y^2+4x=y+8

Solution

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

(h, k) = (\frac{33}{16}, \frac{1}{2}), p = - 1

Solution steps

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

Rewrite

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

in the standard form

4 (- 1) (x - \frac{33}{16}) = (y - \frac{1}{2})^{2}

Therefore parabola properties are:

(h, k) = (\frac{33}{16}, \frac{1}{2}), p = - 1

x^{2} - 16 x - 32 y = - 224

f (x) = x^{2} + 2 x + 3

f (x) = x^{2}

f (x) = - 2 x^{2} - 2 x - 2

f (x) = - 3 x^{2} + 5 x

What is y^2+4x=y+8 ?
The solution to y^2+4x=y+8 is Parabola with (h,k)=(33/16 , 1/2),p=-1