184 Chapter 7 Using Quadratic Functions to Model Data Homework 7.1 1.16 4 4 4 4 2 2 4 = = = =3.8 4 2 4 2 2 2 = = =5.12 4 3 4 3 2 3 = = =7. 4 2 2 2 2 29 3 3 3 3 3= = = 9. 6 6 649 7 49= =11. 5 2 5 2 5 22 2 2 4 = =13. 9 6 9 6 9 6 3 66 2 6 6 36 = = =15. 3 3 3 2 3 2 3 2 3 24 2 8 32 16 2 4 2 2 4 4= = = = = 17. 3 3 2 6 62 2 2 2 4= = =19. 11 11 11 11 5 55 5520 10 20 4 5 2 5 5 2 25= = = = = 21. 236 x =

366xx= = 23. 25 x = 5 x = 25. 23 0 x =

233xx== 27. 232 x =

3216 24 2xxx= = = 29. 2300 x =

300100 310 3xxx= = = 31.248 0 x = 2484816 34 3xxxx== = = 33. 29 x = ; there are no real number solutions. 35. 225 0 x + =means 225 x = ; there are no real number solutions. 37. 24 36 x =

2993xxx== = 39. 25 8 x =

2858 4 2 2 25 5 52 2 5 2 105 5 252 105xxxx== = = = = = SSM: Intermediate AlgebraHomework 7.1 185 41. 23 14 0 x = 223 1414314 143 314 3 423 3 9423xxxxx=== = = = = 43. 2( 4) 25 x =4 254 54 5or 4 54 5or 4 59or 1xxx xx xx x = = = = = + = = = 45. 2(8 3) 36 x + = 8 3 368 3 68 3 6or8 3 68 3or8 93 9or8 8xxx xx xx x+ = + = + = + = = = = = 47. 2(9 7) 0 x =9 7 09 779xxx === 49. 2(8 3) 6 90 x + = 2(8 3) 848 3 848 3 4 218 3 2 218 3 2 213 2 218xxxxxx = = = = = = 51. 23(4 7) 9 x + = 2(4 7) 3 x + = Since the square of a number is nonnegative, there are no real number solutions. 53. 25( 6) 3 23 x + = 225( 6) 20( 6) 46 46 26 2or 6 28or 4xxxxx xx x = = = = = = = = 55. 23(4 9) 5 31 x + = 223(4 9) 36(4 9) 124 9 124 9 4 34 9 2 34 9 2 39 2 34xxxxxxx = = = = = = = 57. Solve for x when f(x) = 0 220 171717xxx= == The x-intercepts are ( )17, 0and ( )17, 0 . 59.Solve for x when h(x) = 0 220 ( 2) 4( 2) 42 42 22 2or 2 24or 0xxxxx xx x= = = = = = = = The x-intercepts are( ) 4, 0and( ) 0, 0 . Homework 7.1SSM: Intermediate Algebra 186 61.Solve for x when f(x) = 0 2220 3( 1) 123( 1) 12( 1) 41 41 21 2or 1 21or 3xxxxxx xx x= + ++ =+ =+ = + = + = + = = = The x-intercepts are( ) 1, 0and( ) 3, 0 . 63.Solve for x when h(x) = 0 2220 2( 5) 12( 5) 11( 5)2xxx= + ++ = + = Since the square of a number is nonnegative, there are no real number solutions. 65.Solve for x when f(x) = 0 22220 ( 5) 9( 5) 9( 5) 9( 5) 35 35 3or 5 38or 2xxxxxx xx x= = = = = = = = = The x-intercepts are( ) 8, 0and( ) 2, 0 . Since these points are symmetric, the average of the x-coordinates,8 2 1052 2+= = , is the x-coordinate of the vertex of f(x). Substitute 5 for x in the function to find the y-coordinate of the vertex: 2( ) (5 5) 9 9 f x = = Therefore, the vertex is (5, -9). 67.Solve for x when h(x) = 0 2220 2( 7) 92( 7) 99( 7)2972972372372xxxxxxx= + ++ =+ =+ = + = + = = Rationalize the denominator: 3 2 3 22 2 2 = 3 2724.88or 9.12xx x= So, the x-intercepts are( ) 4.88, 0 and( ) 9.12, 0 . Since these points are symmetric, the average of the x-coordinates,4.88 ( 9.12)72 += , is the x-coordinate of the vertex of f(x). Substitute -7 for x in the function to find the y-coordinate of the vertex: 2( ) 2( 7 7) 9 9 hx = + + =Therefore, the vertex is (-7, 9). SSM: Intermediate AlgebraHomework 7.2 187 69.Solve for x when g(x) = 0 220 3131315.57or 5.57xxxx x= == So, the x-intercepts are ( )31, 0 and ( )31, 0 . Since these points are symmetric, the average of the x-coordinates,31 ( 31)02+= , is the x-coordinate of the vertex of f(x). Substitute 0 for x in the function to find the y-coordinate of the vertex: 2( ) (0) 31 31 gx = = Therefore, the vertex is (0, -31). 71.No, the student did not solve it correctly. They should have factored the left hand side first. 2210 25 0( 5)( 5) 0( 5) 05 05x xx xxxx + = = = == 73. 112212a a a ab b bb = = = 75.Answers may vary.Example: You can solve quadratic equations by extracting the square root as long as they can be placed in vertex form. (Samples: 281 0 x =or 22( 1) 9 0 x + = ) You can solve quadratic equations by factoring as long as the quadratic expression contained in the equation is not prime.(Samples: 281 ( 9)( 9) 0 x x x = + =or 24 4 0 x x + + = ). Homework 7.2 1. 22126 362c = = = This expression is 212 36 x x + +and its factored form is( )26 x + . 3. 2214( 7) 492c = = = This expression is 214 49 x x +and its factored form is( )27 x . 5. 2227 ( 7) 492 4 2c = = = This expression is 24974x x +and its factored form is 272x . 7. 2223 3 92 4 2c = = = This expression is 2934x x + +and its factored form is 232x + . Homework 7.2SSM: Intermediate Algebra 188 9. 2 2221 1 1 1 12 2 4 16 4c = = = = This expression is 21 12 16x x + +and its factored form is 214x + . 11. 2 2 2224 1 4 2 ( 2) 45 2 10 5 25 5c = = = = = This expression is 24 45 25x x +and its factored form is 225x . 13.Since 2263 92 = = , add 9 to both sides of the equation. ( )2226 16 9 1 93 103 103 10x xx xxxx+ =+ + = ++ =+ = = 15.Since 22126 362 = = , add 36 to both sides of the equation. ( )22212 212 36 2 366 386 386 38x xx xxxx+ =+ + = ++ =+ = = 17.Since 222( 1) 12 = = , add 1 to both sides of the equation. ( )2222 102 1 10 11 111 111 11x xx xxxx = + = + = = = 19.First, change the equation to the form 2 2;18 3 x bx p x x + = = . Since 2218( 9) 812 = = , add 81 to both sides of the equation. ( )22218 318 81 3 819 849 849 2 219 2 21x xx xxxxx = + = + = = = = 21.First, change the equation to the form 2 2;18 9 x bx p x x + = + = . Since 2218(9) 812 = = , add 81 to both sides of the equation. ( )22218 918 81 9 819 909 909 3 109 3 10x xx xxxxx+ = + = ++ =+ = + = = 23.First, change the equation to the form 2 2;4 8 x bx p x x + = + = . Since 224(2) 42 = = , add 4 to both sides of the equation. SSM: Intermediate AlgebraHomework 7.2 189 ( )2224 84 4 8 42 4x xx xx+ = + + = ++ = Since the square of a number is nonnegative, there are no real number solutions. 25.Since 2227 ( 7) 492 4 2 = = , add 494 to both sides of the equation. 22227 349 497 34 47 12 492 4 47 612 47 612 47 612 27 612 27 612x xx xxxxxxx = + = + = + = = = = = 27.Since 2225 5 252 4 2 = = , add 254 to both sides of the equation. 22225 425 255 44 45 16 252 4 45 412 4x xx xxx+ =+ + = + + = + + =

5 412 45 412 25 412 25 412xxxx+ = + = = = 29.First, change the equation to the form 2 2;7 x bx p x x + = + = . Since 2221 1 12 4 2 = = , add 14 to both sides of the equation. 222271 174 41 28 12 4 41 272 4x xx xxx+ = + + = + + = + + = Since the square of a number is nonnegative, there are no real number solutions. 31.Since 2 2225 1 5 ( 5) 252 2 4 16 4 = = = , add 2516 to both sides of the equation. 22225 12 25 25 1 252 16 2 165 8 254 16 165 334 16x xx xxx = + = + = + = Homework 7.2SSM: Intermediate Algebra 190

5 334 165 334 45 334 45 334xxxx = = = = 33.Divide both sides by 2 so that a = 1: 2342x x = . Since( )22 42 42 = = , add 4 to both sides of the equation. ( )( )222234234 4 423 822 2112211221122x xx xxxxx = + = + = + = = = Rationalize the denominator: 11 2 222 2 222224 222 24 222xxx == = = 35.Change the equation so that it is in the form 2ax bx p + = :25 15 9 x x + = . Divide both sides by 5 so that a = 1:2935x x + = . Since 2223 3 92 4 2 = = , add 94to both sides of the equation. 22229359 9 934 5 43 36 452 20 203 92 203 92 203 92 203 32 2 5x xx xxxxxx+ = + + = + + = + + = + = + = + = Rationalize the denominator: 3 5 3 510 2 5 53 3 52 103 3 52 1015 3 510 1015 3 510xxxx =+ = = = = SSM: Intermediate AlgebraHomework 7.2 191 37.Divide both sides by 3 so that a = 1:24 53 3x x + = . Since 2 2224 1 2 2 43 2 3 9 3 = = = , add 49to both sides of the equation. 22224 53 34 4 5 43 9 3 92 15 43 9 92 193 92 193 92 193 92 193 32 193 32 193x xx xxxxxxxx+ =+ + = + + = + + = + = + = + = = = 39.Change the equation so that it is in the form 2ax bx p + = :22 7 x x = . Divide both sides by 2 so that a = 1:21 72 2x x = . Since 2 2221 1 1 ( 1) 12 2 4 16 4 = = = , add 116to both sides of the equation. 22221 72 21 1 7 12 16 2 161 56 14 16 161 574 16x xx xxx = + = + = + = 1 574 161 574 161 574 41 574 41 574xxxxx = = = = = 41.Change the equation so that it is in the form 2ax bx p + = :25 7 3 x x + = . Divide both sides by 5 so that a = 1:27 35 5x x + = . Since 2 2227 1 7 7 495 2 10 100 10 = = = , add 49100to both sides of the equation. 22227 35 57 49 3 495 100 5 1007 60 4910 100 1007 1110 100x xx xxx+ = + + = + + = + + = Since the square of a number is nonnegative, there are no real number solutions. 43.Divide both sides by 8 so that a = 1:25 18 2x x = . Since 2 2225 1 5 ( 5) 258 2 16 256 16 = = = , add 25256

