解：?$(1)$?把?$A(1，$??$-2)$?和?$B(0，$??$-5)$?代入?$y=x^2+bx+c$?
得?$ \begin{cases}{1+b+c=-2}\\{c=-5}\end{cases}，$?解得?$\begin{cases}{ b=2}\\{c=-5}\end{cases}$?
∴二次函數(shù)的表達(dá)式為?$y=x^2+2x-5$?
∵?$y=x^2+2x-5=(x+1)^2-6$?
∴頂點(diǎn)坐標(biāo)為?$(-1，$??$-6) $?
?$(2)$?如圖，∵點(diǎn)?$A(1，$??$-2)$?關(guān)于對(duì)稱軸直線?$x=-1$?的對(duì)稱點(diǎn)為?$C(-3，$??$-2)$?
∴當(dāng)?$y≤-2$?時(shí)，?$x$?的取值范圍是?$-3≤x≤1$?
