What is the axis 6x^2+12y-18=0 ?

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axis 6x^2+12y-18=0

axis

6 x^{2} + 12 y - 18 = 0

x = 0

Solution steps

6 x^{2} + 12 y - 18 = 0

Rewrite

6 x^{2} + 12 y - 18 = 0

in the standard form:

4 (- \frac{1}{2}) (y - \frac{3}{2}) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, \frac{3}{2}), p = - \frac{1}{2}

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 = x^{2} + 6 x + 11

y^{2} = - 12 x

(x - 1)^{2} = 8 (y + 3)

y = 5 x^{2} + 4 x - 3

y = - x^{2} + 10