What is the axis y=-2x^2+11 ?

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axis y=-2x^2+11

axis

y = - 2 x^{2} + 11

x = 0

Solution steps

y = - 2 x^{2} + 11

Rewrite

y = - 2 x^{2} + 11

in the standard form:

4 (- \frac{1}{8}) (y - 11) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 11), 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^{2} - 4 x - 8 y + 12 = 0

y = x^{2} + 10 x

x = (- 20 y + 13)^{2}

(x - 3)^{2} = 12 (y - 1)

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