Answer
(a) See graph.
(b) $(-\infty,\infty)$.
Work Step by Step
(a) Given $f(x)=-x^2+2x-3=-(x^2-2x+1)+1-3=-(x-1)^2-2$, we can determine it opens down with vertex $(1,-2)$, axis of symmetry $x=1$, y-intercept $f(0)=-3$, and x-intercept(s) $none$. See graph.
(b) For $f(x)\le0$, use the graph, we have the solution as $(-\infty,\infty)$.
