Students' knowledge and skills base: starting with a simple quadratic function Y = x2, then Y = AX2, Y = AX2+C, and finally y=a(x-h)2, Y = A (X-H) 2+K, Y = AX2+BX+C. Students have mastered three representations of quadratic functions and the properties of images.
Experience basis of students' activities: By solving the profit maximization with quadratic function, students have experienced the process of transforming practical problems into mathematical problems and have experience in solving such problems.
Second, the analysis of teaching tasks
This lesson will further use quadratic function to solve practical problems, which is the sublimation and improvement of the previous section. The specific teaching objectives are as follows:
Knowledge and skills
Can analyze and express the quadratic function relationship between variables in practical problems under different backgrounds, and can use the knowledge of quadratic function to solve the maximum (minimum) value in practical problems.
(2) Process and method
1. Cultivate students' analytical judgment ability by analyzing and expressing the quadratic function relationship between variables in practical problems under different backgrounds.
2. By using the knowledge of quadratic function to solve practical problems, cultivate students' mathematical application ability.
(3) Emotional attitudes and values
1. By exploring the process of maximum light transmission area of rectangle and window, we can further gain the experience of solving practical problems by using mathematical methods, and further feel the application value of mathematical model ideas and mathematical knowledge.
2. Be able to reflect on the basic strategies to solve problems and form a personal problem-solving style.
3. Further understand the close relationship between mathematics and human society, understand the value of mathematics, enhance the understanding of mathematics and confidence in learning mathematics well, and have certain innovative spirit and practical ability.
Teaching focus
1. By exploring the process of maximum light transmission area of rectangle and window, we can further gain the experience of solving practical problems by using mathematical methods, and further feel the application value of mathematical model ideas and mathematical knowledge.
2. Be able to analyze and express the quadratic function relationship between variables in practical problems under different backgrounds, and use the knowledge of quadratic function to solve practical problems.
Teaching difficulties
Find the expression of quadratic function from the chart.
Third, the analysis of teaching process
This lesson is divided into five parts, which are: creating problem situations to introduce new lessons, induction and sublimation, exploration of classroom practice activities, classroom summary and homework.
The first step is to create problem situations and introduce new courses.
Design Description: By reviewing and analyzing the contents of the last class, we will introduce students into the world of solving problems with quadratic function again, stimulate students' desire to continue exploring and introduce new course contents.
Last class, we solved the problem of profit maximization with quadratic function. As we know, to maximize profit is to find the maximum value of quadratic function, which is actually to solve practical problems with quadratic function. The key to solve this kind of problem is to understand the meaning of the problem, clarify what is to be solved, analyze the relationship between various quantities in the problem, express the problem from a mathematical point of view, clarify what is known and what is sought, and establish a mathematical model. On this basis, using the learned mathematical knowledge, we can gradually get the answers to the questions.
In this lesson, we will continue to use quadratic function to solve the maximum area problem.
Activity content: It consists of four practical questions.
1. Question 1: As shown below, make a rectangle ABCD in a right triangle, where AB and AD are on two right-angled sides respectively.
(1) Let one side of the rectangle AB = x m, then how to express the length of the AD side?
(2) Let the area of a rectangle be y m2, and when X is taken, what is the maximum value of Y? What is the maximum value?
Design purpose of question 1:
Students have seen the same kind of problems when they are studying similarly, so we should leave some time for students to think and communicate in class, so that students can give full play to their dominant position in class. After students give full play to their independent inquiry ability, teachers should cooperate with students to complete the answers to questions. The purpose of doing this is to set an example for students in their future study and help them form a complete solution process in their minds. The specific process is as follows:
Analysis: (1) needs the length of a D side, that is, the length of a BC side, and BC is an edge at △EBC, so BC can be calculated by triangle similarity. It is derived from △EBC∽△EAF, so AD = BC = (40-x).
(2) Finding the maximum value of area y, that is, finding the maximum value of function y = ab ad = x (40-x), is a mathematical problem.
Next, teachers and students will cooperate with each other to complete the answering process.
( 1)∵BC∨AD，
∴△EBC∽△EAF.∴ .
And ab = x, be = 40-x,
∴ .∴BC= (40-x)。
∴ad=bc=(40 x)= 30 x
(2)y=AB AD=x(30- x)=- x2+30x
=- (x2-40x+400-400)
=- (x2-40x+400)+300
=- (x-20)2+300。
When x = 20, the maximum value of y = 300.
That is, when x is 20m, the value of y is the largest, and the maximum value is 300m2.
What have you learned from thinking about solving such problems?
Firstly, analyze the meaning of the problem and find out the relationship between variables. It is found that finding the area is finding the two sides of a rectangle. Secondly, the two sides are represented by an algebraic formula containing X. Finally, the actual problem is brought into the area formula, transformed into a mathematical problem, and solved by mathematical methods. )
2. Question 2: Change the question: "Let the length of the AD side be x m, what will the question be?"
The design purpose of question 2:
The problems that students encounter in life are ever-changing, and they should have the ability to analyze specific problems. Therefore, after making certain changes, students can solve such problems independently through their own analysis. So as to improve students' ability to think and solve problems independently.
Analysis: the area requires the side length of AB, and AB=CD, so it requires the length of DC, which is one side of △FDC, so it can be obtained by triangle similarity.
Solution: ∫DC∨AB,
∴△FDC∽△FAE.
∴ .
AD = x，FD=30-x。
∴ .
∴DC= (30-x)。
∴AB=DC= (30-x)。
y=AB AD=x (30-x)
=- x2+40x
=- (x2-3 0x+225-225)
=- (x- 15 )2+300。
When x = 15, y max = 300.
