Question

二次函数f(x)满足f(x加1)减f(x)=2x,且f(0)=1 (1)求f(x)的解析式(2)在区间[负1,1]上,y

Answer 1

f(x)=ax²+bx+c

f(x+1)=a(x+1)²+b(x+1)+c

f(x加1)减f(x)=2x,且f(0)=1 (1)求f(x)的解析式(

f(x+1)-f(x)=a(x+1)²+b(x+1)+c-(ax²+bx+c)=2x f(0)=1

a=1,b=-1,c=1

f(x)=x²-x+1