Find the function y = x2 + ax + 5, (x belongs to the range of [- 1,2])

Find the function y = x2 + ax + 5, (x belongs to the range of [- 1,2])

If the abscissa of the symmetry axis X = - A / 2 is at x = 0.5, that is, a = - 1, the minimum value of Y is 5-a / 4 = 5.25 at x = 0.5, and the maximum value is 9 + 2A = 7 at x = - 1 or x = 2

Function, y = ax + B, when x = 1, y = 1, when x = 2, y = - 5, (1). Find the value of a and B, (2). When x = 0, find the function value Y (3). When x takes what value, the function value y is 0

(1)1=a+b
-5=2a+b
a=-6
b=7
(2)y=-6x+7
When x = 0
y=7
(3)0=-6x+7
x=7/6

If y = x (1-ax) 2 (a > 2) and y '| x = 2 = 5, then the value of real number a is?

y'=(1-ax)^2-2ax(1-ax)
y'(2)=(1-2a)^2-4a(1-2a)=5
1-4a+4a^2-4a+8a^2-5=0
12a^2-8a-4=0
3a^2-2a-1=0
(3a+1)(a-1)=0
a=-1/3,1
Is this not contradictory to question a > 2? Is it wrong?