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Math Help - VA,HA & Slant Asymptotes

  1. #1
    dto
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    Thumbs up VA,HA & Slant Asymptotes

    I was wondering if this function had a vertical and/or horizontal and/or slant asymptotes?

    Function:

    f(x)= (x^3-2x+5)/(3x^2+4x-1)

    I know how to find the slant asymptotes with long division, but i can't seem to do it with this function.

    Please Help!

    Thanks
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  2. #2
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    Quote Originally Posted by dto View Post

    I know how to find the slant asymptotes with long division, but i can't seem to do it with this function.
    Use (x^3{\color{red}+0x^2}-2x+50)\div (3x^2+4x-1)

    There will plenty of fractions to keep you busy.


    Quote Originally Posted by dto View Post
    I was wondering if this function had a vertical and/or horizontal and/or slant asymptotes?

    Will have vertical asymptotes at 3x^2+4x-1=0

    You can solve this with the quadratic formula.
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  3. #3
    dto
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    so i think the answer for the slant is: y= 1/3x-4/9 R1/9x+4/9)/(1/3x-4/9)

    and is that suppose to be a 50 or a 5?

    Thanks
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  4. #4
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    Quote Originally Posted by dto View Post
    so i think the answer for the slant is: y= 1/3x-4/9 R1/9x+4/9)/(1/3x-4/9)

    and is that suppose to be a 50 or a 5?

    Thanks
    It's supposed to be a 5!
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  5. #5
    dto
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    so is the slant y=1/3x-4/9?
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  6. #6
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    Quote Originally Posted by dto View Post
    so is the slant y=1/3x-4/9?

    Thats what I get!
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  7. #7
    dto
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    and for the va, is it (-2+/- square root of 7)/ 3?

    i know it looks funky, but I don't know how to do the symbols.
    If it is like that, how do i graph the va?

    Thanks

    p.s. about the slant, sweet! what I did wrong was, I did not realize that it was going to be fractions! thanks alot!
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  8. #8
    dto
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    yet another question! 4x/x-2.... if the bottom is more then the top, that means there is no slant right?

    Correct me if I am wrong:
    HA= 4
    VA=2
    Slant= 0 or should it be none?

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