Question

若二次函数f(x)满足f(x+1)-f(x)=2x.f(0)=0,求f(x)的解析式.

Answer 1

设f(x)=ax2+bx+c,f(0)=0,则c=0,

f(x+1)=ax2+(2a+b)x+a+b,

f(x)=ax2+bx,

f(x+1)-f(x)=2ax+a+b=2x,则

a=1,b=-1,即

f(x)=x2-x