$$y=\frac{ax+b}{x+c}$$ find a,b and c given that there are asymptotes at $x=-1$ and $y=-2$ and the curve passes through (3,0)
I know that c=1 but I dont know how to find a and b?
I thought you expressed y in partial fraction so that you end up with something like $y=Ax+B+\frac{C}{x+D}$
You're on the right track; $$\frac{ax+b}{x+c}=\frac{ax+ac-ac+b}{x+c}=a+\frac{-ac+b}{x+c}$$ As $x\to \pm \infty$, the fraction disappears and you're left with $a$. Hence $a=-2$.
The line $x =-1$ is assymptote, in that case we must have:
$$\lim_{x\to -1}\frac{ax+b}{x+c}=\pm\infty$$
I've written $\pm$ because any of them is valid. This is just abuse of notation to make this a little faster. For this to happen we need a zero in the denominator at $x=-1$ since this is just a rational function. In that case we want $c=1$.
The line $y = -2$ is assymptote, so that we have:
$$\lim_{x\to +\infty}\frac{ax+b}{x+c}=-2$$
In that case, our strategy is to calculate the limit and equal it to this value. To compute the limit we do as follows:
$$\lim_{x\to +\infty}\frac{ax+b}{x+c}=\lim_{x\to\infty}\frac{a+\frac{b}{x}}{1+\frac{c}{x}}=a$$
This implies that $a = -2$. Finally, since the curve passes through $(3,0)$ we have (I'll already plug in $a$ and $c$):
$$\frac{-2\cdot3+b}{3+1}=0$$
This implies that $b=6$. So the function you want is $f : \Bbb R \setminus \{1\} \to \Bbb R$ given by:
$$f(x)=\frac{-2x+6}{x+1}$$
$c=1$, and since the curve passes through (3,0), we have $b = -3a$. So $y=a\frac{x-3}{x+1} = a(1-\frac{4}{x+1})$ which converges to a when x goes to infinity. So $a=-2$
