What is the axis 4y=-x^2-12x-48 ?

The axis 4y=-x^2-12x-48 is x=-6

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axis 4y=-x^2-12x-48

axis

4 y = - x^{2} - 12 x - 48

x = - 6

Solution steps

4 y = - x^{2} - 12 x - 48

Rewrite

4 y = - x^{2} - 12 x - 48

in the standard form

4 (- 1) (y - (- 3)) = (x - (- 6))^{2}

Therefore parabola properties are:

(h, k) = (- 6, - 3), p = - 1

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 = - 6

y = - 4 (x + 10)^{2} + 6

(x + 1)^{2} = - 3 (y - 6)

y = 10 x^{2} + 1 - 6 x

y = - 8 x^{2} + 4 x

- 3 y = x^{2} - 8 x + 37