jmc

The graph of the original parabola ($A$) and its final image ($A^{\prime }$) after rotation and translation is shown below:



Rotating the original parabola 180 degrees changes its equation to $y=-(x-2{)}^2+3$. Shifting this last parabola to the left changes its equation to $y=-(x+1{)}^2+3$. Shifting it down changes its equation to $y=-(x+1{)}^2+1$. So the equation of $A^{\prime }$ is $y=-(x+1{)}^2+1$. To find the zeros of this parabola, we set $y=0$ to get $0=-(x+1{)}^2+1$. Expanding the right hand side gives $0=-x^2-2x$. Dividing through by $-1$ and factoring out an $x$ from the right hand side, we get $0=x(x+2)$, so either $x=0$ or $x+2=0$. Thus, $a=0$ and $b=-2$, so $a+b=\boxed{-2}$.