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/
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