# How do you find the vertex, focus and directrix of 4x^2 + 6x -y + 2 = 0?

Nov 9, 2016

The vertex $= \left(- \frac{3}{4} , - \frac{1}{4}\right)$
The focus is $= \left(- \frac{3}{5} , - \frac{3}{16}\right)$
The equation of the directrix is $y = - \frac{5}{16}$

#### Explanation:

Let's rewrite the equation by completing the squares
$4 \left({x}^{2} + \frac{3}{2} x + \frac{9}{16}\right) = y - 2 + \frac{9}{4} = y + \frac{1}{4}$
${\left(x + \frac{3}{4}\right)}^{2} = \frac{1}{4} \left(y + \frac{1}{4}\right)$
This is a parabola
We compare this to the equation of the parabola ${\left(x - a\right)}^{2} = 2 p \left(y - b\right)$
$\therefore$The vertex is $\left(- \frac{3}{4} , - \frac{1}{4}\right)$
$p = \frac{1}{8}$
The focus is $\left(a , b + \frac{p}{2}\right) = \left(- \frac{3}{4} , - \frac{1}{4} + \frac{1}{16}\right) = \left(- \frac{3}{4} , - \frac{3}{16}\right)$

The equation of the directrix is $y = - \frac{1}{4} - \frac{1}{16} = - \frac{5}{16}$
graph{(4x^2+6x-y+2)(y+5/16)=0 [-3.459, 1.409, -1.378, 1.056]}