Graphing Rational Functions: An Example (page 2 of 4) Sections: Introduction, Examples, The special case with the "hole"
First I'll find any vertical asymptotes, by setting the denominator equal to zero and solving: x2
+ 1 = 0
Since this equation has no solutions, then the denominator is never zero, and there are no vertical asymptotes. Copyright © Elizabeth Stapel 2003-2011 All Rights Reserved To find the horizontal or slant asymptote, I look at the degrees of the numerator and denominator. The numerator is linear (that is, it is of degree one) while the denimonator is quadratic (that is, it is of degree two). Since the degree is greater in the denominator, the asymptote will be a horizontal at y = 0; that is, the horizontal asymptote will be the x-axis.
Next I'll find any intercepts, by plugging zero in for each variable, in turn: x = 0: y = (0 + 2)/(0 + 1) = 2 y
= 0: 0 = (x + 2)/(x2
+ 1)
I'm not sure, just from the one asymptote and the two points I have so far, quite what is going on with this graph, so I'll plot a few more points.
I found some of these points just in order to let me know what was happening up near the "top" of the curve, where the graph may get a little messy.
With these points (and the others from the T-chart in between x = 0 and x = 0.5), I can better see what is going on.
I plotted a lot of points so I could see what was going on with this rational, especially near the top of the curve, where the extra points told me that the curve was rounded. Most rationals do not need this many points, but don't be surprised if you do encounter one like this. Just keep plotting points until you're comfortable with your understanding of what the graph should look like. If you have a calculator with a "TABLE" utility, use that (as I did) to help you find all the points you need. << Previous Top | 1 | 2 | 3 | 4 | Return to Index Next >>
|
|
|
|
Copyright © 2003-2012 Elizabeth Stapel | About | Terms of Use |
|
|
|
|
|
|