# What is the the vertex of y = x^2 - x - 6 ?

Mar 15, 2016

$\left(\frac{1}{2} , - \frac{13}{2}\right)$

#### Explanation:

Vertex of a parabola in the form $a {x}^{2} + b x + c$ is given by:
$x = - \frac{b}{2 a}$
Note this is only gives the $x$-coordinate; we will have to evaluate this value to get the $y$-coordinate.

Our parabola ${x}^{2} - x - 6$ has $a = 1$, $b = - 1$, and $c = - 6$. Using the vertex formula above, we see:
$x = - \frac{- 1}{2 \left(1\right)} = \frac{1}{2}$

Evaluating $y$ at this value:
$y = {\left(\frac{1}{2}\right)}^{2} - \left(\frac{1}{2}\right) - 6 = \frac{1}{4} - \frac{1}{2} - 6 = - \frac{13}{2}$

Therefore our vertex occurs at the point $\left(\frac{1}{2} , - \frac{13}{2}\right)$.