Chapter 5: Exercises with Solutions

Section 5.3Zeros of the Quadratic20y20(−1,18)yx10x10f(x)=−2(x+1) 2 +18f(x)=−2(x+1) 2 +1857. First, complete the square:f(x) = 3x 2 + 18x − 48= 3(x 2 + 6x − 16)= 3 ( x 2 + 6x + 9 − 9 − 16 )= 3 (( x 2 + 6x + 9 ) − 9 − 16 )()= 3 (x + 3) 2 − 25= 3 (x + 3) 2 − 75Read off the vertex as (h, k) = (−3, −75).through the vertex with equation x = −3.To find the x-intercepts algebraically, set y = 0 and factor.By the zero product property, either0 = 3x 2 + 18x − 480 = 3(x 2 + 6x − 16)0 = 3(x + 8)(x − 2)Solve these linear equations independently.So the x-intercepts are (−8, 0) and (2, 0).x + 8 = 0 or x − 2 = 0.x = −8 or x = 2The axis of symmetry is a vertical lineLastly, to find the y-intercept, set x = 0 in the equation and solve for y:y = 3x 2 + 18x − 48y = 3(0) 2 + 18(0) − 48y = −48Version: Fall 2007