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May 21, 2025, 05:41:31 pm

Author Topic: Bazza's 3/4 Question Thread  (Read 33492 times)  Share 

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brightsky

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Re: Bazza's 3/4 Question Thread
« Reply #15 on: January 14, 2012, 08:38:35 pm »
+2
ask: how else are you supposed to get x-5 in the numerator?
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pi

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Re: Bazza's 3/4 Question Thread
« Reply #16 on: January 14, 2012, 08:43:28 pm »
+1
What do you mean "resultant factors"? Do you mean after simplification?

has a "linear" part and a hyperbolic part after simplification

How did you simplify  to get a "linear" part?

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Re: Bazza's 3/4 Question Thread
« Reply #17 on: January 14, 2012, 09:21:10 pm »
0
Sorry about my question being unclear, i wrote had to write it on my ipad so i took some shortcuts (obv too many)

It's 2 separate examples



with both denominators in linear form (what do i call this, a hyperbolic part?)

whilst



I can't understand why having as the original denominator produces a "hyperbolic" part and a "truncus" part (1/x^2) whilst having a denominator like (x+3)(x-1) produces two hyperbolic parts

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Re: Bazza's 3/4 Question Thread
« Reply #18 on: January 14, 2012, 09:27:18 pm »
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Well, that's just the nature of the quadratic denominator. Every expression will be different and there in no clear-cut "formula" for finding out how it will simplify (and hence what it's graph will look like). Your best bet would be to approach every problem as new, after a while, you might see patterns:
will always split into two "hyperbolic parts"
will always split into a "hyperbolic part" and an "truncus part"
etc.

Just takes practise to get used to them.

I can't understand why having as the original denominator produces a "hyperbolic" part and a "truncus" part (1/x^2) whilst having a denominator like (x+3)(x-1) produces two hyperbolic parts

That really goes back to each expression being different, its not something that has an "explanation", its just something that just is what it is.

« Last Edit: January 14, 2012, 09:37:06 pm by Rohitpi »

brightsky

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Re: Bazza's 3/4 Question Thread
« Reply #19 on: January 14, 2012, 11:41:41 pm »
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the 'truncus' part (lol) is there because you can expand a fraction with a repeated root in the denominator the way you expand one with, say, (x+1)(x+3). you can try:

(2x-3)/(4x-7)^2 = A/(4x-7) + B/(4x-7) = (A+B)/(4x-7), which is impossible since (2x-3) isn't a factor of (4x-7)^2
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WhoTookMyUsername

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Re: Bazza's 3/4 Question Thread
« Reply #20 on: January 15, 2012, 09:47:30 am »
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i think i semi understand now :) thanks brightsky

uh, is it possible to use "inspection" when dealing with partial fractions that where the numerator has the same or higher degree than the denominator to reduce the numerator power? (i don't think it is, but enwiabe wasn't aware of a place where long division of polynomials was actually necessary, is this it xD)
« Last Edit: January 15, 2012, 09:53:17 am by Bazza16 »

WhoTookMyUsername

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Re: Bazza's 3/4 Question Thread
« Reply #21 on: January 15, 2012, 11:14:35 am »
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I've just been completely mindf*****...

how come when



the resultant graph CROSSES the asymptote? WTF!!! Isn't the whole point of an asymptote that it can't be crossed???

Please explain!!! (firstly, in terms of WHY it can cross the asymptote, and secondly, in what cases / where do you know it will cross (without solving all the time))

thanks

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Re: Bazza's 3/4 Question Thread
« Reply #22 on: January 15, 2012, 11:53:33 am »
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I've just been completely mindf*****...

how come when



the resultant graph CROSSES the asymptote? WTF!!! Isn't the whole point of an asymptote that it can't be crossed???

Please explain!!! (firstly, in terms of WHY it can cross the asymptote, and secondly, in what cases / where do you know it will cross (without solving all the time))

thanks

It crosses the asymptote of the individual functions a/f(x) and b/g(x), but the whole thing does not have the same asymptote, so there is no reason to assume it shouldn't cross this. It may be easier if you plug in actual numbers to see how it works.

Also, I assume you are not talking about the f(x) = 0 or g(x) = 0 asymptote?
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enwiabe

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Re: Bazza's 3/4 Question Thread
« Reply #23 on: January 15, 2012, 11:53:49 am »
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I've just been completely mindf*****...

how come when



the resultant graph CROSSES the asymptote? WTF!!! Isn't the whole point of an asymptote that it can't be crossed???

Please explain!!! (firstly, in terms of WHY it can cross the asymptote, and secondly, in what cases / where do you know it will cross (without solving all the time))

thanks

It CAN be crossed, but not in the infinite limit. The asymptote is what happens as SOMETHING goes to infinity. In the finite realm, an asymptote can certainly be crossed.

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Re: Bazza's 3/4 Question Thread
« Reply #24 on: January 15, 2012, 12:11:15 pm »
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I've just been completely mindf*****...

how come when



the resultant graph CROSSES the asymptote? WTF!!! Isn't the whole point of an asymptote that it can't be crossed???

Please explain!!! (firstly, in terms of WHY it can cross the asymptote, and secondly, in what cases / where do you know it will cross (without solving all the time))

thanks

It CAN be crossed, but not in the infinite limit. The asymptote is what happens as SOMETHING goes to infinity. In the finite realm, an asymptote can certainly be crossed.
yup exactly what enwiabe said, just to reiterate further, it's always good to go back and check the formal definitions to clear up any misunderstandings. Most of the time, misunderstandings come from not knowing the rigorous definition and rather applying your own definition of it.

http://en.wikipedia.org/wiki/Asymptote
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WhoTookMyUsername

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Re: Bazza's 3/4 Question Thread
« Reply #25 on: January 15, 2012, 12:23:28 pm »
0
thanks for the help :)

i think i understand where i went wrong now

WhoTookMyUsername

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Re: Bazza's 3/4 Question Thread
« Reply #26 on: January 15, 2012, 12:55:39 pm »
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For something like




Do f(x) and g(x) become asymptotes as well ? (i think they do but the book doesn't mention it anywhere)

thanks again :)

pi

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Re: Bazza's 3/4 Question Thread
« Reply #27 on: January 15, 2012, 01:12:07 pm »
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I don't think they do actually.

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Re: Bazza's 3/4 Question Thread
« Reply #28 on: January 15, 2012, 04:25:51 pm »
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For the function y = a/f(x) + b/g(x) + c/h(x) + ..., the asymptotes will be at f(x) = 0, g(x) = 0, h(x) = 0 and so on.

No matter how many times you use partial fractions or re-arrange the equation, it is still the same equation, will still look exactly the same and will still have the same asymptotes.
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WhoTookMyUsername

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Re: Bazza's 3/4 Question Thread
« Reply #29 on: January 15, 2012, 06:21:15 pm »
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I don't think they do actually.

m... But because you're essentially adding the y values still, it should never "touch" the original two functions ? Am i missing something here

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« Last Edit: January 03, 2017, 12:45:06 am by pi »