to both sides of the equation. 22225 18 25 25 1 258 256 2 2565 128 2516 256 2565 15316 256x xx xxx = + = + = + = Homework 7.2SSM: Intermediate Algebra 192 5 15316 2565 15316 2565 9 1716 165 3 1716 165 3 1716xxxxx = = = = = 45.First, change the equation so that it is in the form 2ax bx p + = : 222 ( 3) 52 6 52 5 5xx xx x xx x = = = Divide both sides by 2, so that a = 1: 25 52 2x x =Since 2 2225 1 5 ( 5) 252 2 4 16 4 = = = , add 2516

to both sides of the equation. 22225 52 25 25 5 258 16 2 165 40 254 16 165 654 165 654 165 654 165 654 45 654 45 654x xx xxxxxxxx = + = + = + = = = = = = 47.First, change the equation so that it is in the form 2ax bx p + = : 22( 1) 3 5( 1)3 5 56 2xx xx x xx x + = + + = + = Since( )22 63 92 = = , add 9to both sides of the equation. ( )2226 26 9 2 93 113 113 11x xx xxxx = + = + = = = 49.No, the student did not solve the equation correctly. The student should have first divided both sides by 4 and then completed the square and extracted the roots. The correct solution is: 222224 6 13 12 43 9 1 92 16 4 163 4 94 16 163 134 163 134 163 134 163 134 43 134 43 134x xx xx xxxxxxxx+ =+ =+ + = + + = + + = + = + = + = = = SSM: Intermediate AlgebraHomework 7.3 193 51.a. 23 6 13 x x = + + ( )2226 106 9 10 93 1x xx xx+ = + + = ++ = Since the square of a number is nonnegative, there are no real number solutions. b. 24 6 13 x x = + + ( )226 9 03 03 03x xxxx+ + =+ =+ == c. 25 6 13 x x = + + ( )( )26 8 02 4 02 0or 4 02or 4x xx xx xx x+ + =+ + =+ = + == = 53.Solve for x when f(x) = 0: 22220 8 38 38 16 3 16( 4) 134 134 13x xx xx xxxx= + = + = + = = = 55.Solve for x when f(x) = 0: 22220 4 54 54 4 5 4( 2) 1x xx xx xx= + ++ = + + = ++ = Since the square of a number is nonnegative there are no real number solutions. Therefore, there are no x-intercepts. 57. Solve for x when f(x) = 0: 220 10 25( 5) 05 05x xxxx= + ++ =+ == 59.Answers may vary.Homework 7.3 1.2, 5,2 a b c = = = ( )( )( )25 5 4 2 22 25 414xx = = 3.3, 7,2 a b c = = = ( )( )( )27 7 4 3 22 37 736xx = = 5.2, 5,1 a b c = = = ( ) ( ) ( )( )( )25 5 4 2 12 25 174xx == 7.3, 2,8 a b c = = = ( ) ( ) ( )( )( )22 2 4 3 82 32 926xx = = Since the square root of a negative number is not a real number, there are no real solutions. 9.First write 22 5 3 x x + =in the form 20 ax bx c + + = : 22 5 3 0 x x + =2, 5,3 a b c = = = Homework 7.3SSM: Intermediate Algebra 194 ( )( )( )25 5 4 2 32 25 4945 745 7 5 7or4 42 12or4 41or 32xxxx xx xx x = = = + = == == = 11.3, 0,17 a b c = = = ( ) ( )( )( )20 0 4 3 172 320462 516513xxxx == = = 13.First write 22 5 x x = in the form 20 ax bx c + + = : 22 5 0 x x + =2, 5,0 a b c = = = ( )( )( )25 5 4 2 02 25 2545 545 5 5 5or4 40 10or4 450or2xxxx xx xx x = = = + = == == = 15.First write 24 2 3 x x = +in the form 20 ax bx c + + = : 24 2 3 0 x x =4, 2,3 a b c = = = ( ) ( ) ( )( )( )( )22 2 4 4 32 42 5282 2 1382 1 1381 134xxxxx ===== 17.First write( ) 3 1 2 x x + =in the form 20 ax bx c + + = : ( )223 1 23 3 23 3 2 0x xx xx x + = = = 3, 3,2 a b c = = = ( ) ( ) ( )( )( )23 3 4 3 22 33 156xx = =

Since the square root of a negative number is not a real number, there are no real solutions. 19.First write ( )23 2 5 x x =in the form 20 ax bx c + + = : ( )2223 2 53 6 53 5 6 0x xx xx x = = = 3, 5,6 a b c = = = SSM: Intermediate AlgebraHomework 7.3 195 ( ) ( ) ( )( )( )25 5 4 3 62 35 976xx == 21.First write( ) ( ) 3 4 2 1 x x x + = in the form 20 ax bx c + + = : ( ) ( )223 4 2 13 12 2 22 12 0x x xx x xx x+ = + = ++ + = 2, 1,12 a b c = = = ( )( )( )21 1 4 2 122 21 954xx = = Since the square root of a negative number is not a real number, there are no real solutions. 23.First write 23 6x x =in the form 20 ax bx c + + = : 26 3 0 x x + =1, 6,3 a b c = = = ( )( )( )( )26 6 4 1 32 26 4826 4 322 3 323 3xxxxx = = = == 25.2, 5,4 a b c = = = ( ) ( ) ( )( )( )25 5 4 2 42 25 5745 57 5 57or4 43.14or 0.64xxx xx x == + = = 27.First write 22.8 7.1 4.4 x x =in the form 20 ax bx c + + = : 22.8 7.1 4.4 0 x x = ( ) ( ) ( )( )( )22.8, 7.1,4.47.1 7.1 4 2.8 4.42 2.87.1 99.695.67.1 99.69 7.1 99.69or5.6 5.63.05or 0.52a b cxxx xx x= = = ==+ = = 29.First write5.4 ( 9.8) 4.1 3.2 6.9 xx x + + = in the form 20 ax bx c + + = : 225.4 52.92 4.1 3.2 6.95.4 26.02 0.9 0x x xx x + = + = 5.4, 46.02,0.9 a b c = = = ( ) ( ) ( )( )( )246.02 46.02 4 5.4 0.92 5.446.02 2137.280410.846.02 2137.28 46.02 2137.28or10.8 10.88.54or 0.020xxx xx x ==+ 31. 225 x = 225 0( 5)( 5) 05 0or 5 05or 5xx xx xx x =+ =+ = == = Homework 7.3SSM: Intermediate Algebra 196 33.( 8) 16 xx = 228 168 16 0( 4)( 4) 04 04x xx xx xxx = + = = == 35. 24 80 x = 220202 5xxx== = 37. 2(4 3) 2 22 x + + = ( )24 3 204 3 204 3 2 54 3 2 53 2 54xxxxx+ =+ = + = = = 39.2 (3 4) 5 x x x = 2226 8 56 9 5 06, 9, 5( 9) ( 9) 4(6)( 5)2(6)9 3912x x xx xa b cxx = + == = = = = Since the square root of a negative number is not a real number, there are no real solutions. 41. 29 5 0 x x = (9 5) 00or9 5 00or9 550or9x xx xx xx x == == == = 43. 25 3 8 x x = + 25 3 8 0(5 8)( 1) 05 8 0or 1 05 8or 18or 15x xx xx xx xx x = + = = + == = = = 45. 23 27 x x = 23 27 03 ( 9) 0( 9) 00or 9 00or 9x xxxxxx xx x = = == == = 47. 22(2 1) 3 x x + = : 224 2 34 3 2 0x xx x+ = + = 4, 3,2 a b c = = = 2( 3) ( 3) 4(4)(2)2(4)3 238xx = = Since the square root of a negative number is not a real number, there are no real solutions. 49. 212 36 0 x x + + = ( 6)( 6) 06 06x xxx+ + =+ == 51. 224 18 60 0 x x = 226(4 3 10) 04 3 10 0(4 5)( 2) 04 5 0or 2 04 5or 25or 24x xx xx xx xx xx x = =+ =+ = == == = SSM: Intermediate AlgebraHomework 7.3 197 53. 2( 4) 3 7 x = 2( 4) 104 104 10xxx = = = 55.( 1) 7 xx x + = 222272 72 1 7 1( 1) 81 81 2 21 2 2x x xx xx xxxxx+ = + =+ + = ++ =+ = + = = 57. 2 (3 1) 3( 2) x x x = 226 2 3 66 5 6 0x x xx x = + = 6, 5,6 a b c = = = 2( 5) ( 5) 4(6)(6)2(6)5 11912xx = = Since the square root of a negative number is not a real number, there are no real solutions. 59.2 (3 1) 3( 2) x x x = 222222 6 802( 3 ) 803 409 93 404 43 160 92 4 4x xx xx xx xx = = = + = + = + 23 1692 4x = 3 1692 43 132 23 132 23 13 3 13or2 2 2 216 10=8or 52 2xxxx xx x = = = = + = = = = 61.25 35 x x = 25 35 05 ( 7) 05 0or 7 00or 7x xxxx xx x = == == = 63.250 5 105 x x = + 2225 50 105 05( 25 21) 025 21 0( 3)( 7) 03 0or 7 03or 7x xx xx xx xx xx x + = + = + = = = == = 65.2(3 1) 7 x = 3 1 73 1 71 73xxx = = = 67. ( 2)( 5) 1 x x = 222225 2 10 17 949 497 94 47 36 492 4 47 132 4x x xx xx xxx + = = + = + = + = Homework 7.3SSM: Intermediate Algebra 198 7 132 47 132 27 132 27 132xxxx = = = = 69.2 2( 1) ( 2) 0 x x + + = 2 22222222 1 4 4 02 2 5 0502521 5 14 2 41 10 12 4 41 92 41 92 4x x x xx xx xx xx xxxx + + + + =+ + =+ + =+ = + + = + + = + + = + = Since the square root of a negative number is not a real number, there are no real solutions. 71.2 2( 2)( 5) 5 2 10 3 10 x x x x x x x + = + = 73. ( 2)( 5) 3 x x + = 223 10 33 13 0x xx x = = 21, 3, 13( 3) ( 3) 4(1)( 13)2(1)3 612a b cxx= = = == 75.24( 2) 3 1 x + = 2222224( 2)( 2) 3 14( 2 2 4) 3 14( 4 4) 3 14 16 16 3 14 16 12 04( 4 3) 04 3 0( 3)( 1) 03 0or 1 03or 1x xx x xx xx xx xx xx xx xx xx x + = + + = + + = + + = + = + = + = = = == = 77.24( 2) 3 4( 2)( 2) 3 x x x + = + 22224( 2 2 4) 34( 4 4) 34 16 16 34 16 13x x xx xx xx x= + += + += + += + 79. a.Substitute 3 for f(x): 222 23 4 84 5 01, 4, 54 ( 4) 4(1)(5) 16 20 4 0x xx xa b cb ac= + + == = = = = = < So, there are no real number solutions, which means there are no points on f at y = 3. b.Substitute 4 for f(x): 222 24 4 84 4 01, 4, 44 ( 4) 4(1)(4) 16 16 0x xx xa b cb ac= + + == = = = = = So, there is one solution to the equation, which means there is one point on f at y = 4. SSM: Intermediate AlgebraHomework 7.3 199 c.Substitute 5 for f(x): 222 25 4 84 3 01, 4, 34 ( 4) 4(1)(3) 16 12 4 0x xx xa b cb ac= + + == = = = = = > So, there are two real number solutions, which means there are two points on f at y = 5. d. 81. Solve for x when f(x) = 2: 222 6 76 5 0( 5)( 1) 05 0or 1 05or 1x xx xx xx xx x= + + = = = == = Therefore, two points at height 2 are( ) 1, 2and ( ) 5, 2 . Since these points are symmetric and the average of the x-coordinates at these points is 1 532+=the x-coordinate of the vertex is 3. Substitute 3 for x in 2( ) 6 7 f x x x = +to find the y-coordinate of the vertex: 2(3) 3 6(3) 7 2 f = + = So the vertex is( ) 3, 2 . 83. a. Yes, f does model the data well. b.In 2007, t = 17. Solve for f(17) 2(17) 30.32(17) 122.15(17) 109.756795.686796 schoolsf = += c.200 schools per state means 10,000 charter schools. Solve for t when f(t) = 10,000. 22210000 30.32 122.15 109.750 30.32 122.15 9890.25122.15 (122.15) 4(30.32)( 9890.25)2(30.32)16.16or 20.1916 or 20t tt ttt tt t= += = The solution t = -16 is model breakdown. There will be 200 charter schools per state in 2010. 85. a. Yes, f does model the data well. b. 2(13) 23.43(13) 259.14(13) 815.77 f = + 1406.651407 This means that in 2003, there will be1407 convictions of police officers. Homework 7.3SSM: Intermediate Algebra 200 c. 22000 23.43 259.14 815.77 t t = + 222259.14 (259.14) 4(23.43)( 1184.23)2(23.43)0 23.43 259.14 815.770 23.43 259.14 1184.233.48or 14.543 or 15t tt ttt tt t = += = There were 2000 convictions of police officers in 1987 and 2005.87. No, the student did not solve the equation correctly because they did not change the form into 20 ax bx c + + =first. Here is the correct way: 2222 5 12 5 1 0So,2, 5,15 (5) 4(2)( 1)2(2)5 334x xx xa b cxx+ =+ == = = = = 89. Solve for x when f(x) = 0: 220 4 6( 4) ( 4) 4(1)( 6)2(1)4 4024 2 1022 105.16or 1.16x xxxxxx x= ==== The x-intercepts are (5.16, 0) and (-1.16, 0) 91. Solve for x when h(x) = 0: 220 3 2 5( 2) ( 2) 4(3)(5)2(3)2 566x xxx= + = = Since the square root of a negative number is not a real number, there are no real number solutions, and therefore, no x-intercepts. 93. a.0 mx b + = 0 mx b b bmx bmx bm mbxm+ = = = = b.7 21 0 x + =So, m = 7 and b = -21. Using the formula from part a: 2137x = = Solving for x in the usual way: 7 21 07 21 21 0 217 217 217 73xxxxx+ =+ = = = = 95. a = k, b = -12, c = 4 224 0( 12) 4( )(4) 0144 16 016 1449b ackkkk = = = = = 97. a = 2, b = k, c = 8 22224 04(2)(8) 064 0648b ackkkk = = === SSM: Intermediate AlgebraHomework 7.4 201 99. a = 9,b = -6, c = k 224 0( 6) 4(9)( ) 036 36 036 361b ackkkk = = = = = 101. Factoring: 220 0( 5)( 4) 05 0or 4 05or 4x xx xx xx x = + = = + == = Completing the square: 222201 1204 41 812 41 812 41 92 21 9 1 9or2 2 2 210 8or2 25or 4x xx xxxxx xx xx x = + = + = = = = + = = = = = Quadratic formula: 2220 0( 1) ( 1) 4(1)( 20)2(1)1 8121 921 9 1 9or2 25or 4x xxxxx xx x = ===+ = == = 103. Answers may vary. Example:First, put the equation in the form 20 ax bx c + + = . Once you know the value of a, b, and c, substitute these values into the quadratic formula. 242b b acxa =Homework 7.4 1.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2221, 6 : 6 1 12,11 :11 2 23, 8 : 8 3 3a b ca b ca b c= + += + += + +

