Question

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?

Answer 1

The answers are as follows:

d = 8

e = -2

f = 3

To find d and e, we first have to factor the polynomial. You can start by pulling the greatest common factor out, which is 2.

f(x) = 2x^2 - 12x - 32

f(x) = 2(x^2 - 6x - 16)

Now we can factor the inside by finding the two numbers that multiply to the constant (-16) and add up to the middle number (-6). The numbers -8 and 2 satisfy both of these and can be used for the bases of the factoring.

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

Now to find the zeros, all you have to do is set each parenthesis equal to 0 separately.

FIRST ZERO

x - 8 = 0

x = 8

SECOND ZERO

x + 2 = 0

x = -2

Now to find the x value of the vertex, we can simply use the formula for x values of vertex (-b/2a), in which a is the coefficient of x^2 (2) and b is the coefficient of x (-12). Now we'll plug those values in.

-b/2a

-(-12)/2(2)

12/4

3