What is the axis y=x^2-5 ?

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axis y=x^2-5

axis

y = x^{2} - 5

x = 0

Solution steps

y = x^{2} - 5

Rewrite

y = x^{2} - 5

in the standard form:

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

Therefore parabola properties are:

(h, k) = (0, - 5), 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^{2}

y = - \frac{1}{4} x^{2} + 8 x - 4

y = x^{2} - 10 x + 23

y^{2} + 8 x - 4 y + 20 = 0

y = x^{2} - 10 x + 12