What is the relative maximum and minimum of the function?

f (x) = 2x^2 + 28x - 8

A. Minimum Value: - 106 Range y > - 7

B. Minimum Value: - 106 Range y > - 106

C. Minimum Value: 7

Range y > 7

D. Minimum Value: - 7 Range y > - 7

We equate the derivative of the function to 0 to find the x-value at its minimum:

f' (x) = 4x + 28

0 = 4x + 28

x = - 7

Now, we put this value into the original equation to find the minimum value:

f (-7) = - 106

The function has a positive slope so it will increase. Thus, the range will be y > - 106

The answer is B