I have to create a rational function with the characteristics:
3 real zeros(1 of them of multiplicity 2).
y-intercept 1.
vertical asymptote at $x=-2$ and $x=3$.
oblique asymptote $y=2x+1$.
I've tried with various functions but the y-intercept and the oblique asymptote characteristics seem to be impossible to satisfy at the same time.
I had something like:
$$\frac{2(x-a)^2(x-b)}{(x+2)(x-3)}$$
and i tried solving for a and b trying to satisfy the y intercept but the solution isn't rational.
Any help would be great!
Create rational function
algebra-precalculuscalculusrational-functions
Best Answer
You almost did all the work (may be you forgot to add the conditions $(a\neq -2, a\neq 3 , b\neq -2, b\neq 3) $.
Now, using the formula you proposed, perform the long division to get $$y=\frac{2(x-a)^2(x-b)}{(x+2)(x-3)}=2 x-(4 a+2 b-2)+O\left(\frac{1}{x}\right)$$ which makes $$4 a+2 b-2=-1 \implies b=\frac{1}{2}-2 a$$
So $$y=\frac{2(x-a)^2 \left(x+2a-\frac{1}{2}\right)}{(x-3) (x+2)}$$ If $x=0$ then $$\frac{1}{6} \left(a^2-4 a^3\right)=1$$ The cubic equation in $a$ shows only one real root which is $\approx -1.06715 $.
If you want to look fancy, using the hyperbolic solution to get $$a=\frac{1}{12}-\frac{1}{6} \cosh \left(\frac{1}{3} \cosh ^{-1}(1295)\right)$$