因为f(x+3)=f(x-1),所以对称轴x=-1,设f(x)=a(x+1)^2+k,将（0,1）代入,得a+k=1,...(1)令f(x)=a(x+1)^2+k=0,得ax^2+2ax+a+k=0,设两根为x1,x2,则x1+x2=-1,x1*x2=k/a,x1^2+x2^2=(x1+x2)^2-2*x1*x2=(-1)^2-2k/a=10,...(2)...