Algebra lesson notes for 8th grade secondary school
Lesson topic: Function
Objective of the lesson:
· Educational: define the concept of a quadratic function of the form (compare graphs of functions and), show the formula for finding the coordinates of the vertex of a parabola (teach how to use this formula in practice); to develop the ability to determine the properties of a quadratic function from a graph (finding the axis of symmetry, the coordinates of the vertex of a parabola, the coordinates of the points of intersection of the graph with the coordinate axes).
· Developmental: development of mathematical speech, the ability to correctly, consistently and rationally express one’s thoughts; developing the skill of correctly writing mathematical text using symbols and notations; development analytical thinking; development cognitive activity students through the ability to analyze, systematize and generalize material.
· Educational: fostering independence, the ability to listen to others, developing accuracy and attention in written mathematical speech.
Lesson type: learning new material.
Teaching methods:
generalized reproductive, inductive heuristic.
Requirements for students' knowledge and skills
know what a quadratic function of the form is, the formula for finding the coordinates of the vertex of a parabola; be able to find the coordinates of the vertex of a parabola, the coordinates of the points of intersection of the graph of a function with the coordinate axes, and use the graph of a function to determine the properties of a quadratic function.
Equipment:
Lesson Plan
I. Organizational moment(1-2 min)
II. Updating knowledge (10 min)
III. Presentation of new material (15 min)
IV. Consolidating new material (12 min)
V. Summing up (3 min)
VI. Homework assignment (2 min)
Lesson progress
I. Organizational moment
Greeting, checking absentees, collecting notebooks.
II. Updating knowledge
Teacher: In today's lesson we will study a new topic: "Function". But first, let's repeat the previously studied material.
Frontal survey:
1) What is called a quadratic function? (A function where given real numbers, a real variable, is called a quadratic function.)
2) What is the graph of a quadratic function? (The graph of a quadratic function is a parabola.)
3) What are the zeros of a quadratic function? (The zeros of a quadratic function are the values at which it becomes zero.)
4) List the properties of the function. (The values of the function are positive at and equal to zero at; the graph of the function is symmetrical with respect to the ordinate axes; at - the function increases, at - decreases.)
5) List the properties of the function. (If, then the function takes positive values at, if, then the function takes negative values at, the value of the function is 0 only; the parabola is symmetrical about the ordinate axis; if, then the function increases at and decreases at, if, then the function increases at, decreases at.)
III. Presentation of new material
Teacher: Let's start learning new material. Open your notebooks, write down the date and topic of the lesson. Pay attention to the board.
Writing on the board: Number.
Function.
Teacher: On the board you see two graphs of functions. The first graph, and the second. Let's try to compare them.
You know the properties of the function. Based on them, and comparing our graphs, we can highlight the properties of the function.
So, what do you think will determine the direction of the branches of the parabola?
Students: The direction of the branches of both parabolas will depend on the coefficient.
Teacher: Absolutely right. You can also notice that both parabolas have an axis of symmetry. In the first graph of the function, what is the axis of symmetry?
Students: For a parabola, the axis of symmetry is the ordinate axis.
Teacher: Right. What is the axis of symmetry of a parabola?
Students: The axis of symmetry of a parabola is the line that passes through the vertex of the parabola, parallel to the ordinate axis.
Teacher: Right. So, the axis of symmetry of the graph of a function will be called a straight line passing through the vertex of the parabola, parallel to the ordinate axis.
And the vertex of a parabola is a point with coordinates. They are determined by the formula:
Write the formula in your notebook and circle it in a frame.
Writing on the board and in notebooks
Coordinates of the vertex of the parabola.
Teacher: Now, to make it more clear, let's look at an example.
Example 1: Find the coordinates of the vertex of the parabola.
Solution: According to the formula
we have:
Teacher: As we have already noted, the axis of symmetry passes through the vertex of the parabola. Look at the board. Draw this picture in your notebook.
Write on the board and in notebooks:
Teacher: In the drawing: - equation of the axis of symmetry of a parabola with the vertex at the point where the abscissa of the vertex of the parabola is.
Let's look at an example.
Example 2: Using the graph of the function, determine the equation for the axis of symmetry of the parabola.
The equation for the axis of symmetry has the form: , which means the equation for the axis of symmetry of a given parabola.
Answer: - equation of the axis of symmetry.
IV. Consolidating new material
Teacher: The tasks that need to be solved in class are written on the board.
Writing on the board: № 609(3), 612(1), 613(3)
Teacher: But first, let's solve an example not from the textbook. We will decide at the board.
Example 1: Find the coordinates of the vertex of a parabola
Solution: According to the formula
we have:
Answer: coordinates of the vertex of the parabola.
