# How do you find the vertex and axis of symmetry of y=5/4(x+2)^2 -1?

Jun 28, 2017

Vertex is $\left(- 2 , - 1\right)$ and axis of symmetry is $x + 2 = 0$.

#### Explanation:

This is the equation of a horizontal parabola in vertex form i.e.

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

where vertex is $\left(h , k\right)$ and axis of symmetry is $x = h$.

Hence vertex and axis of symmetry of $y = \frac{1}{4} {\left(x + 2\right)}^{2} - 1$

or $y = \frac{1}{4} {\left(x - \left(- 2\right)\right)}^{2} - 1$

are $\left(- 2 , - 1\right)$ and $x = - 2$ or $x + 2 = 0$ respectively

graph{((x+2)^2-4-4y)(x+2)((x+2)^2+(y+1)^2-0.02)=0 [-13, 7, -4.28, 5.72]}