Help creating a rational function

Question

Create a rational function with vertical asymptotes $x=\pm1$ and oblique asymptote of $y=2x-3$ and a $y$-intercept of $4$.

Answer 1

You can go by parts, a la Jack.

First, you want your function to have vertical asymptotes at $x=1,-1$. So we really want $$\tag 1 \frac{1}{(1-x)(1+x)}$$

Now, as $x\to\infty$; this has a "oblique" asymptote of $0$, so we can just add the asymptote we want: $$\frac{1}{(1-x)(1+x)}+2x-3$$

Now, let's evaluate this at $x=0$. It gives $$1-3=-2$$. Since $(1)$ evaluates to $1$; we can multply it by the constant that will give us $4$, i.e. $7$. Then $$\frac{7}{(1-x)(1+x)}+2x-3$$

will evaluate to $4$ on the origin, and have the required conditions. You can try and produce more examples to get the hang of it.

Answer 2

For one family of rational functions $f(x)$ having vertical asymptotes only at $x=\pm 1$, consider those that can be written in the form $$f(x)=p_1(x)+\frac{p_2(x)}{(x+1)^m}+\frac{p_3(x)}{(x-1)^n}$$ for some positive integers $m,n,$ and some polynomials $p_1,p_2,p_3$, with $p_2,p_3$ in particular being non-$0$ polynomials of respective degrees $<m,$ $<n$. Such a function has a slant asymptote if and only if $p_1(x)$ is the line toward which the function tends asymptotically.

Hence, any rational functions $f(x)$ in our family having $y=2x-3$ for a slant asymptotes can be written in the form $$f(x)=2x-3+\frac{p_2(x)}{(x+1)^m}+\frac{p_3(x)}{(x-1)^n},$$ with $m,n,p_2,p_3$ as described above.

We're just looking for one such example, so to make it simpler on ourselves, we may as well assume $m=n=1$, so that $p_2,p_3$ must be constant polynomials, and our function will be $$f(x)=2x-3+\frac{A}{x+1}+\frac{B}{x-1}$$ for some non-zero constants $A,B$. All that remains is to choose non-zero constants $A,B$ so that $f(0)=4$ (i.e.: $f(x)$ has $y$-intercept $4$).