Let a be greater than 0, - 1 ≤ x ≤ 1, the minimum value of function y = - x ^ 2-ax + B + 1 is - 4, and the maximum value is 0 I know that the axis of symmetry is negative, so when x = 1, the minimum is - 4, but the answer is that when x = - 1, the minimum is 0, which I don't understand

Let a be greater than 0, - 1 ≤ x ≤ 1, the minimum value of function y = - x ^ 2-ax + B + 1 is - 4, and the maximum value is 0 I know that the axis of symmetry is negative, so when x = 1, the minimum is - 4, but the answer is that when x = - 1, the minimum is 0, which I don't understand

y=-(x²+ax+a²/4)+a²/4+b+1
=-(x+a/2)²+a²/4+b+1
1. When - 1

Let a > 0, when - 1 ≤ x ≤ 1, the minimum value of the function y = - x2 ax + B + 1 is - 4, and the maximum value is 0

Y = − x2 − ax + B + 1 = − (x + A2) 2 + A24 + B + 1; (1) if − A2 ≤− 1, i.e. a ≥ 2, the function y decreases monotonically on [- 1, 1]; {the minimum value of the function is B-A = - 4; the maximum value is a + B = 0, i.e. a = 2, B = - 2; (2) if − 1 ＜− A2 ＜ 0, i.e. 0 ＜ a ＜ 2, x = − A2, i.e

Given a = - 3, find the maximum and minimum value of the function y = x ^ 3 + ax + 1 on [0,2]

y=x³-3x+1
The derivation of function
y'=3x²-3
Let y '= 3x & sup2; - 3 = 0
Find x = 1 or x = - 1
So the function has only one extremum in the interval [0,2]
When x = 1, y = 1-3 + 1 = - 1
When x = 0, y = 0-0 + 1 = 1
When x = 2, y = 8-6 + 1 = 3
So the maximum value of the function in the interval [0,2] is 3 and the minimum value is - 1

Given the function f (x) = X2 - (2a + 1) x + alnx. (I) when a = 2, find the tangent equation of curve y = f (x) at point (1, f (1)); (II) find the monotone interval of function f (x)

(1) When a = 2, f (x) = X2 - (2a + 1) x + alnx = x2-5x + 2lnx, ｜ f ′ (x) = 2x − 5 + 2x, ｜ f ′ (1) = 2-5 + 2 = - 1, ∵ f (1) = 1-5 = - 4, the tangent equation of the curve y = f (x) at point (1, f (1)) is: x + y + 3 = 0. (II) f ′ (x) = 2x − (2a + 1) + AX = 2x2 − (2

As shown in the figure, if the parabola y = - x2 + 2 (M + 1) x + m + 3 intersects the X axis at two points a and B, and OA: OB = 3:1, then M=______ ．

Let a (3a, 0), B (- A, 0), 2A = 2 (M + 1) 3a · (− a) = − 3

If a = 3x ^ 2-2y ^ 2, B = x ^ 2 + y ^ 2, C = 2XY + x ^ 2, then x = - 1, y = 2, the value of a - [2B - [3a-c]

a-[2b-[3a-c]]=a-2b+3a-c=4a-2b-c=4(3x²-2y²)-2(x²+y²)-(2xy+x²)=9x²-10y²-2xy=9-40+4=-27

How to transform curve equation into curve parameter equation?

Cosine squared plus sine squared equals one. Just put it in your formula

（1)(x²-2x)²+2（x²-2x）+1
(2) 4X to the power of n-2x to the power of n-1, I will not reduce n

① Take (X & sup2; - 2x) as a whole, let it be y, then the formula is Y & sup2; + 2Y + 1 = (y + 1) & sup2;
That's it, and then take X & sup2; - 2x in
② N-1 times of 2x (2x-1)
We should learn the method instead of just knowing the answer

In the plane rectangular coordinate system, if the vertex coordinates of △ ABC are a (3,0), B (- 1,0), C (2,3) respectively, if the quadrilateral whose vertex is a, B, C, D is a parallelogram, then the coordinate of point D is______ ．

When ab ∥ CD, the coordinates of the fourth vertex D are (- 2,3) or (6,3). When ad ∥ BC, the coordinates of the fourth vertex D are (0, - 3). So the answer is (- 2,3) or (6,3) or (0, - 3)

When x tends to - 1, the limit of X is

The limit is 1
If you don't understand this question, you can ask,