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Math Help - Graphing help

  1. #1
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    Graphing help

    so for math, my teacher asked us to graph it, so i was given this
    f(x) = X^2 - 1/ x - 1

    f(x)=X+1 / x^2 -2x- 3

    f(x) = x^2 + 2x / x + 1

    and i was asked to sketch it with the vertical aysmtopes, and x-int and y -int. so i know most of it , but i dunno for some of these equation do u always factor it first, or do u do long divison first. Can u show me a example of through long divison and obtaining the slant aysmtopes. I really dont understand some of these. THANKSS!!
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  2. #2
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    Solution

    The first one you can reduce because you factor the numerator to get (x + 1) * (x - 1). You cancel out the (x -1) from the numerator and denominator to get f(x) = x + 1. The graph of this is the same as f(x) = x, but with the y intercept at y = 1 instead of the origin.
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  3. #3
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    Solution part 2

    For the second equation, you can factor the denominator to get (x+1) * (x-3). Then you can cancel the (x+1) factor to get f(x) = 1/(x-3). The y intercept would be at y = -1/3. The vertical asymptote would be at x = 3. If you substitute some x values into the f(x) equation you could find the trend of the graph. As x approaches 3 from the left, f(x) approaches negative infinity. As x approaches 3 from the right, f(x) approaches positive infinity.
    Last edited by ajj86; October 23rd 2008 at 08:58 PM. Reason: typos
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  4. #4
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    Number 3

    For the last one you can't cancel anything so you have to factor the numerator to get f(x) = x(x+2)/x+1. The vertical asymptote is at x = -1 and the graph intersects at x = -2 and x = 0. As x approaches -1 from the left, f(x) approaches positive infinity. As x approaches -1 from the right, f(x) approaches negative infinity. Again, if you plot some test points by substituting x into the f(x) equation, you can find the trend of the graph. I hope this is helpful.
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