How do you graph # f(x)= x/(x+3)#?

First of all, we see that the function is not defined at #x=-3#. In fact, there is a vertical asymptote at this point since the denominator tends to #0# as #x->-3#, while the numerator is around a constant #-3#.

More than that, as #x->-3# from the right (that is, while it's greater than #-3#), the sign of the function is negative since the numerator is around #-3# and denominator is positive, so the function tends to negative infinity #-oo#.
If #x->-3# from the left (that is, while it's smaller than #-3#), the sign of the function is positive since the numerator is around #-3# and denominator is negative, so the function tends to positive infinity #+oo#.

Continuing analysis, we can see that, as #x->+oo# or as #x->-oo#, the value of the function tends to #1# since both numerator and denominator will increase to #+oo# or decrease to #-oo# with the same speed.

Now about constructing a graph.
The easiest way is to transform our function as follows:
#y=x/(x+3)=(x+3-3)/(x+3)=(x+3)/(x+3)-3/(x+3)=1-3/(x+3)#

So, we have to graph #y=1-3/(x+3)#.
According to the rules of graph transformation (seeUnizor - Algebra - Graph) we can construct this graph in the following steps:

Step 1. Graph #y=1/x#

graph{1/x [-10, 10, -15, 15]}

Step 2. Shift it to the left by #3# to graph #y=1/(x+3)#

graph{1/(x+3) [-10, 10, -15, 15]}

Step 3. Stretch it vertically by a factor of #3# getting the graph of #y=3/(x+3)#

graph{3/(x+3) [-10, 10, -15, 15]}

Step 4. Invert the graph (positive - to negative, negative - to positive), thus getting the graph of #y=-3/(x+3)#

graph{-3/(x+3) [-10, 10, -15, 15]}

Step 5. Finally, shift the graph up by #1# to get #y=1-3/(x+3)#:

graph{1-3/(x+3) [-10, 10, -15, 15]}