Author Topic: Horizontal Asymptote  (Read 9706 times)

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Offline Ergot

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Horizontal Asymptote
« on: May 18, 2006, 01:30:02 am »
Where is it when y = 3x / (x+3)^2 ?
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Offline dark_drake

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Re: Horizontal Asymptote
« Reply #1 on: May 18, 2006, 01:33:45 am »
y = 0

At least if I remember correctly.
errr... something like that...

Offline Ergot

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Re: Horizontal Asymptote
« Reply #2 on: May 18, 2006, 01:41:34 am »
Explain :( ?
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Offline dark_drake

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Re: Horizontal Asymptote
« Reply #3 on: May 18, 2006, 01:49:08 am »
The limit as x approaches +infinity and -infinity is 0.
errr... something like that...

Offline Sidoh

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Re: Horizontal Asymptote
« Reply #4 on: May 18, 2006, 01:50:23 am »
Explain :( ?



Actually, there's a really easy way to do this problem (without graphing or anything).

Since the total powers of x (2) are greater in the denomonater than they are in the numerator, f(x) will approach 0 as x approaches infinity.  This is always the case with any function, no matter how big a constant on either may be.  Conversely, if the powers of x are larger in the numerator than in the demononator, f(x) will approach infinity.
« Last Edit: May 18, 2006, 01:52:04 am by Sidoh »

Offline dark_drake

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Re: Horizontal Asymptote
« Reply #5 on: May 18, 2006, 01:59:41 am »
Explain :( ?
Since the total powers of x (2) are greater in the denomonater than they are in the numerator, f(x) will approach 0 as x approaches infinity.  This is always the case with any function, no matter how big a constant on either may be.  Conversely, if the powers of x are larger in the numerator than in the demononator, f(x) will approach infinity.
This is true; however, learning the limit thing is useful for when the powers are equal.  As always, I reccomend learning the basics and using shortcuts after growing comfortable with the basics.
errr... something like that...

Offline Ergot

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Re: Horizontal Asymptote
« Reply #6 on: May 18, 2006, 02:01:07 am »
Thanks... I have a quiz on this tomorrow, hopefully he'll leave those types out...
drake... explain the basic to me? AFAIK, this will get me pretty far until I learn more in calc or w/e.
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Offline dark_drake

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Re: Horizontal Asymptote
« Reply #7 on: May 18, 2006, 02:06:20 am »
Thanks... I have a quiz on this tomorrow, hopefully he'll leave those types out...
drake... explain the basic to me? AFAIK, this will get me pretty far until I learn more in calc or w/e.
Just the limit as x approaches plus and minus infinity. How you do that, at least how I learned, is about like this:


Basically, you can't evaluate the limit of the function as x approaches infinity because when you plug infinity in, you get infnity/infinity, which is undefined.  So, you have to divide the numerator and denominator by the largest power of x.

Edit: This only works if the powers in the denominator are >= those in the numerator.  If the numerator's are greater, you're dealing with slant asymptotes.
« Last Edit: May 18, 2006, 02:11:07 am by dark_drake »
errr... something like that...

Offline Ergot

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Re: Horizontal Asymptote
« Reply #8 on: May 18, 2006, 02:20:24 am »
I SEE !!!
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Offline rabbit

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Re: Horizontal Asymptote
« Reply #9 on: May 19, 2006, 07:47:07 am »
Or when it's undefined (IE: -3).

Offline Ergot

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Re: Horizontal Asymptote
« Reply #10 on: May 19, 2006, 09:29:49 am »
Or when it's undefined (IE: -3).
That would be x = -3 or the vertical asymptote... that one is easy.
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Offline rabbit

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Re: Horizontal Asymptote
« Reply #11 on: May 19, 2006, 11:52:12 am »
Fuck asymptotes, I'm done with high school!

Offline MyndFyre

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Re: Horizontal Asymptote
« Reply #12 on: May 25, 2006, 11:33:50 pm »
Thanks... I have a quiz on this tomorrow, hopefully he'll leave those types out...
drake... explain the basic to me? AFAIK, this will get me pretty far until I learn more in calc or w/e.
Just the limit as x approaches plus and minus infinity. How you do that, at least how I learned, is about like this:


Basically, you can't evaluate the limit of the function as x approaches infinity because when you plug infinity in, you get infnity/infinity, which is undefined.  So, you have to divide the numerator and denominator by the largest power of x.

Edit: This only works if the powers in the denominator are >= those in the numerator.  If the numerator's are greater, you're dealing with slant asymptotes.

Easier way to do that limit: L'Hospital's Rule. 



Really handy, because it can be applied several times.  For example, you want to find the limit of the following function:


So, as a good, easy mneumonic device:

1.) If the exponent of the numerator is greater, the function will have a limit of +infinity or -infinity.
2.) If the exponent of the denominator is greater, the function will have a limit of 0.  Series defined by this function will be convergent.
3.) If the exponents of the numerator and denominator are the same, the limit will be a constant of the numerator's greatest-power coefficient times the factorial of the power, divided by the denominator's greatest-power coefficient times the factorial of the power (in the example it was 6*5!/12*5!).  I believe that the series will be convergent, but I can't remember for sure.

Also, limits don't need to be taken to infinity.  They can be found at any point along the function.
« Last Edit: May 25, 2006, 11:37:13 pm by MyndFyre[x86] »
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Offline Ergot

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Re: Horizontal Asymptote
« Reply #13 on: May 26, 2006, 01:41:36 am »
Uhh... I don't understand that :( I'm only in Algebra II.
Who gives a damn? I fuck sheep all the time.
And yes, male both ends.  There are a couple lesbians that need a two-ended dildo...My router just refuses to wear a strap-on.
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Offline deadly7

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Re: Horizontal Asymptote
« Reply #14 on: May 26, 2006, 02:23:39 am »
What the hell do factorials have to do with this?  You have weird ways of doing things.. graphing takes about two seconds with a calculator and is way easier.
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