What is the axis y=3x^2+9 ?

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axis y=3x^2+9

axis

y = 3 x^{2} + 9

x = 0

Solution steps

y = 3 x^{2} + 9

Rewrite

y = 3 x^{2} + 9

in the standard form:

4 \cdot \frac{1}{12} (y - 9) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 9), p = \frac{1}{12}

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

\Rightarrow y = 0.5 k x^{2}

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

y = x^{2} + 6 x + 11

y^{2} = - 12 x

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