Question

已知y=f(x)是二次函数,且满足f(0)=1,f(x+1)-f(x)=2X,求f(x)

Answer 1

set f(x)=ax^2+bx+c(a is not equal to 0)

since f(0)=1, so c=1

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

so 2ax+a+b=2x

that is equal to say

a=1,b=-1

so f(x)=x^2-x+1