已知f（x）为二次函数，且f（x+1）+f（x-1）=2x2-4x，则f（x）的解析式是_．

设f（x）=ax2+bx+c，∴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）=2ax2+2bx+2a+2c，∵f（x+1）+f（x-1）=2x2-4x，∴2ax2+2bx+2a+2c=2x2-4x，∴a=1，b=-2，c=-1，则f（x）=...