$\displaystyle { y }^{ 2 }\quad =\quad \frac { (x+1)(x-3) }{ (x+4) } $

okay so,

I know its symmetric about the x axis because of y^2

y = 0 at

x = 3 and x = -1

graph doesn't exist when y^2 < 0

so i did complete the square on the numerator and got

(x - 1)^2 < 4

so x -1 < ±2

so for x < 3 and x < -1 the graph doesn't exist

but apparently there is an asymptote at x = - 4 and the graph does exist between -4 and -1 (I was given the answer)

please help