, 1 b b = =Sketch a dashed rectangle that contains the points( ) ( ) ( ) ( ) 1, 0 , 1, 0 , 0, 1 ,and0,1 , and then sketch the inclined asymptotes. 4 4yx44 39. 2 22 22 281 49 396981 49 39693969 3969 3969149 81x yx yx y+ =+ =+ = Ellipse 249, 7 a a = =x-intercepts:( ) ( ) 7, 0 , 7, 0 281, 9 b b = =y-intercepts:( ) ( ) 0, 9 , 0, 9 888yx844 41. 2 22 2111 1x yx y+ =+ = Circle 21, 1 a a = =x-intercepts:( ) ( ) 1, 0 , 1, 0 21, 1 b b = =y-intercepts:( ) ( ) 0, 1 , 0,1 4 4yx44 43. 2 22 22 29 4 1449 4 144144 144 144116 36y xy xy x = = = Hyperbola 216, 4 b b = =y-intercepts:( ) ( ) 0, 4 , 0, 4 236, 6 a a = =Sketch a dashed rectangle that contains the points( ) ( ) ( ) ( ) 6, 0 , 6, 0 , 0, 4 ,and0, 4 , and then sketch the inclined asymptotes. 888yx8 Homework 11.7SSM: Intermediate Algebra 368 45. 2 2125 25x y =Hyperbola 225, 5 a a = =x-intercepts:( ) ( ) 5, 0 , 5, 0 225, 5 b b = =Sketch a dashed rectangle that contains the points( ) ( ) ( ) ( ) 5, 0 , 5, 0 , 0, 5 ,and0, 5 , and then sketch the inclined asymptotes.888yx8 47. 2 22 216116 16x yx y+ =+ = Circle 216, 4 a a = =x-intercepts:( ) ( ) 4, 0 , 4, 0 216, 4 b b = =y-intercepts:( ) ( ) 0, 4 , 0, 4 22yx22 49. 2 22 22 24 4 644 4 6464 64 64116 16x yx yx y+ =+ =+ = Circle 216, 4 a a = =x-intercepts:( ) ( ) 4, 0 , 4, 0 216, 4 b b = =y-intercepts:( ) ( ) 0, 4 , 0, 4 22yx22 51. 2 22 22 29 9 819 9 8181 81 8119 9x yx yx y = = = Hyperbola 29, 3 a a = =x-intercepts:( ) ( ) 3, 0 , 3, 0 29, 3 b b = =Sketch a dashed rectangle that contains the points( ) ( ) ( ) ( ) 3, 0 , 3, 0 , 0, 3 ,and0, 3 , and then sketch the inclined asymptotes. 4 4yx44 53.a.i. 2 21x yc d+ =2 214 16x y+ =Ellipse 24, 2 a a = =x-intercepts:( ) ( ) 2, 0 , 2, 0 216, 4 b b = =y-intercepts:( ) ( ) 0, 4 , 0, 4 4 4yx44 SSM: Intermediate AlgebraHomework 11.7 369 ii. 2 21x yc d+ =2 214 16x y =Hyperbola 24, 2 a a = =x-intercepts:( ) ( ) 2, 0 , 2, 0 216, 4 b b = =Sketch a dashed rectangle that contains the points( ) ( ) ( ) 2, 0 , 2, 0 , 0, 4 , and ( ) 0, 4 , and then sketch the inclined asymptotes. 4 4yx55 iii. 2 21x yc d+ =2 2116 4y x =Hyperbola 216, 4 b b = =y-intercepts:( ) ( ) 0, 4 , 0, 4 24, 2 a a = =Sketch a dashed rectangle that contains the points( ) ( ) ( ) 2, 0 , 2, 0 , 0, 4 , and ( ) 0, 4 , and then sketch the inclined asymptotes. 4 4yx55 iv. 2 21x yc d+ =2 214 4x y+ =Circle 24, 2 a a = =x-intercepts:( ) ( ) 2, 0 , 2, 0 24, 2 b b = =y-intercepts:( ) ( ) 0, 2 , 0, 2 4 4yx44 b.If0 c >and0 d >andc d , then the graph is an ellipse. If0 c >and0 d < , then the graph is a hyperbola with x-intercepts. If0 c , then the graph is a hyperbola with y-intercepts. Ifc d =and0 c > , then the graph is a circle. 