# How do you find the vertex and intercepts for y=x^2 – 12x + 36?

## Replaced equation (with 2 solutions for $x$) into a corresponding function with vertex and intercepts.

##### 1 Answer
Nov 13, 2015

Vertex at $\left(6 , 0\right)$
y-intercept: $36$
x-intercept: 6#

#### Explanation:

$y = {x}^{2} - 12 x + 36$

$\Rightarrow y = {\left(x - 6\right)}^{2} + 0$
this is the vertex form of a parabola with vertex at $\left(6 , 0\right)$

The y-intercept is the value of $y$ when $x = 0$
$\textcolor{w h i t e}{\text{XXX}} y = {\left(0\right)}^{2} - 12 \left(0\right) + 36 = 36$

The x-intercept is the value of $x$ when $y = 0$
$\textcolor{w h i t e}{\text{XXX}} 0 = {x}^{2} - 12 x + 36 = \left(x - 6\right) \left(x - 6\right)$
$\textcolor{w h i t e}{\text{XXX}} \Rightarrow x = 6$ (single intercept point)
graph{x^2-12x+36 [-25.27, 54.73, -1.6, 38.4]}