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Math Help - vertical asymptote question

  1. #1
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    vertical asymptote question

    for f(x)=(6-3x-3x^2)/(x^2-1)

    I got vertical asymptotes at x=1 and x= -1 but the case where x=1 is wrong

    yet (x^2-1)=0 factors to (x+1)(x-1)=0 which implies that x^2-1)=0 is zero for both x+1 and x-1

    Where is my error?

    thanks!
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: vertical asymptote question

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

    So, your original function has a vertical asymptote at x=-1, but a hole in the graph at x=1.
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