# What is the vertex form of 3y=-5x^2 - x +7?

##### 1 Answer
Aug 5, 2017

$y = - \frac{5}{3} {\left(x - \left(- \frac{1}{10}\right)\right)}^{2} + \frac{141}{60}$

#### Explanation:

Given:

$3 y = - 5 {x}^{2} - x + 7$

Divide both sides by $3$ to get $y$ on the left hand side, then complete the square...

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

$\textcolor{w h i t e}{y} = - \frac{5}{3} \left({x}^{2} + \frac{1}{5} x - \frac{7}{5}\right)$

$\textcolor{w h i t e}{y} = - \frac{5}{3} \left({x}^{2} + 2 \left(\frac{1}{10}\right) x + \frac{1}{100} - \frac{141}{100}\right)$

$\textcolor{w h i t e}{y} = - \frac{5}{3} \left({\left(x + \frac{1}{10}\right)}^{2} - \frac{141}{100}\right)$

$\textcolor{w h i t e}{y} = - \frac{5}{3} {\left(x + \frac{1}{10}\right)}^{2} + \frac{141}{60}$

$\textcolor{w h i t e}{y} = - \frac{5}{3} {\left(x - \left(- \frac{1}{10}\right)\right)}^{2} + \frac{141}{60}$

The equation:

$y = - \frac{5}{3} {\left(x - \left(- \frac{1}{10}\right)\right)}^{2} + \frac{141}{60}$

is in the form:

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

which is vertex form for a parabola with vertex $\left(h , k\right) = \left(- \frac{1}{10} , \frac{141}{60}\right)$ and multiplier $a = - \frac{5}{3}$

graph{3y = -5x^2-x+7 [-4.938, 5.06, -1.4, 3.6]}