If we are given (2x-4)/(x^2-4)2x−4x2−4.
The first step for finding the asymptote is to factor EVERYTHING.
Let's start with 2x-42x−4. We can easily factor out a 22, leaving the expression as 2(x-2)2(x−2). Now le't move on to x^2-4x2−4. This is actually a special case, called a difference of squares. The form for a difference of squares is (x-y)(x+y)(x−y)(x+y). So let's factor x^2-4x2−4 to (x+2)(x-4)(x+2)(x−4). These are as factored as they can be, so we should now take another look at the expression put together.
We now have (2(x-2))/((x+2)(x-2))2(x−2)(x+2)(x−2). There's something intersting going on here; there's an (x-2)(x−2) in both the numerator and denominator. That makes it equal to 11, because (x-2)/(x-2)x−2x−2 is just 11. Now we just have 2/(x-2)2x−2.
The definition of an asymptote is the value that the graph will approach but never touch. The reason for that is that for a certain value, it will make the expression be divide by zero, which cannot happen; in math terms it is illegal. In the case of our expression, when x=2x=2 then we are dividing by zero, which like we said, isn't allowed. We could say x=-2x=−2 except that it ceases to be a value once it becomes 11. x=-2x=−2 is actually a hole, not an asymptote.
Thus the asymptote is x=4x=4.
