# What is the vertex form of y=6x^2+17x+12 ?

Feb 16, 2016

$6 {\left(x + \frac{17}{32}\right)}^{2} + \frac{5277}{512}$ This is the required vertex form. Vertex is $\left(- \frac{17}{32} , \frac{5277}{512}\right)$

#### Explanation:

It is $y = 6 \left({x}^{2} + \frac{17 x}{6}\right) + 12$

= $6 \left({x}^{2} + \frac{17 x}{16} + \frac{289}{1024} - \frac{289}{1024}\right) + 12$

= $6 {\left(x + \frac{17}{32}\right)}^{2} + 12 - 6 \left(\frac{289}{1024}\right)$

=$6 {\left(x + \frac{17}{32}\right)}^{2} + \frac{5277}{512}$ This is the required vertex form. Vertex is $\left(- \frac{17}{32} , \frac{5277}{512}\right)$