# How do you find the vertex for  y=x^2-x-2?

Aug 2, 2017

Vertex is at $\left(\frac{1}{2} , - 2 \frac{1}{4}\right)$

#### Explanation:

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

$y = {\left(x - \frac{1}{2}\right)}^{2} - \frac{9}{4}$ , Comparing with vertex form of equation

y= a (x-h)^2+k ; (h,k) being vertex , we find here

$h = \frac{1}{2} , k = - \frac{9}{4}$ . So vertex is at $\left(\frac{1}{2} , - 2 \frac{1}{4}\right)$

graph{x^2-x-2 [-10, 10, -5, 5]} [Ans]