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Math Help - Curve Sketching

  1. #1
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    Curve Sketching

    Given f(x)=(x^2-x-2)/(x^2+x), find the vertical and horizontal asymptotes and sketch the graph near the asymptotes.

    VA
    x=0 or x=-1

    HA
    as x -> +-infinity, y->1

    So that is my work above and I graphed this all too. But.... my question comes here.
    When I graph the function f(x) using software, there appears to be no vertical asmptote of x=-1, only x=0 is a VA... This is really strange because I do believe there are two asymptotes..

    to find asymptote, set denominator =0 and solve for x.
    x^2 + x = 0
    x(x+1)=0
    x=0 or x=-1
    so?? what did I do wrong now?

    Note: I did not type the eqtn of the function wrong into graphing software.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by skeske1234 View Post
    Given f(x)=(x^2-x-2)/(x^2+x), find the vertical and horizontal asymptotes and sketch the graph near the asymptotes.

    VA
    x=0 or x=-1

    HA
    as x -> +-infinity, y->1

    So that is my work above and I graphed this all too. But.... my question comes here.
    When I graph the function f(x) using software, there appears to be no vertical asmptote of x=-1, only x=0 is a VA... This is really strange because I do believe there are two asymptotes..

    to find asymptote, set denominator =0 and solve for x.
    x^2 + x = 0
    x(x+1)=0
    x=0 or x=-1
    so?? what did I do wrong now?

    Note: I did not type the eqtn of the function wrong into graphing software.
    You did nothing "wrong". You overlooked something...

    Note that \frac{x^2-x-2}{x^2+x}=\frac{\left(x-2\right){\color{red}\left(x+1\right)}}{x{\color{re  d}\left(x+1\right)}}

    Since the term x+1 appears in both the numerator and the denominator, there is a hole in the graph at x=-1, not a vertical asymptote.
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