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

Answer 2

设f(x)=ax²+bx+c
则f(x+1)=a(x+1)²+b(x+1)+c
f(x-1)=a(x-1)²+b(x-1)+c
∴a(x+1)²+b(x+1)+c+a(x-1)²+b(x-1)+c=2x²-4x
2ax²+2bx+2a+2c=2x²-4x
∴2a=2,2b=-4,2a+2c=0
∴a=1,b=-2,c=-1
∴f(x)=x²-2x-1

Answer 3

既然是
二次函数
，那就设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
于是2ax²+2bx+2a+2c=2x^2-4x
2a=2,2b=-4,2a+2c=0
得a=1.b=-2，c=-1
f（x）=x²-2x-1

Answer 4

用待定系数法：
设f（x）=ax^2+bx+c(a不为0)
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=2x^2-4x
a=1,b=-2,c=-1
f(x)=2x^2-2x-1