What is the axis (x+1)^2=12y ?

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axis (x+1)^2=12y

axis

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

x = - 1

Solution steps

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

Rewrite

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

in the standard form:

4 \cdot 3 (y - 0) = (x - (- 1))^{2}

Therefore parabola properties are:

(h, k) = (- 1, 0), p = 3

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 = - 1

x = 2 (y - 5)^{2} + 3

x^{2} + 10 x = - 12 y - 1

2 y^{2} - 8 x - 24 y + 56 = 0

y = \frac{2 x ^{2} + 7 x + 12}{2}

- x^{2} + 4 x - 11