因为y=f(x)是二次函数且满足f(0)=1
y=ax^2+bx+1
又因为f(x=1)-f(x)=2x,(肯定是f(x-1)吧!)
f(x-1)=a(x-1)^2+b(x-1)+1
f(x-1)-f(x)=a(x^2-2x+1)+bx-b+1-ax^2-bx-1
=-2ax+a-b=2x
a=b
-2a=2
a=-1=b
二次函数解析式:y=-x^2-x+1
因为y=f(x)是二次函数且满足f(0)=1
y=ax^2+bx+1
又因为f(x=1)-f(x)=2x,(肯定是f(x-1)吧!)
f(x-1)=a(x-1)^2+b(x-1)+1
f(x-1)-f(x)=a(x^2-2x+1)+bx-b+1-ax^2-bx-1
=-2ax+a-b=2x
a=b
-2a=2
a=-1=b
二次函数解析式:y=-x^2-x+1