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

Question

Benjamin Achtenberg · Accepted Answer

Asymptote is at #x=-2#

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#.

Lily Anderson · Answer

undefined To find the horizontal asymptote for $\frac{{2x - 4}}{{x^2 - 4}}$:
1. Determine the degrees of the numerator and denominator. The degree of the numerator is 1 and the degree of the denominator is 2.
2. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $y = 0$.
3. If the degree of the numerator is equal to the degree of the denominator, divide the leading coefficients of both the numerator and the denominator to find the horizontal asymptote.
4. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.
5. In this case, since the degree of the numerator (1) is less than the degree of the denominator (2), the horizontal asymptote is at $y = 0$.