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

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

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

vertices

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

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

Solution steps

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

Rewrite

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

in the standard form

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

Therefore parabola properties are:

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

24000 x^{2} + 900 x - 150

y = x^{2} + 4 x - 8

y = x^{2} + 4 x - 3

f (x) = x^{2} - 12 x + 27

3 x^{2} + 2

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