Vertical and Horizontal Asymptotes

Date: 6/3/96 at 3:29:21
From: Anonymous
Subject: Vertical and Horizontal Asymptotes

My name is Margot and I am an Algebra II student in 
high school. This is one of the test questions with which I am having 
trouble. Using the function f(x)= the quantity x squared minus 2 times 
x minus 15 divided by the quantity x squared plus 4 times x 
plus 3:

  f(x) = (x^2 - 2x - 15)/(x^2 + 4x + 3)

What is the vertical asymptote(s)?
x-values of any holes?
Real zero(s)
What is the horizontal asymptote?

I understand all but the vertical and horizontal asymptotes.

Date: 6/3/96 at 12:55:21
From: Doctor Darren
Subject:Vertical and Horizontal Asymptotes

If you understand the hole, then you know that we may as well deal 
with the equation f(x)=(x-5)/(x+1). As you can probably tell, if we 
plug in larger and larger values of x, then f(x) approaches 1 from 
below, without ever quite reaching it. Thus, there is an asymptote at 
y=1. 

Similarly, if we plug in largely negative values of x (try -100, 
-1000, etc) you will see that f(x) approaches 1 from above, so we have 
the same asymptote at y=1. This is the only horizontal asymptote.  

Vertical asymptotes will occur when the denominator is 0. You can test
this, by seeing that if you plug in values closer and closer to -1, 
f(x) gets very large (or largely negative) and approaches infinity (or 
negative infinity). Furthermore, if you plug in x=-1, the equation is 
undefined. Thus, we have a vertical asymptote at x=-1.

I hope this helped!

-Doctor Darren,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/