Y=5
(x2 +1)-1

Do you mean this?


y = \frac 5 {x^2+1} -1


What is it that needs to be investigated? One thing to note is that y=0 if \frac 5 {x^2+1} \ =\ 1, which means the roots of the equation are at x^2 = 4, or x = \pm 2.