Question

in the xy coordinate plane, the graph of the equation y = 2x^2 - 12x - 32 has zeros at x = d and x = e, where d > e. The graph has a minimum at (f, -20). what are the values of d, e, and f?

Answer 1

We have the following function:
 y = 2x ^ 2 - 12x - 32
 We match zero:
 2x ^ 2 - 12x - 32 = 0
 We rewrite the function:
 x ^ 2 - 6x - 16 = 0
 (x-8) * (x + 2) = 0
 The zeros of the function are:
 x1 = 8
 x2 = -2
 Where,
 8> -2
 Thus,
 d = 8
 e = -2
 Then, to find the minimum function we derive:
 y '= 4x - 12
 We equal zero and clear x:
 4x-12 = 0
 x = 12/4
 x = 3
 Therefore, the value of f is:
 f = 3
 Answer:
 The values of d, e, and f are:
 d = 8
 e = -2
 f = 3