# How do you find the vertex and intercepts for y = x^2 - 9?

May 30, 2018

Vertex: $\left(0 , - 9\right)$
$y \text{-intercept: } y = - 9$
$x \text{-intercepts: " x=-3" and } x = + 3$

#### Explanation:

The general vertex form is
$\textcolor{w h i t e}{\text{XXX")y=color(green)m(x-color(red)a)^2+color(blue)bcolor(white)("xxx}}$with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

The given equation $y = {x}^{2} - 9$ can be written explicitly in vertex form as
$\textcolor{w h i t e}{\text{XXX")y=color(green)1(x-color(red)0)^2+color(blue)(""(-9))color(white)("xxx}}$with vertex at (color(red)0,color(blue)(""(-9)))

The $y$-intercept is simply the value of $y$ when $x = 0$.
Substituting $0$ for $x$ in the given equation:
$\textcolor{w h i t e}{\text{XXX}} y = {0}^{2} - 9 \Rightarrow y = - 9$

Similarly, the $x$-intercepts are the values of $x$ which solve the equation when $y = 0$.
Substituting $0$ for $y$ in the given equation:
$\textcolor{w h i t e}{\text{XXX}} 0 = {x}^{2} - 9$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow {x}^{2} = 9$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow x = \pm 3$