已知二次函数f(x),满足条件f(0)=(1/2)m和f(x+1)-f(x-1)=4x-2m

二次函数y=f(x)满足条件f(0)=(1/2)m,
∴设f(x)=ax²+bx + m/2,
由f(x+1）-f(x-1)=4x-2m,得
a(x+1)²+b(x+1)+ m/2 -[a(x-1)²+b(x-1) + m/2]=4x-2m,
∴4ax+2b=4x-2m,
即4(a-1)x+2(b+m)=0
要使上式在x取任意值时都成立,只能使
a=1,b=-m.
∴f(x)=x²-mx+m/2