Simplify these equations: ( )( )( )614 2 11 29 3 18 3a b ca b ca b c+ + =+ + =+ + = Eliminate c by multiplying both sides of equation (1) by -1: ( ) 6 4 a b c = Adding the left sides and right sides of equations (2) and (4) gives:( ) 3 5 5 a b + =Adding the left sides and right sides of equations (3) and (4) gives: ( ) 8 2 12 6 a b + =Simplify: ( ) 4 6 7 a b + =Eliminate b by multiplying equation (5) by -1 and add each side to the corresponding side of equation (7): 1 a =Next, substitute 1 for a in equation (5): ( ) 3 1 52bb+ == Then, substitute 1 for a and 2 for b in equation (1): 61 2 63a b ccc+ + =+ + == Therefore, a = 1, b = 2, and c = 3. So, the equation is 22 3 y x x = + + . Homework 7.4SSM: Intermediate Algebra 202 3.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2221, 5 : 5 1 12,11 :11 2 23,19 :19 3 3a b ca b ca b c= + += + += + +

Simplify these equations: ( )( )( )5 14 2 11 29 3 19 3a b ca b ca b c+ + =+ + =+ + = Eliminate c by multiplying both sides of equation (1) by -1: ( ) 5 4 a b c = Adding the left sides and right sides of equations (2) and (4) gives:( ) 3 6 5 a b + =Adding the left sides and right sides of equations (3) and (4) gives: ( ) 8 2 14 6 a b + =Simplify: ( ) 4 7 7 a b + =Eliminate b by multiplying equation (5) by -1 and add each side to the corresponding side of equation (7): 1 a =Next, substitute 1 for a in equation (5): ( ) 3 1 63bb+ == Then, substitute 1 for a and 3 for b in equation (1): 51 3 51a b ccc+ + =+ + == Therefore, a = 1, b = 3, and c = 1. So, the equation is 23 1 y x x = + + . 5.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2221, 9 : 9 1 12, 7 : 7 2 24, 15 : 15 4 4a b ca b ca b c= + += + + = + +

Simplify these equations: ( )( )( )9 14 2 7216 4 15 3a b ca b ca b c+ + =+ + =+ + = Eliminate c by multiplying both sides of equation (1) by -1: ( ) 9 4 a b c = Adding the left sides and right sides of equations (2) and (4) gives:( ) 3 2 5 a b + = Adding the left sides and right sides of equations (3) and (4) gives: ( ) 15 3 24 6 a b + = Simplify: ( ) 5 8 7 a b + = Eliminate b by multiplying equation (5) by -1 and add each side to the corresponding side of equation (7): 2 63aa= = Next, substitute -3 for a in equation (5): ( ) 3 3 27bb + = = Then, substitute -3 for a and 7 for b in equation (1): 93 7 95a b ccc+ + = + + == Therefore, a = -3, b = 7, and c = 5. So, the equation is 23 7 5 y x x = + + . 7. Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2222, 2 : 2 2 23,11 :11 3 34, 24 : 24 4 4a b ca b ca b c= + += + += + +

Simplify these equations: ( )( )( )219 3 11 216 4 24 3a b ca b ca b c+ + =+ + =+ + = Eliminate c by multiplying both sides of equation (1) by -1: ( ) 4 2 2 4 a b c = Adding the left sides and right sides of equations (2) and (4) gives:( ) 5 9 5 a b + =Adding the left sides and right sides of equations (3) and (4) gives: ( ) 12 2 22 6 a b + =Simplify: ( ) 6 11 7 a b + =SSM: Intermediate AlgebraHomework 7.4 203 Eliminate b by multiplying equation (5) by -1 and add each side to the corresponding side of equation (7): 2 a =Next, substitute 2 for a in equation (5): ( ) 5 2 91bb+ == Then, substitute 2 for a and -1 for b in equation (1): 4 2 24(2) 2( 1) 28 2 24a b cccc+ + =+ + = + == Therefore, a = 2, b = -1, and c = -4. So, the equation is 22 4 y x x = . 9.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2221, 3 : 3 1 13, 9 : 9 3 35, 29 : 29 5 5a b ca b ca b c = + += + += + +

Simplify these equations: ( )( )( )3 19 3 9 225 5 29 3a b ca b ca b c+ + = + + =+ + = Eliminate c by multiplying both sides of equation (1) by -1: ( ) 3 4 a b c =Adding the left sides and right sides of equations (2) and (4) gives:( ) 8 2 12 5 a b + =Simplify: ( ) 4 6 6 a b + =Adding the left sides and right sides of equations (3) and (4) gives: ( ) 24 4 32 7 a b + =Simplify: ( ) 6 8 8 a b + =Eliminate b by multiplying equation (6) by -1 and add each side to the corresponding side of equation (8): 2 21aa== Next, substitute 1 for a in equation (6): 4(1) 62bb+ == Then, substitute 1 for a and 2 for b in equation (1): 31 2 36a b ccc+ + = + + = = Therefore, a = 1, b = 2, and c = -6. So, the equation is 22 6 y x x = + . 11.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2223, 7 : 7 3 34, 0 : 0 4 45, 11 : 11 5 5a b ca b ca b c= + += + + = + +

