解:$(1)$由题意可得:$\begin{cases}{ c=1 } \\{ 4a+2b+c=11 }\\{a-b+c=6} \end{cases},$解得:$\begin{cases}{ a=\dfrac {10}{3} } \\{ b=-\dfrac 53 } \\{ c=1 }\end{cases}$
$(2) $由$ (1)$得$ y=\frac {10}{3} x^2-\frac {5}{3} x+1$
∴当$ x=-3$时$, y=\frac {10}{3} ×(-3)^2-\frac {5}{3} ×(-3)+1=36$