Example 2: Find the coordinates of the points of intersection of the parabola with the coordinate axes.
Solution: 1) With axle:
According to Vieta's theorem:
The points of intersection with the x-axis are (1;0) and (2;0).
2) With axle:
The point of intersection with the ordinate axis (0;2).
Answer: (1;0), (2;0), (0;2) – coordinates of the points of intersection with the coordinate axes.
No. 609(3). Find the coordinates of the vertex of the parabola
Solution: Abscissa of the vertex of a parabola:
Parabola vertex ordinate:
Answer: - coordinates of the vertex of the parabola.
No. 612(1). Does the axis of symmetry of the parabola pass through the point (5;10)?
Solution: Equation of the axis of symmetry: .
Find the abscissa of the vertex of the parabola: . So, the equation of the axis of symmetry looks like this. Let us schematically draw this parabola:
Consequently, the axis of symmetry passes through the point (5;10).
No. 613(3). Find the coordinates of the points of intersection of the parabola with the coordinate axes.
Solution: 1) With axle:
We are looking for a discriminant:
This means that there are no intersection points with the abscissa axis.
The point of intersection with the ordinate axis (0;12).
Answer: (0;12) – coordinates of the point of intersection with the ordinate axis; the parabola does not intersect with the abscissa axis.
V. Summing up
Teacher: In today's lesson we studied a new topic: "Function", learned to find the coordinates of the vertex of a parabola, the coordinates of the points of intersection of the parabola with the coordinate axes. In the next lesson we will continue solving problems on this topic.
VI. Homework
Teacher: The homework assignment is written on the board. Write it down in your diaries.
Writing on the board and in the diaries: §38, No. 609(2), 612(2), 613(2).
Literature
1. Alimov Sh.A. Algebra 8th grade
2. Sarantsev G.I. Methods of teaching mathematics in secondary school
3. Mishin V.I. Private technique teaching mathematics in high school
Description of the video lesson
Let's consider some special cases of the quadratic function.
First case. Let's find out what the graph of the function ig is equal to one third x square plus four represents.
To do this, in one coordinate system, we will construct graphs of the functions yrek equals one third x square... and yrek equals one third x square plus four.
Let's make a table of the values of the function yrek equals one third x square. Let's build according to given points function graph.
To get a table of values of the function igrek equals one third x square plus four with the same values of the argument, you should add four to the found values of the function igrek equals one third x square...
Let's make a table of values for the graph of the function igreq equals one third x square plus four. Let's construct points using the specified coordinates and connect them with a smooth line. We get the graph of the function igreq equals one third x square plus four.
It is easy to understand that the graph of the function yrek equals one third x square plus four can be obtained from the graph of the function yrek equals one third x square by parallel translation four units up along the y axis.
Thus, the graph of the function ygr is equal to a x square plus en is a parabola, which is obtained from the graph of the function ygr equals a x square using parallel translation along the y axis by modulus en units upward, if en greater than zero or down if en less than zero.
Second case. Let's consider the function ig is equal to one third of the square of the difference between the numbers x and six and construct its graph.
Let's build a table of values of the function i is equal to one third x square, indicate the obtained points on coordinate plane and connect with a smooth line.
Now let's make a table of values for the function ig is equal to one third of the square of the difference between the numbers x and six. Using the indicated points, we will construct a graph of the function.
It is noticeable that each point of the second graph is obtained from the corresponding point of the first graph using a parallel translation of six units along the x-axis.
The graph of the function ygr is equal to a multiplied by the square of the difference x and em... is a parabola that can be obtained from the graph of the function ygr is equal to a x square by parallel translation along the x axis by the modulus em units to the left, if em is greater than zero or by the modulus em units to the right if um is less than zero.
Let us now consider the graph of the function igreq equals one third times the square of the difference of x and two plus five. Its graph can be obtained from the graph of the function igreq equals one third x squared using two parallel translations - shifting the parabola to the right by two units and upward by five units.
In this case, parallel translations can be made in any order: first along the x-axis, and then along the y-axis, or vice versa.
But why, when adding the number en to a function, its graph moves up by a modulus of en units, if en is greater than zero or down, if en is less than zero, and when adding the number em to an argument, the function moves by a modulus of em units to the right, if em is less than zero or to the left if um is greater than zero?
Let's consider first case. Let it be necessary to construct a graph of the function yrek equals ef from x.. plus en. Note that the ordinates of this graph for all values of the argument are en units greater than the corresponding ordinates of the graph yrek equal to eff of x for positive en and en units less for negative en. Therefore, the graph of the function ygr is equal to ef from x...plus en can be obtained parallel transfer along the ordinate axis of the graph of the function ygr is equal to ef from x by modulus en units up, if en is greater than zero, and by modulus en units down, if en is less than zero.
