Equations with parameters are rightfully considered one of the most difficult problems in school mathematics. It is precisely these tasks that end up year after year on the list of tasks of type B and C in the unified state exam of the Unified State Examination. However, among large number equations with parameters are those that can be easily solved graphically. Let's consider this method using the example of solving several problems.
Find the sum of integer values of the number a for which the equation |x 2 – 2x – 3| = a has four roots.
Solution.
To answer the question of the problem, let’s construct graphs of functions on one coordinate plane
y = |x 2 – 2x – 3| and y = a.
Graph of the first function y = |x 2 – 2x – 3| will be obtained from the graph of the parabola y = x 2 – 2x – 3 by symmetrically displaying with respect to the x-axis that part of the graph that is below the Ox-axis. The part of the graph located above the x-axis will remain unchanged.
Let's do this step by step. The graph of the function y = x 2 – 2x – 3 is a parabola, the branches of which are directed upward. To build its graph, we find the coordinates of the vertex. This can be done using the formula x 0 = -b/2a. Thus, x 0 = 2/2 = 1. To find the coordinate of the vertex of the parabola along the ordinate axis, we substitute the resulting value for x 0 into the equation of the function in question. We get that y 0 = 1 – 2 – 3 = -4. This means that the vertex of the parabola has coordinates (1; -4).
Next, you need to find the intersection points of the parabola branches with the coordinate axes. At the points of intersection of the parabola branches with the abscissa axis, the value of the function is zero. Therefore, we solve the quadratic equation x 2 – 2x – 3 = 0. Its roots will be the required points. By Vieta’s theorem we have x 1 = -1, x 2 = 3.
At the points of intersection of the parabola branches with the ordinate axis, the value of the argument is zero. Thus, the point y = -3 is the point of intersection of the branches of the parabola with the y-axis. The resulting graph is shown in Figure 1.
To obtain a graph of the function y = |x 2 – 2x – 3|, let us display the part of the graph located below the abscissa symmetrically relative to the x-axis. The resulting graph is shown in Figure 2.
The graph of the function y = a is a straight line parallel to the abscissa axis. It is depicted in Figure 3. Using the figure, we find that the graphs have four common points (and the equation has four roots) if a belongs to the interval (0; 4).
Integer values of the number a from the resulting interval: 1; 2; 3. To answer the question of the problem, let’s find the sum of these numbers: 1 + 2 + 3 = 6.
Answer: 6.
Find the arithmetic mean of integer values of the number a for which the equation |x 2 – 4|x| – 1| = a has six roots.
Let's start by plotting the function y = |x 2 – 4|x| – 1|. To do this, we use the equality a 2 = |a| 2 and select the complete square in the submodular expression written on the right side of the function:
x 2 – 4|x| – 1 = |x| 2 – 4|x| - 1 = (|x| 2 – 4|x| + 4) – 1 – 4 = (|x |– 2) 2 – 5.
Then the original function will have the form y = |(|x| – 2) 2 – 5|.
To construct a graph of this function, we construct sequential graphs of functions:
1) y = (x – 2) 2 – 5 – parabola with vertex at point with coordinates (2; -5); (Fig. 1).
2) y = (|x| – 2) 2 – 5 – part of the parabola constructed in step 1, which is located to the right of the ordinate axis, is symmetrically displayed to the left of the Oy axis; (Fig. 2).
3) y = |(|x| – 2) 2 – 5| – the part of the graph constructed in point 2, which is located below the x-axis, is displayed symmetrically relative to the x-axis upward. (Fig. 3).
Let's look at the resulting drawings:
The graph of the function y = a is a straight line parallel to the abscissa axis.
Using the figure, we conclude that the graphs of functions have six common points (the equation has six roots) if a belongs to the interval (1; 5).
This can be seen in the following figure:
Let's find the arithmetic mean of the integer values of parameter a:
(2 + 3 + 4)/3 = 3.
Answer: 3.
