Question

已知二次函数f(x)满足f(x+1)+f(x-1)=2x^2-4x,求f(x)的解析式

Answer 1

设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

∴对应系数相等

{a=1

{b=-2

{c=-1

∴f(x)=x^2-2x-1