b. 5 1 x = −b = = 1 5 1 5 1 f      = − (−∞, domain: ∞) 2a 10 2       [2, range: 2  2 ∞)   5 5 5 5 10 = − = − = − 39. = 3x2 12x 1 f(x) – – 4 2 4 4 4 4 a. = 3. The parabola opens upward and has a a minimum value. at 1 . x = 2 The minimum is −5 b. 12 2 x = −b = =   −5 2a 6 c. domain: (−∞, range:  4 ,  ∞) ∞ = 3(2)2 12(2) 1 = 12 24 1 = f(2)  – – – – –13 The minimum is at 2 . x –13 = c. domain: (−∞, ∞) [−13, range: ∞)
b. x = −b = 5 = 1 domain: (−∞, ∞) 2a f  1  = 5  1 10 2 − 5  1        range: [2, ∞)  2  5  2 5 5 39. f(x) = 3x2 – 12x – 1 = − 5 = − 10 = − 4 2 4 4 4 a. a = 3. The parabola opens upward and has a minimum value. The minimum is −5 4 at x = 1 . 2 b. x = −b = 12 = 2  −5  2a 6 c. domain: (−∞, ∞) range:  4 , ∞  f(2) = 3(2)2 – 12(2) – 1 = 12 – 24 – 1 = –13  The minimum is –13 at x = 2 . c. domain: (−∞, ∞) range: [−13, ∞)
Precalculus-Essentials-5th-Edition-Blitzer-Solutions-Manual.docx