# What are the vertex, focus and directrix of  y=2x^2 +11x-6 ?

Dec 19, 2016

The vertex is $= \left(- \frac{11}{4} , - \frac{169}{8}\right)$
The focus is $= \left(- \frac{11}{4} , - \frac{168}{8}\right)$
The directrix is $y = - \frac{170}{8}$

#### Explanation:

Let rewrite the equation

$y = 2 {x}^{2} + 11 x - 6$

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

$= 2 \left({x}^{2} + \frac{11}{2} x + \frac{121}{16}\right) - 6 - \frac{121}{8}$

$y = 2 {\left(x + \frac{11}{4}\right)}^{2} - \frac{169}{8}$

$y + \frac{169}{8} = 2 {\left(x + \frac{11}{4}\right)}^{2}$

$\frac{1}{2} \left(y + \frac{169}{8}\right) = {\left(x + \frac{11}{4}\right)}^{2}$

This is the equation of the parabola

${\left(x - a\right)}^{2} = 2 p \left(y - b\right)$

The vertex is $= \left(a , b\right) = \left(- \frac{11}{4} , - \frac{169}{8}\right)$

The focus is $= \left(a , b + \frac{p}{2}\right) = \left(- \frac{11}{4} , - \frac{169}{8} + \frac{1}{8}\right)$

$= \left(- \frac{11}{4} , - \frac{168}{8}\right)$

The directrix is $y = b - \frac{p}{2}$

$\implies$, $y = - \frac{169}{8} - \frac{1}{8} = - \frac{170}{8}$

graph{(y-2x^2-11x+6)(y+170/8)=0 [-14.77, 10.54, -21.49, -8.83]}