Question

What is the minimum or maximum of `h(x)=x^2-4x+3`?

Answer 1

Since the coefficient of `x^2` is positive, we're talking about a "valley"-parabola, and so we have a minimum .

Explanation:

We have to set the derivative to zero:
`h'(x)=2x-4=0->x=2`

Put that in `h(x)`:
`h(2)=2^2-4*2+3=-1`

So the minimum is at `(2,-1)`
graph{x^2-4x+3 [-6.58, 13.42, -2.28, 7.72]}