Question

f(x)是二2次函数f(x+4)+f(x-1)=x的平方-2x求f(x)的解析式

Answer 1

设：f(x)=ax²+bx+c
f(x+4)+f(x-1)=a(x+4)²+b(x+4)+c+a(x-1)²+b(x-1)+c
=2ax²+(6a+2b)x+(17a+3b+2c)=x²-2x
2a=1
6a+2b=-2
17a+3b+2c=0
解得：
a=1/2；b=-5/2；c=-1/2
所以：f(x)=(1/2)x²-5x/2-1/2

Answer 2

f(x)=ax^2+bx+c(a不为0)
f(x+4)+f(x-1)=a(x+4)^2+b(x+4)+c+a(x-1)^2+b(x-1)+c
=2ax^2+(8a+b-2a+b)x+16a+4b+2c+a-b
=2ax^2+(6a+2b)x+17a+3b+2c
2a=1
6a+2b=-2
17a+3b+2c=0
a=1/2
b=-5/2
c=-1/2
f(x)=0.5x^2-2.5x-0.5

Answer 3

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