# Thread: Please please help me out its urgent !!!!!!!!!!!!!!!!

1. ## 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

2. Originally Posted by kedhar
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.

3. oooooooooh thank you very much !!

4. and also how do u determine where it is increasing or decreasing?

5. Originally Posted by kedhar
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