It is known that the quadratic function f (x) = ax Λ 2 + BX + 1 (a, B are constants, a > 0) satisfies f (x-3) = f (1-x) and has the maximum value 4 in the interval < - 3,2 > find the value of a, B and find the range of the function f (x) in the interval < - 3,2 >

It is known that the quadratic function f (x) = ax Λ 2 + BX + 1 (a, B are constants, a > 0) satisfies f (x-3) = f (1-x) and has the maximum value 4 in the interval < - 3,2 > find the value of a, B and find the range of the function f (x) in the interval < - 3,2 >


F (x-3) = f (1-x), so the axis of symmetry is [x-3 + 1-x] / 2 = 2, so - B / 2A = 2, and because there is a maximum value of 4 on the interval < - 3,2 >, and in this interval, f (- 3) = 4, so we get 9a-3b + 1 = 4, so we get a = 1 / 7, B = - 4 / 7, because the interval < - 3,2 > is simple



The opening of integral coefficient quadratic function y = ax ^ 2 + BX + C is upward, and the discriminant is B ^ 2-4ac = 5. The values of a, B and C are calculated between the two intersections of X axis and the interval (- 1,0), (1, + ∞)


x1=(-b-√5)/2a.-10
-2a



Given the following equation: √ 9 * 9 + 19 = 10, √ 99 * 99 + 199 = 100, √ 999 * 999 + 1999 = 1000, write the nth equation


√ 99 (n) 9 * 99 (n) 9 + 199 (n) 9 = 10 & # 8319;



According to the above characteristics, write the fourth formula and verify it


√(9999*9999+19999)=10000
Let a = 9999
Then 19999
=9999+9999+1
=2a+1
So the original formula = √ (A & # 178; + 2A + 1)
=√(a+1)²
=a+1
=10000