If the maximum value of the quadratic function y = AX2 + 4x + A is 3, then the value of a is______ .

From the meaning of the title, 4A · a − 42
4a=3，
After sorting out, a2-3a-4 = 0,
A1 = 4, A2 = - 1,
∵ a quadratic function has a maximum value,
∴a＜0，
∴a=-1．
So the answer is: - 1

Let's know that the square + 2aX + 1 of the quadratic function y = ax has the maximum value 4 in the interval [- 3,2]. Find the real number a

y=ax²＋2ax＋1
Its symmetry axis X = - 2A / 2 * a = - 1
① When a < 0, the image opening is downward
∵－1∈[－3,2]
The function has a maximum value of 4 when x = - 1
That is, 4 = a (- 1) 2-2a + 1 = a-2a + 1 = 1-A
=>a=－3
② When a > 0, the opening of the image is upward, and - 3 is the farthest from the axis of symmetry x = 1, so x = - 3 gets the maximum value of 4
∴4=a(－3)²＋2a(－3)＋1
=>9a－6a＋1=4
=>3a=3
=>a=1
To sum up, a = 1 or a = - 3

Given that the maximum value of the quadratic function y = ax ^ 2 + BX + C is equal to - 3a, and its image passes through (- 1, - 2), (1,6) two points, find a, B, C

Because there is a maximum, the opening must be downward
4ac-b squared / 4A = - 3A
Then bring two points into the equation, three equations a for two answers, take the positive tongue

Given the quadratic function f (x) = (4-3a) x ^ 2-2x + A, find the maximum value of F (x) on the interval [0,1]

If f (x) is a quadratic function, then 4-3a ≠ 0, that is, a ≠ 4 / 3
The symmetry axis of F (x) is: x = 1 / (4-3a)
4) / (3 A) / (3 A)
The function f (x) is a decreasing function on [0,1]
When x = 0, f (x) has a maximum value, f (0) = a
If 4-3a > 0, then 1 / (4-3a) > 0
① 1 / (4-3a) ≥ 1, i.e. 0 < 4-3a ≤ 1, 1 ≤ a < 4 / 3
The function f (x) is a decreasing function on [0,1]
When x = 0, f (x) has a maximum value, f (0) = a
② When 4-3a > 1, a < 1
f(0)=a
f(1)=4-3a-2+a=2-2a
Let f (1) - f (0) > 0
2-2a-a>0
a<2/3
Then:
When a < 2 / 3, f (1) > F (0), the maximum value of F (x) in the interval [0,1] is f (1) = 2-2a
When a = 2 / 3, f (1) = f (0) = 2 / 3, and the maximum value of F (x) on the interval [0,1] is 2 / 3
When 2 / 3 says:
When a < 2 / 3, the maximum value of F (x) on the interval [0,1] is f (1) = 2-2a
When a = 2 / 3, the maximum value of F (x) on the interval [0,1] is 2 / 3
When 2 / 34 / 3, the maximum value of F (x) on the interval [0,1] is f (0) = a

It is known that the quadratic function f (x) = AX2 + (2a-1) x + 1 is in the interval [- 3] 2, 2] is 3. Find the value of real number a

Because the maximum value of quadratic function f (x) in the interval [- 32, 2] is 3, there must be f (− 2A − 12a) = 3, or F (2) = 3, or F (− 32) = 3. (1) if f (− 2A − 12a) = 3, i.e. 1 - (2a − 1) 24a = 3, a = - 12 is obtained. At this time, the parabola opening is downward, the equation of symmetry axis is x = - 2, and − 2 ∉ [

Given that the maximum value of quadratic function f (x) = ax square + (2a-1) x + 1 in the interval 〔 - 3 / 2,2] is 3, find the value of real number a In class, the teacher said a little. It seems that we should first find out the axis of symmetry, and then we will discuss something

If the symmetry axis also needs to be divided into different situations in an interval, we should not generalize, or we should not divide them in general
Axis of symmetry (1-2a) / 2a, that interval is symmetric about the point (0.25,0)
1°a>0,
1) If the axis of symmetry is less than 0.25, then f (2) = 3 and a = 0.5
2) When the axis of symmetry is greater than 0.25, f (- 1.5) = 3, and a = - 2 / 3 is obtained

If the function f (x) = AX2 + 2aX + 1 has the maximum value 4 on [- 3, 2], then the real number a=______ .

① When a > 0, because the axis of symmetry is x = - 1, f (2) is the largest, so f (2) = 4, that is, 4A + 4A + 1 = 4, so a = 3
8；
② When a < 0, because the axis of symmetry is x = - 1, f (- 1) is the smallest, so f (- 1) = 4, that is, a-2a + 1 = 4, so a = - 3;
③ When a = 0, f (x) = 1 does not hold
To sum up, a = 3
8 or a = - 3
So the answer is: 3
8 or - 3

Given that the maximum value of quadratic function f (x) = AX2 + 4ax + A2-1 in the interval [- 4,1] is 5, find the value of real number a

Analysis: the symmetric axis equation of F (x) is x = - 2, and the vertex coordinates are. Obviously, the abscissa of its vertex is in the interval [- 4,1). (3 points) (1) if a < 0, the opening of the function image is downward, as shown in Figure (a). When x = - 2, the function obtains the maximum value of 5, that is, f (- 2) = a2-4a-1 = 5, the solution is a = 2-10, (a = 2 + 10 rounding)

It is known that the function f (x) = AX2 + (2a-1) x-3 is in the interval [- 3 If the maximum value of 2,2] is 1, find the value of real number a

When a = 0, f (x) = - x-3, f (x) cannot obtain 1 on [− 32, 2]. Therefore, a ≠ 0, then the symmetric axis equation of F (x) = AX2 + (2a-1) x-3 (a ≠ 0) is x0 = 1 − 2a2a