What is y^2+x+4y+4=0 ?

The solution to y^2+x+4y+4=0 is Parabola with (h,k)=(0,-2),p=-1/4

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y^2+x+4y+4=0

y^{2} + x + 4 y + 4 = 0

(h, k) = (0, - 2), p = - \frac{1}{4}

Solution steps

y^{2} + x + 4 y + 4 = 0

Rewrite

y^{2} + x + 4 y + 4 = 0

in the standard form

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

Therefore parabola properties are:

(h, k) = (0, - 2), p = - \frac{1}{4}

x^{2} + x + 2 m (x) + 1 = 0

2 y = x^{2} - 10

f, f = 6 (x - 2) f (x)^{- 1} (x) f - 1 (x)

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

(y - 3)^{2} = 7 (x - 8)

What is y^2+x+4y+4=0 ?
The solution to y^2+x+4y+4=0 is Parabola with (h,k)=(0,-2),p=-1/4