# What is the vertex form of y= 3x^2+x-55?

Mar 2, 2017

$y = 3 {x}^{2} + x - 55$ has a minimum $- \frac{661}{12}$ at $\left(- \frac{1}{6} , - \frac{661}{12}\right)$

#### Explanation:

$y = 3 {x}^{2} + x - 55$

$y = \left[3 \left({x}^{2} + \frac{x}{3}\right)\right] - 55$

solve using completing a square,
$y = \left[3 {\left(x + \frac{1}{6}\right)}^{2} - 3 \cdot {\left(\frac{1}{6}\right)}^{2}\right] - 55$

$y = 3 {\left(x + \frac{1}{6}\right)}^{2} - 3 \cdot \left(\frac{1}{36}\right) - 55$

$y = 3 {\left(x + \frac{1}{6}\right)}^{2} - \frac{1}{12} - 55$

$y = 3 {\left(x + \frac{1}{6}\right)}^{2} - \frac{661}{12}$

Therefore,
$y = 3 {x}^{2} + x - 55$ has a minimum $- \frac{661}{12}$ at $\left(- \frac{1}{6} , - \frac{661}{12}\right)$