Question

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

Answer 1

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

Explanation:

First, we can simplify the expression.

`(x-2)/(x^2-4)=(x-2)/((x+2)(x-2))=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 `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]}

Answer 2

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: `(x-2)^2`, but that would be wrong. `(x-2)^2` expanded is `(x-2)(x-2)`, which becomes `x^2-4x+4`. If we get an answer and then check it against the original and they are different, we did something wrong. `x^2-4x+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 `(x+2)(x-4)`.

So now the equation looks like this: `(x-2)/((x+2) (x-2))` the `x-2`s will divide out, making a hole. That leaves `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
`(x-2)/((x+2) (x-2))` and seeing where that asymptote is.
graph{(x-2)/((x+2) (x-2))}