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Why are you studing Further/Additional Maths?! watch

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    Trust me, you can't.. lol. I tried P4 for a day and got stuck drawing things like y = (5x - 1)/(2x + 3)

    But now I know how (thanks to people here)... put the denominator equal to zero to find impossible answers, and therefore vertical asymptotes. In this case it would be: x = -3/2

    Then to find the horizontal asymptote, find what happens to y when x tends to infinity. Unfortumately I don't get this quite as much... I tried rearraging in terms of y and got -1/3, but I don't think that's the proper way.

    y = (5x - 1)/(2x + 3)
    2yx + 3y = 5x - 1
    2yx - 5x = -1 -3y
    x (5 - 2y) = (3y + 1)
    x = (5 - 2y)/(3y + 1)

    Anyway, yes you will probably need P2 and 3 (and def. P1) for P4.
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    (Original post by mik1a)

    Then to find the horizontal asymptote, find what happens to y when x tends to infinity. Unfortumately I don't get this quite as much... I tried rearraging in terms of y and got -1/3, but I don't think that's the proper way.
    When x tend to infinity, y tends to 5/2 (I think).
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    i don't need further maths at uni for i am doing medicine...however my unis don't even acknowledge that i do the subject (which makes it worse)...i simply do it for i think it is one of the only subjects which allow you to solve problems...all the rest are spoon fed (even experiments...most of them are in the book and gives a step by step explanation)...and it also makes maths A-Level so much easier!
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    (Original post by Leekey)
    When x tend to infinity, y tends to 5/2 (I think).
    that would make sense, as it would tend to 5*infinity / 2*infinity, because the others are negligable, then cancel infinity! :cool:
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    (Original post by mik1a)
    that would make sense, as it would tend to 5*infinity / 2*infinity, because the others are negligable, then cancel infinity! :cool:
    u can't 'cancel' infinity! because infinity is not a number.
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    (Original post by mik1a)
    Trust me, you can't.. lol. I tried P4 for a day and got stuck drawing things like y = (5x - 1)/(2x + 3)
    The easiest way of finding horizontal asymptotes (IMO) in this type of question is to divide all the terms in the fraction by the highest power of x - in this case just x, and consider what happens when x tends to infinity.

    y = (5 - 1/x)/(2 + 3/x) As x tends to infinity, 1/x and 3/x tend to 0 so y tends to 5/2 as leekey said. He's obviously not a mod for nothing
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    (Original post by IntegralAnomaly)
    u can't 'cancel' infinity! because infinity is not a number.
    well what he meant that when x gets very large the +constants are 'nothing' compared to your very large number, so overall y tends to 5/2... I think he's got the hang of it now..!
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    (Original post by IntegralAnomaly)
    u can't 'cancel' infinity! because infinity is not a number.
    Sure you can.. if it's the same variable tending to infinity... you can cancel! Stop being so pedantic lol
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    If you want a nice neat way to do this the use l'hopitals rule ( google if needed).

    In a nutshell
    lim x tend to Infinity of f(x)/g(x) = lim x tend to infinity f '(x)/g'(x).

    ie f(x)/g(x) has the same limit as f'(x)/g'(x) as x tends to infinity

    now f'(x) = 5
    and g'(x) = 2

    therefor in the limit f'(x)/g'(x) = 5/2

    MrM.
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    IMO the easiest way was just thinking it logically, as x tends to infinity the constants are negligible, remaining coeffients i.e. 5/2 ! However maybe that f '(x) and g '(x) good for using complexer functions?? Like (3x^4+3x^2+6)/(2x^2+1)?
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    (Original post by sweetymango)
    IMO the easiest way was just thinking it logically, as x tends to infinity the constants are negligible, remaining coeffients i.e. 5/2 ! However maybe that f '(x) and g '(x) good for using complexer functions?? Like (3x^4+3x^2+6)/(2x^2+1)?
    My way works for this too:
    Divide top and bottom by x^4, y = (3 + 3/x^2 + 6/x^2)/(2/x^2 + 1/x^4)
    Bottom's going to tend to 0 so no horizontal asymptote! It would have worked if there had been one though! There's also no vertical asymptote as 2x^2 + 1 can't equal 0 for any real values of x! Not a very good example to choose
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    (Original post by Bezza)
    He's obviously not a mod for nothing
    And you guys thought it was because of a chronic addiction to this site!!! I believe there is a maths test to be a mod though as ALL of the mods have done / are doing a maths A-Level, at least I think Expression did (hmmm conspiracy?).
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    (Original post by Bezza)
    My way works for this too:
    Divide top and bottom by x^4, y = (3 + 3/x^2 + 6/x^2)/(2/x^2 + 1/x^4)
    Bottom's going to tend to 0 so no horizontal asymptote! It would have worked if there had been one though! There's also no vertical asymptote as 2x^2 + 1 can't equal 0 for any real values of x! Not a very good example to choose
    lol yeah, i should have done a TOP heavy fraction, doh! So would that l'hopitals rule fail for my silly example?
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    L´Hopitals rule also works for stuff like lim x tends to zero (e^x)/sin(x) ; in fact it works for a whole host of stuff....Check it out .... :°)
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    Guh...we do further maths in two lessons a week. I chose it because I liked the subject, now I'm thinking of taking old reliable up at uni too. But it isn't "old" reliable" anymore as its rock solid!
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    (Original post by Spider)
    Guh...we do further maths in two lessons a week. I chose it because I liked the subject, now I'm thinking of taking old reliable up at uni yoo. But it isn't "old" reliable" anymore as its rock solid!
    hey? whats old reliable?
    and...2 lessons a week?! :eek: how many for regular maths?
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    (Original post by mik1a)
    Sure you can.. if it's the same variable tending to infinity... you can cancel! Stop being so pedantic lol
    What about x²/(x+1) as x-> infinity does it equal infinity/infinity "and the inifinities cancel"?
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    (Original post by It'sPhil...)
    What about x²/(x+1) as x-> infinity does it equal infinity/infinity "and the inifinities cancel"?
    no, the infinities are different infinities... one is infinity, and the other is infinity squared. in this case, you can cances an infinity leaving infinity on the top, the coefficient would be 1, yes?
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    (Original post by Katie Heskins)
    hey? whats old reliable?
    and...2 lessons a week?! :eek: how many for regular maths?

    Just a stupid thing I say...People usually call their favourite guns "old reliable", but I decided to call maths that instead cause it's my favourite subject.

    5 Lessons a week for normal maths, Further maths is an "extra"
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    (Original post by sweetymango)
    I just wondered which people are actually studying Further Maths/Further Additional Maths with the intention of studying it at Uni.
    I'm studying Double Maths, for my passion - but I don't want to study it further (not on its own anyway!), after A-Level that's it!!! I'm off to study something else!

    more work now means more sleep in uni when they reteach it

    otherwise further maths is considered prestigous subject and its 2 subjects for 1 and a halfs work
 
 
 
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