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Math Help - Asymptotes of Rational Functions

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    Asymptotes of Rational Functions

    Find the vertical, horizontal, and oblique asymptotes, if any, of each rational function. Do not graph.

    [1] R(x) = (6x^2 + x + 12)/(3x^2 - 5x - 2)

    [2] F(x) = (x - 1)/(x - x^3)
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  2. #2
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    Quote Originally Posted by fdrhs View Post
    [SIZE="3"]Find the vertical, horizontal, and oblique asymptotes, if any, of each rational function. Do not graph.

    [1] R(x) = (6x^2 + x + 12)/(3x^2 - 5x - 2)
    You need to synthetically divide.
    To get,
    2+\frac{11x+12}{3x^2-5x-2}
    The horizontal asymptotes are the solutions to,
    3x^2-5x-2=0

    And the vertical asymptotes are y=2.
    Because the second part,
    \frac{11x+12}{3x^2-5x-2}\approx 0 for large x.

    There are no slant asymptotes.
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    fdrhs

    How did you get y = 2 as the vertical asymptote?

    Also, I don't understand the second part.
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    Quote Originally Posted by fdrhs View Post
    Find the vertical, horizontal, and oblique asymptotes, if any, of each rational function. Do not graph.

    [1] R(x) = (6x^2 + x + 12)/(3x^2 - 5x - 2)...
    Hello,

    1. To me the vertical asymptotes are perpendicular to the x-axis. You get them if you calculate the zeros of the denominator:

    3x^2-5x-2=0 \Longleftrightarrow x=\frac{-(-5) \pm \sqrt{(-5)^2-4 \cdot 3 \cdot (-2)}}{2 \cdot 3}

    x=\frac{5 \pm 7}{6} \Longleftrightarrow x = 2\ \vee \ x=-\frac{1}{3}

    The vertical asymptotes are x = 2 and x=-\frac{1}{3}


    2. To get the horizontal or oblique asymptotes do long division:

    Code:
     (6x^2 +   x + 12)  (3x^2 - 5x - 2) = 2 
    -(6x^2 - 10x -  4)
    ----------------
             11x + 16    <---- remainder
    Therefore the equation of your function becomes:

    R(x) = \frac{6x^2 + x + 12}{3x^2 - 5x - 2}=2 + \frac{11x+16}{3x^2 - 5x - 2}

    Now calculate the limit if x approaches positive or negative infinity. You see that the fraction approaches zero in both cases. Thus the asymptote is horizantal and has the equation: y = 2

    EB

    PS.: Even though you are not allowed to graph the function it isn't such a bad idea to control the calculations by a sketch. But don't show it around!
    Last edited by earboth; February 1st 2007 at 04:00 AM.
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    Quote Originally Posted by fdrhs View Post
    Find the vertical, horizontal, and oblique asymptotes, if any, of each rational function. Do not graph.
    ...
    [2] F(x) = (x - 1)/(x - x^3)
    Hello,

    first factor the denominator:

    F(x)=\frac{x-1}{x-x^3}=\frac{x-1}{x(1-x)(1+x)}=\frac{-1}{x(x+1)}\ \wedge \ x \neq 1

    Thus you get 2 vertical asymptotes at x = 0 and x = -1 and a hole at x = 1 that means the graph is punctuated(?).

    Because the degree of the nummerator is smaller than the degree of the denominator you get the x-axis as the only horizontal asymptote: y = 0.

    EB
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