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Asymptotes: Examples (page 4 of 4) Sections: Vertical asymptotes, Horizontal asymptotes, Slant asymptotes, Examples In general, you will be given a rational (fractional) function, and you will need to find the domain and any asymptotes. You will need to know what steps to take and how to recognize the different types of asymptotes.
The vertical asymptotes (and any restrictions on the domain) come from the zeroes of the denominator, so I'll set the denominator equal to zero and solve.
4x^{2}
– 9 = 0
Then the domain is all
xvalues
other than ±
^{3}/_{2},
and the two vertical asymptotes are at Since the degrees of
the numerator and the denominator are the same (each being 2),
then this rational has a nonzero (nonxaxis)
horizontal asymptote, and does not have a slant asymptote; the horizontal
asymptote is found by dividing the leading terms: Then the full answer is: domain:
A given rational function may or may not have a vertical asymptote (depending upon whether the denominator ever equals zero), but it will always have either a horizontal or else a slant asymptote. Note, however, that the function will only have one of these two; you will have either a horizontal asymptote or else a slant asymptote, but not both. As soon as you see that you'll have one of them, don't bother looking for the other one.
The vertical asymptotes come from the zeroes of the denominator, so I'll set the denominator equal to zero and solve. x^{2}
+ 9 = 0
Oops! This has no solution. Since the denominator has no zeroes, then there are no vertical asymptotes and the domain is "all x". Since the degree is greater in the denominator than in the numerator, the yvalues will be dragged down to the xaxis, and the horizontal asymptote is therefore "y = 0". Since I have found a horizontal asymptote, I don't have to look for a slant asymptote. Then the full answer is: domain:
all x
The Special Case with a "Hole"
It so happens that this function can be simplified as: The temptation is to say that y equals x + 1 and therefore that this has no vertical asymptote. But the original function does have a zero in the denominator at x = 2. While the graph of y will look very much like x + 1, it will not quite be the same. Since the degree of the numerator is one greater than the degree of the denominator, I'll have a slant asymptote (not a horizontal one), and I'll find that slant asymptote by long division. Hmm... There wasn't any remainder when I divided. Actually, that makes sense: since x – 2 is a factor of the numerator and I'm dividing by x – 2, the division should come out evenly. Then the full answer is: Copyright © Elizabeth Stapel 20032011 All Rights Reserved domain:
When you go to graph the function in this last example, you can draw the line right on the slant asymptote, but you will need to leave a nice open dot where x = 2, to indicate that this point is not actually included in the graph. This last case is not the norm for slant asymptotes, but you should expect to see at least one problem of this type, including perhaps on the test. By the way, different books may treat this case differently: Some books may say that there is no vertical asymptote where there is a "hole". Check for an example in your text, or ask your instructor. In general, the procedure for asymptotes is the following:
When you're working these problems, try to go through these steps in order, so you can remember them on the test. They're not so hard once you get the hang of them, so be sure to do plenty of practice exercises. << Previous Top  1  2  3  4  Return to Index



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