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Math Help - Find Vertical Asymptote

  1. #1
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    Smile Find Vertical Asymptote

    Here is another review question placed on the board by my teacher.

    Find the vertical asymptote(s) of the function

    G(x) = (x^3 - 1)/(x - x^2)

    He said the only vertical asymptote is x = 0.

    However, I disagree.

    I say that x = 1 is another asymptote.

    HERE IS MY WORK:

    I factored the denominator.

    x - x^2 = x(1 - x).

    I then set each factor to 0 and solve for x.

    One factor is x = 0

    I also set 1 - x = 0 and got x = 1 as another asymptote.

    My teacher said that I am wrong but gave me credit for trying.

    Who is right and why?

    Thanks
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by magentarita View Post
    Here is another review question placed on the board by my teacher.

    Find the vertical asymptote(s) of the function

    G(x) = (x^3 - 1)/(x - x^2)

    He said the only vertical asymptote is x = 0.

    However, I disagree.

    I say that x = 1 is another asymptote.

    HERE IS MY WORK:

    I factored the denominator.

    x - x^2 = x(1 - x).

    I then set each factor to 0 and solve for x.

    One factor is x = 0

    I also set 1 - x = 0 and got x = 1 as another asymptote.

    My teacher said that I am wrong but gave me credit for trying.

    Who is right and why?

    Thanks
    Your teacher is correct. What you found was where the graph has point discontinuity. There's a "hole" in the graph where x=1.

    When you simplify the rational expression, one of the factors divides out. When that happens, it is not an asymptote.

    Point of Discontinuity
    Definition of Point of Discontinuity

    A function is said to have a point of discontinuity at x = a or the graph of the function has a hole at x = a, if the original function is undefined for x = a, whereas the related rational expression of the function in simplest form is defined for x = a.

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

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

    The above function in its original state is not defined for x=0 nor x=1.

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

    In the simplified form above, the function is still not defined for x = 0.

    Therefore, x = 0 is an asymptote. However, it is defined for x = 1.
    Plugging x=1 into the simplified function, we find f(1)=-3.
    Thus we can say that the function f(x) has a point of discontinuity at (1, -3).
    Last edited by masters; October 20th 2008 at 12:46 PM.
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  3. #3
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    Your Teacher is Right, because, the curve has Vertical Asymptote x = 0. It also has one Linear Oblique Asymptote, y = - x - 1

    f(x)=\frac{x^3-1}{x-x^2}=\frac{(x-1)(x^2+x+1)}{-x(x-1)}=\frac{x^2+x+1}{-x}, \;\;x\ne1

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

    So, Linear Oblique Asymptote,

    y = -x-1

    The value x = 1 is excluded from the graph, because it will make function undefined.

    See the attached graph. Green line x = 0 is vertical Asymptote VA(y-axis)
    Green dotted line (y = -x-1) is Linear Oblique Asymptote (LOA).
    Attached Thumbnails Attached Thumbnails Find Vertical Asymptote-graph8.jpg  
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