The maximum value of the function y x 2 ax 3 0 a 2 on 1,1 is the minimum value is

The axis of symmetry is x=-a2 (1,0).

Since the parabolic opening is upward, the axis of symmetry is within the specified interval [-1,1].

Therefore, the function value corresponding to the vertex is the minimum value.

The formula yields y=x2+ax+3 =(x+a2) +3-a4, so when x=-a2, the minimum value is 3-a4, because the opening is upward, and the axis of symmetry is to the left of the midpoint of the interval [-1,1], so x=1 is farther away from the axis of symmetry than x=-1.

Therefore, when x=1, the maximum value is a+4

y=x2+ax+3 (0, the axis of symmetry is x=-a2 (1,0).)

y=y=x 2+ax+3 =(x+a 2) +3-a 4 on [-1,1].

So when x=-a2, there is a minimum value3- a4, and when x=1, there is a maximum.

The maximum value is A+4

Calculated with software.

The minimum is 2 and the maximum is 6

**：clear,clc;

n=1;for x=-1:1

for a=0:2

y(n)=x.^2+a*x+3;

n=n+1;

endend

min(y)

max(y, where:

min(y)……Minimum.

max(y)……Maximum.

y = x 2 + 2x + 1 is a quadratic function chain pose.

To find the maximum and minimum values of a quadratic function over a particular interval, you can find the vertices within the interval and determine whether the function value of that point is the maximum or minimum value of the interval.

First, find the vertices of the quadratic lead call function:

x = b 2a = 2) 2) =1 Then, calculate the function value at the vertice:

y = 1) 2 + 2(1) +1 = 2 Therefore, the vertex is (1, 2).

Because the quadratic function is a convex function, the vertex is the maximum value of the interval.

So, a ranges from 0 to 1.

Answer]: bSince y=ax must be a monotonic function, the maximum and minimum values of the circle of the known number on [0,1] must be obtained at x=0 and x=1. That is, the solution of a0+a1=3 is obtained by lifting the collapse a=2.

y=x is an increment function.

y= (x-1) is also an additive function.

Then y=x+ refers to God or (x-1) is an increasing function.

The defined domain satisfies x-1 0

x 1 is the only way to do it when x = 1

Time. The minimum value of the function is y=1 + 0=1

The axis of symmetry of the function is x=-a 2

Because of 0, when x=-a2, ymin=3-a2 4x=1, ymax=a+4

First, determine the axis of symmetry of the quadratic function: i.e., x=-2a b. This question is -a 2.

Because 0

The maximum is A+4

The minimum is 3-quarter a square.

1. If a>1, then the maximum is a, and the minimum is 1, that is, a 1=3, getting: a=2;

2. If 0 is synthetic, it gets: a=2

01, at this time, the maximum value is taken at a=1, which is a, and the minimum value is taken at a=0, which is 1

Thus there is 1+a=3

So a=2

Test Points: Maxima of Functions and Their Geometric Meanings Special Topic: Calculation Questions; Number Combination Analysis: Function y=x

The axis of symmetry of 2+ax+3(0 a 2) is x=-a2 (-1,0), and its image opening is upward, so the maximum value is y(1) and the minimum value is .

y(-a2) solution: solution: function y=x

The axis of symmetry of 2+ax+3(0 a 2) is x=-a2 (-1,0), and its image is open upward, so the maximum value is taken at x=1, its value is 4+a, and the minimum value is x=-

a2 with a value of .

3-a24, so the answer is: 4+a, 3-a24 Comments: The test point of this question is the maximum value of the function and its geometric significance, the test is judged by the image features and the maximum and minimum value of the function is calculated, and the maximum value of the quadratic function in the closed interval is the hot spot of the college entrance examination

The axis of symmetry is x=-a2

Since the parabolic opening is upward, the axis of symmetry is within the specified interval [-1,1], so the function value corresponding to the vertex is the minimum value.

The recipe yields y=x 2+ax+3

x+a/2）²+3-a²/4

Therefore, when x=-a 2, the minimum value is 3-a4, because the opening is upward, and the axis of symmetry is to the left of the midpoint of the interval [-1,1], so x=1 is farther away from the axis of symmetry than x=-1.

Therefore, when x=1, the maximum value is a+4

The function y=x 2+ax+3 can be deformed into y=(x+a 2) 2+3-(a 2) 4, according to the functional analysis (you can also draw a graph to help you understand), it gets: 1. The function image opens upward, and when x=-a 2, the minimum value 3-(a 2) 4 is obtainedThe 0 function is on [1,1], and when x=1 achieves the maximum: a+4.

Let x1 and x2 be the values of the two independent variables of the original letter bridge base number, and x10, x2-1>0, so 2(x1-x2) (x1-1)(x2-1)<0, so the original function is a subtraction function on the zone front space [2,6].

The maximum value is f(2)=2, and the minimum value is f(6)=2 5

What grade questions?

This function is y=2 x to shift the right of the friend by one unit.

So in the interval of 2-6, it is still a monotonically decreasing function.

So the maximum value is y(2)=2

The minimum value y(6)=2 disturbs Xiaohuai and is cautious 5