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]