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Asymptotes: Examples (page 4 of 4) Sections: Vertical asymptotes, Horizontal asymptotes, Slant asymptotes, Examples In general, you will be given rational (fractional) expressions, and you will have 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. 4x2 – 9 = 0
Then the domain is all x-values
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 non-zero (non-x-axis) horizontal asymptote, not a slant asymptote, and the horizontal asymptote is found by dividing the leading terms:
Then the full answer is: domain: Note: You may or may not have a vertical asymptote (depending upon whether the denominator ever equals zero), but you will always have a horizontal or slant asymptote in these problems. However, you will only have one of the two. That is, 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, don't bother looking for the other.
The vertical asymptotes come from the zeroes of the denominator, so I'll set the denominator equal to zero and solve. x2 + 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 y-values will be dragged down to the x-axis, 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"
Note that this 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, so 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 the 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 2006-2008 All Rights Reserved domain: Note: When you go to graph the function in this last example, you will draw the line right on the slant asymptote, but you will leave a nice open dot where x = 2, to indicate that this point is not included. 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. << Previous Top | 1 | 2 | 3 | 4 | Return to Index
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Copyright © 2006-2008 Elizabeth Stapel | About | Terms of Use |
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