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

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Answer 1 · 2016-05-29T03:09:13

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

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.

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