(1)设f(x)=ax2+bx+c,
a(x+1)2+b(x+1)+c+a(x-1)2+b(x-1)+c=-2x2+4x,
2ax2+2bx+2a+2c=-2x2+4x,
,∴f(x)=?x2+2x+1.
a=?1 b=2 c=1
(2)f(x)=-(x-1)2+2,
①a+2<1即a<-1,当x=a+2,f(x)max=-a2-2a+1;
②a≤1≤a+2即-1≤a≤1,当x=1,f(x)max=2;
③a>1,当x=a,f(x)max=-a2+2a+1;
故f(x)max=
?a2?2a+1,a<?1 2,?1≤a≤1 ?a2+2a+1,a>1