因为是二次函数,则设为y=ax^2+bx+c
因为过点(0,1)
代入得 1=a*0^2+b*0+c c=1
y=ax^2+bx+1
f(x+1)=a(x+1)^2+b(x+1)+1
f(x)=ax^2+bx+1
f(x+1)-f(x)=a(x+1)^2+b(x+1)+1 - (ax^2+bx+1)
=a(x+1)^2-ax^2+b(x+1)-bx
=a(x+1+x)(x+1-x)+b
=a(2x+1)+b
=2ax+a+b=2x
2a=2 a+b=0
a=1 b=-1
则y=x^2-x+1