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

Mar 27, 2016

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:

$\frac{25 - {x}^{2}}{5 - x}$

This factors into:

$\frac{\left(5 + x\right) \left(5 - x\right)}{5 - x}$

Now you notice that the $\left(5 - x\right)$'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 $\left(5 , 10\right)$).

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 $\frac{4 {x}^{2}}{{x}^{2}}$

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 $\frac{4}{1}$, 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