# What is the vertex form of y= x^2 - 10x - 9 ?

Apr 19, 2018

$y = {x}^{2} + 10 x - 9$

First, we need to complete the square

$y = \textcolor{g r e e n}{\left({x}^{2} + 10 x\right)} - 9$

What would make $\textcolor{g r e e n}{t h i s}$ $\left({x}^{2} + 10 x\right)$ a perfect square? Well, $5 + 5$ equals $10$ and $5 \times 5$ equals $25$ so let's try adding that to the equation:

${x}^{2} + 10 x + 25$

As a perfect square:

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

Now let's look at our original equation.

$y = {\left(x + 5\right)}^{2} - 9 \textcolor{red}{- 25}$

NOTE that we subtracted $25$ after we added it. That's because we added $25$, but as long as we later subtract it, we haven't changed the value of the expression

$y = {\left(x + 5\right)}^{2} - 34$

To check our work, let's graph our original function and what we have. If we did it right, they should be the same

graph{y=x^2+10x-9}

graph{y=(x+5)^2-34}

Looks like we were right!