# What is the vertex form of y= 3x^2-11x+6 ?

Aug 11, 2017

$\left(\frac{11}{6} , - \frac{49}{12}\right)$

#### Explanation:

The $x$ value of the axis of symmetry is the same as the $x$ value of the vertex.

Use the axis of symmetry formula $x = - \frac{b}{2 a}$ to find the $x$ value of the vertex.

$x = \frac{- \left(- 11\right)}{2 \left(3\right)}$

$x = \frac{11}{6}$

Substitute $x = \frac{11}{6}$ into the original equation for the $y$ value of the vertex.

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

$y = - \frac{49}{12}$

Therefore, the vertex is at $\left(\frac{11}{6} , - \frac{49}{12}\right)$.