It is known that the quadratic function f (x) = ax ^ 2 + BX satisfies the condition 1. For any x belonging to R, f (x-4) = f (2-x) 2. The image of function f (x) is tangent to the line y = X The quadratic function f (x) = ax ^ 2 + BX satisfies the condition 1. For any x belonging to R, f (x-4) = f (2-x) 2. The image of function f (x) is tangent to the line y = X Find: if and only if x belongs to [4, M], f (x-t) ≤ x is constant, try to find the value of T M

It is known that the quadratic function f (x) = ax ^ 2 + BX satisfies the condition 1. For any x belonging to R, f (x-4) = f (2-x) 2. The image of function f (x) is tangent to the line y = X The quadratic function f (x) = ax ^ 2 + BX satisfies the condition 1. For any x belonging to R, f (x-4) = f (2-x) 2. The image of function f (x) is tangent to the line y = X Find: if and only if x belongs to [4, M], f (x-t) ≤ x is constant, try to find the value of T M

1. Find f (x) first
f(x-4)=f(2-x)
a(x-4)^2+b(x-4)=a(2-x)^2+b(2-x)
(b-2a) x = 3b-6a
Because the above formula holds for any x, b-2a = 0, that is, B = 2A
Because f (x) and y = x are tangent (there is only one intersection) and both f (x) and y = x cross the origin, the origin is the tangent point
F '(x) = 2aX + B = 2aX + 2A. The slope of the tangent point is 1, so 1 = 2a0 + 2a, that is, a = 1 / 2
So f (x) = 1 / 2 x ^ 2 + X
2、f(x-t)≤x
1/2(x-t)^2+(x-t)≤x
It is reduced to x ^ 2-2tx + T ^ 2-2t ≤ 0
Because if and only if x belongs to [4, M], f (x-t) ≤ X
So 4, m are two parts of the equation x ^ 2 - 2tx + T ^ 2 - 2T = 0
Substituting 4, T ^ 2-10t + 16 = 0, t = 2, t = 8
When t = 2, we substitute m into m to get m ^ 2-4m = 0, and the solution is m = 0 (rounding off), M = 4. So t = 2, M = 4
When t = 8, substituting m, m ^ 2-16m + 48 = 0, M = 4, M = 12
(M = 4 also holds for f (x-t) ≤ x, but the interval becomes a point, so I don't know if M = 4 should be discarded.)

Let the quadratic function f (x) = ax & # 178; + BX + C satisfy the conditions f (0) = 2, f (1) = - 1, and the length of the line cut by the image on the x-axis is 2 times longer than the sign 2

Then f (x) = ax & # 178; - (a + 3) x + 2. If the abscissa of the intersection of F (x) and X-axis x1, X2.. then X1 + x2 = (a + 3) / A, x1x2 = 2 / A, so (x2-x1) &# 178; = (x1 + x2) &# 178; - 4x1x2. That is, (a + 3) &# 178; - 8A = 8A & # 178;, it is reduced to 7a & # 178

Let the quadratic function f (x) = ax ^ 2 + BX (a ≠ 0) satisfy the condition 1. F (x-4) = f (2-x); 2. The image of F (x) is tangent to the line y = x, and the analytic expression of F (x) is obtained

By taking x-4 and 2-x into f respectively, a (x-4) ^ 2 + B (x-4) = a (2-x) ^ 2 + B (2-x)
The results show that: - 4ax + 12a + 2bx-6b = 0
2[(b-2a)x+(2a-b)]=0
(b-2a)x+(2a-b)=0
2a(1-x)-b(1-x)=0
Because the domain of X is r, so 2A = B. so f (x) = ax ^ 2 + 2aX
Because the image of F (x) is tangent to the line y = x, ax ^ 2 + 2aX = x, ax ^ 2 + 2ax-x = 0
So Δ = (2a-1) ^ 2 = 0, so 2a-1 = 0, a = 0.5, so B = 1, f (x) = 0.5x ^ 2 + X