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

Dec 16, 2015

This equation for a parabola is already in vertex form ...

$y = a {\left(x - h\right)}^{2} + k$ where vertex $= \left(h , k\right)$

#### Explanation:

Using the vertex form above,

vertex $= \left(h , k\right) = \left(- 1 , 4\right)$

hope that helped