The graph of the parabola defined by the equation $y=-(x+1)^2+1$ is shifted 1 unit to the right, then shifted 5 units down, then rotated 180 degrees about its vertex. The resulting parabola has zeros at $x=a$ and $x=b$, where $b\ge a$. What is $b-a$?
The graph of the original parabola ($A$) and its final image ($A'$) after rotation and translation is shown below:

[asy]

Label f;

f.p=fontsize(4);

xaxis(-4,4,Ticks(f, 2.0));

yaxis(-6,5,Ticks(f, 2.0));

real f(real x)

{

return x^2-4;

}

draw("$A'$", graph(f,-3,3), linewidth(1));

real g(real x)

{

return -(x+1)^2+1;

}

draw("$A$", graph(g,-3.5,1.5), linewidth(1));

[/asy]

Shifting the original parabola 1 unit to the right changes its equation to $y=-x^2+1$. Shifting this last parabola 5 units down changes its equation to $y=-x^2-4$. Rotating it by 180 degrees changes its equation to $y=x^2-4$. So the equation of $A'$ is $y=x^2-4$. To find the zeros of this parabola, we set $y=0$ to get $0=x^2-4$. Factoring the right hand side, we get $0=(x-2)(x+2)$, so either $x-2=0\Rightarrow x=2$ or $x+2=0 \Rightarrow x=-2$. Thus, $a=-2$ and $b=2$, so $b-a=\boxed{4}$.