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]

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