Simplify these equations: ( )( )( )9 3 7116 4 0225 5 11 3a b ca b ca b c+ + =+ + =+ + = Eliminate c by multiplying both sides of equation (1) by -1: ( ) 9 3 7 4 a b c = Adding the left sides and right sides of equations (2) and (4) gives:( ) 7 7 5 a b + = Adding the left sides and right sides of equations (3) and (4) gives: ( ) 16 2 18 6 a b + = Simplify: ( ) 8 9 7 a b + = Eliminate b by multiplying equation (5) by -1 and add each side to the corresponding side of equation (7): 2 a = Next, substitute -2 for a in equation (5): ( ) 7 2 77bb + = = Then, substitute -2 for a and 7 for b in equation (1): 9 3 79( 2) 3(7) 718 21 74a b cccc+ + = + + = + + == Therefore, a = -2, b = 7, and c = 4. So, the equation is 22 7 4 y x x = + + . Homework 7.4SSM: Intermediate Algebra 204 13.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2223, 2 : 2 3 34,16 :16 4 45, 36 : 36 5 5a b ca b ca b c= + += + += + +

Simplify these equations: ( )( )( )9 3 2 116 4 16 225 5 36 3a b ca b ca b c+ + =+ + =+ + = Eliminate c by multiplying both sides of equation (1) by -1: ( ) 9 3 2 4 a b c = Adding the left sides and right sides of equations (2) and (4) gives:( ) 7 14 5 a b + =Adding the left sides and right sides of equations (3) and (4) gives: ( ) 16 2 34 6 a b + =Simplify: ( ) 8 17 7 a b + =Eliminate b by multiplying equation (5) by -1 and add each side to the corresponding side of equation (7): 3 a =Next, substitute 3 for a in equation (5): ( ) 7 3 147bb+ == Then, substitute 3 for a and -7 for b in equation (1): 23 7 24a b ccc+ + = + == Therefore, a = 3, b = -7, and c = -4. So, the equation is 23 7 4 y x x = . 15.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2222, 5 : 5 2 24, 3 : 3 4 45,13 :13 5 5a b ca b ca b c = + += + += + +

Simplify these equations: ( )( )( )4 2 5116 4 3225 5 13 3a b ca b ca b c+ + = + + =+ + = Eliminate c by multiplying both sides of equation (1) by -1: ( ) 4 2 5 4 a b c =Adding the left sides and right sides of equations (2) and (4) gives:( ) 12 2 8 5 a b + =Simplify: ( ) 6 4 6 a b + =Adding the left sides and right sides of equations (3) and (4) gives: ( ) 21 3 18 7 a b + =Simplify: ( ) 7 6 8 a b + =Eliminate b by multiplying equation (6) by -1 and add each side to the corresponding side of equation (8): 2 a =Next, substitute 2 for a in equation (6): ( ) 6 2 48bb+ == Then, substitute 2 for a and -8 for b in equation (1): 4 2 54(2) 2( 8) 58 16 53a b ccc+ + = + + = = = Therefore, a = 2, b = -8, and c = 3. So, the equation is 22 8 3 y x x = + . 17.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2221, 1 : 1 1 12, 1 : 1 2 23, 3 : 3 3 3a b ca b ca b c = + + = + += + +

Simplify these equations: ( )( )( )1 14 2 1 29 3 3 3a b ca b ca b c+ + = + + = + + = Eliminate c by multiplying both sides of equation (1) by -1: ( ) 1 4 a b c =Adding the left sides and right sides of equations (2) and (4) gives:( ) 3 0 5 a b + =Adding the left sides and right sides of equations (3) and (4) gives: ( ) 8 2 4 6 a b + =Simplify: SSM: Intermediate AlgebraHomework 7.4 205 ( ) 4 2 7 a b + =Eliminate b by multiplying equation (5) by -1 and add each side to the corresponding side of equation (7): 2 a =Next, substitute 2 for a in equation (5): ( ) 3 2 06bb+ == Then, substitute 2 for a and -6 for b in equation (1): 12 6 13a b ccc+ + = + = = Therefore, a = 2, b = -6, and c = 3. So, the equation is 22 6 3 y x x = + . 19.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2220, 4 : 4 0 02, 8 : 8 2 23,1 :1 3 3a b ca b ca b c= + += + += + +

Simplify these equations: ( )( )( )4 14 2 829 3 1 3ca b ca b c=+ + =+ + = Since c = 4, substitute 4 for c in equations (2) and (3): 4 2 4 89 3 4 1a ba b+ + =+ + = Simplifying these equations gives: ( )( )4 2 4 49 3 3 5a ba b+ =+ = To eliminate b, multiply both sides of equation (4) by -3 and both sides of equation (5) by 2: ( )( )12 6 12 618 6 6 7a ba b = + = Adding the left and the right sides of equations (6) and (7) gives: 6 183aa= = Next, substitute -3 for a in equation (4): 4( 3) 2 42 168bbb + === Therefore, a = -3, b = 8, and c = 4. So, the equation is 23 8 4 y x x = + + . 21.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2220, 1 : 1 0 01, 3 : 3 1 12,13 :13 2 2a b ca b ca b c = + += + += + +

Simplify these equations: ( )( )( )113 24 2 13 3ca b ca b c= + + =+ + = Since c = -1, substitute -1 for c in equations (2) and (3): ( 1) 34 2 ( 1) 13a ba b+ + =+ + = Simplifying these equations gives: ( )( )444 2 14 5a ba b+ =+ = To eliminate b, multiply both sides of equation (4) by -2 and add each side to the corresponding side in equation (5): 2 63aa== Next, substitute 3 for a in equation (4): 3 41bb+ == Therefore, a = 3, b = 1, and c = -1. So, the equation is 23 1 y x x = + . 23.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2220,17 :17 0 02,11 :11 2 23, 2 : 2 3 3a b ca b ca b c= + += + += + +

Simplify these equations: ( )( )( )17 14 2 1129 3 2 3ca b ca b c=+ + =+ + = Since c = 17, substitute 17 for c in equations (2) and (3): 4 2 17 119 3 17 2a ba b+ + =+ + = Simplifying these equations gives: ( )( )4 2 6 49 3 155a ba b+ = + = To eliminate b, multiply both sides of equation (4) by -3 and both sides of equation (5) by 2: Homework 7.4SSM: Intermediate Algebra 206 ( )( )12 6 18 618 6 30 7a ba b =+ = Adding the left and the right sides of equations (6) and (7) gives: 6 122aa= = Next, substitute -2 for a in equation (4): 4( 2) 2 62 21bbb + = == Therefore, a = -2, b = 1, and c = 17. So, the equation is 22 17 y x x = + + . 25.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2221,1 :1 1 12, 4 : 4 2 23, 9 : 9 3 3a b ca b ca b c= + += + += + +

Simplify these equations: ( )( )( )1 14 2 429 3 93a b ca b ca b c+ + =+ + =+ + = Eliminate c by multiplying both sides of equation (1) by -1: ( ) 1 4 a b c = Adding the left sides and right sides of equations (2) and (4) gives:( ) 3 3 5 a b + =Adding the left sides and right sides of equations (3) and (4) gives: ( ) 8 2 8 6 a b + =Simplify: ( ) 4 4 7 a b + =Eliminate b by multiplying equation (5) by -1 and add each side to the corresponding side of equation (7): 1 a =Next, substitute 1 for a in equation (5): ( ) 3 1 30bb+ == Then, substitute 1 for a and 0 for b in equation (1): 11 0 10a b ccc+ + =+ + == Therefore, a = 1, b = 0, and c = 0. So, the equation is 2y x = . 27.Substitute the given points into 2( ) f x ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2221,1 :1 1 12, 2 : 2 2 23, 3 : 3 3 3a b ca b ca b c= + += + += + +

Simplify these equations: ( )( )( )114 2 2 29 3 3 3a b ca b ca b c+ + =+ + =+ + = Eliminate c by multiplying both sides of equation (1) by -1: ( ) 1 4 a b c = Adding the left sides and right sides of equations (2) and (4) gives:( ) 3 1 5 a b + =Adding the left sides and right sides of equations (3) and (4) gives: ( ) 8 2 2 6 a b + =Simplify: ( ) 4 1 7 a b + =Eliminate b by multiplying equation (5) by -1 and add each side to the corresponding side of equation (7): 0 a =Next, substitute 0 for a in equation (5): ( ) 3 0 11bb+ == Then, substitute 0 for a and 1 for b in equation (1): 10 1 10a b ccc+ + =+ + == Therefore, a = 0, b = 1, and c = 0. So, the equation is( ) f x x = , which is a linear function. 30.Solve for a in 2( ) y ax h k = +by substituting 5 for h and -7 for k since (5, -7) is the vertex, and substitute 8 for x and 11 for y since (8, 11) lies on the parabola:SSM: Intermediate AlgebraHomework 7.4 207 22( )11 (8 5) ( 7)11 9 718 92y ax h kaaaa= += += == Since a = 2, 22( 5) 7 y x = or, expanding the solution: 2222( 5)( 5) 72( 10 25) 72 20 50 72 20 43y x xx xx xx x= = + = + = +. 31.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2220, 4 : 4 0 01, 0 : 0 1 12, 0 : 0 2 2a b ca b ca b c= + += + += + +

Simplify these equations: ( )( )( )4 1024 2 03ca b ca b c=+ + =+ + = Since c = 4, substitute 4 for c in equations (2) and (3): 4 0 4 2 4 0 a ba b+ + =+ + = Simplify these equations: ( )( )444 2 4 5a ba b+ = + = Eliminate b by multiplying equation (4) by -2 and add each side to the corresponding side of equation (5): 2 42aa== Next, substitute 2 for a in equation (4): 2 46bb+ = = Therefore, a = 2, b = -6, and c = 4. So, the equation is 22 6 4 y x x = + . 33.Linear: y = 2x+2 since the slope is 2 and the y-intercept is (0l, 2); Quadratic: answers may vary. Example: 22 2 y x = + ; Exponential:2(2)xy = . 35.Three possible points are (2, 8), (3, 4), and (6, 4). Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2222, 8 : 8 2 23, 4 : 4 3 36, 4 : 4 6 6a b ca b ca b c= + += + += + +

