Question

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

Answer 1

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

f(0)=1

则:c=1

∴f(x)=ax^2+bx+1

f(x+1)=a(x+1)^2+b(x+1)+1

f(x+1)-f(x)

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

=2ax+a+b=2x

∴系数对应相等

{2a=2

{a+b=0

解得

a=1,b=-1

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