What is the minimum value for the function y=2x^2-32x+256

The answer would be: 128

You can find minimum value by looking at the vertex. First, differentiate the equation and find the value of x.

y=2x^2-32x+256

4x - 32=0

4x=32

x=8

If you put x=8, the value would be:

y=2x^2-32x+256

y=2 (8) ^2-32 (8) + 256

y = 128 - 256 + 256 = 128