Simplify these equations: ( )( )( )4 2 8 19 3 4 236 6 4 3a b ca b ca b c+ + =+ + =+ + = Eliminate c by multiplying both sides of equation (1) by -1: ( ) 4 2 8 4 a b c = Adding the left sides and right sides of equations (2) and (4) gives:( ) 5 4 5 a b + = Adding the left sides and right sides of equations (3) and (4) gives: ( ) 32 4 4 6 a b + = Simplify: ( ) 8 1 7 a b + = Eliminate b by multiplying equation (5) by -1 and add each side to the corresponding side of equation (7): 3 31aa== Next, substitute 1 for a in equation (5): ( ) 5 1 49bb+ = = Then, substitute 1 for a and -9 for b in equation (1): 4 2 84(1) 2( 9) 84 18 822a b cccc+ + =+ + = + == Therefore, a = 1, b = -9, and c = 22. So, the equation is29 22 y x x = + . 37.Solve for a in 2( ) y ax h k = +by substituting 5 for h and 8 for k since (5, 8) is the vertex, and substitute 4 for x and 6 for y since (4, 6) lies on the parabola: 22( )6 (4 5) 86 82y ax h kaaa= += += += Homework 7.5SSM: Intermediate Algebra 208 Since a = -2, 22( 5) 8 y x = +orexpanding the solution: 2222( 5)( 5) 422( 10 25) 422 20 50 422 20 42y x xx xx xx x= = + = + = + . 39.Answers may vary. Example:( ) 0, 2 ,( ) 1,1 , and6, 4 . The equation of the parabola through these points is22 2 y x x = + . Homework 7.5 1.a.A quadratic function would be reasonable. b.A linear function would be reasonable. c.An exponential function would be reasonable. d.None of the mentioned types of functions would be reasonable for this scattergram. 3. The data does not suggest a quadratic relationship based on the scattergram above. A quadratic function is not a reasonable function. 5.First, draw a scattergram of the data. The quadratic regression equation, which is 2( ) 0.66 21.49 95.85 f t t t = + , fits the data well. This equation can be verified by graphing it with the scattergram: 7.a.First draw a scattergram of the data. The linear regression equation is : ( ) 5.09 11.75 f t t = +The exponential regression equation is : ( ) 16.63(1.14)tf t =The quadratic regression equation is : 2( ) 0.277 1.976 16.493 f t t t = + +b.The exponential model gives the best estimate. 9.First draw a scattergram of the data: SSM: Intermediate AlgebraHomework 7.5 209 The quadratic regression equation, which is 2( ) 0.013 1.188 28.239 f t t t = + , fits the data well. This equation can be verified by graphing it with the scattergram: 11.First draw a scattergram of the data: The quadratic regression equation is, which is 2( ) 0.0067 0.12 6.39 f t t t = + , fits the data well. This equation can be verified by graphing it with the scattergram:. 13.a., b.The quadratic function 2( ) 0.000333 0.152 4 f d d d = + + , is the best model for the three points (0, 4), (60.5, 12) and (127.3, 18). The quadratic function is the best model since the value of f(d) will eventually return to 0 as the horizontal distance increases. 15.a.First draw a scattergram for C. The quadratic regression equation, 2( ) 0.90 19.94 121.38 Ct t t = + , is the best model for this data (as can be seen from the scattergram below. Next, draw a scattergram of the data for H. The quadratic regression equation, ( ) 2.48 1.30 Ht t = , is the best model for this data (as can be seen from the scattergram below. b.Graph the two regression equations on the same coordinate axes on your calculator. Use your calculator to find the points of intersection. Homework 7.6SSM: Intermediate Algebra 210 The intersection points for Cand Hare ( ) 8.1,18.8and( ) 16.8, 40.5 .These points mean that the sales of Corona and Heineken beers were equal in 1988 and in 1997, when the sales were at 18.8 million cases, and 40.5 million cases, respectively for each company. Homework 7.6 1.a.Solve for t when( ) 0 f t = : 220 0.66 21.49 95.8521.49 (21.49) 4( 0.66)( 95.85)2( 0.66)5.33or 27.23t ttt t= + = The t-intercepts are( ) 5.33, 0and( ) 27.23, 0 . So, according to this model, no firms performed drug tests on employees in 1985 or 2027.b.Any value of t that cause the percentage of firms performing drug tests to be negative. This would occur in years before 1985 ( ) 5.33 t 19.94) as the percent of schools with Internet access can not be negative. e. Better model SSM: Intermediate AlgebraChapter 7 Review Exercises 213 17.a.Linear:( ) 1.86 1.61 f t t = Exponential:( ) 1.63(1.26)tf t =Quadratic: 2( ) 0.25 0.92 2.62 f t t t = + The quadratic and exponential models fit the data well. b.The exponential model, as it gives values lower than 1.9 (the 1990 capacity) for years before 1990, whereas the quadratic model gives values higher than 1.9 for years before 1990. c.Since 1 MW meets the needs of 1000 people. 1 thousand MW meets the needs of (1000)(1000) = 1 million people. There are 6.2 billion people in the world (or, 6200 million people). So, 6200 thousand MW are needed. Let f(t) = 6200 and solve for t: 6200 1.63(1.26)62001.261.636200log(1.26 ) log1.636200log(1.26) log1.636200log1.63log(1.26)35.7tttttt== = = = So, according to the exponential model, there will be enough wind energy to meet the needs of the world in 2026. d.Let f(t) = 6200 and solve for t: 2226200 0.25 0.92 2.620 0.25 0.92 6197.380.92 ( 0.92) 4(0.25)( 6197.38)2(0.25)155.6 (model breakdown) or159.3t tt ttt t= += = So, the quadratic model predicts that there will be enough wind energy to meet the needs of the world in 2049. e.Answers may vary. Example: An exponential model gets larger faster than a quadratic model, so the exponential function will take less time to obtain enough power to supply the world. 19.Answers may vary. Chapter 7 Review Exercises 1.72 36 2 36 2 6 2 = = =2. 3 3 3 5 155 5 5 5 5= = =3. 50 50 25 2 5 249 7 7 49= = =4. 49 49 7100 10 100= =5. 26 3 2 0 x x = 2( 3) ( 3) 4(6)( 2)2(6)3 5712xx == 6. 25 7 x = 27575757 55 5355x == = = = Chapter 7 Review ExercisesSSM: Intermediate Algebra 214 7. 25( 3) 4 7 x + = 225( 3) 33( 3)53353 535 51535xxxxx = = = = = 153515 155xx= = 8. 298 x = 9849 27 2x = = = 9.( 1)( 7) 4 x x + = 2227 7 46 11 0( 6) ( 6) 4(1)( 11)2(1)6 8026 4 523 2 5x x xx xx + = = ==== 10. 22( 4) 9 x + = 29( 4)2942xx+ = + = Since the square of a negative number is not a real number, there are no real solutions. 11. 23 6 x x = 23 6 03 ( 2) 03 0or 2 00or 2x xxxx xx x = == == = 12. 22 24 0 x x = ( 6)( 4) 06 0or 4 06or 4x xx xx x + = = + == = 13. 24 25 0 x = (2 5)(2 5) 02 5 0or2 5 02 5or2 55 5or2 2x xx xx xx x + = = + == = = = 14. 23 7 2 0 x x + = (3 1)( 2) 03 1 0or 2 03 1or 21or 23x xx xx xx x = = == == = 15. 225 9 0 x = (5 3)(5 3) 05 3 0or5 3 05 3or5 33 3or5 5x xx xx xx x + = = + == = = = 16. 2 23 1 5 1 x x x = + 222 5 2 05 5 4(2)( 2)2(2)5 414x xxx+ = = = 17. 22 4 5 x x = 222 5 4 05 5 4(2)( 4)2(5)5 5710x xxx+ = = = SSM: Intermediate AlgebraChapter 7 Review Exercises 215 18. 24 1 x x = 2224 14 4 1 4( 2) 32 32 3x xx xxxx = + = + = = = 19. 2 23 5 5 x x x x + = + + 2222 5 02 55252525 22 2102xxxxxxx ==== = = = 20. 22 (2 5) 13 3 5 x x x + = + 2 2224 10 13 3 510 8 0( 10) ( 10) 4(1)(8)2(1)10 68210 2 1725 17x x xx xxxxx + = + + = ==== 21. 23 ( 7) 13 2 7 xx x + = 2 223 21 13 2 721 20 0( 20)( 1) 020 0or 1 020or 1x x xx xx xx xx x + = + = = = == = 22. 2 22( ) 3( 2 ) 5 x x x x = + 2 2222 2 3 6 54 5 0( 4) ( 4) 4(1)(5)2(1)4 42x x xx xxx = + + = = = Since the square of a negative number is not a real number, there are no real solutions. 23. 2 2( 2) ( 3) 2 x x + + = 2 22224 4 6 9 22 2 13 22 2 11 0( 2) ( 2) 4(2)(11)2(2)2 844x x x xx xx xxx+ + + + = + = + = = = Since the square of a negative number is not a real number, there are no real solutions. 24. 25(5 8) 0 x = 22225 40 925 494925492575xxxxx ==== = 25. 22.7 5.1 9.4 x x = 222.7 5.1 9.4 0( 5.1) ( 5.1) 4(2.7)( 9.4)2(2.7)1.15or 3.04x xxx x = = Chapter 7 Review ExercisesSSM: Intermediate Algebra 216 26. 21.7( 2.3) 3.4 2.8 x x = 2221.7 3.91 3.4 2.81.7 2.8 7.31 02.8 2.8 4(1.7)( 7.31)2(1.7)3.05or 1.41x xx xxx x = + = = 27. 26 4 0 x x + = 2226 46 9 4 9( 3) 133 133 13x xx xxxx+ =+ + = ++ =+ = = 28. 28 12 8 0 x x + + = 22231 023 9 912 16 163 74 163 74 16x xx xxx+ + =+ + = + + = + = Since the square of a negative number is not a real number, there are no real solutions. 29. 22 3 6 x x = + 22222 3 63323 9 932 16 163 574 163 574 163 574 43 574 43 574x xx xx xxxxxx+ =+ =+ + = + + = + = + = = = 30.( 3) 2 2 ( 1) 1 xx xx + = + + 2 22223 2 2 2 11 525 255 14 45 292 45 292 45 292 25 292 25 292x x x xx xx xxxxxx + = + += ++ + = + + = + = + = = = 31.Solve for x when( ) 0 f x = : 2223 10 03 1010310310 33 3303xxxxxx ==== = = The x-intercepts are 30, 03 and 30, 03 . 32.Solve for x when( ) 0 gx = : 2222( 3) 5 02( 3) 55( 3)2532xxxx + + = + = + =+ = SSM: Intermediate AlgebraChapter 7 Review Exercises 217 5 232 2103210326 102xxxx+ = + = = = The x-intercepts are 6 10, 02 + and 6 10, 02 . 33.Solve for x when( ) 0 hx = : 223 2 2 02 2 4(3)( 2)2(3)2 2862 2 761 73x xxxxx+ = = = == The x-intercepts are 1 7, 03 and 1 7, 03 + . 34.Solve for x when( ) 0 kx = : 225 3 1 03 3 4( 5)( 1)2( 5)3 1110x xxx + = = = Since the square root of a negative number is not a real number, there are no real number solutions. Therefore, there are no x-intercepts. 35.Solve for x when( ) 0 f x = : 222 3 4 0( 3) ( 3) 4(2)( 4)2(2)3 414x xxx = == The x-intercepts are 3 41, 04 + and 3 41, 04 . 36.Solve for x when( ) 0 gx = : 223 5 7 05 5 4( 3)(7)2( 3)5 10961(5 109)65 1096x xxxxx + + = = = == The x-intercepts are 5 109, 06 + and 5 109, 06 . 37.Factoring: 22 8 0( 4)( 2) 04 0or 2 04or 2x xx xx xx x = + = = + == = Completing the square: 22222 8 02 82 1 8 1( 1) 91 91 31 3or 1 34or 2x xx xx xxxxx xx x = = + = + = = = = = = = Chapter 7 Review ExercisesSSM: Intermediate Algebra 218 Quadratic Formula: 222 8 0( 2) ( 2) 4(1)( 8)2(1)2 3622 622 6 8 2 6 4= or2 2 2 24or 2x xxxxx xx x = ===+ = = == = 38.Find k when the discriminant equals 0: 22224 04(3)(12) 0144 014414412b ackkkxx = = === = 39.a. 23 6 7 3 x x + = 223 6 4 0( 6) ( 6) 4(3)(4)2(3)6 126x xxx + = = = Since the square root of a negative is not a real number, there are no real number solutions. There is no such value for x. b. 23 6 7 4 x x + = 223 6 3 0( 6) ( 6) 4(3)(3)2(3)6 06616x xxxx + = === = c. 23 6 7 5 x x + = 223 6 2 0( 6) ( 6) 4(3)(2)2(3)6 1266 2 363 33x xxxxx + = ==== d.Answers may vary. 40.Using the equation 2( ) y ax h k = + , substitute -4 for h and 3 for k since the vertex is (-4, 3). Also, substitute -3 for x and 6 for y since (-3, 6) lies on the parabola. 222( )6 ( 3 ( 4)) 36 ( 3 4) 36 (1) 33y ax h kaaaa= += += + += += So, the equation is 23( 4) 3 y x = + +41.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2221, 3 : 3 1 12, 6 : 6 2 23,13 :13 3 3a b ca b ca b c= + += + += + +

