Question

二次函数f(x)满足f(x+1)-f(x)=2x且f(0)=1,(1)求f(x)的解析式

Answer 1

f(x+1)-f(x)=2x,

f(1)-f(0)=0,f(1)=f(0)=1.

f(2)-f(1)=2,

f(3)-f(2)=4,

f(4)-f(3)=6,

f(5)-f(4)=8.

.

f(x)-f(x-1)=2(x-1).

f(x)-f(1)=[2+2(x-1)]*(x-1)/2,

f(x)=x(x-1)+1=x^2-x+1.

2.f(x)=x^2-x+1,y=2x+m,

x^2-x+1=2x+m,

x^2-3x+1+m=0,

⊿