已知二次函数f(x)=ax^2+bx(a,b为常数,且a≠0)满足f(-x+5)=f(x-3)且方程f(x)=x有等

（1）f（-x+5）=f（x-3）,说明函数图像有对称轴x=1,所以-b/(2a)=1,b=-2a。方程f（x）=x有等根,即ax^2+(b-1)x=0有等根。判别式△=0,所以b=1,从而a=-1/2。f（x）=-1/2x^2+x。（2）f（x）=-1/2x^2+x=-1/2(x-1)^2+1/2。①n≤1时,[m,n]上函数递增,所以f(m)=3m,f(n)=3n,即-1/2m^2+m=3m,-1/2n^2+n=3n,解得m=-4,n=0。②m