A good way to start with any of these questions is to guess what type of curve this is, and if that is too difficult then the next best thing is to find some points. In this case it is obvious that it is a hyperbola from the question (and form of the equation).
Next is to find the asymptotes. We can take an educated guess that they will be parallel to the x and y axes. To find the x-parallel-asymptote we substitute x=inf into the equation and by inspection, this gives y=2 (we can use l’hopitals rule to prove this). To find the y-parallel-asymptote we make the fraction tend to infinity by making the denominator 0, therefore x=-1 is the other asymptote.
Lastly, which way round does the hyperbola go? If we have already plotted a few points then this is obvious, and if not then just plot a few points. We can merge this step and finding the intersection points into one by setting y=0 gives x=0.5 and setting x=0 gives y=-1 and so, we have our solution.
For the extension, we take our original equation and use polynomial division to get y = 2 – 3/(x+1). In this form it is obvious that there is a shift by 1 to the left, a flip in the y-axis, a stretch in the y-axis by a factor of 3 and then finally a shift of the curve up by 2 (Can explain better with graphs).