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

Feb 3, 2016

Asymptote is at $x = - 2$

#### Explanation:

If we are given $\frac{2 x - 4}{{x}^{2} - 4}$.

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

Let's start with $2 x - 4$. We can easily factor out a $2$, leaving the expression as $2 \left(x - 2\right)$. 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 $\left(x - y\right) \left(x + y\right)$. So let's factor ${x}^{2} - 4$ to $\left(x + 2\right) \left(x - 4\right)$. These are as factored as they can be, so we should now take another look at the expression put together.

We now have $\frac{2 \left(x - 2\right)}{\left(x + 2\right) \left(x - 2\right)}$. There's something intersting going on here; there's an $\left(x - 2\right)$ in both the numerator and denominator. That makes it equal to $1$, because $\frac{x - 2}{x - 2}$ is just $1$. Now we just have $\frac{2}{x - 2}$.

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 $x = 4$.