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Angie
science forum beginner
Joined: 09 Dec 2005
Posts: 4
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 Posted: Sat May 20, 2006 2:25 pm Post subject:
Calculus Question - Curve Sketching
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Hello,
I need to sketch the following curve and find asymptotes, cusps, etc. Typically, I haven't had difficulties sketching curves of functions, however, I am stumped on this one because it has a root in the numerator. Fom my analysis, it looks like there are no vertical asymptotes or cusps, and a horizontal asymptote at x=0, however, I'm very unsure as to whether or not this is correct.
The function is f(x) = x^(3/4) / (x+2).
Can anyone help with this one?
Thank You. |
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Frederick Williams
science forum addict
Joined: 19 Nov 2005
Posts: 97
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| Posted: Sat May 20, 2006 3:04 pm Post subject:
Re: Calculus Question - Curve Sketching
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AMC wrote:
| Quote: |
Hello,
I need to sketch the following curve and find asymptotes, cusps, etc. Typically, I haven't had difficulties sketching curves of functions, however, I am stumped on this one because it has a root in the numerator. Fom my analysis, it looks like there are no vertical asymptotes or cusps, and a horizontal asymptote at x=0, however, I'm very unsure as to whether or not this is correct.
The function is f(x) = x^(3/4) / (x+2).
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If it had an asymptote at x = 0 it would be a vertical one--but it
doesn't.
--
Remove "antispam" and ".invalid" for e-mail address. |
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matt271829-news@yahoo.co.
science forum Guru
Joined: 11 Sep 2005
Posts: 846
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| Posted: Sat May 20, 2006 4:31 pm Post subject:
Re: Calculus Question - Curve Sketching
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AMC wrote:
| Quote: | Hello,
I need to sketch the following curve and find asymptotes, cusps, etc. Typically, I haven't had difficulties sketching curves of functions, however, I am stumped on this one because it has a root in the numerator. Fom my analysis, it looks like there are no vertical asymptotes or cusps, and a horizontal asymptote at x=0, however, I'm very unsure as to whether or not this is correct.
The function is f(x) = x^(3/4) / (x+2).
Can anyone help with this one?
Thank You.
|
You mean a horizontal asymptote at y = 0, right? If you haven't done so
already, you might find if helpful to roughly sketch the graph of y =
x^(3/4), and superimpose a sketch of y = x + 2. Then go along the
x-axis "mentally dividing" one by the other so as to see what the
overall shape of x^(3/4) / (x+2) will be. |
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Angie
science forum beginner
Joined: 09 Dec 2005
Posts: 4
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| Posted: Sat May 20, 2006 9:13 pm Post subject:
Re: Calculus Question - Curve Sketching
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Thanks for your help.
Yes, I did actually mean a horizontal asymptote at y=0, not x=0.
I will take your advice and go from there. If anyone else has any suggestions, I will be glad to hear them.
Thanks again. |
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