Given that the image of quadratic function y = x2 + BX + C passes through point a (C, 0) and is symmetric with respect to line x = 2, the analytic expression of the quadratic function may be______ Just write a possible analytical expression

Given that the image of quadratic function y = x2 + BX + C passes through point a (C, 0) and is symmetric with respect to line x = 2, the analytic expression of the quadratic function may be______ Just write a possible analytical expression

According to the meaning of C2 + BC + C = 0 (1), B = - 4A = - 4 (2) (1) (2), the solution of simultaneous equations is b = - 4, C = 0 or 3, then the analytic expression of quadratic function is y = x2-4x or y = x2-4x + 3

The solution set of F (x) > 2x is 1

It is known that the quadratic coefficient of quadratic function f (x) is a, and the solution set of inequality f (x) > - 2x is (1,3)
Find (1) if the equation f (x) + 6A = 0 has two equal roots, find the analytic formula
(2) If the maximum value of F (x) is a positive number, find the range of A
Let y = ax ^ 2 + BX + C, because the solution set of F (x) > - 2x is (1,3)
That is, the solution set of ax ^ 2 + (B + 2) x + C > 0 is (1,3)
So the solution of the equation AX ^ 2 + (B + 2) x + C = 0 is X1 = 1, X2 = 3, and a < 0
So a + B + C + 2 = 0 and 9A + 3 (B + 2) + C = 0
(1) Because the equation f (x) + 6A = 0 has two equal roots
So △ = B ^ 2 - 4a (c + 6a) = 0
Three equations are solved simultaneously: a = - 1 / 5, B = - 6 / 5, C = - 3 / 5
So the analytic formula is: y = - 1 / 5 * (x ^ 2 + 6x + 3)
(2) Because a + B + C + 2 = 0 and 9A + 3 (B + 2) + C = 0
So B = - 4a-2, C = 3A
So the analytic formula is: y = ax ^ 2 - 2 (2a + 1) x + 3a
Because the maximum value y = [12a ^ 2 - 4 (2a + 1) ^ 2] / 4A = (a ^ 2 + 4A + 1) / a > 0
So a ＜ - 2 - √ 3 or - 2 + √ 3 ＜ a ＜ 0

It is known that the coefficient of quadratic term of quadratic function f (x) is a, and the solution set of F (x) > - 2x is (1,3)
1: If the equation f (x) + 6A = 0 has two equal roots, find the analytic expression of F (x)
2. If f (x) + 3 ＜ 0 is r constant for any x, the range of a is obtained

The solution set of F (x) + 2x > 0 is (1,3)
Therefore, f (x) + 2x = a (x-1) (x-3), and a