The graphs of a function $f(x)=3x+b$ and its inverse function $f^{-1}(x)$ intersect at the point $(-3,a)$. Given that $b$ and $a$ are both integers, what is the value of $a$?
Since the graph of $f$ contains the point $(-3,a)$, we know that  \[a=f(-3)=3(-3)+b=b-9.\]Since the graph of $f^{-1}$ also contains this point we know that $f^{-1}(-3)=a$ or $-3=f(a)$.  Therefore \[-3=f(a)=3a+b.\]Substitution for $a$ gives \[-3=3(b-9)+b=4b-27.\]Therefore $b=\frac14(27-3)=6$. This forces \[a=b-9=6-9=\boxed{-3}.\]One could also recall that the graph of $f$ is a line and the graph of $f^{-1}$ is that line reflected through $y=x$.  Since the slopes of these lines are not 1, the lines both intersect $y=x$ at a single point and that point is also the point of intersection of the graphs of $f$ and $f^{-1}$.  Therefore the intersection point must be $(-3,-3)$, giving $a=\boxed{-3}$.