Simplify these equations: ( )( )( )314 2 6 29 3 13 3a b ca b ca b c+ + =+ + =+ + = Eliminate c by multiplying both sides of equation (1) by -1: ( ) 3 4 a b c = Adding the left sides and right sides of equations (2) and (4) gives:( ) 3 3 5 a b + =Adding the left sides and right sides of equations (3) and (4) gives: ( ) 8 2 10 6 a b + =Simplify: ( ) 4 5 7 a b + =SSM: Intermediate AlgebraChapter 7 Review Exercises 219 Eliminate b by multiplying equation (5) by -1 and add each side to the corresponding side of equation (7): 2 a =Next, substitute 2 for a in equation (5): ( ) 3 2 33bb+ == Then, substitute 2 for a and -3 for b in equation (1): 32 ( 3) 34a b ccc+ + =+ + == Therefore, a = 2, b = -3, and c = 4. So, the equation is 22 3 4 y x x = + . 42.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2221, 4 : 4 1 13, 2 : 2 3 34, 11 : 11 4 4a b ca b ca b c= + + = + + = + +

Simplify these equations: ( )( )( )419 3 2 216 4 11 3a b ca b ca b c+ + =+ + = + + = Eliminate c by multiplying both sides of equation (1) by -1: ( ) 4 4 a b c = Adding the left sides and right sides of equations (2) and (4) gives:( ) 8 2 6 5 a b + = Simplify: ( ) 4 3 6 a b + = Adding the left sides and right sides of equations (3) and (4) gives: ( ) 15 3 15 7 a b + = Simplify: ( ) 5 5 8 a b + = Eliminate b by multiplying equation (6) by -1 and add each side to the corresponding side of equation (8): 2 a = Next, substitute -2 for a in equation (6): ( ) 4 2 35bb + = = Then, substitute -2 for a and 5 for b in equation (1): 42 5 41a b ccc+ + = + + == Therefore, a = -2, b = 5, and c = 1. So, the equation is 22 5 1 y x x = + + . 43.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2222, 9 : 9 2 23,18 :18 3 35, 48 : 48 5 5a b ca b ca b c= + += + += + +

Simplify these equations: ( )( )( )4 2 919 3 18 225 5 48 3a b ca b ca b c+ + =+ + =+ + = Eliminate c by multiplying both sides of equation (1) by -1: ( ) 4 2 9 4 a b c = Adding the left sides and right sides of equations (2) and (4) gives:( ) 5 9 5 a b + =Adding the left sides and right sides of equations (3) and (4) gives: ( ) 21 3 39 6 a b + =Simplify: ( ) 7 13 7 a b + =Eliminate b by multiplying equation (5) by -1 and add each side to the corresponding side of equation (7): 2 42aa== Next, substitute 2 for a in equation (5): ( ) 5 2 91bb+ == Then, substitute 2 for a and -1 for b in equation (1): 4 2 94(2) 2( 1) 98 2 93a b cccc+ + =+ + = + == Therefore, a = 2, b = -1, and c = 3. So, the equation is 22 3 y x x = + . Chapter 7 Review ExercisesSSM: Intermediate Algebra 220 44.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2220, 5 : 5 0 02, 3 : 3 2 24, 15 : 15 4 4a b ca b ca b c= + += + + = + +

Simplify these equations: ( )( )( )514 2 3 216 4 153ca b ca b c=+ + =+ + = Since c = 5, substitute 5 for c in equations (2) and (3): 4 2 (5) 3 16 4 (5) 15a ba b+ + =+ + = Simplifying these equations gives: ( )( )4 2 2416 4 20 5a ba b+ = + = To eliminate b, multiply both sides of equation (4) by -2 and add each side to the corresponding side in equation (5): 8 162aa= = Next, substitute -2 for a in equation (4): 4 2 24( 2) 2 22 63a bbbb+ = + = == Therefore, a = -2, b = 3, and c = 5. So, the equation is 22 3 5 y x x = + + . 45.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2220, 7 : 7 0 01, 8 : 8 1 13, 8 : 8 3 3a b ca b ca b c= + += + + = + +

Simplify these equations: ( )( )( )718 29 3 83ca b ca b c=+ + =+ + = Since c = 7, substitute 7 for c in equations (2) and (3): (7) 89 3 (7) 8a ba b+ + =+ + = Simplifying these equations gives: ( )( )1 49 3 15 5a ba b+ =+ = To eliminate b, multiply both sides of equation (4) by -3 and add each side to the corresponding side in equation (5): 6 183aa= = Next, substitute -3 for a in equation (4): 1( 3) 14a bbb+ = + == Therefore, a = -3, b = 4, and c = 7. So, the equation is 23 4 7 y x x = + + . 46.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2220, 3 : 3 0 02,13 :13 2 23, 24 : 24 3 3a b ca b ca b c= + += + += + +

Simplify these equations: ( )( )( )314 2 1329 3 243ca b ca b c=+ + =+ + = Since c = 3, substitute 3 for c in equations (2) and (3): 4 2 (3) 13 9 3 (3) 24a ba b+ + =+ + = Simplifying these equations gives: ( )( )4 2 1049 3 215a ba b+ =+ = To eliminate b, multiply both sides of equation (4) by -3and multiply both sides of equation (5) by 2 ( )( )12 6 30618 6 427a ba b = + = To solve for a, add the corresponding sides of equations (6) and (7) together: 6 122aa== Next, substitute 2 for a in equation (4): 4 2 104(2) 2 102 21a bbbb+ =+ === Therefore, a = 2, b = 1, and c = 3. So, the equation is 22 3 y x x = + + . SSM: Intermediate AlgebraChapter 7 Review Exercises 221 47.Linear: 2 4slope 21 0intercept (0, 4)So,2 4myy x= = = == + Exponential: As the value of x increases by 1, the value of y is multiplied by ,, so the base 12b = . The y-intercept is (0, 4).Therefore, the exponential function is 142xy = Quadratic: Answers may vary. Example: 22 4 y x = +48.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2220, 7 : 7 0 02,1 :1 2 25, 7 : 7 5 5a b ca b ca b c= + += + += + +

