Question

设二次函数y=f(x）满足f（0）=1,且f(x+1)-f(x-1)=4x,求f(x)的解析式？

Answer 1

解：设二次函数的解析式为f(x)=ax^2+bx+c
f(0)=1
所以c=1 f(x)=ax^2+bx+1
又f(x+1)-f(x-1)=4x
f(1+1)-f(1-1)=4
f(2)=5
f(-1+1)-f(-1-1)=-4
f(-2)=5
4a+2b+1=5
4a-2b+1=5
b=0
a=5/4
f(x)=x^2+1

Answer 2

设f(x）=ax^2+bx+c，因为f（0）=1，所以C=1
f(x+1)=a(x+1)^2+b(x+1)+1
=a(x^2+2x+1)+bx+b+1
=ax^2+（2a+b）x+a+b+1
f(x-1)=a(x-1)^2+b(x-1)+1
=a(x^2-2x+1)+bx-b+1
=ax^2+（b-2a）x+a-b+1
所以f(x+1)-f(x-1)=4ax+2b
又因为f(x+1)-f(x-1)=4x
所以a=1,b=0
所以f(x)=x^2+1

Answer 3

令f（x）=ax^2+bx+c带进去就求出来了