Given the quadratic function f (x) = x & # 178; - 4x, its domain of definition is d. if d = [a, a + 2], find the minimum value of function f (x)

Given the quadratic function f (x) = x & # 178; - 4x, its domain of definition is d. if d = [a, a + 2], find the minimum value of function f (x)

F (x) = (X-2) ^ 2-4, the opening is upward, the axis of symmetry is x = 2, and there is a minimum f (2) = - 4 at x = 2
Discussion a:
If 0=

Given the quadratic function y = 2x2-4ax + A2 + 2A + 2 (1), through the formula, when x takes what value, y has the maximum or minimum value, what is the maximum or minimum value? (2) When - 1 ≤ x ≤ 2, the function has the minimum value 2

(1) Y = 2x2-4ax + A2 + 2A + 2, y = 2 (x-a) 2-A2 + 2A + 2, when x = a, y has the minimum value of 3 - (A-1) 2; (2) when - 1 ≤ x ≤ 2, 3 - (A-1) 2 = 2, the solution is a = 0 or a = 2, when x ＜ - 1, when x = - 1, y = 2, the solution is a = - 3 − 7, when x ＞ 2, when x = 2, y = 2, the solution is a = 4

Given that the minimum value of quadratic function y = x2 + (2a + 1) x + A-1 is zero, what is the value of a

To put it simply, if I remember correctly, the coefficient in front of x2 of quadratic function means that the opening of function is upward, then this function has a minimum value ~ the coefficient in front of X can adjust the left and right movement of function ~ the last number C is to adjust the up and down movement of function ~ if C = 0 is A-1 = 0, when a = 1, the function is y = x2 + 3x, then the function has a minimum value of 0