設(shè)f(x)=ax2 +bx+c
那么f(x+1)+f(x-1)=2ax2+2bx+2a+2c=2x^2-4x,
得a=1,b=-2,c=-1
所以f(x)=x2 -2x-1
若f(x)為二次函數(shù),且滿足f(x+1)+f(x-1)=2x^2-4x,求f(x)的解析式
若f(x)為二次函數(shù),且滿足f(x+1)+f(x-1)=2x^2-4x,求f(x)的解析式
數(shù)學(xué)人氣:500 ℃時(shí)間:2020-06-17 23:11:06
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