What is the axis x^2=-16y ?

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axis x^2=-16y

axis

x^{2} = - 16 y

x = 0

Solution steps

x^{2} = - 16 y

Rewrite

x^{2} = - 16 y

in the standard form:

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

Therefore parabola properties are:

(h, k) = (0, 0), p = - 4

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 = 6 x^{2} - 12 x

y = + x^{2} + 3 x + 14

4 x^{2} - 12 x - 16 y + 41 = 0

x = (y - 12)^{2} - 156

x = \frac{1}{4} y^{2}