Chapter 9 Radical Functions

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Homework 9.5 SSM: Intermediate Algebra 67. f ( x) = 3 x − 7 3 x − 7 = −1 3 x = 6 x = 2 ( ) 2 2 x = 2 x = 4 Check x = 4 ( ) 3 4 − 7 =−1 ? − 1 = −1 true When x = 4 , f ( x ) = − 1. 69. f ( x) = −2 x − 4 + 5 −2 x − 4 + 5 = 7 −2 x − 4 = 2 x − 4 = −1 ( x − 4) = ( −1) 2 2 x − 4 = 1 x = 5 Check x = 5 ( ) −2 5 − 4 − 5= 7 ? ? − 7 = 7 false No real number solutions. There is no value of x that would make f ( x ) = 7 . 71. a. Somewhere during the 1970’s average test scores declined. During the 1980’s and 1990’s test scores have been improving to near the 1970 level. b. 3.9 t + 280 = 305 3.9 t = 25 ( t ) t = 2 25 3.9 ⎛ 25 ⎞ = ⎜ ⎟ ⎝ 3.9 ⎠ 2 2 ⎛ 25 ⎞ t = ⎜ ≈ 41.09 3.9 ⎟ ⎝ ⎠ According to the model, average test scores will return to 305 in 2023. c. 3.9 t + 280 = 500 3.9 t = 220 ( t ) t = 2 220 3.9 ⎛ 220 ⎞ = ⎜ ⎟ ⎝ 3.9 ⎠ 2 2 ⎛ 220 ⎞ t = ⎜ ⎟ ≈ 3182.12 ⎝ 3.9 ⎠ According to the model, the average test score will reach the maximum score of 500 in 5164. Model breakdown has occurred. The model assumes that the average score will continue to rise. 73. No. When the student squared both sides of the equation, the work on the left side was incorrect. The student did not square the factor 5 on the left side. ( 5 2 x + 3 ) = 25 ( x + 3 ) 75. y = 3 x − 4 y = − 2 x + 6 Since the left hand sides are equal, set the right hand sides equal to each other and solve the resulting equation. 3 x − 4 = − 2 x + 6 5 x = 10 x = 2 ( ) 2 2 x = 2 x = 4 Substitute this value into either original equation and solve for y. y = 3 ( 4) − 4 = 2 The solution is ( 4, 2 ) . 77. The left hand side was not squared properly in the second line. 296
SSM: Intermediate Algebra Homework 9.6 Homework 9.6 1. ( 0,3 ) and ( 4,5 ) Substitute the point ( 0,3 ) into the equation y = a x + b . 3 = a 0 + b b = 3 Substitute the point ( ) y = a x + 3 and solve for a. 5 = a 4 + 3 2a = 2 a = 1 The equation is y = x + 3 . 4,5 into the equation 3. ( 0, 2 ) and ( 9,6 ) Substitute the point ( 0, 2 ) into the equation y = a x + b . 2 = a 0 + b b = 2 Substitute the point ( ) y = a x + 2 and solve for a. 6 = a 9 + 2 3a = 4 a = 4 3 The equation is 9,6 into the equation 4 y = x + 2 . 3 5. ( 0, 2 ) and ( 3,5 ) Substitute the point ( 0, 2 ) into the equation y = a x + b . 2 = a 0 + b b = 2 Substitute the point ( ) y = a x + 2 and solve for a. a 5 = a 3 + 2 3 = 3 a = = 3 3 3 3,5 into the equation The equation is y = 3 ⋅ x + 2 or y = 3x + 2 . 7. ( 1,2 ) and ( 4,3 ) Substitute the points into the equation y a x b = + . 2 = a 1 + b 3 = a 4 + b Rewrite as: a + b = 2 2a + b = 3 Solve the resulting system. Multiply the first equation by − 1 and add to the second equation. −a − b = −2 2 a + b = 3 a = 1 Substitute the point ( ) y = x + b and solve for b. 2 = 1 + b b = 1 The equation is y = x + 1 . 9. ( 2, 4 ) and ( 3,5 ) 1,2 into the equation Substitute the points into the equation y a x b = + . 4 = a 2 + b 5 = a 3 + b Rewrite as: 1.4142a + b = 4 1.7321a + b = 5 Solve the resulting system. Multiply the first equation by − 1 and add to the second equation. −1.4142a − b = −4 1.7321 a + b = 5 0.3179a = 1 a ≈ 3.15 2, 4 into the equation Substitute the point ( ) y = 3.15 x + b and solve for b. 4 = 3.15 2 + b b ≈ −0.45 The equation is roughly y = 3.15 x − 0.45 . 297
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Chapter 9 Radical Functions

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