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

 

[Fonz]

oops


y=1/2x2+2x+2

Looks simple enough.

First make n equation with just the Vertical asymptotes.
Forget the horizontal ones.


X cannot be smaller/equal to 1 or bigger than/equal to 5

So,

Y= 1/((X-1)(X-5))

Y= 1/(x2-6x+5)

NOW, put in the horizontal asymptotes.

We know Y cannot be greater than/equal to 2.

So, 2(x2-6x+5)=1
2x2-12x+10=1

So,


2x2-12x+9=0

Roots are:

12+-(root of)(144-4(2)(9))/ 2(2)

= 12+-(root of)(72)/4

= 12+-(root of 72)/4
= 12+-2exp3*3exp2/4
=12+-6(root of(2))/4

=1/2 * 6+-3root(2)

So, x=3+3root(2) or X=3-3root(2)

So, X=7.2 or X=-1.2

Since the roots are less than X=1 and greater than X=5,
the equation remains the same as before.

Y=1/(X-5)(X-1)

or

Y=1/(x2-6x+5)

Godspeed

You could use Newton-Ralphson for faster root approximation
but not really necessary.

 

[bigmag]

you're also wrong....to give it a horizontal asymptote, there has to be the same coefficient in the numerator as the denominator, and the numerator would also have to have a 2 infront of the variable to give it a horizantal asymptote of 2. Then for the denominator xhas to equal 1 and 5, so it would have to be a linear equation. so it would be like this, if there's no intercepts

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

 

[May1010]

Why don't you just ask a fat girl?

 

[Fonz]

you're also wrong....to give it a horizontal asymptote, there has to be the same coefficient in the numerator as the denominator, and the numerator would also have to have a 2 infront of the variable to give it a horizantal asymptote of 2. Then for the denominator xhas to equal 1 and 5, so it would have to be a linear equation. so it would be like this, if there's no intercepts

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

I don't think you fully understand what I did.

The function f(x) does not go above Y=2 for the
Range X=1 to X=5 so the Horizontal Y-asymptote becomes
irrelevant.

Its only when X becomes smaller than X=1 and greater than
X=5 that the Y-Value becomes greater than 2 and
nullifies the condition hat Ycannot be greater than 2.

Since the actual function f(x) has parameters that set it
in 1<X<5, my equation is perfectly justified since it doesn't violate
the horizontal Y=2 asymptototal premise.

If X had no vertical asymptotes at X=1 and X=5, you'd be right.

Godspeed

 

[-SD-]

It should be a variation of 1/X^2. The only info Ive been given is the asymptotes.

 

[guards]

I tried......flaky psych/law major here

 

[Juicer Jr.]

Unless I forgot ALL my calculus...

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

Vertical assymptotes at 1 & 5, horizontals at 2