What is the axis y=3x^2-12x+8 ?

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axis y=3x^2-12x+8

axis

y = 3 x^{2} - 12 x + 8

x = 2

Solution steps

y = 3 x^{2} - 12 x + 8

Rewrite

y = 3 x^{2} - 12 x + 8

in the standard form

4 \cdot \frac{1}{12} (y - (- 4)) = (x - 2)^{2}

Therefore parabola properties are:

(h, k) = (2, - 4), p = \frac{1}{12}

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

x^{2} = 8 y

y = x^{2} + 4 x - 45

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

(x + 6)^{2} = 12 (y + 1)

y = x^{2} - 6 x - 3