What is the axis y>2x^2+5 ?

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axis y>2x^2+5

axis

y > 2 x^{2} + 5

x = 0

Solution steps

y > 2 x^{2} + 5

Rewrite

y > 2 x^{2} + 5

in the standard form:

4 \cdot \frac{1}{8} (y - 5) = (x - 0)^{2}

Therefore parabola properties are:

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

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

y = (x + 5)^{2} + 3

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

x = (2 y)^{2}

x = 4 y^{2}