Simplify these equations: ( )( )( )714 2 1225 5 73ca b ca b c=+ + =+ + = Since c = 7, substitute 7 for c in equations (2) and (3): 4 2 (7) 125 5 (7) 7 a ba b+ + =+ + = Simplifying these equations gives: ( )( )4 2 6 425 5 05a ba b+ = + = Simplify these equations: ( )( )2 3 65 0 7a ba b+ = + = To eliminate b, multiply both sides of equation (6) by -1 and add each side to the corresponding side in equation (7): 3 31aa== Next, substitute 1 for a in equation (5): 2 32(1) 35a bbb+ = + = = Therefore, a = 1, b = -5, and c = 7. So, the equation is 25 7 y x x = + . 49.a.First, draw a scattergram of the data. The quadratic regression equation,2( ) 0.0123 1.85 0.64 f t t t = + + , is the best model for this data. b.Find f(18) 2(18) 0.0123(18) 1.85(19) 0.64 29.9 f = + + So, about 30% of 18-year-old Americans voted. c.Solve for t when f(t) = 50 22250 0.0123 1.85 0.640 0.0123 1.85 49.361.85 1.85 4( 0.0123)( 49.36)2( 0.0123)34.7or 115.7 t tt ttt t= + += + = 115.7t is model breakdown.So, according to the model, 50% of people at age 35 vote.d.Using a graphing calculator, find that the maximum point of the parabola is approximately (75.3, 70.2). So the age of Americans who are most likely to vote is age 75.Chapter 7 TestSSM: Intermediate Algebra 222 e.Answers may vary. Example:Candidates tend to focus on issues for older people because they are the most likely to vote. 50.a. b.Using quadratic regression: 2( ) 0.037 0.47 31.84 f t t t = +c. 2(37) 0.037(37) 0.47(37) 31.84 65.1 f = + This means that in 2001, 65.1% of the grades at Princeton will be As. d. 2100 0.037 0.47 31.84 t t = + 220 0.037 0.47 68.160.47 ( 0.47) 4(0.037)( 68.16)2(0.037)37.04or 49.74t ttt t= = In 1933 and in 20020, all grades at Princeton will be As. Model breakdown has occurred. Chapter 7 Test 1.32 16 2 16 2 4 2 = = =2. 7 7 2 7 22 2 2 2= =3. 20 20 2 5 2 5 3 2 1575 15 75 5 3 5 3 3= = = =4. 23 10 0 x x = ( 5)( 2) 05 0or 2 05or 2x xx xx x + = = + == = 5. 26 100 x = 2210065035035035 2 33 35 63xxxxxx=== = = = 6. 23 4 0 x x + = 21 1 4( 3)( 4)2( 3)1 476x = = Since the square of a negative number is not a real number, there are no real solutions. 7. 24( 3) 1 7 x + = 2224( 3) 66( 3)43( 3)2xxx = = = 3323 232 26326326 62xxxxx = = = = = 8. 23 21 0 x x = 3 ( 7) 03 0or 7 00or 7xxx xx x == == = SSM: Intermediate AlgebraChapter 7 Test 223 9. 281 0 x = ( 9)( 9) 09 0or 9 09or 9x xx xx x + = = + == = 10.( 3)( 5) 6 x x + = 222225 3 15 62 21 02 212 1 21 1( 1) 221 221 22x x xx xx xx xxxx+ =+ =+ =+ + = ++ =+ = = 11. 25( 4) 10 x + = 2( 4) 24 2xx+ = + = Since the square of a negative number is not a real number, there are no real solutions. 12.2 ( 5) 4 3 xx x + = 2222 10 4 32 6 3 06 6 4(2)(3)2(2)x x xx xx+ = + + = = 6 1246 2 343 32xxx = = = 13. 29 16 24 x x = + 229 24 16 0( 24) ( 24) 4(9)( 16)2(9)24 11521824 24 2184 4 33x xxxxx = ==== 14. 23.7 2.4 5.9 x x = 223.7 5.9 2.4 05.9 5.9 4(3.7)( 2.4)2(3.7)1.93or 0.34x xxx x+ = = 15. 28 2 0 x x = 2228 28 16 2 16( 4) 184 184 3 24 3 2x xx xxxxx = + = + = = = = 16. 22( 4) 3 x x = 222222 8 32 3 83423 9 942 16 163 64 94 16 16x xx xx xx xx = + =+ =+ + = + + = + 23 734 163 734 163 734 43 734 43 734xxxxx + = + = + = = = Chapter 7 TestSSM: Intermediate Algebra 224 17.Solve for x when( ) 0 f x = : 223 8 1 0( 8) ( 8) 4(3)(1)2(3)8 5268 2 1364 133x xxxxx + = ==== The x-intercepts are 4 13, 03 + and 4 13, 03 . 18.Solve for x when f(x) = 0 2220 2( 3) 52( 3) 55( 3)2532532Rationalize the denominator:5 232 2xxxxxx= + = = = = = Solve for:103210321.42or 4.58xxxx x = = So, the x-intercepts are approximately( ) 1.42, 0and( ) 4.58, 0 . Since these points are symmetric, the average of the x-coordinates,1.42 4.5832+= , is the x-coordinate of the vertex of f(x). Substitute 3for x in the function to find the y-coordinate of the vertex: 2( ) 2(3 3) 5 5 hx = + =Therefore, the vertex is (3, 5). 19.Solve for a when the discriminant equals 0: 222224 0( 4) 4( )(4 ) 016 16 016 1611b aca aaaaa = = = = == 20.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2221, 4 : 4 1 12, 9 : 9 2 23,16 :16 3 3a b ca b ca b c= + += + += + +

Simplify these equations: ( )( )( )4 14 2 9 29 3 163a b ca b ca b c+ + =+ + =+ + = Eliminate c by multiplying both sides of equation (1) by -1: ( ) 4 4 a b c = Adding the left sides and right sides of equations (2) and (4) gives:( ) 3 5 5 a b + =Simplify: ( ) 8 2 12 6 a b + =Adding the left sides and right sides of equations (3) and (4) gives: ( ) 4 6 7 a b + =Eliminate b by multiplying equation (6) by -1 and add each side to the corresponding side of equation (7): SSM: Intermediate AlgebraChapter 7 Test 225 1 a =Substitute 1 for a in equation (5): ( ) 3 1 52bb+ == Then, substitute 1 for a and 2 for b in equation (1): 41 2 41a b ccc+ + =+ + == Therefore, a = 1, b = 2, and c = 1. So, the equation is 22 1 y x x = + + . 21.Using the equation 2( ) y ax h k = + , substitute 5 for h and 3 for k since the vertex is (5, 3). Also, substitute 3 for x and 11 for y since (3,11) lies on the parabola. 2222( )11 (3 5) 311 (3 5) 311 ( 2) 311 4 38 42y ax h kaaaaaa= += += += += +== So, the equation is 22( 5) 3 y x = +22.a. 26 11 1 x x + = 226 10 0( 6) ( 6) 4(1)(10)2(1)6 42x xxx + = = = Since the square root of a negative is not a real number, there are no real number solutions. There is no such value for x. b. 26 11 2 x x + = 226 9 0( 6) ( 6) 4(1)(9)2(1)6 02632x xxxx + = === = c. 26 11 3 x x + = 226 8 0( 6) ( 6) 4(1)(8)2(1)6 426 228 4or2 24or 2x xxxxx xx x + = ==== == = 23.a.First, draw a scattergram of the data. The quadratic regression equation,2( ) 1.14 25.26 81.2 f t t t = + , is the best model for this data. b. 2(15) 1.14(15) 25.26(15) 81.2 41.2 f = + So, 41.2% of television shows will be at least an hour long in 2005. c.Solve for t when f(t) = 50 22250 1.14 25.26 81.20 1.14 1.85 131.225.26 25.26 4( 1.14)( 131.2)2( 1.14)8.31or 13.8t tt ttt t= + = + = So, according to the model, 50% of television shows will be at least an hour long in 1996 and in 2004. Cumulative Review Chapters 1-7SSM: Intermediate Algebra 226 d.Solve for t when f(t) = 0 220 1.14 25.26 81.225.26 25.26 4( 1.14)( 81.2)2( 1.14)3.90or 18.26t ttt t= + = So, the t-intercepts are( ) 3.90, 0and ( ) 18.26, 0 . This means that according to the model no television shows were at least an hour in 1994 and none will be at least an hour in 2008. (Model breakdown has probably occurred. e.Model breakdown will definitely occur for years before 1994 (t < 3.90) and for years after 2008 (t > 18.26) because there cant be less than zero shows that last at least an hour. 24.Using a graphing calculator, find that the maximum point of the parabola is approximately (2.50, 103). So, the maximum height reached by the ball is 103 feet at 2.5 seconds. Cumulative Review of Chapters 1-7 1. 25 24 x = 22452452452 6 55 52 305x == = = = 2. 5 1 76 3 2x + = 5 7 16 2 35 21 26 6 65 196 66 5 19 65 6 6 5195xxxxx= = = = = 3. 4log ( 3) 2 x + = 23 43 1613xxx+ =+ == 4.3 5 49xe = 3 5454354ln( ) ln354ln( ) ln354ln 2.89043xxxeeex ex== = = = 5. 65 4 82 b + = 66165 78785781.58075bbb== = 6. 23 5 4 0 x x = 2( 5) ( 5) 4(3)( 4)2(3)5 736xx == SSM: Intermediate AlgebraCumulative Review Chapters 1-7 227 7. 4 53(2) 95x= 4 54 595(2)395log(2) log395(4 5) log(2) log395log34 5log(2)95log34 5log(2)95log35log(2)2.49624xxxxxx= = = = = + += 8. 22(3 10) 7 x x = 226 20 76 7 20 0(3 4)(2 5) 03 4 0or2 5 03 4or2 54 5or3 2x xx xx xx xx xx x = + = + = = + == = = = 9.log (65) 4b= 41465652.8394bbb== 10. 2 23 5 12 20 x x x + = + 2228 12 20 04(2 3 5) 02 3 5 0(2 5)( 1) 02 5 0or1 02 5or15or12x xx xx xx xx xx xx x = = = + = = + == = = = 11.2(5 2) 1 9( 3) x x + = 10 4 1 9 2710 3 9 2730x xx xx+ = + = = 12. 4 32 23log (4 ) 2log (4 ) 5 x x = ( )4 3 3 22 212 62 2122 6626 56616log (4 ) log (4 ) 5log (64 ) log (16 ) 564log 516log 4 54 24 328(8)1.4142x xx xxxxxxxxx = = = ===== 13. 22 (3 4) 5 3 x x x + = 2 2226 8 5 37 8 2 0( 8) ( 8) 4(7)(2)2(7)8 8148 2 2144 27x x xx xxxxx + = + = ==== 14. 4 73ln(2 ) 2ln(3 ) 8 x x + = 4 3 7 212 142626 88261826ln(2 ) ln(3 ) 8ln(8 ) ln(9 ) 8ln(72 ) 872721.154072x xx xxx eexex+ =+ ==== = Cumulative Review Chapters 1-7SSM: Intermediate Algebra 228 15. 25( 3) 4 13 x + = 225( 3) 99( 3)5935335Rationalize the denominator:3 535 53 5353 53515 3 55xxxxxxxx = = = = = = = = 16. 22 3 6 0 x x + = 222222 3 63323 9 932 16 163 48 94 16 163 574 163 574 16x xx xx xxxx+ =+ =+ + = + + = + + = + = 3 574 43 574 43 574xxx+ = = = 17.In order to eliminate y, multiply4 7 10 x y = by 4 and3 4 11 x y + =by 7. This gives: 16 28 4021 28 77x yx y = + = Add corresponding sides to obtain x: 37 371xx== Substitute 1 for x in one of the original equations to solve for y: 4(1) 7 107 142yyy = = = The solution is (1, 2). 18.Substitute 3 1 y x = in2 3 11 x y = 2 3(3 1) 112 9 3 117 142x xx xxx = + = = = Substitute 2 for x in one of the original equations to solve for y: 3 13(2) 15y x = = = The solution is (2, 5). 19.In order to eliminate y, multiply 1 52 2x y =by 35 . This gives 3 3 1510 5 10x y + = . Add the left sides and the right sides of the equations: 3 3 1510 5 102 3 65 5 5x yx y + = = This yields: 3 2 15 610 5 10 53 4 15 1210 10 10 101 310 1010 1 3 101 10 10 13x xx xxxx + = + + = += = = Substitute -3 for x in one of the original equations to solve for y: SSM: Intermediate AlgebraCumulative Review Chapters 1-7 229 1 5( 3)2 23 52 23 52 2842yyyy = == = = So the solution is (-3, -4). 20.2(3 4) 5 3(6 5) x x < + 2(3 4) 5 3(6 5)6 8 5 18 156 8 18 1024 21121,12x xx xx xxx < + < < < < 21. 4 5 3 1 2 4(2 ) (3 ) b c b c 3 12 15 4 4 812 4 15 88 23823(2 )(3 )(8 81)( )( )648648b c b cb b c cb cbc == == 22. 2 35 43 15 21015b cb c 3 1 2 34 2 5 51 3 25 4 41 15 41 15 423232323b cb cb cbc + ==== 23. 38 34 168b cb c 38 34 138 ( 4) 3 ( 1)312 23122366683434342764b cb cb cb cbcbc = = = = 24. 7 53ln(2 ) 2ln(4 ) x x + 7 3 5 221 1021 1031ln(2 ) ln(4 )ln(8 ) ln(16 )ln(8 16 )ln(128 )x xx xx xx= += += = 25. 8 33log (4 ) 4log (2 )b bx x ( )8 3 3 424 12241212log (4 ) log (2 )log (64 ) log (16 )64log16log 4b bb bbbx xx xxxx= = = = 26. 2(3 4) (3 4)(3 4) x x x = 229 12 12 169 24 16x x xx x= += + 27.(5 7)(5 7) x x + 2225 35 35 4925 49x x xx= + = 28. 2 33 ( 5)( 8) xx x + 5 2 35 3 26 4 33 ( 8 5 40)3 ( 5 8 40)3 15 24 120xx x xxx x xx x x x= + = + = + + x 112 Cumulative Review Chapters 1-7SSM: Intermediate Algebra 230 29. 2(2 3)( 4 5) x x x + 3 2 23 22 8 10 3 12 152 5 22 15x x x x xx x x= + += + + 30. 2( ) 2( 5) 3 f x x = + 22222( 5)( 5) 32( 5 5 25) 32( 10 25) 32 20 50 32 20 47x xx x xx xx xx x= += + += + += + += + 31. 2 2 281 25 (9 ) 5 (9 5)(9 5) x x x x = = +32. 3 213 40 x x x + 2( 13 40)( 8)( 5)xx xxx x= += 33. 28 22 21 (2 7)(4 3) x x x x + = + 34. 3 24 9 36 x x x + : 22( 4) 9( 4)( 4)( 9)( 4)( 3)( 3)x x xx xx x x= + += + = + + 35.Since the y-intercept is (0, 20) and as x increases by 1, f(x) decreases by 3: ( ) 3 20 f x x = + 36.As x increases by 1, g(x) increases by a factor of 3, so the base b = 3. Substitute a point from the table into( )xgx ab =and solve for a. 212 (3)12 943So, the equation is:4( ) (3)3xaaagx==== 37.For the function k, as x increases by 1, k(x) increases by 4, so the slope is 4. 38.(4) 108 g = 39.1 x =since(1) 4 g = 40. 1(7) 4 h=since(4) 7 h = 41.The x-intercept of k is (2, 0) 42. 43. 44.First, change the form intoy mx b = + 2 5 205 2 20245x yy xy x = = += SSM: Intermediate AlgebraCumulative Review Chapters 1-7 231 45. 46. 47. 48. 49. 2 12 14 ( 2) 633 ( 5) 2y ymx x = = = = 50.First, change the form of the equation given to y mx b = + . 3 4 54 3 53 54 4x yy xy x = = += The slope of the perpendicular line will be 43 . So the equation of the perpendicular line will be 43y x b = + . Substitute the given point into this equation to solve for b. 46 ( 2)386386318 83 3103bbbbb= += += = = So, the equation is 4 103 3y x = +or 1.33 3.33 y x = + . 51.The equation is of the formxy ab = . Since (0, 78) is on the curve, a = 78.So, the equation is of the form:78xy b = . Now substitute (9, 13) into the equation and solve for b. 991913 781378130.8278bbb== = So, the equation is78(0.82)xy = . 52.Both points satisfy the equation xy ab = ; we have the system of equations: 528327abab== Cumulative Review Chapters 1-7SSM: Intermediate Algebra 232 Combining these two equations yields 5231383278327831.4527ababbb== = So the equation is of the form:(1.45)xy a =To find a, substitute (2, 27) into(1.45)xy a = 227 (1.45)27 2.102512.84aaa=== So, the equation is12.84(1.45)xy = 53.Substitute the given points into 2y ax bx c = + + . ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2221, 1 : 1 1 12, 4 : 4 2 24, 20 : 20 4 4a b ca b ca b c = + += + += + +

