Question

若二次函数f(x)的图像过点(0,0),且f(x+!)=f(x)+x+1,求f（x）的解析式

Answer 1

设f(x）=ax^2+bx+c∵图像过（0,0)∴c=0

f(x)=ax^2+bx

f(x+1)=a(x+1)^2+b(x+1)...(1)

f(x)+x+1=ax^2+bx+x+1...(2)

(1)=(2)a(x^2+2x+1)+bx+b=ax^2+(b+1)x+1

ax^2+（2a+b)x+(a+b)=ax^2+(b+1)x+1

2a+b=b+1 a=1/2

a+b=1 b=1/2

解析式为f(x)=x^2/2+x/2