What is the vertices 2y^2-8x-24y+56=0 ?

The vertices 2y^2-8x-24y+56=0 is (h,k)=(-2,6),p=1

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vertices 2y^2-8x-24y+56=0

vertices

2 y^{2} - 8 x - 24 y + 56 = 0

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

Solution steps

2 y^{2} - 8 x - 24 y + 56 = 0

Rewrite

2 y^{2} - 8 x - 24 y + 56 = 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 + 4)^{2}

f (x) = x^{2} + 3 x + \frac{1}{4}

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

f (x) = x^{2} + 6 x + 2

f (x) = x^{2} + 6 x + 3

What is the vertices 2y^2-8x-24y+56=0 ?
The vertices 2y^2-8x-24y+56=0 is (h,k)=(-2,6),p=1