Question

已知f(x)是二次函数且满足f(x+4)+f（x-1)=X的平方-2x 求f(x)

Answer 1

设f(x)=ax²+bx+c
f(x+4)=a(x+4)²+b(x+4)+c
=ax²+(8a+b)x+16a+4b+c
f(x-1)=a(x-1)²+b(x-1)+c
=ax²+(b-2a)x+a-b+c
f(x+4)+f(x-1)=ax²+(8a+b)x+16a+4b+c+ax²+(b-2a)x+a-b+c
=2ax²+(6a+2b)x+17a+3b+2c
f(x+4)+f(x-1)=x²-2x
2ax²+(6a+2b)x+17a+3b+2c=x²-2x
2a=1
a=1/2
6a+2b=-2
6*1/2+2b=-2
b=-5/2
17a+3b+2c=0
17*1/2+3(-5/2)+2c=0
c=-1/2
f(x)=1/2x²-5/2x-1/2

Answer 2