Let's consider second case. Let it be necessary to construct a graph of the function yreq equals ef from the sum of x and em. Let's consider the function yrek equals ef from x, which at some point x equal to x the first takes on the value yk first is equal to ef from x first. Obviously, the function ygr is equal to ef from the sum of x and em will take the same value at the point x-second, the coordinate of which is determined from the equality x-second plus em is equal to x-first, that is, x-first is equal to x-first minus em. Moreover, the equality under consideration is valid for all values of x from the domain of definition of the function. Consequently, the graph of the function can be obtained by parallel moving the graph of the function igreq equals ef from x along the abscissa axis to the left by the module em units to the left, if em is greater than zero, and by the module em to the right, if em is less than zero. Parallel movement of the graph of a function along the x-axis by em units is equivalent to moving the y-axis by the same number of units, but in the opposite direction.
When a parabola rotates around its axis, a figure is obtained that is called a paraboloid. If inner surface make a mirror paraboloid and direct a beam of rays parallel to the axis of symmetry of the parabola at it, then the reflected rays will converge at a point called the focus. At the same time, if the light source is placed in focus, then reflected from mirror surface paraboloid, the rays will be parallel and not scattered.
The first property allows us to obtain at the focus of a paraboloid high temperature. According to legend, this property was used by the ancient Greek scientist Archimedes. During the defense of Syracuse in the war against the Romans, he built a system of parabolic mirrors, which made it possible to focus the reflected sun rays on the ships of the Romans. As a result, the temperature at the foci of the parabolic mirrors turned out to be so high that a fire broke out on the ships and they burned down. This property is also used in the manufacture of parabolic antennas.
The second property is used in the manufacture of spotlights and car headlights.
Determining the values of the coefficients of a quadratic function from a graph.
Methodological development by Sagnaeva A.M.
MBOU secondary school No. 44, Surgut, Khanty-Mansi Autonomous Okrug-Yugra .
Ι. Finding the coefficient A
- Using the graph of a parabola, we determine the coordinates of the vertex (m,n)
2. Using the graph of a parabola, we determine the coordinates of any point A (X 1 ;y 1 )
3. Substitute these values into the formula of a quadratic function specified in a different form:
y=a(x-m)2+n
4. solve the resulting equation.
Oh 1 ;y 1 )
parabola
ΙΙ. Finding the coefficient b
1. First we find the value of the coefficient a
2. In the formula for the abscissa of a parabola m= -b/2a substitute the values m And a
3. Calculate the value of the coefficient b .
Oh 1 ;y 1 )
parabola
ΙΙΙ. Finding the coefficient c
1. We find the ordinate of the point of intersection of the parabola graph with the Oy axis, this value is equal to the coefficient With, i.e. dot (0;s)-the point of intersection of the parabola graph with the Oy axis.
2. If it is impossible to find the point of intersection of the parabola with the Oy axis from the graph, then we find the coefficients a,b
(see steps Ι, ΙΙ)
3. Substitute the found values a, b ,A(x 1; at 1 ) into the equation
y=ax 2 +bx+c and we find With.
Oh 1 ;y 1 )
parabola
Tasks
clue
Ιx 2 Ι , a x 1 0, because a The ordinate of the point of intersection of the parabola with the OY axis is the coefficient c Answer: 5 c x 1 x 2 "width="640" - The branches of the parabola are directed downwards,
- The roots have different signs,Ι x 1 ΙΙх 2 Ι , and x 1 0, because a
- The ordinate of the point of intersection of the parabola with the OY axis is the coefficient With
X 1
X 2
P Clue
0 x 1 +x 2 = - b/a 0. a 0. Answer: 5 "width="640" 1.The branches of the parabola are directed downwards, which means a
- x 1 +x 2 = - b/a 0. a 0.
0 because the branches of the parabola are directed upward; 2. c=y(0)3. The vertex of the parabola has a positive abscissa: in this case a 0, therefore b4. D0, because the parabola intersects the OX axis at two different points. "width="640" The figure shows a graph of the function y=ax 2 +bx+c. Specify signs coefficients a,b,c and discriminant D.
Solution:
1. a0, because the branches of the parabola are directed upward;
3. The vertex of the parabola has a positive abscissa:
in this case a 0, therefore b
4. D0, because the parabola intersects the OX axis at two different points.
The picture shows a parabola
Specify values k And t .
Find the coordinates of the vertex of the parabola and write the function whose graph is shown in the figure.
Find where are the abscissas of the intersection points
parabolas and horizontal straight lines (see figure).
A bad teacher presents the truth, a good teacher teaches how to obtain it.
