What is the axis y=x^2+8 ?

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axis y=x^2+8

axis

y = x^{2} + 8

x = 0

Solution steps

y = x^{2} + 8

Rewrite

y = x^{2} + 8

in the standard form:

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

Therefore parabola properties are:

(h, k) = (0, 8), 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} + 10 x - 2 y + 21 = 0

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

x = (y + 1) (y + 2)

4 x^{2} - 128 y = 0

(y - 3)^{2} = - 8 (x - 4)