设f（x）=ax2+bx+c，∴a（x+1）2+b（x+1）+c=2ax2+2bx+2c-x2，∴ax2+2ax+a+bx+b+c=2ax2+2bx+2c-x2，∴（a-1）x2+（b-2a）x+c-a-b=0，∴a-1=0，b-2a=0，c-a-b=0，解得：a=1，b=2，c=3∴f（x）=x2+2x+3
