What is the axis x^2=-125(y-20) ?

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axis x^2=-125(y-20)

axis

x^{2} = - 125 (y - 20)

x = 0

Solution steps

x^{2} = - 125 (y - 20)

Rewrite

x^{2} = - 125 (y - 20)

in the standard form:

4 (- \frac{125}{4}) (y - 20) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 20), p = - \frac{125}{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 = 3 x^{2} - 6 x + 4

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

y = 3 x^{2} (1.3)

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

36 y^{2} + 72 x - 132 y - 239 = 0