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Math Help - asymptotes and discontinuitites

  1. #1
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    asymptotes and discontinuitites

    (x^2 +5x+4)/(x^3 +x^2 -9x -9)

    just trying to find the vertical and horizontal asymptotes of this function...

    hortizontal at 0.

    the vertical asymptotes at -1, 3 and -3..just wondering if this is correct..

    and how do i go on about finding where the discontinuities are in this function by looking at the graph...how would i work it out??

    thanks
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  2. #2
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    Quote Originally Posted by b00yeah05 View Post
    (x^2 +5x+4)/(x^3 +x^2 -9x -9)

    just trying to find the vertical and horizontal asymptotes of this function...

    hortizontal at 0. Mr F says: Strictly speaking, y = 0.

    the vertical asymptotes at -1, 3 and -3.. Mr F says: Strictly speaking, x = -1, x = 3 and x = -3.

    just wondering if this is correct.. Mr F says: Now you know.

    and how do i go on about finding where the discontinuities are in this function by looking at the graph...how would i work it out??

    thanks
    Imagine trying to run a pen along the graph. Every time you have to lift your pen off the page, you've met a discontinuity. Clearly the function is discontinuous at the vertical asymptotes, that is, at x = -1, 3 and -3.

    I spoke too soon. There's a mistake in the above .... See my next reply.
    Last edited by mr fantastic; February 11th 2008 at 02:25 AM. Reason: There's a mistake in this reply - I wonder who can spot it >before< reading my next reply .....
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  3. #3
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    Quote Originally Posted by b00yeah05 View Post
    (x^2 +5x+4)/(x^3 +x^2 -9x -9)

    just trying to find the vertical and horizontal asymptotes of this function...

    hortizontal at 0. Mr F says: Yes, at y = 0.

    the vertical asymptotes at -1, 3 and -3..just wondering if this is correct.. Mr F says: NO! There's vertical asymptotes at x = -3 and x = 3. But NOT x = -1. See my main reply below.

    and how do i go on about finding where the discontinuities are in this function by looking at the graph...how would i work it out??

    thanks
    \frac{x^2 +5x+4}{x^3 +x^2 -9x -9} = \frac{(x + 1)(x + 4)}{(x + 1)(x + 3)(x - 3)} = \frac{x+4}{(x+3)(x - 3)} , provided x \neq -1 .....

    So the function is undefined at x = 3 and x = -3 => vertical asympotes. But the function does not exist (different from undefined) at x = -1. At x = -1 you have an open circle (to indicate the point is NOT included).

    The function is still discontinuous though at x = -3, x = 3 AND x = -1 ..... you still have to left your pen off the page at x = -1 ......
    Last edited by mr fantastic; February 11th 2008 at 02:31 AM. Reason: Replaced "not defined" with "does not exist"
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