55.No. The graph of 2 214 81x y+ =is an ellipse. The graph fails the vertical line test. 57.a. 2542y x = The square root of a nonnegative number is a nonnegative real number and the square root of a negative number is an imaginary number. Since 52 is positive, 2542x is nonnegative for real values of y. Thus, 0 y for real number values of y. b. 4 4yx44 Section 11.7 QuizSSM: Intermediate Algebra 370 59.Answers may vary. Possible answer: 2 22 21x ya b+ =To keep the ellipses from intersecting, we increase the value of a and b for each equation. Some possible equations: 2 2 2 22 2 2 22 21, 1,1 4 4 91, 1,9 16 16 25125 36x y x yx y x yx y+ = + =+ = + =+ = 888yx8 61. 2 2 22 2 22 2 22 22 21x y rx y rr r rx yr r+ =+ =+ = Ellipse Since 2 2 2a b r = = , it is also a circle. (Note that all circles are ellipses just as all squares are rectangles.) Section 11.7 Quiz 1. 2 219 25x y+ =29, 3 a a = =x-intercepts:( ) ( ) 3, 0 , 3, 0 225, 5 b b = =y-intercepts:( ) ( ) 0, 5 , 0, 5 4 4yx44 2. 2 2149 9y x =249, 7 b b = =y-intercepts:( ) ( ) 0, 7 , 0, 7 29, 3 a a = =Sketch a dashed rectangle that contains the points( ) ( ) ( ) ( ) 3, 0 , 3, 0 , 0, 7 ,and0, 7 , and then sketch the inclined asymptotes. 8yx844 3. 2 22 22 24 164 1616 16 1614 16x yx yx y = = = 24, 2 a a = =x-intercepts:( ) ( ) 2, 0 , 2, 0 216, 4 b b = =Sketch a dashed rectangle that contains the points( ) ( ) ( ) ( ) 2, 0 , 2, 0 , 0, 4 ,and0, 4 , and then sketch the inclined asymptotes. 4 4yx55 SSM: Intermediate AlgebraSection 11.7 Quiz 371 4. 2 22 22 216 3 4816 3 4848 48 4813 16x yx yx y+ =+ =+ = 23, 3 a a = =x-intercepts: ( ) ( )3, 0 , 3, 0 216, 4 b b = =y-intercepts:( ) ( ) 0, 4 , 0, 4 4 4yx55 5. 2 218 3x y =28, 2 2 a a = =x-intercepts: ( ) ( )2 2, 0 , 2 2, 0 23, 3 b b = =Sketch a dashed rectangle that contains the points ( ) ( ) ( )2 2, 0 , 2 2, 0 , 0, 3 , and ( )0, 3 , and then sketch the inclined asymptotes. 4 4yx44 6. 2 22 22 24 4 164 4 1616 16 1614 4y xy xy x = = = 24, 2 b b = =y-intercepts:( ) ( ) 0, 2 , 0, 2 24, 2 a a = =Sketch a dashed rectangle that contains the points( ) ( ) ( ) ( ) 2, 0 , 2, 0 , 0, 2 ,and0, 2 , and then sketch the inclined asymptotes. 4 4yx44 7. 2 215 14x y+ =25, 5 a a = =x-intercepts: ( ) ( )5, 0 , 5, 0 214, 14 b b = =y-intercepts: ( ) ( )0, 14 , 0, 14 4 4yx44 8. 2 22 22 29 819 8181 81 81181 9x yx yx y+ =+ =+ = 281, 9 a a = =x-intercepts:( ) ( ) 9, 0 , 9, 0 29, 3 b b = =y-intercepts:( ) ( ) 0, 3 , 0, 3 888yx8Homework 11.8SSM: Intermediate Algebra 372 9.No. The graph of 2 219 4x y =is a hyperbola. The graph fails the vertical line test. 10.Answers will vary. The points( ) 0, 3and( ) 0, 3 are y-intercepts. An ellipse with these y-intercepts has the form 2 2219x ya+ = . Three possibilities: let1, 2,and 4 a = . 2 211 9x y+ =2 214 9x y+ =2 2116 9x y+ =Homework 11.8 1. 