Question

in the xy-coordinate plane, the graph of the equation y=2x^2-12x-32 has zeros x=d and x=e, where d>e. the graph minimum at (f, -50). what are the values of d,e, and f?

Answer 1

d=8 e=-2 and f=3


hope this helps

Answer 2

Answer:

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

Step-by-step explanation:

To find the zeros, we factor the quadratic.  We see that all of the terms are even, so we can factor out a 2:

y=2x²-12x-32

y=2(x²-6x-16)

Now we want factors of -16 that sum to -6; -8(2) = -16 and -8+2 = -6:

y=2(x-8)(x+2)

Using the zero product property, we know that x-8=0 or x+2=0; this gives us x=8 or x=-2.  Since d and e are zeros, and we know that d>e, this means d = 8 and e = -2.

To find the x-coordinate of the minimum, we find the axis of symmetry; this is given by the formula x=-b/2a:

x = -(-12)/2(2) = 12/4 = 3

This means f=3.