How do you find the horizontal asymptote for #(2x-4)/(x^2-4)#?

1 Answer
Feb 3, 2016

Answer:

Asymptote is at #x=-2#

Explanation:

If we are given #(2x-4)/(x^2-4)#.

The first step for finding the asymptote is to factor EVERYTHING.

Let's start with #2x-4#. We can easily factor out a #2#, leaving the expression as #2(x-2)#. Now le't move on to #x^2-4#. This is actually a special case, called a difference of squares. The form for a difference of squares is #(x-y)(x+y)#. So let's factor #x^2-4# to #(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))#. There's something intersting going on here; there's an #(x-2)# in both the numerator and denominator. That makes it equal to #1#, because #(x-2)/(x-2)# is just #1#. Now we just have #2/(x-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=2# then we are dividing by zero, which like we said, isn't allowed. We could say #x=-2# except that it ceases to be a value once it becomes #1#. #x=-2# is actually a hole, not an asymptote.

Thus the asymptote is #x=4#.