What is the axis 4x^2-128y=0 ?

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axis 4x^2-128y=0

axis

4 x^{2} - 128 y = 0

x = 0

Solution steps

4 x^{2} - 128 y = 0

Rewrite

4 x^{2} - 128 y = 0

in the standard form:

4 \cdot 8 (y - 0) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 0), p = 8

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 - 3)^{2} = - 8 (x - 4)

y = (x - 13)^{2}

y = x^{2} - 5 x - 84

y = + \frac{1}{4} x^{2} + \frac{2}{3} x - \frac{1}{6}

x = + (y + 1)^{2} - 9