# How do you write the quadratic in vertex form given f(x)=x^2 + 6x + 12?

May 12, 2015

$f \left(x\right) = {x}^{2} + 6 x + 12$

$= {\left(x + 3\right)}^{2} - 9 + 12$

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

In general, $a {x}^{2} + b x + c$ can be written in vertex form as

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

In the case $a = 1$, this simplifies to

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