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, -50). What are the values of d, e, and f? Select the correct answer from the following. Show your work?
a) d=2, e =8, and f = -1/8
b) d = 8, e = -2, and f = 3
c) d = 2, e = 8, and f = 2
d) d = 2, e = 8, and f = -3
To find the values of d and e, we can set y = 0 in the equation y = 2x^2 - 12x - 32 and solve for x:
2x^2 - 12x - 32 = 0
x^2 - 6x - 16 = 0
(x - 8)(x + 2) = 0
Therefore, x = 8 or x = -2. Since d > e, d = 8 and e = -2.
To find the x-coordinate of the minimum, we use the formula x = -b / 2a (for the vertex of a parabola y = ax^2 + bx + c). In this case, a = 2, b = -12:
f = -(-12) / (2*2)
f = 12 / 4
f = 3
Therefore, the correct answer is:
b) d = 8, e = -2, and f = 3