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Math Help - Asymptote Question...

  1. #1
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    Unhappy Asymptote Question...

    I have found the slant asymptote of \frac{x^4-6x^3-11x^2+60x+100}{x^3-9x^2+27x-27} to be y=x+3. Somehow, on it's graph, the slant looks totally off. What is the problem with my math? The factored form that was given is \frac{(x-5)^2(x+2)^2}{(x-3)^4}. Thanks!
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  2. #2
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    Quote Originally Posted by nivek516 View Post
    I have found the slant asymptote of \frac{x^4-6x^3-11x^2+60x+100}{x^3-9x^2+27x-27} to be y=x+3. Somehow, on it's graph, the slant looks totally off. What is the problem with my math? The factored form that was given is \frac{(x-5)^2(x+2)^2}{(x-3)^{\color{red}4}}. Thanks!
    Do you mean (x-3)^3 = x^3 - 9x^2 + 27x-27? If so then your expansions are correct and your slant asymptote is correct. The graph you're looking at might be wrong .... Is it a graph you've drawn using technology? Check how you entered the equation.
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  3. #3
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    Hello nivek:

    Your math looks fine; the equation of the slant asymptote is y = x + 3.

    (There is a typographical error in your factored form; the exponent in the denominator should be 3.)

    What do you mean by saying that the slant on your graph is off?

    The slope is one; if the scales on both axes are the same, then the graph of y = x + 3 should make a 45-degree angle with the horizontal axis.

    Cheers,

    ~ Mark
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  4. #4
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    Yes, I typed in the wrong equation in my post. When I say the asymptote is off, I mean that the graph in quadrant 1 begins at around x=3 and avoids the asymptote. As x\rightarrow \infty, the graph starts to acknowlege the aymptote. Take a look. Could you explain why this happens?
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