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

Answer 1

Savannah Anderson

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.

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Answer 2

Hazel Anglin

Vertex: (-2, 8) Focus: (-2, 9) Directrix: y = 7

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.