Question

若二次函数f(x)=ax²+bx+c(c≠0)满足f(x+1)-f(x)=2x,且f(0)=1.(1)若在区间

Answer 1

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

f(0)=1

1=c

f(1)=f(0)

a+b+1=1==>b=-a

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

在f(x+1)=f(x)+2x中,令x= - 1

f(0)=f(-1)-2

f(-1)=3

2a+1=3

a=1

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

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