If the minimum value of quadratic function y = x & # 178; - MX + 3 is - 3, then M=

If the minimum value of quadratic function y = x & # 178; - MX + 3 is - 3, then M=

By y = x & # 178;; - MX + M & # 178;; / 4 + 3-m & # 178;; / 4
=（x-m/2)²;+3-m²;/4,
When the minimum value is 3-m & # / 4 = - 3,
m²=24
m=±2√6

It is known that the quadratic functions y = x & # 178; - MX + Half M & # 178; + 1 and y = x & # 178; - MX - half M & # 178; + 2 of X are two quadratic functions
One of the images intersects the X axis at two different points a and B
1) It is to determine which secondary image may pass through two points a and B
2) If the coordinate of point a is (- 1,0), try to find the coordinate of point B
3) Under the condition of (2), for the quadratic function passing through a and B, when x takes any value, the value of Y decreases with the increase of X

According to the vertex coordinates: (- B / 2a, (4ac-b & # 178;) / 4A) for y = x & # 178; - MX + (M & # 178; + 1) / 2, the ordinate of the lowest point y = (4ac-b & # 178;) / 4A = (M & # 178; + 1) / 2 + M & # 178;; / 4 = 3 / 4m & # 178; + 1 / 2 > 0, so

It is known that the image of quadratic function y = (M2-2) x2-4mx + n is symmetric with respect to the line x = 2, and its highest point is on the line y = & frac12; X + 1
(1) Find the analytic expression of the quadratic function;
(2) If the opening direction of the parabola remains unchanged and the vertex moves to the point m on the straight line y = & frac12; X + 1, the image and X-axis intersect at two points a and B, and s △ ABM = 8, the analytic expression of the quadratic function at this time is obtained

1) On the symmetry of the line x = 2,
x=-b/2a=4m/2(m^2-2)=2,m1=-1,m2=2,
Because there is the highest point, so m = - 1,
Substituting x = 2 into y = x / 2 + 1, y = 2,
Substitute M = - 1, (2,2) into n = - 2
Analytical formula: y = - x ^ 2 + 4x-2
2) Because the vertex moves to the point m on the line y = & frac12; X + 1, let m (h, H / 2 + 1),
Because the opening direction of the parabola is constant, a = - 1,
Let y = - (X-H) ^ 2 + H / 2 + 1
=-x^2+2hx-h^2+h/2+1,
AB=√△=√(2h+4),
From s △ ABM = 8,
So: (1 / 2) * [√ (2H + 4)] * (H / 2 + 1) = 8,
Let √ (2H + 4) = t,
t^3=64,
t=4,
h=6,
Analytical formula: y = - x ^ 2 + 12x-32

Given the image of quadratic function y = AX2 + BX + C (a ≠ 0), we have the following conclusions: (1) A-B + C > 0; (2) ABC > 0; (3) 4a-2b + C > 0; (4) b2-4ac > 0; (5) 3A + C > 0; (6) a-c > 0
A. 2B. 3C. 4D. 5

When x = - 1, y ＜ 0, then A-B + C ＜ 0, so ① wrong; when the opening of parabola is upward, then a ＞ 0; when the symmetry axis is on the right side of Y axis, x = - B2A ＞ 0, then B ＜ 0; when the intersection coordinate of parabola and Y axis is below X axis, then C ＜ 0, then ABC ＞ 0, so ② correct; when x = - 2, y ＞ 0, then 4a-2b + C ＞ 0, so ③ positive