# What is the vertex form of 5y = 11x^2-15x-9?

Apr 10, 2017

$y = \frac{11}{5} {\left(x - \frac{15}{22}\right)}^{2} - \frac{621}{220}$

#### Explanation:

Vertex form of such equation is $y = a {\left(x - h\right)}^{2} + k$, with $\left(h , k\right)$ as vertex.

Here we have $5 y = 11 {x}^{2} - 15 x - 9$

or $y = \frac{11}{5} {x}^{2} - 3 x - \frac{9}{5}$

or $y = \frac{11}{5} \left({x}^{2} - 3 \times \frac{5}{11} x\right) - \frac{9}{5}$

$= \frac{11}{5} \left({x}^{2} - 2 \times \frac{15}{22} x + {\left(\frac{15}{22}\right)}^{2} - {\left(\frac{15}{22}\right)}^{2}\right) - \frac{9}{5}$

$= \frac{11}{5} {\left(x - \frac{15}{22}\right)}^{2} - {\left(\frac{15}{22}\right)}^{2} \times \frac{11}{5} - \frac{9}{5}$

$= \frac{11}{5} {\left(x - \frac{15}{22}\right)}^{2} - \frac{45}{44} - \frac{9}{5}$

$= \frac{11}{5} {\left(x - \frac{15}{22}\right)}^{2} - \frac{45 \times 5 + 44 \times 9}{220}$

$= \frac{11}{5} {\left(x - \frac{15}{22}\right)}^{2} - \frac{225 + 396}{220}$

$= \frac{11}{5} {\left(x - \frac{15}{22}\right)}^{2} - \frac{621}{220}$

and vertex is $\left(\frac{15}{22} , - \frac{621}{220}\right)$

graph{5y=11x^2-15x-9 [-4.667, 5.333, -4.12, 0.88]}