Question

二次函数f(x)满足f(x+1)+f(x-1)=2x^2+4x,求f(x)？

Answer 1

f(x)=ax²+bx+c
f(x+1)=a(x+1)²+b(x+1)+c
=ax²+2ax+a+bx+b+c
f(x-1)=a(x-1)²+b(x-1)+c
=ax²-2ax+a+bx-b+c
ax²+2ax+a+bx+b+c +ax²-2ax+a+bx-b+c
=2ax²+2a+2bx+2c
=2x^2+4x
2a=2
a=1
2b=4
b=2
2a+2c=0
c=-1
f(x)=x²+2x-1,3,设函数=ax2+bx+c，然后代入，与后式进行比较，求出系数abc，既得函数解析式,2,令u=x-1，f(x+1)+f(x-1)=2x^2+4x =》
f(u)+f(u-2)= 2(u-1)^2 + 4(u-1)
=u^2+(u-2)^2 +4u-2
= u^2+(u-2)^2 + 2u +2(u-2)+2
= u^2+2u+1 +(u-2)^2 +2(u-2)+1
=> f(u)= u^2+2u+1
f(x)=x^2+2x+1,2,设二次函数f(x)=a²x+bx+c
f(x+1)+f(x-1=a（x+1）²+b（x+1）+c+a（x-1）²+b（x-1）+c=2ax²+2bx+2a+2c=2x^2+4x
a=1，b=2，c=-1
f(x)=x²+2x-1,2,f(x)=x的平法+2x-1
具体算法如下：既然是二次函数，不妨设f（x）=ax平房+bx+c
再化简，最后会出来个三元一次方程，很简单的,2,设函数=ax2+bx+c，然后代入，与后式进行比较，求出系数abc，既得函数解析式,1,