What is the axis y=-12x^2+156x ?

The axis y=-12x^2+156x is x= 13/2

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axis y=-12x^2+156x

axis

y = - 12 x^{2} + 156 x

x = \frac{13}{2}

Solution steps

y = - 12 x^{2} + 156 x

Rewrite

y = - 12 x^{2} + 156 x

in the standard form

4 (- \frac{1}{48}) (y - 507) = (x - \frac{13}{2})^{2}

Therefore parabola properties are:

(h, k) = (\frac{13}{2}, 507), p = - \frac{1}{48}

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 = \frac{13}{2}

y^{2} + 2 x - 2 y - 3 = 0

y = 2 x^{2} - x

(x + 3)^{2} = 4 (y - 1)

y = x^{2} + 2 x + 1

y = - x^{2} - 2 x + 7