What is y^2+4y+4x+8=0 ?

The solution to y^2+4y+4x+8=0 is Parabola with (h,k)=(-1,-2),p=-1

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y^2+4y+4x+8=0

y^{2} + 4 y + 4 x + 8 = 0

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

Solution steps

y^{2} + 4 y + 4 x + 8 = 0

Rewrite

y^{2} + 4 y + 4 x + 8 = 0

in the standard form

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

Therefore parabola properties are:

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

e^{l n (y)} = e^{2 \cdot l n (x)}

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

\frac{s ^{2} - 2 p + 6 p}{( s + 2 )} = 1

f, f (x)^{- 1} (x) = \frac{0.75}{10 - x}

T (n) = T (n - 1) + n^{2}

What is y^2+4y+4x+8=0 ?
The solution to y^2+4y+4x+8=0 is Parabola with (h,k)=(-1,-2),p=-1