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

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axis y=2x^2-8

axis

y = 2 x^{2} - 8

x = 0

Solution steps

y = 2 x^{2} - 8

Rewrite

y = 2 x^{2} - 8

in the standard form:

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

Therefore parabola properties are:

(h, k) = (0, - 8), p = \frac{1}{8}

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 = x^{2} - 2 x - 9

x^{2} - 6 x - 5 y - 1 = 0

y = + (x + 4)^{2} - 6

y = (x + 8)^{2}

(x + 3)^{2} = 24 (y - 1)