Find polynomials p(x) and q(x) such that f(x) = p(x)/q(x) has vertical asymptotes x=-2, x=5, horizontal asymptote y=1 and f(-1) = f(4)=0.

Any hints on how to do this?

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- Aug 2nd 2009, 02:31 PMPTL[SOLVED] Finding polynomials from a diagram...
Find polynomials p(x) and q(x) such that f(x) = p(x)/q(x) has vertical asymptotes x=-2, x=5, horizontal asymptote y=1 and f(-1) = f(4)=0.

Any hints on how to do this? - Aug 2nd 2009, 02:50 PMpickslides
as then

A horizontal asymptote could imply a constant term of +1 in

The points will be used to solve any arbitrary constants. - Aug 2nd 2009, 03:48 PMPTL
Does that mean we have f(x) =

So p(-1) = 0

p(4) = 0

and p(x)/q(x) has a local max at (1.5, almost(1))?

So the derivative of at 1.5 = 0? (1.5 because it's half-way between -2 and 5..

Is that how I'm supposed to solve for p(x)? - Aug 2nd 2009, 03:56 PMpickslides
There's a lot of Latex errors there my friend.

- Aug 3rd 2009, 02:18 PMPTL
- Aug 3rd 2009, 03:19 PMDeMath
- Aug 3rd 2009, 03:26 PMSoroban
Hello, PTL!

Quote:

Find polynomials and such that: . has vertical asymptotes ,

horizontal asymptote and

. . Hence: .

So we have: .

For the horizontal asymptote, the function goes to 1 as

. . Hence, has the same degree as , and has a leading coefficient of 1.

That is: .

And we have: .

Since .[1]

Since .[2]

Add [1] and [2]: .

Therefore: .

Edit: Ah ... DeMath beat me to it.

Oh well, my explanation is*slightly*different.

. - Aug 3rd 2009, 03:50 PMDeMath
Also see picture

http://i066.radikal.ru/0908/73/fcc543c1795a.png