已知y=f(x)是二次函数,且满足f(0)=1,f(x+1)-f(x)=2x,求函数y=f(x)的解析式

设f(x)=ax^2+bx+c
f(0)=1 ==> c=1

f(x+1)-f(x)=2x
==> a(x+1)^2+b(x+1)+1-ax^2-bx-1=2x
==> 2ax+(a+b)=2x
==> 2a = 2 ==> a = 1
a+b=0, a=1 ==> b = -1

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

f(0)==1
设解析式为:
f(x)=ax^2+bx+1
2x=f(x+1)-f(x)=a(x+1)^2+b(x+1)-ax^2-bx=ax^2+2ax+a+bx+b-ax^2-bx=2ax+b
2ax+(a+b)=2x
2a=2
a+b=0

算出来a,b放入f(x)=ax^2+bx+1中/