# What is the vertex form of y=6x^2-13x-5 ?

Mar 2, 2016

$y = 6 {\left(x - \frac{13}{12}\right)}^{2} - \frac{289}{24}$

#### Explanation:

The standard form of the quadratic function is $a {x}^{2} + b x + c$

the function here $y = 6 {x}^{2} - 13 x - 5 \text{ is in this form }$

by comparison , a = 6 , b = -13 and c = -5

The vertex form is : $y = a {\left(x - h\right)}^{2} + k$

where (h,k) are the coords of the vertex.

the x-coord of the vertex (h)$= \frac{- b}{2 a} = - \frac{- 13}{12} = \frac{13}{12}$

and y-coord (k) $= 6 {\left(\frac{13}{12}\right)}^{2} - 13 \left(\frac{13}{12}\right) - 5 = - \frac{289}{24}$

here $\left(h , k\right) = \left(\frac{13}{12} , - \frac{289}{24}\right) \mathmr{and} a = 6$

$\Rightarrow y = 6 {\left(x - \frac{13}{12}\right)}^{2} - \frac{289}{24} \text{ is the equation }$