# How do you find the vertex of  f(x)= -x^2-6x-5?

Jun 2, 2015

First, find the x value of the vertex with the following formula:

${x}_{v} = \frac{- b}{2 a}$

In your function, the values are:

$a = - 1$
$b = - 6$
$c = - 5$

Plugging those values:
${x}_{v} = \frac{- \left(- 6\right)}{2 \left(- 1\right)} = \frac{6}{- 2} = - 3$

Now you do $f \left({x}_{v}\right)$, which is equal to the y value of the vertex (${y}_{v}$):
$f \left({x}_{v}\right) = - {\left(- 3\right)}^{2} - 6 \left(- 3\right) - 5$
$f \left({x}_{v}\right) = - 9 - \left(- 18\right) - 5$
$f \left({x}_{v}\right) = 4$

$\left({x}_{v} , {y}_{v}\right)$
$\left(- 3 , 4\right)$