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  1. #1
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    graph the function...

    graph the function. label all intercepts and asymptotes.

    y = x-2 / (x+1)(x-3)

    i got for the intercepts: y= 2/3
    x= -2

    y-asymptote = 2

    how will this graph look?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ohhhh View Post
    graph the function. label all intercepts and asymptotes.

    y = x-2 / (x+1)(x-3)

    i got for the intercepts: y= 2/3
    x= -2

    y-asymptote = 2

    how will this graph look?
    Check your y asymptote. I get x = -1 and x = 3 as asymptotes.

    -Dan
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  3. #3
    Super Member wingless's Avatar
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    I think you tried to solve x-2 (numerator) to find the horizontal asymptote. No, horizontal asymptotes are not found that way. You have to solve
    \lim_{\text{    }x\to \infty } \frac{x-2}{(x+1)(x-3)}
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  4. #4
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    yeah i did do the numerator.. woops. so, asymptotes at -1 and 3? my main concern is how this graph will look. i know it'll resemble the graph of 1/x .. what else do i need to do to figure our how the graph will look?
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  5. #5
    Super Member wingless's Avatar
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    \lim_{\text{    }x\to \mp \infty } \frac{x-2}{(x+1)(x-3)}

    \lim_{\text{    }x\to \mp \infty } \frac{x-2}{x^2 - 2x -3}

    To calculate this limit, we divide both numerator and denominator by x^n, n is the highest power of x in the denominator. In x^2 - 2x -3, the highest power of x is 2. So n = 2. We'll divide both numerator and denominator by x^2.

    \lim_{\text{    }x\to \mp \infty } \frac{\frac{x-2}{x^2}}{\frac{x^2 - 2x -3}{x^2}}

    \lim_{\text{    }x\to \mp \infty } \frac{\frac{1}{x}-\frac{2}{x^2}}{1 - \frac{2}{x} - \frac{3}{x^2}}

    When x goes to \infty, the fractions \frac{1}{x}, \frac{1}{x^2}, \frac{2}{x}, \frac{3}{x^2}... etc will approach 0.

    So, the limit becomes \frac{0}{1} = 0. That means the horizontal asymptote is the line y=0, which is also called the x axis.
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