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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: captainoats on June 29, 2009, 08:10:05 pm

Title: Question regarding asymptotes with a composite function?
Post by: captainoats on June 29, 2009, 08:10:05 pm: Evening all,

I have the equation:

$y=\frac{x-2}{x+3}$

So I know that there is an asymptote at x=-3, as you can't divide by zero.

But, how do you go about working out the y asymptote manually? I know that you can plug it into the calulator and work it out, but I don't understand how I would tackle this if I don't have my calculator.

Thanks! :)
Title: Re: Question regarding asymptotes with a composite function?
Post by: TrueTears on June 29, 2009, 08:13:29 pm: Long divide it, this yields $1 - \frac{5}{x+3}$

Hence y asy at y = 1

code for fractions

\frac{ }{ }
Title: Re: Question regarding asymptotes with a composite function?
Post by: captainoats on June 29, 2009, 08:23:12 pm: Quote from: TrueTears on June 29, 2009, 08:13:29 pm
Long divide it, this yields $1 - \frac{5}{x+3}$

Hence y asy at y = 1

code for fractions

\frac{ }{ }

Awesome! Thankyou! :)
Title: Re: Question regarding asymptotes with a composite function?
Post by: dcc on June 29, 2009, 08:48:07 pm: Quote from: captainoats on June 29, 2009, 08:10:05 pm
$y=\frac{x-2}{x+3}$

A trick which I picked up off VN a long, long time ago was to consider rewriting the equation, as follows:

$y = \frac{x-2}{x+3} = \frac{\left(x + 3\right) - 5}{x + 3} = 1 - \frac{5}{x + 3}$

It's very simple. Another example would be:

$y = \frac{x + 2}{2x + 3} = \frac{\frac{1}{2} \left(2x + 3\right) + \frac{1}{2}}{2x + 3} = \frac{1}{2} + \frac{\frac{1}{2}}{2x + 3}$

Very useful, in my opinion :)
Title: Re: Question regarding asymptotes with a composite function?
Post by: captainoats on June 29, 2009, 08:51:04 pm: Quote from: dcc on June 29, 2009, 08:48:07 pm
Quote from: captainoats on June 29, 2009, 08:10:05 pm
$y=\frac{x-2}{x+3}$

A trick which I picked up off VN a long, long time ago was to consider rewriting the equation, as follows:

$y = \frac{x-2}{x+3} = \frac{\left(x + 3\right) - 5}{x + 3} = 1 - \frac{5}{x + 3}$

It's very simple. Another example would be:

$y = \frac{x + 2}{2x + 3} = \frac{\frac{1}{2} \left(2x + 3\right) + \frac{1}{2}}{2x + 3} = \frac{1}{2} + \frac{\frac{1}{2}}{2x + 3}$

Very useful, in my opinion :)

Wow! That's sweet! It took me a while to see what was happening.. (long day!)... but that is really cool!
Title: Re: Question regarding asymptotes with a composite function?
Post by: kamil9876 on June 29, 2009, 09:27:42 pm: Quote from: dcc on June 29, 2009, 08:48:07 pm
Quote from: captainoats on June 29, 2009, 08:10:05 pm
$y=\frac{x-2}{x+3}$

A trick which I picked up off VN a long, long time ago was to consider rewriting the equation, as follows:

$y = \frac{x-2}{x+3} = \frac{\left(x + 3\right) - 5}{x + 3} = 1 - \frac{5}{x + 3}$

It's very simple. Another example would be:

$y = \frac{x + 2}{2x + 3} = \frac{\frac{1}{2} \left(2x + 3\right) + \frac{1}{2}}{2x + 3} = \frac{1}{2} + \frac{\frac{1}{2}}{2x + 3}$

Very useful, in my opinion :)

Very cool indeed, you can use this to prove the long division algorithm (ie: if you avoid rewriting the redundant parts such as the denominator etc. and just extract the numbers that you are actually manipulating you get that algorithm :) )

