# What is the vertex form of y= 36x^2+132x+121 ?

${\left(x + \frac{11}{6}\right)}^{2} = \frac{1}{36} y$

#### Explanation:

Given equation:

$y = 36 {x}^{2} + 132 x + 121$

$y = 36 \left({x}^{2} + \frac{11}{3} x\right) + 121$

$y = 36 \left({x}^{2} + \frac{11}{3} x + {\left(\frac{11}{6}\right)}^{2}\right) - 36 {\left(\frac{11}{6}\right)}^{2} + 121$

$y = 36 {\left(x + \frac{11}{6}\right)}^{2}$

${\left(x + \frac{11}{6}\right)}^{2} = \frac{1}{36} y$

The above is the vertex form of parabola with vertex at

$\left(x + \frac{11}{6} = 0 , y = 0\right) \setminus \equiv \left(- \frac{11}{6} , 0\right)$