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

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

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vertices 2y^2-8x=0

vertices

2 y^{2} - 8 x = 0

(h, k) = (0, 0), p = 1

Solution steps

2 y^{2} - 8 x = 0

Rewrite

2 y^{2} - 8 x = 0

in the standard form:

4 \cdot 1 \cdot (x - 0) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 0), p = 1

x^{2} + 3 x + 1

x^{2} + 3 x - 1

x^{2} - 7 x - 8

0 = (y - 1)^{2} - 16 (x + 4)

y = + 8 - 9 x^{2}