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]}