How do you find the vertex, focus and directrix of $y-2=(-1/8)(x+2)^2$ ?

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Answer 1 · 2018-01-06T12:51:12

Vertex is at $(-2,2)$ , directrix is $y=4$ and focus is at $(-2,0)$

$y-2=(-1/8)(x+2)^2 or y=(-1/8)(x+2)^2 +2$

Comparing with vertex form of equation

$y = a(x-h)^2+k ; (h,k)$ being vertex we find

here $h=-2 , k=2 ,a= -1/8 :.$ Vertex is at $(-2,2)$

$a$ is negative,so parabola opens downward and

directrix is above the vertex . Vertex is at midway

between focus and directrix. we know distance of vertex

from directrix is $d = 1/(4|a|):. d=1/(4|-1/8|)=2$ .

So directrix is $y=(2+2) or y=4$ and focus is at

${-2, (2-2)}$ or at $(-2,0)$

graph{-1/8(x+2)^2+2 [-10, 10, -5, 5]} [Ans]

How do you find the vertex, focus and directrix of y-2=(-1/8)(x+2)^2y-2=(-1/8)(x+2)^2?