Question

How do you find the vertical, horizontal or slant asymptotes for `(4x^2+5)/( x^2-1)`?

Answer 1

Vertical asymptotes: `x = 1` and `x = -1`
Horizontal asymptote: `y = 4`

Explanation:

Ok, let's start with the vertical asymptotes. You know that a function is not defined when a denominator equals `0`. In this case, that would be when `x^2 - 1 = 0`. Using basic equation rules, can can change that into `x^2 = 1`. Then, we take the square root of both sides, making sure to note the positive and negative possibilities.

This gives us:
`x = 1` or `x = -1`

I don't know how advanced your problems will get, but there is now a possibility of either a vertical asymptote or a removable discontinuity. If you have no idea what a removable discontinuity is, you can probably skip this next part.

Removable discontinuities:

In order to know if you have a removable discontinuity, try factoring both the top and the bottom. If any of the factors cancel, then you have a removable discontinuity at that point. Take for example:

`(25 - x^2)/(5-x)`

This factors into:

`((5 + x)(5-x))/(5-x)`

Now you notice that the `(5-x)`'s cancel leaving:

`5 + x` or `x + 5` (if you prefer the `x` first)

This makes a line, which is pretty easy to graph. However, because `5-x = 0` when `x` is `5`, it cannot be in the domain. Therefore, you are left with the line `y = x + 5` with a domain of all real numbers except for `5`. At that point, there is simply a hole in the graph notated by an empty circle where the point would be if it were in the domain (in my example, that point would be at `(5, 10)`).

However, the top part of your problem doesn't even factor, so there cannot be a removable discontinuity giving you vertical asymptotes at `x = 1` and `x = -1`

End of Removable Discontinuities

Now on to the horizontal and oblique (or slant as you called it) asymptotes.

This is done by doing the following:
First take the highest powered variables on the top and the bottom and remove the rest. In your case this would be `(4x^2)/(x^2)`

We can do this because horizontal and oblique asymptotes are when `x` is really big, meaning that only this term is going to be relevant. For example, is `1,000,001` that different from `999,998`?
Not really.

Now, determine which power is greater; `x^3` is bigger than `x^2` no matter how big or small the coefficients (the numbers in front of the variables) are.

If the top is bigger:
Simplify! What you end up with is your oblique asymptote

If the bottom is bigger:
You got off easy. The horizontal asymptote is just `y = 0`

If they are equal (like your problem):
Divide the coefficients (in your case it's `4/1`, which is just `4`) and the horizontal asymptote is `y =`whatever you got.

That means that your horizontal asymptote is `y = 4`.

Hope this helped!

Jonathan 'JMoney' Moore