# How do you find the vertex of y = x^2 - 4x + 4?

Jun 10, 2018

$\left(2 , 0\right)$

#### Explanation:

Method 1:
$y = {x}^{2} - 4 x + 4$
complete the square to get:
$y = {\left(x - 2\right)}^{2}$

pick out the x value (2), and the y value (o)
to get:
$\left(2 , 0\right)$

method 2:
$y = {x}^{2} - 4 x + 4$
$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - 4$
let $\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = 0$ as the vertex is when the gradient of the curve is 0.

to get $x = 2$
substitute that into the original equation to get $\left(2 , 0\right)$