What is the axis x^2+8y=0 ?

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axis x^2+8y=0

axis

x^{2} + 8 y = 0

x = 0

Solution steps

x^{2} + 8 y = 0

Rewrite

x^{2} + 8 y = 0

in the standard form:

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

Therefore parabola properties are:

(h, k) = (0, 0), p = - 2

Parabola is of the form

4 p (y - k) = (x - h)^{2}

and is symmetric around the

y

-axis

Axis of symmetry is a line parallel to the

y

-axis which intersects the vertex:

x = 0

y = x^{2} + 12 x + 28

y = 6 x^{2} - 4

y = 3 x^{2} + 15 x

y = - 2 x^{2} + 11

x^{2} - 4 x - 8 y + 12 = 0