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

Feb 17, 2016

The vertex is $\left(- 3 , 4\right)$

#### Explanation:

I found the vertex by changing the equation, $y = - {x}^{2} - 6 x - 5$, into vertex form. The first step for me at least is to factor out the $- 1$ from the equation. I like to have a coefficient of $1$, and this time it's an easy fix. So now we have $y = - 1 \left({x}^{2} + 6 x + 5\right)$. I think that we should rewrite the equation into vertex form, and the easiest way to do that is by "completing the square".

Completing the square is where we find a number which will make $y = - 1 \left({x}^{2} + 6 x + 5\right)$ a perfect square. To find that, we take the middle term, $6 x$, and (only looking at the coefficient), we divide it by two, giving us $3$. Now, we square the quotient, giving us $9$. We put that into the equation, which gives us $y = - 1 \left({x}^{2} + 6 x + 9 + 5\right)$.
WAIT! We just stuck a random number into an equation! We have to either add $9$ to both sides, or we can just add it and then immediately subtract it. I prefer the second method, which gives us $y = - 1 \left({x}^{2} + 6 x + 9 - 9 + 5\right)$.

So, we perfect went to all this work to make the equation a perfect square, so let's right it like one. Now we have $y = - 1 \left({\left(x + 3\right)}^{2} - 4\right)$, and if we distribute the $- 1$, we have $y = - 1 {\left(x + 3\right)}^{2} + 4$. To get the vertex from vertex form we just change the sign of the value in the parentheses $\left(x + 3\right)$ and keep the constant the same $\left(4\right)$. That gives us $\left(- 3 , 4\right)$, and we can check this by graphing the original equation and seeing what the vertex of that is. Hopefully, if we did our math right, they should have the same vertex.
graph{y=-1^2-6x-5}.

And it is! Good job!