2 22 2254 25 100x yx y+ =+ = 442yx2(5, 0) (5, 0) The two intersections points( ) 5, 0 and( ) 5, 0are the solutions to the system. Solve using elimination: 2 22 2224 4 1004 25 100 21 000x yx yyyy = + ==== Let0 y =in 2 225 x y + =and solve for x. 2 220 25255xxx+ === The solutions are( ) 5, 0 and( ) 5, 0 . 3. 213y xy x= += + 84yx4 2 2(2, 5)(1, 2) The two intersection points( ) 2, 5 and( ) 1, 2are the solutions of the system.Solve using substitution.Substitute 21 x +for y in the second equation. ( )( )2231 32 02 1 0y xx xx xx x= ++ = ++ =+ = 2 0or 1 02or 1x xx x+ = == = Let2 x = and1 x =in3 y x = +and solve for y. ( ) 2 32 35y = += += ( ) 1 31 32y = += += The solutions are( ) 2, 5 and( ) 1, 2 . 5. 2226y xy x= = + 444yx4(2, 2) (2, 2) The two intersection points( ) 2, 2 and( ) 2, 2are the solutions of the system. Solve using substitution. Substitute 22 x for y in the second equation. SSM: Intermediate AlgebraHomework 11.8 373 22 22262 62 842y xx xxxx= + = +=== Let2 x = and2 x =in 22 y x = and solve for y. ( )22 24 22y = = = ( )22 24 22y = = = The solutions are( ) 2, 2 and( ) 2, 2 . 7. 2 22 24916x yx y+ =+ = 888yx8 The graphs do not intersect. The solution set is the empty set,. Solve using elimination. 2 22 2 49160 33 Falsex yx y+ = = = There is no solution. 9. 2 2251x yy x+ == 444yx4(4, 3)(3, 4) The two intersection points( ) 4, 3 and( ) 3, 4 are the solutions of the system.Solve using substitution. Substitute1 x for y in the first equation. ( )( )( )2 2222 222251 252 1 252 2 24 012 04 3 0x yx xx x xx xx xx x+ =+ =+ + + =+ =+ =+ = 4 0or 3 04or 3x xx x+ = == = Let4 x = and3 x =in1 y x = and solve for y. ( ) 4 14 13y = = = ( ) 3 13 14y = = = The solutions are( ) 4, 3 and( ) 3, 4 . 11. 2 22164y xy x =+ = 244yx4(3, 5)(0, 4)(3, 5) The three intersection points( ) 3, 5 ,( ) 0, 4 , and( ) 3, 5 are the solutions to the system. Solve using elimination. 2 22216420y xy xy y =+ =+ = ( )( )220 05 4 0y yy y+ =+ = 5 0or 4 05or 4y yy y+ = == = Let5 y = and4 y =in 24 y x + =and solve for x. 225 493xxx + ===

224 400xxx+ === The solutions are( ) 3, 5 ,( ) 3, 5 , and( ) 0, 4 . Homework 11.8SSM: Intermediate Algebra 374 13. 2 22 225 9 2254 9 36x yx y =+ = 444yx4(3, 0) (3, 0) The two intersection points( ) 3, 0 and( ) 3, 0are the solutions to the system. Solve using elimination. 2 22 22225 9 225 4 9 36 29 26193x yx yxxx =+ ==== Let3 x = and3 x =in 2 24 9 36 x y + =and solve for y. ( )222224 3 9 3636 9 369 000yyyyy + =+ ==== ( )222224 3 9 3636 9 369 000yyyyy+ =+ ==== The solutions are( ) 3, 0 and( ) 3, 0 . 15. 2 29 93 3x yy x+ == + 444yx4(1, 0)(0, 3) The two intersection points( ) 1, 0 and( ) 0, 3are the solutions to the system. Solve using substitution. Substitute3 3 x +for y in the first equation. ( )( )2 2222 229 99 3 3 99 9 18 9 918 18 018 1 0x yx xx x xx xx x+ =+ + =+ + + =+ =+ = 18 0or 1 00or 1x xx x= + == = Let0 x =and1 x = in3 3 y x = +and solve for y. ( ) 3 0 30 33y = += += ( ) 3 1 33 30y = += += The solutions are( ) 0, 3and( ) 1, 0 . 