What is the vertices y^2-2y-8x-31=0 ?

The vertices y^2-2y-8x-31=0 is (h,k)=(-4,1),p=2

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vertices y^2-2y-8x-31=0

vertices

y^{2} - 2 y - 8 x - 31 = 0

(h, k) = (- 4, 1), p = 2

Solution steps

y^{2} - 2 y - 8 x - 31 = 0

Rewrite

y^{2} - 2 y - 8 x - 31 = 0

in the standard form

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

Therefore parabola properties are:

(h, k) = (- 4, 1), p = 2

x^{2} - x - 6

x = 4 y^{2}

2 x^{2} - 3

x^{2} = - 12 y

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

What is the vertices y^2-2y-8x-31=0 ?
The vertices y^2-2y-8x-31=0 is (h,k)=(-4,1),p=2