What is the vertex of # x = 1/12 (y/4 - 4)^2 - 5 #?

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Answer 1 · 2017-05-28T12:13:13

**Vertex is at #** (-5 , 16) #

#x = 1/12 (y/4 -4)^2 -5 or 1/12 (y/4 -4)^2 = x+5 # or

#1/12*1/16 (y -16)^2 = x+5 or 1/192 (y -16)^2 = x+5# or

# (y -16)^2 = 192 (x+5) or (y -16)^2 = 4*48 (x+5)#.

Comparing with standard equation of parabola #(y-k)^2=4a(x-h)#.

Vertex is at # (h,k) :. h= -5 , k= 16 #

Vertex is at # (-5,16) #

graph{x= 1/12(y/4-4)^2-5 [-320, 320, -160, 160]} [Ans]