How do you find the vertex, directrix and focus of #y + 12x - 2x^2 = 16#?

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Answer 1 · 2017-07-03T12:40:52

Vertex is at # (3,-2)# , directrix is # y = -17/8# and
focus is at # (3, -15/8)#

#y+12x-2x^2=16 or y = 2x^2-12x+16 # or

#2(x^2-6x) +16 or 2(x^2-6x+9) -18 +16# or

#2(x-3)^2 -2 # . Comparing withe vertex form of equation #y=a(x-h)^2+k ; (h,k)# being vertex , we find here #h =3 , k= -2#

Vertex is at #(h,k) or (3,-2) ; a =2# ,

Vertex is at mid point between directrix and focus. The parabola opens upward as #a>0#, so diirectrix is below the vertex.

The distance of directrix from vertex is #d=1/(4|a|) =1/(4*2)=1/8#

So directrix is #y= (-2-1/8) or y = -17/8#

Focus is at #3, (-2+1/8) or (3, -15/8)#

graph{2x^2-12x+16 [-10, 10, -5, 5]} [Ans]