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

Nov 3, 2017

vertex at $\left(x , y\right) = \left(0 , - 2\right)$
y-intercept: $y = 0$
x-intercepts: $x = - \sqrt{2}$ and $x = + \sqrt{2}$

#### Explanation:

One way to find the vertex is to convert the given equation into an explicit vertex form
$\textcolor{w h i t e}{\text{XXX}} y = {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$ with vertex at color(red)a,color(blue)b)

$y = {x}^{2} - 2$
is equivalent to
$\textcolor{w h i t e}{\text{XXX}} y = {\left(x - \textcolor{red}{0}\right)}^{2} + \textcolor{b l u e}{\left(- 2\right)}$
the vertex form with vertex at $\left(\textcolor{red}{0} , \textcolor{b l u e}{\left(- 2\right)}\right)$

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

In this case $y = {\textcolor{g r e e n}{x}}^{2} - 2$ with $\textcolor{g r e e n}{x} = \textcolor{g r e e n}{0}$
becomes $y = {\textcolor{g r e e n}{0}}^{2} - 2 = - 2$

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The x-intercepts are the values of $x$ which are possible when $y = 0$

In this case $\textcolor{m a \ge n t a}{y} = {x}^{2} - 2$ with $\textcolor{m a \ge n t a}{y} = \textcolor{m a \ge n t a}{0}$
becomes $\textcolor{m a \ge n t a}{0} = {x}^{2} - 2$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow {x}^{2} = 2$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow x = \pm \sqrt{2}$