Question

What are the vertex, focus, and directrix of `y=8 - (x + 2) ^2`?

Answer 1

The vertex is at `(h, k)=(-2, 8)`
Focus is at `(-2, 7)`
Directrix: `y=9`

Explanation:

The given equation is `y=8-(x+2)^2`

The equation is almost presented in the vertex form

`y=8-(x+2)^2`

`y-8=-(x+2)^2`

`-(y-8)=(x+2)^2`

`(x--2)^2=-(y-8)`

The vertex is at `(h, k)=(-2, 8)`

`a=1/(4p)` and `4p=-1`

`p=-1/4`

`a=1/(4*(-1/4))`

`a=-1`

Focus is at `(h, k-abs(a))=(-2, 8-1)=(-2, 7)`

Directrix is the horizontal line equation

`y=k+abs(a)=8+1=9`

`y=9`

Kindly see the graph of `y=8-(x+2)^2` and the directrix `y=9`

graph{(y-8+(x+2)^2)(y-9)=0[-25,25,-15,15]}

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