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Math Help - Finding Verticle asymptotes, holes and, and horizontal asymptotes

  1. #1
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    Finding Verticle asymptotes, holes and, and horizontal asymptotes

    im having a really difficult time finding out how to find the verticle asymptote and holes and horizontal asymptote. its really frustrating

    for exmaple here are some questions:









    thank you in advance
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  2. #2
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    Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes

    let's start with something simple to see what you know ...

    y = \frac{(x-1)(x+2)(x+3)}{2(x+1)(x+2)(x-3)}

    where are the vertical asymptotes?

    what is the horizontal asymptote?

    where is the point discontinuity (hole) ?
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    Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes

    out of my little understanding of it (my instructor sucks)
    i can say that
    horizontal asymptote at 1/2
    vertical asymptotes at -1,-2, and 3
    holes at........i dont even know. LOL
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    Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes

    Quote Originally Posted by Recyclop View Post
    out of my little understanding of it (my instructor sucks)
    i can say that
    horizontal asymptote at 1/2
    vertical asymptotes at -1,-2, and 3
    holes at........i dont even know. LOL
    horizontal asymptote at y = 1/2 ... it's the equation of a line

    vertical asymptotes at x = -1 and x = 3 , "hole" at x = -2 ... note the common factor in the numerator and denominator.

    OK ... using one of your problems

    g(x) = \frac{6x^2}{x^4 - 4x^2} = \frac{6x^2}{x^2(x^2 - 4)} = \frac{6x^2}{x^2(x-2)(x+2)}

    horizontal asymptote at y = ?

    two vertical asymptotes ... where?

    one "hole" ... where?
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    Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes

    so horizontal asymptote @ y=6
    verticle asymptotes at x=+2 and x=-2
    the hole would be at.....X^2 would be the only common factor i see. but that would be sqrt0.... so there is no hole?
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    Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes

    horizontal asymptote is y = 0 ... the degree of the numerator < degree of the denominator

    vertical asymptotes correct

    hole at x = 0
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    Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes

    so after class today i got a good understanding of asymptotes and what not.
    only thing that still doesnt make sense is holes. i understand that a hole occurs when you have common factors but what do i do to find the point and what does a hole mean?
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    Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes

    A "hole" is really a point discontinuity ... the discontinuity occurs at the x-value that makes the common factor in the numerator and denominator equal to zero.

    for example, the function y = \frac{x^2-1}{x-1} = \frac{(x+1)(x-1)}{x-1} has a "hole" at x = 1. the graph of the function will look like the line y = x+1 except for the "hole" at x = 1 ...

    note the graph made on my TI calculator ... see the "hole" ?
    Attached Thumbnails Attached Thumbnails Finding Verticle asymptotes, holes and, and horizontal asymptotes-graph1.gif  
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    Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes

    ya i see, so how do we determine the y value? is it just whatever number happens to be there if we plug in 1 for the complete function?
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    Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes

    sub in x = 1 after dividing out the common factor ... in this case, y = x+1 = 1 + 1 = 2 ... the "hole" is at (1,2).
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    Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes

    so if there are holes, are the common factors canceled first? then there would be no Vertical asymptotes?
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    Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes

    the function y = \frac{(x+1)(x-2)}{(x-3)(x+1)} has both a hole at one x-value and a vertical asymptote at another x-value ... where?
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    Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes

    hole at 1/2. but would 3 and -1 still be verticle asymptotes and would -1 still be a root?
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    Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes

    sorry, but no ...

    there is a hole at x = -1 (the value of x that makes the factor common to the numerator and denominator 0), the y-value of the hole is 3/4

    there is one vertical asymptote at x = 3
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    Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes

    ok i see now, thank you very much!
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