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

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axis y=-x^2+12

axis

y = - x^{2} + 12

x = 0

Solution steps

y = - x^{2} + 12

Rewrite

y = - x^{2} + 12

in the standard form:

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

Therefore parabola properties are:

(h, k) = (0, 12), p = - \frac{1}{4}

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 - 3)^{2} = 20 (y + \frac{119}{20})

x > - (y + 4)^{2}

y = - x^{2} + 2 x + 10

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

y = + \frac{17}{4} (x + 2)^{2}