How do you sketch the parabola ${(x + 2)}^{2} = - 20 (y - 5)$ $(x+2)^2=-20(y-5)$ and find the vertex, focus, and directrix?

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Answer 1 · 2017-11-05T20:29:14

V(-2;5)
F(-2;0)
D: y=10 graph{(x+2)^(2)=-20 (y-5) [-12.375, 7.625, -4.68, 5.32]}

${(x - h)}^{2} = 4 p (y - k)$ $(x-h)^2=4p(y-k)$
from here we know
vertex is V(h,k)
focus is F(h,k+p)
and the directrix is y=k-p

in
${(x + 2)}^{2} = - 20 (y - 5)$ $(x+2)^(2)=-20 (y-5)$

$h = - 2 : k = 5 and 4 p = - 20 \to p = - 5$ $h=-2: k=5 and 4p=-20 -> p=-5$

then
V(h,k)=V(-2,5)

F(h,k+p) = F(-2,5-5) then F(-2,0)

and

directrix
$y = k - p \to y = 5 - (- 5) \to y = 10$ $y=k-p -> y=5-(-5) -> y=10$

How do you sketch the parabola (x+2)^2=-20(y-5)(x+2)2=−20(y−5)(x+2)^2=-20(y-5) and find the vertex, focus, and directrix?