MATH SOLVE

2 months ago

Q:
# 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?

Accepted Solution

A:

The answers are as follows:d = 8e = -2f = 3To 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 - 32f(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 ZEROx - 8 = 0 x = 8SECOND ZERO x + 2 = 0 x = -2Now 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)12/43