What is the vertices y^2-12y+4x+4=0 ?

The vertices y^2-12y+4x+4=0 is (h,k)=(8,6),p=-1

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vertices y^2-12y+4x+4=0

vertices

y^{2} - 12 y + 4 x + 4 = 0

(h, k) = (8, 6), p = - 1

Solution steps

y^{2} - 12 y + 4 x + 4 = 0

Rewrite

y^{2} - 12 y + 4 x + 4 = 0

in the standard form

4 (- 1) (x - 8) = (y - 6)^{2}

Therefore parabola properties are:

(h, k) = (8, 6), p = - 1

2 x - y^{2} = 4

f (x) = - 2 x^{2} + 5

f (x) = 3 (x - 4)^{2} + 2

2 x^{2} - 12 x + 3

y = - 3 x^{2} + 6 x - 1

What is the vertices y^2-12y+4x+4=0 ?
The vertices y^2-12y+4x+4=0 is (h,k)=(8,6),p=-1