Question

若二次函数y=fx满足f(x-2)-fx=16x,且f(0)=2

Answer 1

1）由f(0)=2,设f(x)=ax^2+bx+2

则f(x-2)-f(x)=a(-4x+4)-2b=-4ax+4a-2b=16x

对比系数得：-4a=16, 4a-2b=0,

解得：a=-4, b=-8

因此f(x)=-4x^2-8x+2

2) 不等式为m