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21 October, 22:00

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

+3
Answers (1)
  1. 22 October, 01:41
    0
    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
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