已知f(x)是二次函数,若f(0)=0,且f(1+x)=f(x)+x+1,试求f(x)的解析式.要有详细部骤.

Question

已知f(x)是二次函数,若f(0)=0,且f(1+x)=f(x)+x+1,试求f(x)的解析式.
要有详细部骤.

Answer 1

f(x)=ax^2+bx+c
f(0)=0
c=0
f(x)=ax^2+bx
f(x+1)=a(x+1)^2+b(x+1)
=ax^2+2ax+a+bx+b
=ax^2+(2a+b)x+a+b
f(x)+x+1=ax^2+(b+1)x+1
2a+b=b+1
a=1/2
a+b=1
b=1/2
f(x)=x^2/2+x/2