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

1 Answer
Mar 27, 2016

Answer:

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