f(x)=ax²+bx=x
ax²+(b-1)x=0有等跟
判別式等于0
(b-1)²-4a*0=0
(b-1)²=0
b=1
f(x)=ax²+x
f(-x+5)=f(x-3)
所以 a(-x+5)²+(-x+5)=a(x-3)²+(x-3)
ax²-10ax+25a-x+5=ax²-6ax+9a+x-3
16a+8=4ax+2x=x(4a+2)
此式恒成立則只有4a+2=16a+8=0
a=-1/2
所以f(x)=-x²/2+x