解析
因为f(0)=0
所以c=0
所以二次函数f(x)=ax²+bx
f(x-1)-f(x)=a(x-1)²+b(x-1)-ax²-bx
=a(x²-2x+1)+bx-b-ax²-bx
=ax²-2ax+a+b-ax²
=-2ax+a+b
=3x
所以
-2a=3 a=-3/2
b=3/2
f(x)=-3/2x²+3/2x
解析
因为f(0)=0
所以c=0
所以二次函数f(x)=ax²+bx
f(x-1)-f(x)=a(x-1)²+b(x-1)-ax²-bx
=a(x²-2x+1)+bx-b-ax²-bx
=ax²-2ax+a+b-ax²
=-2ax+a+b
=3x
所以
-2a=3 a=-3/2
b=3/2
f(x)=-3/2x²+3/2x