There is an asymptote at x=-1 and f(x)=x-1-(1/(x+1)). The range is negative infinity to plus infinity. For x<-1 f(x) drops from plus infinity, crossing the x axis to continue heading to negative infinity gradually as x becomes more negative. For x>1 f(x) rises from negative infinity to plus infinity, crossing the x axis. As x becomes large negative or large positive the graph approximates to the line f(x)=x, which is a sloping asymptote.