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 is greater than e. The graph has a minimum at (f,-50). What are the values of d, e, and f

Answer choices:

A. d=2, e=8, f=-1/8

B. d=8, e=-2, and f=3

C. d=-2, e=8 and f=2

D. d=2, e=8, f=-3

Given:

y = 2x^2 - 12x - 32

the zeros of the equation:

x = d

x = e

where d > e

minimum = (f, - 50)

To determine the values of d and e, substitute the zeros in the equation

0 = 2d^2 - 12d - 32

solve for d and e:

d = 8

e = - 2

To determine the Minimum:

y = 4x - 12

0 = 4x - 12

x = 3 = f

Therefore, the answer is B) d = 8; e = - 2; f = 3