Expert: Paul Klarreich - 2/11/2008

Question
In the rational funtion F(x)= x-1/x-x^3

I am attempting to locate the asymptotes

Do I rearrange the demnominator and make it -x(x^2-1) and therefore the vertical asymptotoes would be 0, 1,-1

I am unsure of if there are horizontal or oblique asympt.
since n<m would the horizontal be y=0 ?


Answer
Questioner:   Jessica
Category:  Advanced Math
Private:  No
 
Subject:  Asymptotes
Question:  In the rational funtion F(x)= x-1/x-x^3

I am attempting to locate the asymptotes

Do I rearrange the demnominator and make it -x(x^2-1) and therefore the vertical asymptotoes would be 0, 1,-1

I am unsure of if there are horizontal or oblique asympt.
since n<m would the horizontal be y=0 ?
..........................................
Hi, Jessica,

Comments:
1. Write your function either:

F(x)= (x-1)/(x-x^3)   << parenthesize.

or
         x - 1
F(x) = ---------
       x - x^3

2. Use proper vocabulary. (It will definitely raise your own understanding level.)

x - x^3 = -x(x^2 -1) is FACTORING, not 'rearranging'.

Nevertheless, that's usually a good idea, and after doing that, you can go a little further.

         x - 1
F(x) = ------------
      - x(x^2 - 1)

         x - 1
F(x) = -----------------
      - x(x - 1)(x + 1)

Now you will note that a factor of (x - 1) cancels (but further comment will be necessary.)

             1
F1(x) = -----------
      - x(x + 1)

Now you will conclude that  x = 0 and x = -1 are vertical asymptotes.  YOu will also note that as  x --> +- infinity,  F(x) --> 0, so  y = 0 is a horizontal asymptote.

[You can't have BOTH horizontal and oblique asymptotes.]

What about  x = 1?  YOu see I wrote  F1(x) after canceling (x - 1).  The cancelling (sometimes I like two l's, sometimes I like one.)   is technically illegal, because if x = 1, x-1 is zero and you cannot cancel a factor that is zero.  So

F1(x) = F(x), provided x /= 1.  How does it affect the graph?  F1(1) = 1/(-2) = -1/2

So (1, -1/2) is on the graph of F1(x), but not on F(x), so if you draw the graph, put a tiny 'hole' at that point.