A.Disterweg
Teacher: Netikova Margarita Anatolyevna, mathematics teacher, GBOU school No. 471 Vyborg district St. Petersburg.
Lesson topic: “Graph of a functiony= ax 2 »
Lesson type: lesson in learning new knowledge.
Target: teach students to graph a function y= ax 2 .
Tasks:
Educational: develop the ability to construct a parabola y= ax 2 and establish a pattern between the graph of the function y= ax 2
and coefficient A.
Educational: development cognitive skills, analytical and comparative thinking, mathematical literacy, ability to generalize and draw conclusions.
Educators: nurturing interest in the subject, accuracy, responsibility, demandingness towards oneself and others.
Planned results:
Subject: be able to use a formula to determine the direction of the branches of a parabola and construct it using a table.
Personal: be able to defend your point of view and work in pairs and in a team.
Metasubject: be able to plan and evaluate the process and result of their activities, process information.
Pedagogical technologies: elements of problem-based and advanced learning.
Equipment: interactive whiteboard, computer, handouts.
1.Formula of roots quadratic equation and decomposition quadratic trinomial by multipliers.
2. Reduction of algebraic fractions.
3.Properties and graph of the function y= ax 2 , dependence of the direction of the branches of the parabola, its “stretching” and “compression” along the ordinate axis on the coefficient a.
Lesson structure.
1.Organizational part.
2.Updating knowledge:
Examination homework
Oral work based on finished drawings
3.Independent work
4.Explanation of new material
Preparing to study new material (creating a problem situation)
Primary assimilation of new knowledge
5. Fastening
Application of knowledge and skills in a new situation.
6. Summing up the lesson.
7.Homework.
8. Lesson reflection.
Technological map of an algebra lesson in 9th grade on the topic: “Graph of a functiony=
ax 2
»
| Lesson steps | Stage tasks | Teacher activities | Student activities | UUD |
| 1.Organizational part 1 minute | Creating a working mood at the beginning of the lesson | Greets students checks their preparation for the lesson, notes those absent, writes the date on the board. | Getting ready to work in class, greeting the teacher | Regulatory: organization of educational activities. |
| 2.Updating knowledge 4 minutes | Check homework, repeat and summarize the material learned in previous lessons and create conditions for successful independent work. | Collects notebooks from six students (selectively two from each row) to check homework for assessment (Appendix 1), then works with the class on interactive whiteboard (Appendix 2). | Six students hand in their homework notebooks and then answer questions. frontal poll (Appendix 2). | Cognitive: bringing knowledge into the system. Communicative: the ability to listen to the opinions of others. Regulatory: evaluating the results of your activities. Personal: assessing the level of mastery of the material. |
| 3.Independent work 10 minutes | Test your ability to factor a quadratic trinomial and reduce algebraic fractions and describe some properties of functions based on its graph. | Hands out cards to students with individual differentiated task (Appendix 3). and solution sheets. | Execute independent work, independently choosing the difficulty level of exercises based on points. | Cognitive: Personal: assessing the level of mastery of the material and one’s capabilities. |
| 4.Explanation of new material Preparing to study new material Primary assimilation of new knowledge | Creating a favorable environment for getting out of a problematic situation, perception and comprehension of new material, independent coming to the right conclusion | So, you know how to graph a function y= x 2 (graphs are pre-built on three boards). Name the main properties of this function: 3. Vertex coordinates 5. Periods of monotony What's in in this case equal to the coefficient at x 2 ? Using the example of the quadratic trinomial, you saw that this is not at all necessary. What sign could he be? Give examples. You will have to find out for yourself what parabolas with other coefficients will look like. The best way study something is to discover for yourself. D.Poya We divide into three teams (in rows), choose captains who come to the board. The task for the teams is written on three boards, the competition begins! Construct function graphs in one coordinate system 1 team: a)y=x 2 b)y= 2x 2 c)y= x 2 Team 2: a)y= - x 2 b)y=-2x 2 c)y= - x 2 Team 3: a)y=x 2 b)y=4x 2 c)y=-x 2 Mission accomplished! (Appendix 4). Find functions that have identical properties. Captains consult with their teams. What does this depend on? But how do these parabolas differ and why? What determines the “thickness” of a parabola? What determines the direction of the branches of a parabola? We will conventionally call graph a) “initial”. Imagine a rubber band: if you stretch it, it becomes thinner. This means that graph b) was obtained by stretching the original graph along the ordinate. How was graph c) obtained? So, when x 2 there can be any coefficient that affects the configuration of the parabola. This is the topic of our lesson: "Graph of a functiony= ax 2 » | 1. R 4. Branches up 5. Decreases by (- Increases by ) Did you like the article? Share with your friends!
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