# What is the vertex of  y= 1/3(x/5+1)^2+4/15 ?

Jan 2, 2016

The vertex form is $y = a {\left(x - h\right)}^{2} + k$ where $\left(h , k\right)$ is the vertex. For our problem the vertex is $\left(- 5 , \frac{4}{15}\right)$

#### Explanation:

$y = \frac{1}{3} {\left(\frac{x}{5} + 1\right)}^{2} + \frac{4}{15}$

$y = \frac{1}{3} {\left(\frac{x + 5}{5}\right)}^{2} + \frac{4}{15}$

$y = \frac{1}{75} {\left(x + 5\right)}^{2} + \frac{4}{15}$

Compare with
$y = a {\left(x - h\right)}^{2} + k$

$h = - 5$ and $k = \frac{4}{15}$

The vertex $\left(h , k\right)$ is $\left(- 5 , \frac{4}{15}\right)$