Answer
$a=3,b=1,c=-1$
Work Step by Step
Step 1. Given $y=\frac{x^2+a}{bx+c}$, we have $y'=\frac{2x(bx+c)-b(x^2+a)}{(bx+c)^2}=\frac{bx^2+2cx-ab}{(bx+c)^2}$
Step 2. The extrema can be found by letting $y'=0$ which gives $bx^2+2cx-ab=0$
Step 3. Using the given condition at $x=3$, we have $b(3)^2+2c(3)-ab=0$ or $9b+6c-ab=0$
Step 4. Using the given condition at $x=-1$, we have $b(-1)^2+2c(-1)-ab=0$ or $b-2c-ab=0$
Step 5. As the point $(-1,-2)$ is on the curve, we have $\frac{(-1)^2+a}{b(-1)+c}=-2$ or $2b-2c=a+1$
Step 6. Combine the results of steps 3-5 to get the following system of equations:
$\begin{cases} 9b+6c-ab=0\\b-2c-ab=0\\ 2b-2c=a+1 \end{cases}$
Step 7. Subtract the second equation from the first one to get $c=-b$ and the last equation gives $a=4b-1$
Step 8. Substitute the above relations in the second equation to get $b-2(-b)-(4b-1)b=0$ which gives $3b(b-1)=0$
Step 9. As $b\ne0$ (otherwise $c=0$ and the function will be undefined), we have $b=1$, thus $c=-1, a=3$