Simplify these equations: ( )( )( )114 2 4216 4 20 3a b ca b ca b c+ + = + + =+ + = Eliminate c by multiplying both sides of equation (1) by -1: ( ) 1 4 a b c =Adding the left sides and right sides of equations (2) and (4) gives:( ) 3 5 5 a b + =Adding the left sides and right sides of equations (3) and (4) gives: ( ) 15 3 21 6 a b + =Simplify: ( ) 5 7 7 a b + =Eliminate b by multiplying equation (5) by -1 and add each side to the corresponding side of equation (7): 2 21aa== Next, substitute 1 for a in equation (5): ( ) 3 1 52bb+ == Then, substitute 1 for a and 2 for b in equation (1): 11 2 14a b ccc+ + = + + = = Therefore, a = 1, b = 2, and c = -4. So, the equation is 22 4 y x x = + . 54.a.Linear: 6 3slope 31 0intercept (0, 3)So,3 3myy x= = = == + Exponential: As the value of x increases by 1, the value of y is multiplied by 2,, so the base b = 2. The y-intercept is (0, 3).Therefore, the exponential function is( ) 3 2xy =Quadratic: Answers may vary. Example: 23 3 y x = + . b. 55. 26 5 3 x x + = 26 8 0( 4)( 2) 04 0or 2 04or 2x xx xx xx x + = = = == = 56. 26 5 4 x x + = 226 9 0( 3)( 3) 0( 3) 03 03x xx xxxx + = = = == SSM: Intermediate AlgebraCumulative Review Chapters 1-7 233 57. 26 5 5 x x + = 226 10 06 ( 6) 4(1)(10)2(1)6 42x xxx + = = = Since the square root of a negative is not a real number, there are no real number solutions. There is no such value for x. 58.To find the x-intercepts, let f(x) = 0 and solve for x: 226 5 06 5 0( 5)( 1) 05 0or 1 05or 1x xx xx xx xx x + = + = = = == = The x-intercepts are (5, 0) and (1, 0). 59.To find the y-intercept, find f(0) 2(0) (0) 6(0) 5 5 f = + = The y-intercept is (0. -5) 60. 61. 2log (16) 4 =since 42 16 = 62. 3log ( ) 3bb =since 3 3b b = 63. ( )25 5 5 21 1log log log 5 225 5 = = = 64. 9log(2)log (2) 0.3155log(9)= 65. 66.( ) 3xgx = 13( ) log ( ) g x x= 67. 2( ) 15f x x = + 111Replace( ) with:215Solve for:2155( 1)25 52 2Replace with( ) :5 5( )2 2Write in terms of:5 5( )2 2f x yy xxy xy xx yx g yg y yxg x x= + = == = = 68.a.First, draw a scattergram of the data. The linear regression equation, ( ) 3.97 3.02 f t t = + , is the best model for this data. Cumulative Review Chapters 1-7SSM: Intermediate Algebra 234 b.(18) 3.97(18) 3.02 74.48 f = + = . In 2008, 74.48%of companies will offer stock options to at least half of their employees. c.100 3.97 3.02 t = + 96.98 3.9724.43tt= All companies will offer stock options to at least half of their employees in 2014. d.Set f(t) = 0 and solve for t 0 3.97 3.023.02 3.973.020.763.97ttt= + == So, the t-intercept is (-0.76, 0). This means that no companies offered stock options to at least half of their employees in 1989. Model breakdown has occurred. e.t < -0.76 (since there cant be less than 0 percent offering stock options) or t> 24.43 (because there cant be more than 100 percent of companies offering stock options). 69.a.Find the different models using the regression feature on a graphing calculator: Linear:( ) 0.18 0.61 f t t = Exponential:( ) 0.15(1.22)tf t =Quadratic: 2( ) 0.0216 0.19 0.88 f t t t = + b.Answers may vary. Example: both the quadratic and exponential functions model the data well. c.Quadratic model d.Using a graphing calculator, find that the minimum point of the parabola is approximately (4.41, 0.46). This means that the minimum sales revenues was 0.46 billion dollars. e.Since the base of the exponential model is b = 1.22, the rate of growth is approximately 1.22-1 = 0.22 or 22% per year. f.Quadratic model: 2225 0.0216 0.19 0.880 0.0216 0.19 4.120.19 ( 0.19) 4(0.0216)( 4.12)2(0.0216)10.1or 18.9t tt ttt t= += = t = -10.1 is model breakdown, so the year that the quadratic model predicts that sakes will reach $5 billion in 2009. Exponential model: 5 0.15(1.22)33.3333 1.22log(1.22) log(33.33333)log(1.22) log(33.33333)log(33.33333)log(1.22)17.6tttttt===== So the year that the exponentialmodel predicts that sakes will reach $5 billion in 2008. Answers may vary. Example: It makes sense that the year predicted by the exponential model is before the year predicted by the quadratic model, as the exponential model is a steeper curve, and rises quicker than the quadratic model. 70.a.First, draw a scattergram of the data for h. The linear regression equation,( ) 2.4 10.7 h t t = + , is the best model for this data. SSM: Intermediate AlgebraCumulative Review Chapters 1-7 235 First, draw a scattergram of the data for c. The linear regression equation,( ) 53 c t t = + , is the best model for this data. b.The rates of change are the slopes of the respective models. Rate of change at Hartford: 2.4 percent per year. Rate of change in Connecticut: 1 percent per year. c.(18) 2.4(18) 10.7 53.9 h = + = . This means that in 2008, the percent of students who will score above the goals in Hartford will be 53.9%. (18) 18 53 71 c = + = . This means that in 2008, the percent of students who will score above the goals in Connecticut will be 71%. d.Solve for t when( ) ( ) h t c t = 2.4 10.7 531.4 42.330.21t ttt+ = += So, the percentage of Hartford and Connecticut students who will score above the goals will be equal in 2020.