I came across this problem:
Find a formula for the function $f(x)$ such that:
I can get any 4 out of the 5 criteria, but I cannot get the last one. I suspect that the function is something like $\displaystyle \frac{x^2}{x^2-16} + 1$.
Any tips or advice? Thank you.
The function you found is very close you just need a root on $3$ and on $-3$ (cause f is even). You can use the even polynomial $x^2-9$ for that. This will do the trick for $-4<x<4$. Now when you get $|x|>4$ you can use a different formula for $x$. Try to find a function $g(x)$ such $$ f(x)= \begin{cases} g(-x),x<-4\\\\ \displaystyle \frac{x^2(x^2-9)}{x^2-16},-4<x<4\\\\ g(x), x>4 \end{cases} $$ And $g(x)$ has to meet the requirment of having $y=2$ as a horizontal asymptote.
You had a good start. Let's write $$f(x)=\frac{P(x)}{Q(x)}$$ You noticed that since it has asymptotes at $\pm4$ you might want to write $Q(x)=x^2-4^2$. But keep in mind that's not the only solution. I can write $Q(x)=(x^2-16)^2$ or something like $(x^2-16)(x^2+1)$, in case I want to make it higher order. Why would I want to make it higher order? Because the problem says you have roots at $0$ and $3$, and since the function is even, you have at $-3$ as well. So the numerator is at least 3rd degree polynomial. 4th degree is easier, to keep it even. So I can write $$P(x)=x^2(x^2-9)$$ and $$Q(x)=(x^2-16)^2$$ I want to have both the same degree, so I have horizontal asymptote at $y\ne 0$. To make it at $y=2$, you just multiply everything by $2$: $$f(x)=\frac{2x^2(x^2-9)}{(x^2-16)^2}$$
