What is the axis x^2+12x+y+40=0 ?

The axis x^2+12x+y+40=0 is x=-6

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axis x^2+12x+y+40=0

axis

x^{2} + 12 x + y + 40 = 0

x = - 6

Solution steps

x^{2} + 12 x + y + 40 = 0

Rewrite

x^{2} + 12 x + y + 40 = 0

in the standard form

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

Therefore parabola properties are:

(h, k) = (- 6, - 4), p = - \frac{1}{4}

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^{2} + 4 x = 7

y = 8 x - 2 x^{2}

y = - 6 x^{2} + 600 x - 5726

y > 2 x^{2} + 5

(x + 4)^{2} = - 8 (y - 3)