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Math Help - help finding asymptotes and local extrema?

  1. #1
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    help finding asymptotes and local extrema?

    This question was on my extra credit test, and I'm having problems with parts of it:
    Graph the rational function below:

    Solve ALGEBRAICALLY X and Y intercepts, all asymptotes, and local extrema (min and max) points and any intersection point with slant asymptote. Find domain, range, intervals where the function is <0 and >0, increasing and decreasing using interval notation.



    I understand how to do most of this but not how to find the local extrema algebraically. I've tried completing the square and I can't quite get it to work.

    So far I've gotten that the y int is 1.5, there's no x int, the slant asymptote is y = x - 4, the vertical asymptote is 2, and there's no horizontal asymptote. I'm not sure if all that information is correct but it's what I've gotten so far, and now I'm sort of stuck on the max and min points.

    Thank you for any help with any of the aspects of this problem!
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  2. #2
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    I think you made some mistakes.
    Vertical should be at x = -2.

    So you have to find min and max algebraically, without using derivatives?
    Last edited by snowtea; December 30th 2010 at 11:40 AM.
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  3. #3
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    yes, no derivatives. we haven't even learned them yet.

    I'm almost positive the asymptote is x + 4, I checked it on my graphing calculator and x + 4 lined up perfectly with the slant asymptote in the graph of the function. I got x + 4 by doing long division and finding the quotient, which is the method my teacher taught us to use.
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  4. #4
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    Yes, you are correct about the slant asymptote (I've fixed that).

    f(x) = \frac{x^2 - 2x + 3}{x + 2} = x + \frac{-4x + 3}{x + 2} = x - 4 + \frac{11}{x + 2}

    Now for the local max and min, you can find that they are at x = 2 \pm\sqrt{11} with derivatives.
    I still cannot see a simple way algebraically. What techniques were you taught in class?

    Do you know how to find the max and min for functions like f(x) = x + a + \frac{b}{x}?
    If so, you can substitute x'=x+2 for your problem to simplify it into this form.
    Last edited by snowtea; December 30th 2010 at 12:26 PM.
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