What is the minimum value for h(x)=x2−16x+60?

Minimum value is equal to x=8, y=-4First find the derivative of the original equation which equals= d/dx(x^2-16x+60) = 2x - 16at x=8, f'(x), the derivative of x equals zero, so therefore, at point x = 8, we have a minimum value.Just plug in 8 to the original equation to find the answer for the minimum value.