( ) 35. If the student’s average in the class is directly = 0 x 11. proportional to the amount of time the student − 9 = x2 x studies, then the relationship between the 0 f average and time studying be linear. That ( − 9) = 0 x x = is, we could express the relationship as A kT where is the time spent studying and is the T A = − = 0 or 9 0 x x student’s average. = 9 x The zeros of ( ) = − 9 are 0 and 9. The f x x2 x x- intercepts of the of are 0 and 9. f Section 2.3 12. ( ) = 0 f x (2 ( 2 − − 3 = − 3) + 1) 2. x2 x x x ( ( − 5x − 6 = − 6) 1. x2 x x + + 4x = 0 x2 ( + 4) = 0 x x = + = 0 or 4 0 x x 3. 2x − 3 = −4 = x 11 2 = x The zeros of are −4 and 0. The ( ) 14 = + 4 f x x2 x of the graph of are −4 and 0. f = 7 x The solution set is {7}. ( ) = 0 g x 13. ( (3x − 4. 3) + 5) x − = 25 x2 = 0 + − 0 ( 5)( x x − 3 = 0 or 3x + 5 = x = 5) 0 0 = 3 3x = −5 x + = − = 5 0 or 5 0 x x 5 x =− = − = 5 x x 5 3 The solution set is ⎧− 5 ,3 . ( ) The zeros of = − 25 are −5 and 5. The g x x2 −5 of the graph of are and 5. x-intercepts g ⎨ 3 ⎬ ⎩ ⎭ ( ) = 0 5. add; ⎛ 1 ⋅ 6 ⎞ = 9 G x 14. − 9 = 0 x2 ⎜ ⎟ 2 ⎝ ⎠ ( + 3)( − 3) = 0 x x
35. If the student’s average in the class is directly proportional to the amount of time the student studies, then the relationship between the 11. f ( x ) = 0 x2 − 9 x = 0 average and time studying be linear. That x ( x − 9) = 0 is, we could express the relationship as A = kT where T is the time spent studying and A is the student’s average. x = 0 or x − 9 = 0 x = 9 The zeros of f ( x ) = x2 − 9 x are 0 and 9. The x- Section 2.3 12. intercepts of the of f are 0 and 9. f ( x ) = 0 1. x2 − 5x − 6 = ( x − 6) ( x + 2. 2 x2 − x − 3 = (2 x − 3) ( x + 1) x2 + 4x = 0 x ( x + 4) = 0 3. 2x − 3 = 11 x = 0 or x + 4 = 0 x = −4 2 x = 14 The zeros of f ( x ) = x2 + 4 x are −4 and 0. The x = 7 The solution set is {7}. 4. ( x − 3) (3x + 5) = 0 x − 3 = 0 or 3x + 5 = 0 13. of the graph of f are −4 and 0. g ( x ) = 0 x2 − 25 = 0 ( x + 5)( x − 5) = 0 x = 3 3x = −5 x 5 =− x + 5 = 0 or x = − 5 x − 5 = 0 x = 5 3 The solution set is ⎧− 5 ,3 . The zeros of g ( x ) = x2 − 25 are −5 and 5. The x-intercepts of the graph of g are −5 and 5. ⎨ 3 ⎬ ⎩ ⎭ 5. add; ⎛ 1 ⋅ 6 ⎞ = 9 14. G ( x ) = 0 x2 − 9 = 0 ⎜ 2 ⎟ ⎝ ⎠ ( x + 3)( x − 3) = 0
Downloadable-Solution-Manual-for-Precalculus-Concepts-Through-1.docx