Ask a question to find the range of function y = x2-ax + 1 (x belongs to (- 1,1) a is a constant) You know what? Come on

Ask a question to find the range of function y = x2-ax + 1 (x belongs to (- 1,1) a is a constant) You know what? Come on

Y = (x-a / 2) ^ 2 + 1-A ^ 2 / 4, opening upward, axis of symmetry x = A / 2
Therefore, there are:
When 0=

Find the range of function y = x & # 178; - ax + 2 (a is constant) x ∈ [- 1,1]
Why consider [- 1,0] [0, - 1] instead of [- 1,1]

y=x^2-ax+2
=(x-a/2)^2 -a^2 /4 +2
(1) When x = A / 2 (x = A / 2 is the axis of symmetry) = 1
F (- 1) is the maximum 3 + A, f (1) is the minimum 3-A range [3-A, 3 + a]
(3) When - 1

Find the range of function f (x) = x2-ax + 1 (a is constant), X ∈ [- 1,1]

F (x) = x2 ax + 1 = (x − A2) 2 + 1 − A24; ① when A2 ≤− 1, i.e. a ≤ - 2, the function f (x) increases monotonically on [- 1, 1]; the range of function f (x) is [f (- 1), f (1)] = [2 + A, 2-A]; ② when - 1 < A2 ≤ 0, i.e. - 2 < a ≤ 0, x = A2, the function f (x) takes the minimum value of 1 − A24

The range of function y = ax + B on [1,2] is [0,1], then the value of a + B is ()
A. 0b. 1C. 0 or 1D. 2

Because the function y = ax + B is a monotone function, if x = 1, then y = 0, so a + B = 0; if x = 1, then y = 1, so a + B = 1, so a + B = 0, or a + B = 1, then C