It is known that the quadratic function f (x) = ax ^ 2 + BX (AB ∈ R, a ≠ 0) satisfies f (- x + 5) = f (x-3) and the equation f (x) = x has equal roots; It is known that the quadratic function f (x) = ax ^ 2 + BX (AB ∈ R, a ≠ 0) satisfies f (- x + 5) = f (x-3) and the equation f (x) = x has equal roots. (1) find the analytic expression of F (x); (2) whether there are real numbers m, n (m < n), so that the domain of definition and value of F (x) are [M, n] and [3M, 3N]? If there are, find the value of M, N; if not, explain the reason

It is known that the quadratic function f (x) = ax ^ 2 + BX (AB ∈ R, a ≠ 0) satisfies f (- x + 5) = f (x-3) and the equation f (x) = x has equal roots; It is known that the quadratic function f (x) = ax ^ 2 + BX (AB ∈ R, a ≠ 0) satisfies f (- x + 5) = f (x-3) and the equation f (x) = x has equal roots. (1) find the analytic expression of F (x); (2) whether there are real numbers m, n (m < n), so that the domain of definition and value of F (x) are [M, n] and [3M, 3N]? If there are, find the value of M, N; if not, explain the reason

(1)
f(-x+5)=f(x-3)
a(x-5)^2+b(5-x)=a(x-3)^2+b(x-3)
a[(x-5)^2-(x-3)^2]+b[(5-x)-(x-3)]=0
a(x-5+x-3)(x-5-x+3)+b(8-2x)=0
2a(8-2x)+b(8-2x)=0
(2a+b)(8-2x)=0
If x takes any value in the domain, then only 2A + B = 0, B = - 2A
Equation f (x) = x
ax^2-2ax=x
ax^2-(2a+1)x=0
x[ax-(2a+1)]=0
X = 0 or x = (2a + 1) / A
Two equal
(2a+1)/a=0 a=-1/2 b=1
The analytic expression of F (x) is f (x) = - x ^ 2 / 2 + X
(2)
f(x)=-x^2/2+x=(-1/2)(x-1)^2+1/2
The axis of symmetry of the function is x = 1
When n ≤ 1, the function increases monotonically
Let f (m) = 3M
3M = - m ^ 2 / 2 + m, m (M + 4) = 0, M = - 4 or M = 0
f(n)=3n
N (n + 4) = 0, n = 0 or n = - 4
M = - 4, n = 0
m> 1, the function decreases monotonically
Let f (m) = 3N, f (n) = 3M
3n=-m^2/2+m
3m=-n^2/2+n
6(n-m)=(n+m)(n-m)-(n-m)
It's time to tidy up
(n-m)(n+m-7)=0
n+m-7=0 n=7-m>1 1