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.
$(x-1)^2<4$ is equivalent to $-2<x-1<2$ and not to $x-1<\pm2$.
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
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.
$(x-1)^2<4$ is equivalent to $-2<x-1<2$ and not to $x-1<\pm2$.