What is the axis x=-8y^2 ?

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axis x=-8y^2

axis

x = - 8 y^{2}

y = 0

Solution steps

x = - 8 y^{2}

Rewrite

x = - 8 y^{2}

in the standard form:

4 (- \frac{1}{32}) (x - 0) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 0), p = - \frac{1}{32}

Parabola is of the form

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

and is symmetric around the

x

-axis

Axis of symmetry is a line parallel to the

x

-axis which intersects the vertex:

y = 0

(x + 3)^{2} = 16 (y - 9)

y = - 0.1 x^{2} + 1.4 x + 98.4

y = 4 x^{2} + 2

y = x^{2} + 4 x

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