設f(x)=ax^2+bx+c(a≠0)
∴f(x+1)=a(x+1)^2+b(x+1)+c
f(x-1)=a(x-1)^2+b(x-1)+c
∴f(x+1)+f(x-1)
=a(x+1)^2+b(x+1)+c+a(x-1)^2+b(x-1)+c
=2ax^2+2bx+2a+2c
∵f(x+1)+f(x-1)=2x^2-4x
∴2ax^2+2bx+2a+2c=2x^2-4x
∴對應系數(shù)相等
{a=1
{b=-2
{c=-1
∴f(x)=x^2-2x-1