jmc

The graphs of $y=a{x}^2+3x+1$ and $y=-x-1$ intersect at exactly one point when the equation $$a{x}^2+3x+1=-x-1$$has only one solution. This equation simplifies to $a{x}^2+4x+2=0$, which has only one solution when the discriminant is $0$, in other words, $$4^2-4(a)(2)=0.$$Solving for $a$ gives $a=\boxed{2}$.