That is, when the AD length is 15m, the rectangular area is the largest, and the largest area is 300m2.
Perception: the maximum problem in the same practical problem has nothing to do with the set independent variables, and it exists objectively.
Teaching instructions:
In class, students are required to complete the complete answer to this question independently. Ask one or two students to act it out, and then the other students will comment to find out the perfect answer process. It embodies the consciousness and ability of students' independent exploration and cooperation, and also fully embodies the incentive function of life evaluation.
3. Question 3: Change the question repeatedly
The design purpose of question 3:
The answer to the second question will make some students completely follow the format of the first question, which will be a bit unskilled at this time, but the third question reinterprets the problem with the largest area from another angle. That is, let students re-examine this problem and thoroughly understand the way of thinking of this kind of problem. Let students see real math problems in class, feel that math is closely related to life, and let students really understand the value of math.
As shown in the figure, make a rectangle ABCD in a right triangle, where point A and point D are on two right angles and BC is on the hypotenuse.
(1). Let one side of the rectangle BC=xm, then how to express the length of the AB side?
(2) Let the area of the rectangle be ym2. What is the maximum value of y when x is taken?
Analysis: This question has been sublimated on the basis of the first two questions. The relationship between the two variables cannot be found directly from the graph, but only depends on the height on the hypotenuse of the right triangle. Therefore, this question needs to add an auxiliary line-the height on the hypotenuse.
Activity purpose:
With the first two questions as the basis, the teacher can lead the students to analyze this problem first, and then leave it as an exercise for the students to solve.
4. Question 4:
The design purpose of question 4:
Basic training on the problem of maximum area has been involved before. Here, the improvement questions aim at improving students' ability to solve problems.
The windows of the building are shown in the following figure. Its upper part is semi-circular and its lower part is rectangular. The total length of the materials used to make the window frame (the sum of the lengths of all black lines in the figure) is 15m. When x is equal to what, the window passes the most light (the result is accurate to 0.0 1m)? At this time, what is the area of the window?
Analysis: X is the radius of a semicircle, and 2x is the long side of a rectangle, so X is related to both the semicircle area and the rectangular area, which requires the maximum light passing through the window, that is, to find the maximum sum of the rectangle and semicircle area, that is, 2xy+ x2, y = because 4y+4x+3x+π x = 7x+4y+π x = 15. The area s = π x2+2xy = π x2+2x = π x2+=-3.5x2+7.5x. At this time, it has been transformed into a mathematical problem, that is, a quadratic function, as long as it is converted into a vertex or substituted into a vertex coordinate formula,
Solution: ∫7x+4y+πx = 15,
∴y=。
Let the area of the window be S(m2), then
S= πx2+2xy
= πx2+2x
= πx2+
=-3.5x2+7.5x
=-3.5(x2- x)
=-3.5(x- )2+。
When x = ∴ 1.07,
S max = ≈ 4.02。
That is, when x≈ 1.07m, the maximum S ≈ 4.02m2, at which time the window passes the most light.
Actual teaching effect:
Compared with the first three questions, the quantitative relationship in the fourth question is more complicated to handle, and teachers should give students timely guidance and help.
The second link is induction and sublimation.
Activity content:
Can students summarize according to the previous examples, and what is the basic idea to solve this kind of problem? Communicate with your peers.
Activity purpose:
Through the study and feelings of the previous examples, the students discuss and communicate, and with the help of teachers, draw a conclusion:
The basic process is: understanding the topic, analyzing the known and unknown quantities, and transforming them into mathematical problems.
The basic idea to solve this kind of problem is:
(1) Understanding the problem;
(2) Analyze the variables and constants in the problem and their relationships;
(3) The relationship between variables is expressed by quadratic function;
(4) determining the maximum value or minimum value;
(5) Test the rationality of the results and expand the application.
The third part is classroom exercises and activities.
Activity content:
Build a rectangular chicken farm with a 48-meter bamboo fence. One side of the chicken farm is brick, the other three sides are surrounded by bamboo fences, and a 2-meter-wide door (no fence) is opened on the opposite side of the brick wall. When the side length of a chicken farm is several meters, which chicken farm occupies the largest area? What is the maximum area?
Design description:
Through the study of a class, let students further feel the idea of finding the largest area with quadratic function. In order to let more students experience success, we should use this relatively simple question to consolidate in time and help students build confidence.
Section 4 Summary of Link Classes
In this lesson, we further learned how to solve the maximum area problem with quadratic function knowledge, enhanced the consciousness of applying mathematical knowledge, gained the experience of solving practical problems with mathematical methods, and further felt the application value of mathematical modeling thought and mathematical knowledge.
Description: It aims to cultivate students' plastic thinking and awareness of cooperation and communication. Ask students to summarize the content of this lesson independently in class. Teachers should encourage, praise and affirm.
The fifth link homework after class
Exercise 2.8 1, 2
Fourth, evaluation and reflection.
The purpose of this lesson is to let students experience the process of maximum light transmission of rectangles and windows, further gain the experience of solving practical problems with mathematical methods, and further feel the application value of mathematical model ideas and mathematics.
In teaching, students should be given as much time as possible to think and reflect, so that they can translate what the teacher has taught into their own understanding, and students can solve problems with their own cognition. Teachers only need to give timely guidance. Classroom is a classroom for students, and students' creativity is infinite. Students should be allowed to play and create in class.
Because students' mathematical language expression ability is still lacking, the training of logical thinking ability needs to be strengthened, so writing training in problem-solving process should be strengthened in class.