# What is the axis of symmetry and vertex for the graph y=2x^2 +8x-3?

$x = - 2 \text{ }$the axis of symmetry
$\left(- 2 , - 11\right) \text{ }$ the Vertex

#### Explanation:

From the given
$y = 2 {x}^{2} + 8 x - 3$
from the first two terms of the right side of the equation, factor out the number 2 then complete the square.

$y = 2 \left({x}^{2} + 4 x + 4 - 4\right) - 3$

$y = 2 {\left(x + 2\right)}^{2} - 8 - 3$

$y = 2 {\left(x - - 2\right)}^{2} - 11$

$2 {\left(x - - 2\right)}^{2} = y + 11$

${\left(x - - 2\right)}^{2} = \frac{1}{2} \left(y - - 11\right) \text{ " }$The vertex form

by inspection , we can see that $x = - 2$ is the axis of symmetry and the vertex is at $\left(h , k\right) = \left(- 2 , - 11\right)$

graph{(x--2)^2=1/2(y--11)[-20,20,-12,10]}

God bless...I hope the explanation is useful.