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

Jan 20, 2017

$\left(\frac{1}{2} , - \frac{47}{4}\right)$

#### Explanation:

$y = - {x}^{2} + x - 12$ => complete the square to convert to vertex form:
$y = - \left({x}^{2} - x\right) - 12$
$y = - \left({x}^{2} - x + \frac{1}{4}\right) - 12 + \frac{1}{4}$
$y = - {\left(x - \frac{1}{2}\right)}^{2} - \frac{47}{4}$
=> in the vertex form of${\left(x - h\right)}^{2} + k$ where$\left(h , k\right)$ is the vertex:
so in this case the vertex is:
$\left(\frac{1}{2} , - \frac{47}{4}\right)$