What is the axis y=+x^2+81 ?

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axis y=+x^2+81

axis

y = + x^{2} + 81

x = 0

Solution steps

y = x^{2} + 81

Rewrite

y = x^{2} + 81

in the standard form:

4 \cdot \frac{1}{4} (y - 81) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 81), 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

y = - \frac{1}{16} (x - 1)^{2} - 1

y = x^{2} + 7 x + 30

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

y = 6 x - x^{2} - 5

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