What is the vertices y^2-12y-4x+28=0 ?

The vertices y^2-12y-4x+28=0 is (h,k)=(-2,6),p=1

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vertices y^2-12y-4x+28=0

vertices

y^{2} - 12 y - 4 x + 28 = 0

(h, k) = (- 2, 6), p = 1

Solution steps

y^{2} - 12 y - 4 x + 28 = 0

Rewrite

y^{2} - 12 y - 4 x + 28 = 0

in the standard form

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

Therefore parabola properties are:

(h, k) = (- 2, 6), p = 1

f (x) = (x - 2)^{2} - 9

(3 x - 4) (4 x - 3) - 2 x

4 x^{2}

y = x^{2} - 2

x^{2} + 2 x - 3

What is the vertices y^2-12y-4x+28=0 ?
The vertices y^2-12y-4x+28=0 is (h,k)=(-2,6),p=1