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

Question

3535 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-06T10:39:34

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

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]}

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