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