Lol i always forget that algorithm so i stick to this, even for cubics over quadratics, which can get tedious
Title: Re: Question regarding asymptotes with a composite function?
Post by: Mao on June 30, 2009, 01:48:44 am: A less rigorous way can get you the answer faster:

as $x\to \infty$ , $x-2\to x$ , $x+3\to x$ , $\implies \frac{x-2}{x+3}\to \frac{x}{x} = 1$

as $x\to \infty$ , $x+2 \to x$ , $2x+3\to 2x$ , $\implies \frac{x+2}{2x+3} \to \frac{x}{2x} = \frac{1}{2}$

(basically ignore the constant)
Title: Re: Question regarding asymptotes with a composite function?
Post by: dcc on June 30, 2009, 12:06:25 pm: That might get you the constant term, however to find the other term its still equivalent to solving:

$1 + \frac{R}{x+3} = \frac{x-2}{x+3}$

Which is essentially the process I outlined above, done in a semi-reversed fashion.
Title: Re: Question regarding asymptotes with a composite function?
Post by: Gloamglozer on June 30, 2009, 02:02:49 pm: Quote from: dcc on June 29, 2009, 08:48:07 pm
Quote from: captainoats on June 29, 2009, 08:10:05 pm
$y=\frac{x-2}{x+3}$

A trick which I picked up off VN a long, long time ago was to consider rewriting the equation, as follows:

$y = \frac{x-2}{x+3} = \frac{\left(x + 3\right) - 5}{x + 3} = 1 - \frac{5}{x + 3}$

It's very simple. Another example would be:

$y = \frac{x + 2}{2x + 3} = \frac{\frac{1}{2} \left(2x + 3\right) + \frac{1}{2}}{2x + 3} = \frac{1}{2} + \frac{\frac{1}{2}}{2x + 3}$

Very useful, in my opinion :)

With the second example, can you please explain how you got to $\frac{1}{2}$ ?
Title: Re: Question regarding asymptotes with a composite function?
Post by: TrueTears on June 30, 2009, 02:41:19 pm: Because the denominator is 2x+3 and the number is x+2 you need ask yourself "How do I get a 2x+3 on the top so when I split the fraction they can cancel"

So to get a coefficient of 2 in front of the x, we put a 2 in front of it but then we must multiply (x+2) by $\frac{1}{2}$ , the net result is still the same for the coefficient of x.

But now we have $\frac{3}{2}$ instead of 2, so what do we need to do to $\frac{3}{2}$ to make it into a 2? We add $\frac{1}{2}$ hence net result is still 2.

EDIT: too fast shinny :P
Title: Re: Question regarding asymptotes with a composite function?
Post by: shinny on June 30, 2009, 02:43:19 pm: Quote from: TrueTears on June 30, 2009, 02:41:19 pm
EDIT: too fast shinny :P

Nah deleted my post =P I actually didn't realise he 'artificially' made the 2x+3 and just explained how he cancelled it out to be 1/2. Read TT's explanation =T
Title: Re: Question regarding asymptotes with a composite function?
Post by: kamil9876 on June 30, 2009, 02:53:08 pm: Another way to do it if you aren't sure:

$<br />\frac{x+2}{2x+3}$
$=\frac{1}{2}(\frac{2x+4}{2x+3})$
$=\frac{1}{2}(\frac{2x+3}{2x+3}+\frac{1}{2x+3}))$
$=\frac{1}{2}(1+\frac{1}{2x+3})$
Title: Re: Question regarding asymptotes with a composite function?
Post by: /0 on June 30, 2009, 03:44:34 pm: It's good for simple examples but with complicated examples I would just go with long division, which is a sure thing, instead of wasting time thinking of how to split it up.
Title: Re: Question regarding asymptotes with a composite function?
Post by: Gloamglozer on June 30, 2009, 04:18:13 pm: Thanks for the explanation TT and kamil (and shinny even though I didn't read your post lol).

I understand it now. :D
Title: Re: Question regarding asymptotes with a composite function?
Post by: shinny on June 30, 2009, 04:19:13 pm: Quote from: Gloamglozer on June 30, 2009, 04:18:13 pm
Thanks for the explanation TT and kamil (and shinny even though I didn't read your post lol).

I understand it now. :D

Heh ignore mine. Misunderstood what you were misunderstanding and answered something retardedly basic =P
Title: Re: Question regarding asymptotes with a composite function?
Post by: Gloamglozer on June 30, 2009, 04:26:06 pm: Either way, still karma'red you for trying. :D