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 (- \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 - 3)^{2} = 8 (y - 5)

y = \frac{x ^{2} 5 x 8}{x 3}

x^{2} - 4 y = 0

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

y^{2} = - x