If the increasing range of quadratic function y = AXX + BX + C is (- ~, 2), what is the decreasing range of quadratic function y = Bxx + ax + C? Please give me a detailed explanation. Thank you very much Functions are really hard,

From the meaning of the title, A0
The symmetry axis of y = Bxx + ax + C is - A / 2B = 1 / 8, and the opening is upward, so the decreasing interval of y = Bxx + ax + C is (- ~, 1 / 8)

How to determine the values of a, B and C in quadratic function, such as y = ax ^ 2 + BX + C?

Three sets of (x, y) values are given
And then it's a system of three equations
For example:
When x = 1, y = 2
When x = 0, y = 4
When x = 2, y = 8
Then the numerical value is brought in and the ternary equations are set up
2=a+b+c
4=c
8=4a+2b+c
Then, the value of ABC can be calculated by the method of bringing in or other methods
The above formula: a = 4, B = - 6, C = 4

X1 and X2 are the two real roots of the quadratic equation x ^ 2-2 (m-1) x + m + 1 = 0 with respect to X. see the description for other contents
X1 and X2 are the two real roots of the quadratic equation x ^ 2-2 (m-1) x + m + 1 = 0 with respect to x, and y = X1 + x2. Find the analytic expression of y = f (m) and the domain of definition of this function

X1 and X2 are the two real roots of the quadratic equation x ^ 2-2 (m-1) x + m + 1 = 0,
According to Weida theorem: X1 + x2 = 2 (m-1)
So y = f (m) = 2m-2
Discriminant 4 (m-1) ^ 2-4 (M + 1) > = 0
m^2-2m+1-(m+1)>=0
m^2-3m>=0
m=3
So the domain: M > = 3 or m

Use square root to solve the following equation 1. The square of x = - 2. The square of 2.3 (x + 1) = 5

1. The square of x = - 2
∵ the square of any number cannot be negative,
The equation has no solution
2.3 (x + 1) = 5
Divide the two sides by three at the same time
(x+1)²=5/3
Squared on both sides
x+1=±√(5/3)=±(1/3)√15
∴x1=-1+(1/3)√15,x2=-1-(1/3)√15.

If the increasing range of quadratic function y = AXX + BX + C is (- ~, 2), what is the decreasing range of quadratic function y = Bxx + ax + C? Please give me a detailed explanation. Thank you very much Functions are really hard,