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DAGESTAN INSTITUTE OF ADVANCED PROFESSIONALS
TEACHING STAFF
DEPARTMENT OF PHYSICS AND MATHEMATICS AND ICT
Project
on the topic:
"Construction and reformation
function graphs
V school course mathematicians »
Rabadanova P.A.
math teacher
MBOU "Kochubeyskaya Secondary School"
Tarumovsky district
2015
1. Introduction…………………………………………………………….….3
2. Chapter I. Literature review on the project topic………………………….….5
3. Chapter II. Empirical part:
3.1. Basic methods for converting function graphs……….….7
3.2. Plotting evenAndodd functions…………….. 10
3.3. Plotting a graph inverse function………………………... 11
3.4. Deformation (compression and stretching) of graphs………………….12
3.5.Combination of transfer, reflection and deformation………………......13
4.Tasks for independent solution………………………..…...14
5. Conclusion………………………………………………………………………………15
6. Conclusions………………………………………………………..………17
INTRODUCTION
Transformation of function graphs is one of the fundamental mathematical concepts directly related to practical activities. The graphs reflect the variability and dynamism of the real world, the mutual relationships of real objects and phenomena.
The functional line is a basic topic covered in the Main and Unified State Exams.Also, many mathematical concepts are examined using graphical methods. For example, toquadraticthe function is introduced and studied in close connection with quadratic equations and inequalities.It follows thatTeaching students how to construct and transform graphs of functions is one of the main tasks of teaching mathematics at school.
Studying the function makes it possible to find aboutdomain of definition and domain of value of a function, domainDecreasing or increasing rates, asymptotes, intervalssign constancy, etc. However, to construct a graphyou can use many functionsuse a number of methodsmaking it easierconstruction. Therefore, students must have the competence to construct graphs using methodological schemes.
The above determinesrelevance research topics.
Object of study is the study of converting function line graphs into school math.
Subject of research – the process of constructing and transforming function graphs in a secondary school.
Purpose of the study: educational - consists in identifying a methodological scheme for constructing and transforming function graphs;developing - development of abstract, algorithmic, logical thinking, spatial imagination;educational – nurturing the graphic culture of schoolchildren, developing mental work skills.
The goals led to the decision of the followingtasks:
1. Analyze educational and methodological materials on the problem under study.
2. Identify methodological schemestransforming graphs of functions in a school mathematics course.
3. Select the most effective methods and meansconstruction and transformation of function graphs in secondary school, promoting: meaningful learning educational material; increasing the cognitive activity of students; development of their creative abilities.
HYPOTHESIS research: the formation of graphic skills in the process of studying functions and nurturing the graphic culture of students will effective if students know the methodological scheme for constructing and transforming graphs of functions in a school mathematics course.
CHAPTER I . REVIEW OF LITERATURE ON THE TOPIC OF THE PROJECT.
In preparation for the project, we studied the following literature:
Sivashinsky, I. Kh. Theorems and problems in algebra, elementary functions - M., 2002. - 115 p.
Gelfand, I. M., Glagoleva, E. G., Shnol, E. E. Functions and graphs (basic techniques) - M., 1985. - 120 s.
V.Z.Zaitsev, V.V. Ryzhkov, M.I. Scanavi. Elementary mathematics - M., 2010 (reprint). - 590 s.
Kuzmin, M. K. Graphing a function - J. Mathematics at school. - 2003. - No. 5. - P. 61-62.
Shilov G.E. How to build graphs? - M., 1982.
Isaac Thanatar. Geometric transformations of function graphs - MCNMO, 2012
INIt is noted that the ability to “read” the behavior of a function on a certain set using a graph is used not only in mathematics courses, but also in any practical human activity in which he has to deal with certain graphic images dependencies. Therefore, students should be able to determine some of its properties from the graph of a function.
