# What is the vertex form of y= x^2+x-12?

Dec 19, 2015

Complete the square to find:

$y = 1 {\left(x - \left(- \frac{1}{2}\right)\right)}^{2} + \left(- \frac{49}{4}\right)$

in vertex form

#### Explanation:

Complete the square as follows:

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

$= {x}^{2} + x + \frac{1}{4} - \frac{1}{4} - 12$

$= {\left(x + \frac{1}{2}\right)}^{2} - \frac{49}{12}$

That is:

$y = 1 {\left(x - \left(- \frac{1}{2}\right)\right)}^{2} + \left(- \frac{49}{4}\right)$

This is in vertex form:

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

with $a = 1$, $h = - \frac{1}{2}$ and $k = - \frac{49}{4}$

so the vertex is at $\left(h , k\right) = \left(- \frac{1}{2} , - \frac{49}{4}\right)$