解：?$(1)$?依題意，得?$\begin{cases}{c=-1}\\{ 1+b+c=2}\end{cases}，$?解得?$\begin{cases}{b=2}\\{c=-1}\end{cases}$?
∴該函數(shù)的表達式為?$y=x^2+2x-1$?
∵?$y=x^2+2x-1=(x+1)^2-2$?
∴該函數(shù)圖像的頂點坐標(biāo)為?$(-1，$??$-2)$?
?$(2)$?∵在函數(shù)?$y=x^2+bx+c $?中，二次項系數(shù)為?$1，$?該函數(shù)圖像的頂點坐標(biāo)為?$(m，$??$k)$?
設(shè)拋物線對應(yīng)的函數(shù)表達式為?$y=(x-m)^2+k$?
∵?$y=(x-m)^2+k$?的圖像經(jīng)過另一點?$(k，$??$m)$?
∴?$m=(k-m)^2+k$?
∴?$m-k=(m-k)^2，$?解得?$m-k=0$?或?$m-k=1$?
?$(3)$?∵函數(shù)?$y=x^2+bx+c $?的圖像經(jīng)過?$A(x_{1}，$??$y_{1})、$??$B(x_{1}-t，$??$y_{2})、$??$C(x_{1}-2t，$??$y_{3})$?三個不同點
∴?$y_{1}=x_{1}^2+ bx_{1}+c，$??$t≠0，$??$y_{2}=(x_{1} -t)^2+b(x_{1}-t)+c=x_{1}^2 -2 x_{1}t +t^2+ bx_{1} - bt +c，$?
?$y_{3}=(x_{1} -2t)^2+b(x_{1} -2t)+c=x_{1}^2-4x_{1}t +4t^2+ bx_{1} -2\ \mathrm {bt} +c$?
∴?$M=y_{2}-y_{1}=x_{1}^2-2x_{1}t +t^2+ bx_{1} - bt +c-(x_{1}^2+ bx_{1}+c)=-2x_{1}t +t^2- bt，$?
?$N= y_{3} -y_{2}=x_{1}^2-4x_{1}t +4t^2+ bx_{1}-2\ \mathrm {bt} + c-(x_{1}^2-2x_{1}t +t^2+ bx_{1} - bt +c)=-2 x_{1}t +3t^2- bt$?
?$N-M=-2 x_{1}t +3t^2- bt -(-2 x_{1}t +t^2-bt)=2t^2> 0$?
∴?$M< N$?
