How do you find the asymptotes for y = (x-2)/( x^2-4)?

Jan 28, 2016

Vertical asymptote at $x = - 2$, horizontal asymptote at $y = 0$

Explanation:

First, we can simplify the expression.

$\frac{x - 2}{{x}^{2} - 4} = \frac{x - 2}{\left(x + 2\right) \left(x - 2\right)} = \frac{1}{x + 2}$

Here, the $x - 2$ has been cancelled out. That means that there is a removable discontinuity at $x = 2$. A removable discontinuity is also known as a hole.

Our new function is $\frac{1}{x + 2}$ with a hole at $x = 2$.

Vertical asymptotes:

The vertical asymptotes will occur when the denominator of the function equals $0$.

$x + 2 = 0$

$x = - 2$

There is a vertical asymptote at $x = - 2$.

Horizontal asymptotes:

When the degree of the denominator is greater than the denominator of the numerator, the $x$ axis or the line $y = 0$ is the function's horizontal asymptote.

We can check a graph:

graph{(x-2)/(x^2-4) [-10, 10, -5, 5]}

Jan 29, 2016

The asymptote is $x = - 2$

Explanation:

This is how I got that asymptote.
First, I factored everything as much as I can. It's like taking apart a puzzle to see every piece of it.
$x - 2$ is already as factored as it gets. However, ${x}^{2} - 4$ can be factored some more. Your first thought might be to factor it into this: ${\left(x - 2\right)}^{2}$, but that would be wrong. ${\left(x - 2\right)}^{2}$ expanded is $\left(x - 2\right) \left(x - 2\right)$, which becomes ${x}^{2} - 4 x + 4$. If we get an answer and then check it against the original and they are different, we did something wrong. ${x}^{2} - 4 x + 4$ does not equal ${x}^{2} - 4$, so we have to figure something else out.

Now, ${x}^{2} - 4$ is actually a special case called "difference of squares", and from that I know that is can be factored to $\left(x + 2\right) \left(x - 4\right)$.

So now the equation looks like this: $\frac{x - 2}{\left(x + 2\right) \left(x - 2\right)}$ the $x - 2$s will divide out, making a hole. That leaves $\frac{1}{x + 2}$. Now, an asymptote is a line that the graph will get closer and closer to but never touch. This is because there is a value which will make the denominator equal to zero, and you cannot divide by zero, ever.

So let's find the value that makes $x + 2 = 0$. Subtract $2$ on both sides and we have $x = - 2$. The line that the graph will never touch --the asymptote-- is $x = - 2$.

We can always check our work by graphing
$\frac{x - 2}{\left(x + 2\right) \left(x - 2\right)}$ and seeing where that asymptote is.
graph{(x-2)/((x+2) (x-2))}