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 - 1)^{2} = - 16 (x - 1)

y = \frac{- 7 + 2 x ^{2} - 4 x}{3}

y = x^{2} - 10 x - 7

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

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