# What is the vertex form of 5y = -3x^2-2x+2?

Jul 8, 2018

$y = - \frac{3}{5} {\left(x + \frac{1}{3}\right)}^{2} + \frac{7}{15}$

#### Explanation:

Given: $5 y = - 3 {x}^{2} - 2 x + 2$

Vertex form: $y = a {\left(x - h\right)}^{2} + k$,

where the vertex is $\left(h , k\right)$ and $a$ is a constant.

Complete the square to find the vertex form.

First divide by $5$ to make $y =$:

$y = \left(- \frac{3}{5} {x}^{2} - \frac{2}{5} x\right) + \frac{2}{5}$

Factor out $- \frac{3}{5}$ so we only have an ${x}^{2}$:

$y = - \frac{3}{5} \left({x}^{2} + \frac{2}{3} x\right) + \frac{2}{5}$

To complete the square we need to multiply $\frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$. We also need to subtract the squared term that we added:

$- \frac{3}{5} \left(x + \frac{1}{3}\right) \left(x + \frac{1}{3}\right) = \textcolor{red}{- \frac{3}{5}} \left({x}^{2} + \frac{2}{3} x \text{ } \textcolor{red}{+ \frac{1}{9}}\right)$

$y = - \frac{3}{5} {\left(x + \frac{1}{3}\right)}^{2} + \frac{2}{5} - \left(\textcolor{red}{- \frac{3}{5} {\left(\frac{1}{3}\right)}^{2}}\right)$

$y = - \frac{3}{5} {\left(x + \frac{1}{3}\right)}^{2} + \frac{3}{3} \cdot \frac{2}{5} + \frac{1}{15}$

$y = - \frac{3}{5} {\left(x + \frac{1}{3}\right)}^{2} + \frac{7}{15}$