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

Aug 6, 2015

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:
$h ' \left(x\right) = 2 x - 4 = 0 \to x = 2$

Put that in $h \left(x\right)$:
$h \left(2\right) = {2}^{2} - 4 \cdot 2 + 3 = - 1$

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