# Math Help - Graph Sketching ARRGHH!!

1. ## Graph Sketching ARRGHH!!

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

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)

2. ## Re: Graph Sketching ARRGHH!!

Originally Posted by Applestrudle
graph doesn't exist when y^2 < 0

so i did complete the square on the numerator and got

(x - 1)^2 < 4
You need to determine the sign of the right-hand side as a whole and not just its numerator. Please review how to solve rational inequalities.

Just one remark.

Originally Posted by Applestrudle
so x -1 < ±2
$(x-1)^2<4$ is equivalent to $-2<x-1<2$ and not to $x-1<\pm2$.