jmc

Since we know that $A,B,C$ are integers, we know that the vertical asymptotes occur at the vertical lines $x=-2$ and $x=3$. Also, since the degree of the numerator and denominator of $f$ are the same, it follows that the horizontal asymptote of $f$ occurs at the horizontal line $y=1/A$.

We see from the graph that $1/A<1.$ Also, we are told that for sufficiently large values of $x,$ $f(x)>0.4,$ so $$0.4\le \frac{1}{A}<1.$$As $A$ is an integer, it follows that $A=2$.

Hence, the denominator of the function is given by $A{x}^2+Bx+C=2(x+2)(x-3)=2x^2-2x-12$. Then, $A+B+C=2-2-12=\boxed{-12}$.