To know where x intercepts, we must solve for x from the equation and see what values y can take:
[tex]\begin{gathered} y=-(x-3)^2-1 \\ y=-(x-3)^2-1 \\ y+1=-(x-3)^2 \\ \sqrt[]{(x-3)^2}=\sqrt[]{-y-1} \\ x-3=\sqrt[]{-y-1} \\ x=3+\sqrt[]{-y-1} \\ \end{gathered}[/tex]In this case we have to make the root positive then:
[tex]\begin{gathered} -y-1\ge0 \\ \\ y\leq-1 \end{gathered}[/tex]that is, y can take any value less than -1 so that x intercepts
