What is the axis x=-3y^2+5 ?

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axis x=-3y^2+5

axis

x = - 3 y^{2} + 5

y = 0

Solution steps

x = - 3 y^{2} + 5

Rewrite

x = - 3 y^{2} + 5

in the standard form:

4 (- \frac{1}{12}) (x - 5) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (5, 0), p = - \frac{1}{12}

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

y = - x^{2} + 5 x - 7

y = x^{2} - 2 x - 3

y = - x^{2} - 6 x - 8

y = 3 x^{2} - 5 x + 7

y^{2} = - 28 x