Question

待定系数法求函数解析式 已知f（x)是二次函数,且f(x+1)+f(x-1)=2x^2-4x+4,求f(x)设f(x)=

Answer 1

f(x)=ax^2+bx+c

f(x+1)=a(x+1)^2+b(x+1)+c=ax^2+a+2ax+bx+b+c

f(x-1)=a(x-1)^2+b(x-1)+c=ax^2+a-2ax+bx-b+c

f(x+1)+f(x-1)= 2ax^2+2a+2bx+2c

f(x+1)+f(x-1)=2ax^2+2bx+2a+2c=2x^2-4x+4

2a=2 a=1

2b=-4 b=-2

2a+2c=4

a=1,b=-2,c=1

明白了吗