The theoretical material for converting graphs is strictly presented. The technique is accompanied by illustrations, drawings, examples of varying complexity and their solutions, which makes it possible to in-depth expand knowledge and construct graphs of complex functions.
Represents electronic training course, the volume and content of which correspond to the requirements for a high school mathematics course high school. The theoretical material is supported by graphic animation illustrations that provide a visual representation of the topic being studied. The course includes three modules: study module theoretical material, self-test module and knowledge control module.
From , , methodological schemes for constructing graphs and examples for independent work were used for the empirical part of the project.
Conclusions to Chapter 1
The study of educational and methodological literature allowed:
1. Identify the methodological schemestudying, constructing and transforming graphs of functions in a school mathematics course.
2. Select the most effective methods and meansconstruction and transformation of function graphs in school mathematics,contributing:
meaningful learning of educational material;
increasing the cognitive activity of students;
development of their creative abilities.
3. show that The function line has a significant impact when learning various concepts in mathematics.
Chapter 2. EMPIRICAL PART
In this chapter, we will look at the main methods for transforming function graphs and provide methodological schemes for constructing various combinations of graphs for various functions.
2.1. BASIC METHODS FOR TRANSFORMING GRAPHICS OF A FUNCTION
Translation along the ordinate axis
f ( x ) f ( x )+ b .
Forgraphing a functiony = f( x) + bfollowinget:
1. graph the functiony= f( x)
2. move the axisabscissa to| b| units up atb>0 or on| b| eatprostrate down atb < 0. Received in new system coordinat graph is a graph of a functiony = f( x) + b.
2. Transfer along axes abscissa
f ( x ) f ( x + a ) .
y = f( x+ a) followinget:
3. Plotting a function of the form y = f (- x )
f (x ) f (- x ).
To graph a functiony = f( - x) follows:
graph a functiony = f( x)
reflect it onrelative to the y-axis
the resulting graph isfunction graphy = f( - X).
4. Plotting a function of the form y = - f ( x )
f ( x ) - f ( x )
- f( x) follows:
graph a functiony= f( x)
reflect it relative to the x-axis
2.2. Plotting even And odd functions
When plottingeven and not even function convenient to use the following properties:
1. Graph of an even symmetry functionric relative to the ordinate axis.
2. Schedule odd function symmetrical about the origin.
To plot graphs of an even and odd function, it is enough to plot only right branch graphics for positive values argument. Left branch is completed symmetrically relative to the origin of coordinates for an odd function and relative to the ordinate axis for an even function.
To graph an even function y = f ( x ) next blows:
build a branch of the graph of this function only in volumerange of positive values of the argument x≥O.
ABOUTtrace this branch relative to the ordinate axis
To graph an odd function y = f ( x ) follows:
build a branch of the graph of this function only inareas of positive argument values (x≥0).
ABOUTtrace this branch relative to the originto the region of negative x values.
2.3. Graphing the Inverse Function
As already noted, direct and inverse functions youreflect the same relationship between variablesx and y, with the only difference that in the inverse function thesethe variables have swapped roles, which is tantamount to changingchanging the designations of coordinate axes. Therefore the schedulethe inverse function is symmetrical to the graph of the direct functionrelative to the bisectorIAndIIIcoordinate angles,i.e. relatively straighty = x. Thus, we getnext rule.
To plot the function y = (x), inverse to the functiony = f( x), should be builtscheduley = f( x) and reflect it relative to the straight line y = x.
2.4. Deformation (compression and stretching) of graphs
1. Compression (stretching) of the graph along the ordinate axis
f ( x ) A ∙ f ( x ).
To graph a functiony= A∙ f( x) follows:
8. Compression (stretching) of the graph along the x-axis
f( x)
To plot the function y= f( x) follows:
2.5. Combination of transfer, reflection and warp
Very often when constructing graphs of functions whenchange the combination of techniques.
Consecutive Application a number of such posture techniquesallows you to significantly simplify the construction of a graph usingrunning function and often reduce it in the end toconstruction of one of the simplest elementary functionstions. Let us consider how, taking into account the above, it followsbuild graphs of functions.
