What is the vertices y^2-4y+4x=-16 ?

The vertices y^2-4y+4x=-16 is (h,k)=(-3,2),p=-1

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vertices y^2-4y+4x=-16

vertices

y^{2} - 4 y + 4 x = - 16

(h, k) = (- 3, 2), p = - 1

Solution steps

y^{2} - 4 y + 4 x = - 16

Rewrite

y^{2} - 4 y + 4 x = - 16

in the standard form

4 (- 1) (x - (- 3)) = (y - 2)^{2}

Therefore parabola properties are:

(h, k) = (- 3, 2), p = - 1

x^{2} + y^{2} + 6 x - 4 y - 23 = 0

y^{2} = - 8 x

16 y^{2} - 32 y - 9 x^{2} - 18 x = 137

4 x^{2} + 4 y^{2} - 4 x + 2 y - 1 = 0

25 x^{2} + 16 y^{2} = 400

What is the vertices y^2-4y+4x=-16 ?
The vertices y^2-4y+4x=-16 is (h,k)=(-3,2),p=-1