Question

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

Answer 1

differentiate the function to give you 2x-16
let 2x-16=0 and solve. Then sub value for x into your origional equation to find the y value

Answer 2

When you use the quadratic formula to solve
$$
h(x)=0
$$
you'll get
$$
x=-\cfrac{b}{2}\pm\cfrac{\sqrt{b^2-4c}}{2}
$$
But h(x) is a symmetric function around the point \(x=-\cfrac{b}{2}\). Since the term in front of the \(x^2\) term is positive, the curve (i.e. parabola) will point upward. So the point \(x=-\cfrac{b}{2}\) is where the minimum of h(x) will occur.

The minimum thus is
$$
h\left (-\cfrac{b}{2}\right)
$$
where \(b=-16\) .
Thus, the minimum of h(x) is
$$
h\left (-\cfrac{b}{2}\right)=h(8)=8^2-16(8)+60=-4
$$

http://www.wolframalpha.com/input/?i=minimize+x%5E2-16x%2B60

Make sense?