17. 2 22 24 9 3616 25 225x yx y+ =+ = 444yx4 The graphs do not intersect. The solution set is the empty set,. Solve using elimination. 2 22 22216 36 14416 25 22511 818111x yx yyy = + = == Since 2ycannot be negative (in the real number system), there is no solution. 19.31y xy x= = 442yx2(1, 2) SSM: Intermediate AlgebraHomework 11.8 375 The intersection point( ) 1, 2 is the solution to the system. Solve using substitution. Substitute3 x for y in the second equation. ( )( )( )( )222213 1224 45 4 04 1 0y xx xx xx xx x xx xx x= = = += += + + = = 4 0or 1 04or1x xx x = == = Let4 x =and1 x = in3 y x = and solve for y. 4 32 31y = = = 1 31 32y = = = Check each result in1 y x = . ( ) 1 4 11 5False = =

2 1 12 2True = = The only solution is( ) 1, 2 . 21. 222 52y xy x= = 44yx43 ( , 1) 3 ( , 1) The two intersection points ( )3,1 and ( )3,1are the solutions to the system. Solve using substitution. Substitute 22 5 x for y in the second equation. 22 2222 5 233y xx xxx= = == Let3 x = and3 x =in 22 y x = and solve for y. ( )23 23 21y = = =

( )23 23 21y = = = The solutions are ( )3,1 and ( )3,1 . 23. 2 22 225 4 1009 9y xx y =+ = 44yx4(0.74, 2.02)(0.74, 2.02)(0.74, 2.02)(0.74, 2.02) The four intersection points( ) 0.74, 2.02 , ( ) 0.74, 2.02 ,( ) 0.74, 2.02 , and( ) 0.74, 2.02are the solutions of the system. Solve using substitution. 2 22 29 99 9x yy x+ == Substitute 29 9x for 2yin the first equation. ( )2 22 22 22225 4 10025 9 9 4 100225 225 4 100229 1251252291250.74229y xx xx xxxx = = = = == Let0.74 x = and0.74 x =in 2 29 9 x y + =and solve for y. ( )2229 0.74 94.074.07 2.02yyy + === ( )2229 0.74 94.074.07 2.02yyy+ === The solutions are( ) 0.74, 2.02 ,( ) 0.74, 2.02 , ( ) 0.74, 2.02 , and( ) 0.74, 2.02 . Homework 11.8SSM: Intermediate Algebra 376 25. 2 22 225 9 22516x yx y+ =+ = 445yx522(2.23, 3.31) (2.25, 3.31)(2.25, 3.31) (2.25, 3.31) The four intersection points( ) 2.25, 3.31 , ( ) 2.25, 3.31 ,( ) 2.25, 3.31 , and( ) 2.25, 3.31are the solutions to the system. Solve using elimination. 2 22 222 25 9 2259 9 14416 8181

169

42.25x yx yxxxx+ = = === = Let2.25 x = and2.25 x =in 2 216 x y + =and solve for y. ( )2222.25 1610.93753.31yyy + === ( )2222.25 1610.93753.31yyy+ === The solutions are( ) 2.25, 3.31 ,( ) 2.25, 3.31 , ( ) 2.25, 3.31 , and( ) 2.25, 3.31 . 27. 2 22 29 852 3 6x yx y+ = = Solve using elimination. 2 22 22227 3 255 2 3 629 261 9 3x yx yxxx+ = ==== Let3 x = and3 x =in 2 29 85 x y + =and solve for y. ( )22229 3 8581 8542yyyy + =+ === ( )22229 3 8581 8542yyyy+ =+ === The four solutions are( ) 3, 2 ,( ) 3, 2 , ( ) 3, 2 , and( ) 3, 2 . Each result satisfies both equations. 29. 