# How do you find the axis of symmetry for a quadratic equation y=x^2+8x+12?

Jun 6, 2015

$y = {x}^{2} + 8 x + 12$

$= {\left(x + 4\right)}^{2} - {4}^{2} + 12$

$= {\left(x + 4\right)}^{2} - 16 + 12$

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

So the vertex is given by $x = - 4$, $y = - 4$

... the axis of symmetry being the vertical line $x = - 4$

In general, if $y = a {x}^{2} + b x + c$ then

$y = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$

which has vertex at $\left(- \frac{b}{2} a , c - {b}^{2} / \left(4 a\right)\right)$

and axis of symmetry $x = - \frac{b}{2 a}$