# What are the vertex, focus and directrix of  y=3 -8x -4x^2 ?

Vertex $\left(h , k\right) = \left(- 1 , 7\right)$

Focus $\left(h , k - p\right) = \left(- 1 , 7 - \frac{1}{16}\right) = \left(- 1 , \frac{111}{16}\right)$

Directrix is an equation a horizontal line

$y = k + p = 7 + \frac{1}{16} = \frac{113}{16}$
$y = \frac{113}{16}$

#### Explanation:

From the given equation $y = 3 - 8 x - 4 {x}^{2}$

Do a little rearrangement

$y = - 4 {x}^{2} - 8 x + 3$

factor out -4

$y = - 4 \left({x}^{2} + 2 x\right) + 3$

Complete the square by adding 1 and subtracting 1 inside the parenthesis

$y = - 4 \left({x}^{2} + 2 x + 1 - 1\right) + 3$

$y = - 4 {\left(x + 1\right)}^{2} + 4 + 3$

$y = - 4 {\left(x + 1\right)}^{2} + 7$

$y - 7 = - 4 {\left(x + 1\right)}^{2}$

${\left(x - - 1\right)}^{2} = - \frac{1}{4} \left(y - 7\right)$ The negative sign indicates that the parabola opens downward

$- 4 p = - \frac{1}{4}$

$p = \frac{1}{16}$

Vertex $\left(h , k\right) = \left(- 1 , 7\right)$

Focus $\left(h , k - p\right) = \left(- 1 , 7 - \frac{1}{16}\right) = \left(- 1 , \frac{111}{16}\right)$

Directrix is an equation a horizontal line

$y = k + p = 7 + \frac{1}{16} = \frac{113}{16}$
$y = \frac{113}{16}$

Kindly see the graph of $y = 3 - 8 x - 4 {x}^{2}$

graph{(y-3+8x+4x^2)(y-113/16)=0[-20,20,-10,10]}

God bless...I hope the explanation is useful.