# What is the axis of symmetry and vertex for the graph y=x^2+8x+12?

May 18, 2018

$x = - 4 \text{ and vertex } = \left(- 4 , - 4\right)$

#### Explanation:

"given a parabola in standard form "color(white)(x);ax^2+bx+c

$\text{then the x-coordinate of the vertex which is also the }$
$\text{equation of the axis of symmetry is}$

•color(white)(x)x_(color(red)"vertex")=-b/(2a)

$y = {x}^{2} + 8 x + 12 \text{ is in standard form}$

$\text{with "a=1,b=8" and } c = 12$

$\Rightarrow {x}_{\text{vertex}} = - \frac{8}{2} = - 4$

$\text{substitute this value into the equation for y}$

${y}_{\text{vertex}} = {\left(- 4\right)}^{2} + 8 \left(- 4\right) + 12 = - 4$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(- 4 , - 4\right)$

$\text{axis of symmetry is } x = - 4$
graph{(y-x^2-8x-12)(y-1000x-4000)=0 [-10, 10, -5, 5]}