What is the vertex of x = 1/12 (y/4 - 4)^2 - 5 x=112(y4−4)2−5?

1 Answer
Binayaka C.
May 28, 2017

**Vertex is at ** (-5 , 16) ∗(−5,16)

Explanation:

x = 1/12 (y/4 -4)^2 -5 or 1/12 (y/4 -4)^2 = x+5 x=112(y4−4)2−5or112(y4−4)2=x+5 or

1/12*1/16 (y -16)^2 = x+5 or 1/192 (y -16)^2 = x+5112⋅116(y−16)2=x+5or1192(y−16)2=x+5 or

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

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