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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

asked
User Aymen
by
7.2k points

1 Answer

2 votes
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
answered
User Songyuanyao
by
8.5k points

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