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Math Help - find the asymptote and holes in the graph

  1. #1
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    find the asymptote and holes in the graph

    This is my last item in my assignment and all I know is the vertical asymptote. Can you guys help me with this problem.

    Given f(x) = x^3 +8 / x +2

    Find the asymptotes and holes of the graph. Find the x and Y axis.

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by cjru View Post
    This is my last item in my assignment and all I know is the vertical asymptote. Can you guys help me with this problem.

    Given f(x) = x^3 +8 / x +2

    Find the asymptotes and holes of the graph. Find the x and Y axis.

    Thanks in advance.
    I assume that you mean:

    f(x) = \dfrac{x^3+8}{x+2}

    If so:
    1. Use brackets
    2. Re-write the equation of the function:

    f(x) = \dfrac{(x+2)(x^2-2x+4)}{x+2}

    Since you can cancel out the factor (x + 2) the graph of the function is a parabola with a hole at H(-2, 12).
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  3. #3
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    How did you get (-2, 12)? So what is the indication a hole if the graph? Is it the cancelation of the numerator and denominator?

    Thanks for the help.
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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by cjru View Post
    How did you get (-2, 12)? So what is the indication a hole if the graph? Is it the cancelation of the numerator and denominator?

    Thanks for the help.
    The rational function in its original state is not defined at x=-2 because that would cause the denominator to be 0.

    After Earboth simplified the function to f(x)=x^2-2x+4, x=-2 is now ok. If it were still undefined at x=-2, you would have a vertical asymptote at x = -2. But that's not the case. Therefore, you have a hole in the curve where x=-2. The corresponding y coordinate is found by using your simplified function and seeing that f(-2) = 12. This function has 'point discontinuity' at (-2, 12)
    Attached Thumbnails Attached Thumbnails find the asymptote and holes in the graph-parabola.bmp  
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  5. #5
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    Thank you very much. Now I inderstand why it became like that.
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