# What is the vertex of  y=1/5(x/2-15)^2-2 ?

Feb 13, 2016

vertex: $\left(30 , - 2\right)$

#### Explanation:

Our "target will be to convert the given equation into "vertex form":
$\textcolor{w h i t e}{\text{XXX}} y = m {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$ with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

Given
$\textcolor{w h i t e}{\text{XXX}} y = \frac{1}{2} {\left(\frac{x}{2} - 15\right)}^{2} - 2$

$y = \frac{1}{2} {\left(\frac{x - 30}{2}\right)}^{2} - 2$

$y = \frac{1}{2} \left(\frac{{\left(x - 30\right)}^{2}}{{2}^{2}}\right) - 2$

$y = \frac{1}{8} {\left(x - \textcolor{red}{30}\right)}^{2} + \textcolor{b l u e}{\text{("-2")}}$

which is the vertex form with a vertex at $\left(\textcolor{red}{30} , \textcolor{b l u e}{- 2}\right)$

The graph below may help to indicate that our answer is (at least approximately) correct:
graph{1/5(x/2-15)^2-2 [9.41, 49.99, -10.61, 9.69]}