Line Crossing Horizontal and Vertical AsymptotesDate: 03/16/99 at 02:58:11 From: Kristin Subject: Horizontal Asymptotes--why would a line cross the horizontal asymptote? I encountered a function that had two vertical asymptotes and one horizontal asymptote. The function is 5x + 1 ------- x^2 - 1 The line between the two vertical asymptotes crosses the horizontal asymptote. Why does it do that? I have thought about the concept of range vs. domain. Does this function really not have a horizontal asymptote? Or is there a hole? Does y ever equal 0? Or, maybe the function has a horizontal asymptote that does not count between -1 and 1 (the vertical asymptotes). It does not seem that the denominator and the numerator have a common factor, so that cannot be the problem. These are some of my speculations. Neither my teacher nor can I figure this out. Thanks! Date: 03/16/99 at 08:26:43 From: Doctor Jerry Subject: Re: Horizontal Asymptotes--why would a line cross the horizontal asymptote? I think you must be assuming something that just is not true. Namely, you believe that horizontal asymptotes cannot share points with the curve. As this graph (which you appear to have graphed correctly) shows, the curve can cross a horizontal asymptote. A horizontal asymptote is a line or a curve that the curve approaches for large values of |x|. For small |x|, the horizontal asymptote has nothing much to do with the curve. Now, to answer the other questions. It happens that the range of this function is the set of all real numbers. Let y be any real number. Look at the equation y = (5x + 1)/(x^2 - 1) If y = 0, we can choose x = -1/5. So, 0 is in the range. Now assume that y is not zero. Expand the above equation: y*x^2 - y = 5x + 1 y*x^2 - 5x + (-y - 1) = 0 Solve the quadratic formula: y = [5+/- sqrt(25 + 4(y^2 + y))]/(2y) Since y^2 + y = y(y + 1), which is positive outside of [-1, 0], there are two solutions for all such y. For y in [-1,0) (y is not zero in this part of the argument), 25 + 4(y^2 + 1) is positive. So, there is at least one root here too. There is no hole in the graph. Of course, x cannot be -1 or 1; that is, the domain is all real except for these two values. Here is something that may help explain. In most textbook graphing problems, the graph does not cross any horizontal asymptote. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ |
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