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    Q. How many horizontal and vertical asymptotes can the graph of a given rational function have?
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    as many vertical as you want to my knowledge
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    how ?
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    well say if on the demoninator you had (x-1)(x-2) it would give you 2 asymptotes wouldnt it?

    cant you just add more factors to the denominator?
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    what about the horizontal ones ...i think the answer is one for that cuz you don't seem to get more than one answers when you substitute infinity ..
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    yer that makes sense, when you go to inifinty it can only tend to 1 answer, but i think you can have an infinite amount of vertical asymptotes though.

    So in answer to your question:

    its infinity + 1

    so an infinite amount lol
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    f(x) = \frac{g(x)}{h(x)} where g,h are polynomial and share no common roots.

    A vertical asymptote means either g->infinity or h->0. Since a polynomial only goes to infinity if it's argument does, h->0. Thus each asymptote equates to a zero of h(x). If you didn't cancel common roots from g and h, then h could have a zero which isn't a vertical asymptote if g has the same zero.

    A vertical asymptote is, by definition, when x->infinity or -infinity and g/h -> constant. Thus a max of only 2 is possible.

    Hence the answer is that 0, 1 or 2 vertical asymptotes are possible and an arbitrary (depending on the specific h(x)) but finite (since h is polynomial) number of vertical asymptotes are possible.
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    you mentioned vertical in both cases

    but surely you can add as many factors to the polynomial as you want

    so their is no upper limit on the number you can add
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    (Original post by thedemon13666)
    you mentioned vertical in both cases

    but surely you can add as many factors to the polynomial as you want

    so their is no upper limit on the number you can add
    Correct. That doesn't mean it's an "infinite" amount, it must be some finite amount. But it's an arbitrarily large amount, meaning it can have as many as you want.
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    thats what i meant haha
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    (Original post by thedemon13666)
    thats what i meant haha
    Yeah, course. I think we all knew what you meant, just that "it can have infinitely many" is wrong, because infinitely many is the only amount it can't have.
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    Surely Tan(x) has an infinite amount?
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    ...
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    (Original post by HPSH)
    Also, for example (forgive me - I've not learnt Latex)

    F(x)=(x-1)/(product between n=2 & n=infinity)(x-n)

    So we can have an infinite amount in the polynomial case
    Sorry, but \displaystyle \prod_{n=2}^\infty (x-n) is not normally considered a polynomial.
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    (Original post by HPSH)
    Surely Tan(x) has an infinite amount?
    Yep, but that's not a polynomial.
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    (Original post by thedemon13666)
    you mentioned vertical in both cases

    but surely you can add as many factors to the polynomial as you want

    so their is no upper limit on the number you can add
    The 0, 1 or 2 case is horizontal. Sorry, I have a tendency to be thinking a bit ahead or behind what my fingers are typing and end up typing something completely different.

    As for upper limits, not there's no upper limit but it must be finite.
 
 
 

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