Math bros, quick question

[-SD-]

I need to write an equation for a function that has verticle asymptotes at X=1 and X=5 and has a horizontal asymptote at Y=2. i know it is a variation of the 1/X function with an x squared thrown in. Thanks

 

[bigmag]

I could help you out....but I'd half to know the x-intercept and the y-intercept.
pm me if you want me to help!!!

 

[Dr. E]

something like this:

y = (x-2)/(x-1)(x-5)

 

[The Varnsen]

That has to be a parabole (sp?)...

 

[bigmag]

no....that wouldn't work....then you would have a horizantal asymptote.....from the info you gave me, I'd say it would be something like this.....

y=(2x-squared+1)/x-squared+6x+5)

but like I said....you'll have to tell me the x-intercept and y-intercept

 

[The Varnsen]

Hey man can´t you read? It hasn´t intercepts!!! Only if one assimptote was negative and the other positive... The assimptotes are all positive so the equation only takes place in the positive area... x positive and y positive...

 

[The Varnsen]

It has to have some sort of fractions in it... But i dunno, i have stoped working with those fuckers sometime ago...

 

[guards]

y=x2+2x+2

 

[The Varnsen]

hey guards that hasn´t a fraction...

 

[guards]

oops


y=1/2x2+2x+2