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Math Help - asymptote question

  1. #1
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    asymptote question

    y=\frac{(x^2-4)}{(x+2)(x-3)}

    Find all vertical, horizontal, and oblique asymptotes, and state if the function has any holes.

    My first step was to simplify:

    y=\frac{(x-2)(x+2)}{(x+2)(x-3)}

    and cancel

    y=\frac{(x-2)}{(x-3)}

    My answer is:
    Vertical asymptote at x=3. Horizontal asymptote at y=1. I don't think there is an oblique asymptote, because the degree of the numerator and denominator are equal. I believe there is a hole at -2,\frac{4}{5}

    Is this correct? I guess my main question is, because I cancelled, does that eliminate the vertical asymptote x=-2 and replace it with the hole mentioned above?
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  2. #2
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    Quote Originally Posted by absvalue View Post
    y=\frac{(x^2-4)}{(x+2)(x-3)}

    Find all vertical, horizontal, and oblique asymptotes, and state if the function has any holes.

    My first step was to simplify:

    y=\frac{(x-2)(x+2)}{(x+2)(x-3)}

    and cancel

    y=\frac{(x-2)}{(x-3)}

    My answer is:
    Vertical asymptote at x=3. Horizontal asymptote at y=1. I don't think there is an oblique asymptote, because the degree of the numerator and denominator are equal. I believe there is a hole at -2,\frac{4}{5}

    Is this correct? I guess my main question is, because I cancelled, does that eliminate the vertical asymptote x=-2 and replace it with the hole mentioned above?
    Your work is OK. However, I'd format the coordinates of the hole as \left( -2,\frac{4}{5}\right).
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  3. #3
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     y=\frac{(x-2)}{(x-3)}=\frac{1}{(x-3)}+1

    Therefore asymptotes are at x=3 and y = 1
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  4. #4
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    Thanks!
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