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Math Help - Oblique asymptote

  1. #1
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    Oblique asymptote

    Not sure if this is the correct section, but in the following function, \displaystyle f(x)=\frac{2x^5-3x^2+5}{3x^4+5x-4}, is the oblique asymptote \displaystyle y=\frac{2}{3}x? I ask because the answer in the book is different, but I can't see my mistake.
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  2. #2
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    Quote Originally Posted by youngb11 View Post
    Not sure if this is the correct section, but in the following function, \displaystyle f(x)=\frac{2x^5-3x^2+5}{3x^4+5x-4}, is the oblique asymptote \displaystyle y=\frac{2}{3}x? I ask because the answer in the book is different, but I can't see my mistake.
    divide every term by x^4 ...

    f(x) = \dfrac{2x - \frac{3}{x^2} + \frac{5}{x^4}}{3 + \frac{5}{x^3} - \frac{4}{x^4}}

    looks like y = \dfrac{2x}{3} to me
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  3. #3
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    Looks that way to me as well.

    -Dan
    Attached Thumbnails Attached Thumbnails Oblique asymptote-asymptote.jpg  
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  4. #4
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    Even a "long division" gives
    \frac{2x^5- 3x^2+ 5}{3x^4+ 5x- 4}= \frac{2}{3}x- \frac{\frac{13}{3}x^2- \frac{8}{3}x- 5}{3x^4+ 5x- 4}
    so the oblique asymptote is, indeed, y= \frac{2}{3}xj.
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