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

1 Answer
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:

(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