222 5 113 4y x xy x x= = + Solve using substitution. Substitute 22 5 11 x x for y in the second equation. ( )( )22 223 42 5 11 3 42 15 05 3 0y x xx x x xx xx x= + = + = + = 5 0or 3 05or 3x xx x = + == = Let5 x =and3 x = in 23 4 y x x = +and solve for y. ( ) ( )25 3 5 425 15 414y = += += ( ) ( )23 3 3 49 9 422y = += + += The solutions are( ) 5,14and( ) 3, 22 . Each result satisfies both equations. 31. 23 22 4y x xy x= += Solve using substitution. Substitute2 4 x for y in the first equation. ( )( )2223 22 4 3 25 6 03 2 0y x xx x xx xx x= + = + + = = 3 0or 2 03or 2x xx x = == = Let3 x =and2 x =in2 4 y x = and solve for y.SSM: Intermediate AlgebraSection 11.8 Quiz 377 ( ) 2 3 46 42y = = =( ) 2 2 44 40y = = = The solutions are( ) 3, 2and( ) 2, 0 . Each result satisfies both equations. 33. 2 22 22 2254 25 1004 25 100x yx yx y+ = =+ = 44yx(5,0)(5,0) The two intersection points for all three graphs are( ) 5, 0 and( ) 5, 0 . These are the solutions to the system. Each result satisfies all three equations. 35.Answers may vary. One possible answer: 2 22 21616x yx y+ = = 37. 2 222 825x cyy x dx+ == + + Substitute( ) 1, 4into each equation. ( ) ( )2 22 1 4 822 16 8216 805cccc+ =+ === ( ) ( )24 1 1 54 1 52ddd= + += + += 39. Written response. Answers may vary. Section 11.8 Quiz 1. 2 22 29 819x yx y+ =+ = 888yx8(3,0) (3,0) The two intersection points( ) 3, 0 and( ) 3, 0are the solutions to the system. Solve using elimination. 2 22 2229 819 9 81 8 0 0 0x yx yyyy+ = = === Let0 y =in 2 29 x y + =and solve for x. 2 22 2290 993x yxxx+ =+ === The solutions are( ) 3, 0 and( ) 3, 0 . 2. 222 1y xy x= = + 2 284yx4(3,7)(1, 1) The two intersection points( ) 1, 1 and( ) 3, 7 are the solutions of the system. Solve using substitution. Substitute 22 x for y in the second equation. Section 11.8 QuizSSM: Intermediate Algebra 378 ( )( )222 12 2 12 3 03 1 0y xx xx xx x= + = ++ =+ = 3 0or 1 03or 1x xx x+ = == = Let3 x = and1 x =in2 1 y x = +and solve for y. ( ) 2 3 16 17y = += += ( ) 2 1 12 11y = += += The solutions are( ) 3, 7 and( ) 1, 1 . 3. 2236 9y xy x x= += + 2 284yx4(1, 4) The intersection point( ) 1, 4is the solution to the system. Solve using substitution. Substitute 23 x +for y in the second equation. 22 26 93 6 96 6 01 01y x xx x xxxx= ++ = + = == Let1 x =in 23 y x = + and solve for y. ( )21 31 34y = += += The solution is( ) 1, 4 . 4. 2 22 225 4 1009 9x yx y =+ = 444yx4 The graphs do not intersect. There are no solutions to the system. Solve using elimination. 2 22 222225 36 900225 25 22561 675675 61x yx yyy = = == Since 2ymust be nonnegative (in the real number system), there are no solutions to the system. The solution set is the empty set,. 5. ( )2 22 2216164x yx yy x =+ == + 88yx8 (4, 0) The intersection point of all three graphs is ( ) 4, 0 . This is the solution of the system. 6.Answers may vary. One possible answer: 2 22255x yy x+ == +