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Math Help - vertical asymptote help!!!!!

  1. #1
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    vertical asymptote help!!!!!

    the problem is:

    h(x)=x^2/x^2-3x+2

    i need to identity the vertical and horizontal asymptote as well as the domain

    so i got, horizontal asympt y=1
    vertical asymptote = 2,1
    and domain is all numbers except 2 and 1

    to obtain the vertical asymptote i set the denominator to equal zero, but apprently i did something wrong somewhere because the answer says that the vertical asymptote CANNOT be 2. is there another way to obtain the vertical asymptote? can i use this method for all problems?

    why can the vertical asymptote NOT be 2?

    thanks!

    thanks!
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  2. #2
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    Re: vertical asymptote help!!!!!

    Quote Originally Posted by noork85 View Post
    the problem is:

    h(x)=x^2/x^2-3x+2

    i need to identity the vertical and horizontal asymptote as well as the domain

    so i got, horizontal asympt y=1
    vertical asymptote = 2,1
    and domain is all numbers except 2 and 1

    to obtain the vertical asymptote i set the denominator to equal zero, but apprently i did something wrong somewhere because the answer says that the vertical asymptote CANNOT be 2. is there another way to obtain the vertical asymptote? can i use this method for all problems?

    why can the vertical asymptote NOT be 2?



    thanks!

    thanks!
    Since :

    \displaystyle \lim_{x \to \2^{-}}f(x)=-\infty ~\text{and}~ \displaystyle \lim_{x \to \2^{+}}f(x)=+\infty

    the line x=2 is a vertical asymptote .
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  3. #3
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    Re: vertical asymptote help!!!!!

    im confused....
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  4. #4
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    Re: vertical asymptote help!!!!!

    Quote Originally Posted by noork85 View Post
    im confused....
    Vertical asymptotes
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  5. #5
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    Re: vertical asymptote help!!!!!

    (x^2)/(x^2-3x+2) Vertical asymptotes occure where the function is undefined. A function is undefined when the denominator is zero. Where is the denominator zero? Factor the denominator (x-2)(x-1) and set them to zero. x=2 and x=1 Those are your vertical. For horizontal, the degree of the numerator and the degree of the denominator are equal, so long division on the leading expressions gives us y=1. Your horizontal. I hope that helps.
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  6. #6
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    Re: vertical asymptote help!!!!!

    x=1 and x=2 are vertical asymptotes. Go to 'drawing graphs' on Google to see the graph.
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  7. #7
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    Re: vertical asymptote help!!!!!

    Are you sure you have copied the problem correctly? x= 2 certainly is a vertical asymptote for x^2/(x^2- 3x+ 2).

    On the other hand, x= 2 is not a vertical asymptote for (x- 2)/(x^2- 3x+ 2) even though the denominator is 0 there. Do you see why?
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