For quadratic function y = - 2x ^ 2, find the value range of y when - 1 ≤ x ≤ 2

For quadratic function y = - 2x ^ 2, find the value range of y when - 1 ≤ x ≤ 2

y=-2x^2
This function is even and symmetric about the Y axis
When x = 0, the maximum value is 0
When x = 2, the minimum value is - 8
So the range is [- 8,0]

It is known that the derivative of the quadratic function f (x) equals ax & # 178; + BX + C is f '(x) = 2x-1, and f (1) = 2. The analytic expression of the quadratic function is obtained

Solution
f(x)=ax²+bx+c
f‘(x)=2ax+b=2x-1
∴2a=2,b=-1
∴a=1
∴f(x)=x²-x+c
∵f(1)=2
∴1²-1+c=2
∴c=2
∴f(x)=x²-x+2

The method of collocation is used to find the maximum value of the quadratic function y = 2 / 3x square - 3x-2 / 1

y=3/2(x²-2x)-1/2
=3/2(x²-2x+1-1)-1/2
=3/2(x²-2x+1)-3/2-1/2
=3/2(x-1)²-2
Minimum = - 2

Is the square + 2x + 1 of y = 1 / 3x a quadratic function?
I'm very confused. I just learned quadratic function. I don't understand it,
If so, please write down the coefficients and constants of quadratic and quadratic,

Yes, 1 / 9, 2, 1