# What is the vertex form of y=- 2x^2 + 3x -6 ?

Jan 28, 2016

$- 2 {\left(x - \frac{3}{4}\right)}^{2} - \frac{39}{8} = y$

#### Explanation:

We start with $- 2 {x}^{2} + 3 x - 6$.
The way I would solve this is by completing the square. The first step for that is to make the coefficient of ${x}^{2}$ 1. We do that by factoring out a $- 2$. The equation now looks like this:
$- 2 \left({x}^{2} - \frac{3}{2} x + 3\right)$.

From here, we need to find a term which will make the equation factorable. We do that by taking the middle factor, $- \frac{3}{2}$, and dividing it by $2$, making it $- \frac{3}{4}$. Then we square this, changing it to $\frac{9}{16}$.

Now that we found the number which will make the${x}^{2} - \frac{3}{2}$part of the equation factorable, what do we do with it? I'll tell you what we do with it; we plug it in . But, we can't just put a random number into the equation. I'll show you how we resolve this in a minute.

First we rewrite the equation as $- 2 \left({x}^{2} - \frac{3}{2} \textcolor{red}{+ \frac{9}{16}} \textcolor{red}{- \frac{9}{16}} + 3\right)$. NOTE we resolved the issue of sticking in a number by subtracting it so that it actually has no effect on the value of the equation.

Anyway, we can now condense $- 2 \left({x}^{2} - \frac{3}{2} + \frac{9}{16} - \frac{9}{16} + 3\right)$ into $- 2 \left({\left(x - \frac{3}{4}\right)}^{2} + \frac{39}{16}\right)$.

We are very nearly done, except that we can stil simplify further by multiplying $- 2$ to the $\frac{39}{16}$, making it $- \frac{39}{8}$.

The final answer is $- 2 {\left(x - \frac{3}{4}\right)}^{2} - \frac{39}{8} = y$