已知二次函数f（x）=ax²+bx（x≠0）满足f（-x+5）=f（x-3）,且方程f（x）=x有等根,①求函数解析式 ⑥当x【-1/2,1】时,函数f（x）的图像总在函数y=2x+m的下方,求m的范围,

∵f(x)=ax^2+bx=x(ax+b)∴f(-x+5)=(-x+5)[a(-x+5)+b]=ax^2+(-b-10a)x+25a+5bf(x-3)=(x-3)[a(x-3)+b]=ax^2+(-6a+b)x+9a-3b又f(-x+5)=f(x-3)∴ax^2+(-b-10a)x+25a+5b=ax^2+(-6a+b)x+9a-3b整理得：b=-2a又f(x)=x有等根,...