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Math Help - Please please help me out its urgent !!!!!!!!!!!!!!!!

  1. #1
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    Exclamation Please please help me out its urgent !!!!!!!!!!!!!!!!

    Hi guys
    My rational funtcion is f(x)= 3x/(x^2-x-6)
    Now for this I got the horizontal asymptote as y=0 because the degree of the denominator is greater than that of the numerator ... but how come the x intercept is 0 so it has to pass through zero ... and here is the link for the graph can anyone please help me out here ??

    Link for graph - http://my.hrw.com/math06_07/nsmedia/tool...

    Thank you very much
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  2. #2
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    Quote Originally Posted by kedhar View Post
    Hi guys
    My rational funtcion is f(x)= 3x/(x^2-x-6)
    Now for this I got the horizontal asymptote as y=0 because the degree of the denominator is greater than that of the numerator ... but how come the x intercept is 0 so it has to pass through zero ... and here is the link for the graph can anyone please help me out here ??

    Link for graph - http://my.hrw.com/math06_07/nsmedia/tool...

    Thank you very much
    Brace yourself for this ......

    Graphs are allowed to cross their horizontal asymptote. It is only vertical asymptotes that graphs can never cross. If you weren't taught this then you were probably the victim of lazy teaching.
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    oooooooooh thank you very much !!
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  4. #4
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    and also how do u determine where it is increasing or decreasing?
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  5. #5
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    Quote Originally Posted by kedhar View Post
    Hi guys
    My rational funtcion is f(x)= 3x/(x^2-x-6)
    Now for this I got the horizontal asymptote as y=0 because the degree of the denominator is greater than that of the numerator ... but how come the x intercept is 0 so it has to pass through zero ... and here is the link for the graph can anyone please help me out here ??

    Link for graph - http://my.hrw.com/math06_07/nsmedia/tool...

    Thank you very much
    1.

    f(x)=\dfrac{3x}{x^2-x-6} = \dfrac{3x}{(x+2)(x-3)}

    2. The vertical asymptotes are at x = -2 and x = 3

    3. The non-vertical asymptote can be calculated as:

    as: y = \lim_{|x|\to \infty}(f(x)) = 0

    That means the asymptote is an approximate function of f for very large values of x
    Attached Thumbnails Attached Thumbnails Please please help me out its urgent !!!!!!!!!!!!!!!!-asympt_horvert.png  
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