What is (x+1)^2=-8(y+4) ?

The solution to (x+1)^2=-8(y+4) is Parabola with (h,k)=(-1,-4),p=-2

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(x+1)^2=-8(y+4)

(x + 1)^{2} = - 8 (y + 4)

(h, k) = (- 1, - 4), p = - 2

Solution steps

(x + 1)^{2} = - 8 (y + 4)

Rewrite

(x + 1)^{2} = - 8 (y + 4)

in the standard form:

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

Therefore parabola properties are:

(h, k) = (- 1, - 4), p = - 2

(x - 3)^{2} = 12 (y + 2)

3 y^{2} + 6 y - 4 x + 15 = 0

(x - \frac{3}{2})^{2} = - 7 (2) (y - 1)

x^{2} + 4 x + 12 y - 56 = 0

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

What is (x+1)^2=-8(y+4) ?
The solution to (x+1)^2=-8(y+4) is Parabola with (h,k)=(-1,-4),p=-2