Curve sketching

$1 = x(2+x)$
So x= 1 or x = -2

Well, vertical asymptotes occur when there's division by zero. Where do you see division in y? When can the divisor be zero?

The next asymptotes you need to consider are when x ---> infinity and when x ---> -infinity. With algebraic fractions, the way to do it is to make sure the degree of the denominator is greater than the degree of the numerator (by degree we mean the highest power of x). Notice in (1-x)/(1+x) the degrees are the same, which means we can actually divide the expression through (long division style). You'll end up with something like

(1-x)/(1+x) = A + B/(1+x) (where A and B are constants you need to work out via long division or whatever method you've been taught).

Now as x tends to infinity, what does A + B/(1 + x) tend to? But that's not the whole story, remember that what we're actually dealing with is

y = x + A + B/(1 + x)

So if A + B/(1+x) tends to L (L is something I've left for you to work out), then y tends to x + L - note that this is the equation of a line, that's what the function tends to (the function may well cut these "oblique" asymptotes, by the way, but as we tend towards our good friend infinity, the line x + L behaves similarly to a vertical asymptote).

Post back if anything's unclear, but don't be vague.

*Edit*

One more thing, I made it sound like we can only divide algebraic fractions through when the degrees are equal, but in reality we can do it when the degree of the numerator is greater than or equal to that of the denominator's (e.g. (x^3 + 4x - 5)/(7x - 2)).