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

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axis y=2x^2-8x+8

axis

y = 2 x^{2} - 8 x + 8

x = 2

Solution steps

y = 2 x^{2} - 8 x + 8

Rewrite

y = 2 x^{2} - 8 x + 8

in the standard form

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

Therefore parabola properties are:

(h, k) = (2, 0), 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 = 2

y = + \frac{1}{7} x^{2}

y = - 8 x^{2} + 4

x = + 2 (y + 1)^{2} - 5

y = x^{2} + 2 x + 3

y = 3.1 + 6.6 x - 2.3 x^{2}