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Math Help - What values of a makes f(x)= ax^2 + (a+1)/x have a horizontal asymptote when x->+/-∞?

  1. #1
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    What values of a makes f(x)= ax^2 + (a+1)/x have a horizontal asymptote when x->+/-∞?

    If you could explain how to do this I would be SOOOO grateful! Thanks!!!

    I have no idea where to start...
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  2. #2
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    Quote Originally Posted by natashabu View Post
    If you could explain how to do this I would be SOOOO grateful! Thanks!!!

    I have no idea where to start...
    Is it f(x) = ax^2 + \frac{a+1}{x} ? Then you require a+1 \neq 0 \, ....
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    Quote Originally Posted by mr fantastic View Post
    Is it f(x) = ax^2 + \frac{a+1}{x} ? Then you require a+1 \neq 0 \, ....

    Thanks I understand now (after thinking about what you wrote)
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  4. #4
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    Quote Originally Posted by steven_smith View Post
    F(x)=ax^2+(a+1)/x

    First you need to find your critical points to do this. take the dervitive of that function and set it = to 0.

    Can't remember the quotent rule right now i think it's if F(x)=a/b then f'(s) = (b')a-(a')b/(b^2)

    f'(x)= 2ax .....= 0
    then just solve for all values a and x that make it = 0 that will tell you where you'r critical points are. put those in the equation and see what it does.
    When answering a question from a poster on Math Help Forum it is generaly considered good form to attempt to answer that question not some other question. You will also find this a usefull exam technique in its own right.

    CB
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