如果函数f(x)=x2+bx+c对任意实数t都有f(2+t)=f(2-t),那么f(1),f(2),f(4)的大小关系

f(1) = 1+b+c
f(3) = 9+2b+c
f(1) = f(3)
1+b +c = 9+2b+c
b = -8
f(2)= 4+2b-8
f(0) = -8
f(2) = f(0)
4+2b-8 = -8
b = -2
f(x) = x^2-2x-8
f(1) = -9
f(2) = -12
f(4) = 0
f(4) > f(1)> f(2)