as then
A horizontal asymptote could imply a constant term of +1 in
The points will be used to solve any arbitrary constants.
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)?
Hello, PTL!
For vertical asymptotes, the denominator must go to zero forFind 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.
.