Can't find the way to solve this problem related to limits.

Question

I've been trying for a long while, and I can't find how to methodically (not guessing) solve this problem:

Find $a$, $b$, $c$ and $d$ for a function $f(x)=\frac{ax+b}{cx+d}$ such that

• $f(2) = 0$
• There is a vertical asymptote as $x \rightarrow 3$
• There is a horizontal asymptote at $y=-2$

I've tried with a system of equations, but I end up looking for four (or five, with $x$) variables to find, and only three equations, four equations if I consider $\lim_{x\to\infty}$ as two different $\lim_{x\to+\infty}$ and $\lim_{x\to-\infty}$, all of which seems quite absurd and has led me virtually nowhere.

The only (stupidly basic) idea I have is that $3c+d = 2a + b = 0$ and that $ax+b=-2(cx+d)$.

Thanks in advance.

Answer 1

To create a vertical asymptote x=k you would insert a factor of (x-k) in the den. w/o a common factor in the num[which would create a "hole"]. For a horizontal asymptote you wish the ratio of the highest powers of x in num & den to be j for y=j to be your horizontal asymptote.

So for(ax+b)/(cx+d) let a=-2; b=0;c=1;d=-3. If you feel it bad form to have a negative 1st term in the num, change all the signs.

Note that going to right or left, the function will approach the same asymptote.

Answer 2

You're searching for a homographic function, so

• there's a vertical asymptote at the root of the denominator. If this vertical asymptote has equation $x=3$, this means $\;3c+d=0$.
• the equation of the horizontal asymptote is the ratio of the leading coefficients $a/c$, so $a=-2c$.
• $f(2)=0$ means $2a+b=0$.

Can you proceed? (remember the coefficients are defined up to a non-zero factor).