What is the maximum or minimum value of the function? what is the range?

y = –2x2 + 32x –12 (1 point)
maximum: 116
range: y 116
maximum: –116
range: y –116
maximum: 116
range: y 116
maximum: –116
range: y –116

You can use the formula
f(x) = ax^2+bx+c
f(x) = a(x-xv)^2+yv
where xv = -b/2a = -32/-4=8
yv = f(xv) = 116

f(x) = -2(x-8)^2+116

The maximum point is (8, 116).  The answer to your question is C. I hope that this is the answer that you were looking for and it has helped you.