What is the maximum or minimum value of the function? what is the range? y = -2x^2 + 32x - 12

Y= -2x² + 32x -12
Take the derivative, and set it equal to zero. We are finding where the slope equals zero (the peak of the parabola)
y'= -4x + 32
0 = -4x + 32
4x=32
x=8
The maximum is at the point x=8. Plugging into the original equation:
y= -2x² + 32x -12
y= -2(8)² + 32(8) -12
y= 116

The maximum is at point y=116
Keep in mind what maximum means. It is the largest value for y that the function has. That means that the range, or all possible y-values, is 
y ≤ 116.

Therefore, the answer is A) Max: 116, range: y ≤ 116