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

Jan 15, 2016

Vertex: $\left(2 , 0\right)$
y-intercept: $4$
x-intercept: $2$

#### Explanation:

The general vertex form for a parabola is:
$\textcolor{w h i t e}{\text{XXX}} y = {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$
for a parabola with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

$y = {x}^{2} - 4 x + 4$

$\textcolor{w h i t e}{\text{X}} = {\left(x - \textcolor{red}{2}\right)}^{2} + \textcolor{b l u e}{0}$
is therefore a parabola with vertex at $\left(\textcolor{red}{2} , \textcolor{b l u e}{0}\right)$

$\overline{\textcolor{w h i t e}{\text{XXXXXXXXXXXXXXXXXXXXXXXXXXXX}}}$

The y intercept is the value of $y$ when #x=0

Given $y = {x}^{2} - 4 x + 4$
when $x = 0$
$\textcolor{w h i t e}{\text{XXX}} y = 4$

$\overline{\textcolor{w h i t e}{\text{XXXXXXXXXXXXXXXXXXXXXXXXXXXX}}}$

The x intercept is the value(s) of $x$ when $y = 0$

Given $y = {x}^{2} - 4 x - 4$
when $y = 0$
$\textcolor{w h i t e}{\text{XXX}} 0 = {x}^{2} - 4 x + 4$

$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 4 x + 4 = 0$

$\textcolor{w h i t e}{\text{XXX}} {\left(x - 2\right)}^{2} = 0$

$\textcolor{w h i t e}{\text{XXX}} \left(x - 2\right) = 0$

$\textcolor{w h i t e}{\text{XXX}} x = 2$