Question 108981
Let's do some analysis. 
When you look at the function,{{{(x^2+5)/(x+1)}}} and you're interested in asymptotes, you need to ask a couple of questions. 
1. When does the denominator go to zero? 
2. What happens when x gets very large in the positive direction?
3. What happens when x gets very large in the negative direction?
Let's look at each question.
1. The denominator goes to zero at x=-1. 
So at x=-1, your function is undefined. 
Just to the right of it, the function goes near positive infinity (x+1>0).
Just to the left of it, the function goes near negative infinity (x+1<0). 
So at x=-1, you have a vertical asymptote.
2. As x gets very large positively, 
{{{(x^2+5)/(x+1)=(x^2)/(x)=(x)}}}
So for very large x, the function looks like y=x. 
As an example, at x=100, 
{{{(x^2+5)/(x+1)=(100^2+5)/(100+1)=99.05}}}
So you have another asymptote that looks like the function y=x.
3. As x gets very large negatively, 
{{{(x^2+5)/(x+1)=(x^2)/(x)=(x)}}}
So for very large x, the function also looks like y=x for large positive x
As an example, at x=-100, 
{{{(x^2+5)/(x+1)=((-100)^2+5)/(-100+1)=-101.061}}}
So you have another asymptote that looks like the function y=x for large negative x.
But it all together and you can graph the function.

{{{ graph( 300, 200, -20, 20, -20, 20,(x^2+5)/(x+1),x,100000(x+1)) }}}