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

Mar 29, 2017

minimum vertex at $- \frac{49}{24}$ and symetry at $x = - \frac{1}{12}$

#### Explanation:

it can be solve by using completing a square.

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

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

$y = 6 {\left(x + \frac{1}{12}\right)}^{2} - \frac{1}{24} - \frac{48}{24}$

$y = 6 {\left(x + \frac{1}{12}\right)}^{2} - \frac{49}{24}$

since coefficient of ${\left(x + \frac{1}{12}\right)}^{2}$ is +ve value, it has a minimum vertex at $- \frac{49}{24}$ and it symetry at $x = - \frac{1}{12}$