What is the axis x^2+10x=-12y-1 ?

The axis x^2+10x=-12y-1 is x=-5

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axis x^2+10x=-12y-1

axis

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

x = - 5

Solution steps

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

Rewrite

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

in the standard form

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

Therefore parabola properties are:

(h, k) = (- 5, 2), 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 = - 5

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

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

- x^{2} + 4 x - 11

y = 5 x^{2} - 9

4 x^{2} - 24 x - 40 y - 4 = 0