# How do you find the vertex of f(x)= x^2+10x-8?

##### 1 Answer
Jun 7, 2015

To find the vertex you can complete the square:

$f \left(x\right) = {x}^{2} + 10 x - 8$

$= \left({x}^{2} + 10 x + 25\right) - 25 - 8$

$= {\left(x + 5\right)}^{2} - 33$

The vertex is where the term ${\left(x + 5\right)}^{2}$ has its minimum possible value, $0$, which is when $x = - 5$ and $f \left(x\right) = - 33$

That is, the vertex is at $\left(- 5 , - 33\right)$

In general:

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