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Math Help - rational function need help: asymptotes, intercepts, holes

  1. #1
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    rational function need help: asymptotes, intercepts, holes

    Please help me with this problem

    f(x)= 3x^2+3x-6 / x^2-6x+5

    1. Determine if the graph of f(x) has any holes.
    2. Find all asymptotes for f(x).
    3. Find atleast one point between any verticle asymptote(s) and x-intercept(s).

    Thank u so much for any help.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tpwkbs View Post
    Please help me with this problem

    f(x)= 3x^2+3x-6 / x^2-6x+5

    1. Determine if the graph of f(x) has any holes.
    find the zeros of the numerator and the denominator. if you have a zero that is the same for both, it is a hole

    2. Find all asymptotes for f(x).
    these occur at x-values (that are not holes) where the denominator is zero.

    3. Find atleast one point between any verticle asymptote(s) and x-intercept(s).
    answer part 2 and you will be able to answer this
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  3. #3
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    Quote Originally Posted by tpwkbs View Post
    Please help me with this problem

    f(x)= 3x^2+3x-6 / x^2-6x+5

    1. Determine if the graph of f(x) has any holes.
    2. Find all asymptotes for f(x).
    3. Find atleast one point between any verticle asymptote(s) and x-intercept(s).

    Thank u so much for any help.
    Factor the numerator and the denominator of f:

    f(x)=\frac{3x^3+3x-6}{x^2-6x+5}= \frac{3(x-1)(x+2)}{(x-1)(x-5)}

    to #1.: The graph is punctured(?) at x = 1

    to #2.: Since \lim_{|x|\mapsto \infty}f(x) = 3 there exist a horizontal asymptote y = 3

    Since f(x) is not defined at x = 5 there exist a vertical asymptote x = 5

    to #3.: I don't understand your question - but because the x-intercept is at x = -2 maybe you mean a point like (0, 0) ?
    Last edited by earboth; April 21st 2008 at 10:03 PM.
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  4. #4
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    thank you

    thank you
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