# What is the vertex form of 6y=-x^2 + 9x ?

Aug 29, 2016

$y = - \frac{1}{6} {\left(x - \frac{9}{2}\right)}^{2} + \frac{27}{8}$

#### Explanation:

Divide both sides by $6$ to get:

$y = - \frac{1}{6} \left({x}^{2} - 9 x\right)$

$= - \frac{1}{6} \left({\left(x - \frac{9}{2}\right)}^{2} - {9}^{2} / {2}^{2}\right)$

$= - \frac{1}{6} {\left(x - \frac{9}{2}\right)}^{2} + \frac{1}{6} \cdot \frac{81}{4}$

$= - \frac{1}{6} {\left(x - \frac{9}{2}\right)}^{2} + \frac{27}{8}$

Taking the two ends together, we have:

$y = - \frac{1}{6} {\left(x - \frac{9}{2}\right)}^{2} + \frac{27}{8}$

which is in vertex form:

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

with multiplier $a = - \frac{1}{6}$ and vertex $\left(h , k\right) = \left(\frac{9}{2} , \frac{27}{8}\right)$

graph{(6y+x^2-9x)((x-9/2)^2+(y-27/8)^2-0.02) = 0 [-5.63, 14.37, -3.76, 6.24]}