Please note that it is timeIt is advisable to carry out the simplification doc in the following sequenceness.
Using parity orodd function.
Transfer of axes.
Reflection and deformation.
The construction of the graph is performed in the reverse order.
Example.
Graph the function
We will carry out the construction in the following steps:
1. build a graph natural logarithm :
2. squeezeto the axisOY2 times:;
3.
display symmetricallyrelative to the axisOY:
;
4. move along the axisOXon(!!!) to the right:
:
5. display symmetrically about the axisOX:
;
6. let's movealong the axisOYup 3 units:
:
EXAMPLES OF CONSTRUCTION AND TRANSFORMATION OF GRAPHICS OF A FUNCTION
Example 1. Graph the function.
First, let's draw a sine graph; its period is equal to:
graph of a functionobtained by compressing the graphto the ordinate axis twice. log .
Graph the functionat = 2 cosX.
Graph the functiony = sinx .
CONCLUSION
While working on project work Various educational and methodological literature on this problem were analyzed. The results of the study made it possible to identify the most characteristic positive aspects studying, construction and transformation of graphs of functions in a school mathematics course
The main goal of the project is to develop students’ skills in reading and completing drawings, and to develop rational methods of independent activity in them.
The need to improve graphic education in general is dictated not only by modern production requirements, but also by the role of graphics in the development of technical thinking and cognitive abilities students. A person’s ability to process graphic information is one of the indicators of his mental development. Therefore, graphic training should become an integral element of general education training.
Conclusions
Thus, the developed project “Construction and transformation of graphs of functions”, dedicated to one of the central concepts of mathematics - functional dependence, is aimed at systematizing and expanding students’ knowledge. The study of specific methods for transforming function graphs is carried out analytically and graphically according to strict methodological schemes. Collected material can be used in lessons and for students’ self-study. To conduct classes, various forms and methods of organization and training can be used.
Student research work on the topic:
"Application linear function in problem solving"
MKOU "Bogucharskaya secondary secondary school No. 1"
Research work in mathematics.
Topic: “Application of the graph of a linear function to solving problems”
7 "B" class
Head: Olga Mikhailovna Fomenko
Boguchar city
1.Introduction………………………………………………………………………………… 2
2. Main part……………………………………………………………3-11
2.1 Methodology for solving word problems using linear function graphs
2.2 Solving word problems on movement using graphs
3. Conclusion………………………………………………………………………………11
4. Literature……………………………………………………………….12
INTRODUCTION
"Algebra.7 grade" examines problems in which, according to a given schedule, it is necessary to answer a number of questions.
For example:
No. 332 The summer resident went from home by car to the village. First he drove along the highway, and then along a country road, slowing down. The summer resident's movement schedule is shown in the figure. Answer the questions:
a) how long did the summer resident drive on the highway and how many kilometers did he travel; what was the speed of the car on this section of the route;
b) how long did the summer resident drive along the country road and how many kilometers did he travel; what was the speed of the car in this section;
c) how long did it take the summer resident to travel all the way from his house to the village?
While searching for material on this topic in the literature and the Internet, I discovered that in the world many physical, and even social and economic phenomena and processes are in a linear relationship, but I settled on movement, as it is the most familiar and popular among us. In the project, I described word problems and ways to solve them using linear function graphs.
Hypothesis: Using graphs, you can not only get a visual representation of the properties of a function, get acquainted with the properties of a linear function and its particular form, direct proportionality, but also solve word problems.
The purpose of my research was the study of the use of linear function graphs in solving word problems on motion. In connection with the implementation of these goals, the following were put forward: tasks:
Study the technique of solving text problems on movement using linear function graphs;
Learn to solve motion problems using this method;
Draw comparative conclusions about the advantages and disadvantages of solving problems using linear function graphs.
Object of study: graph of a linear function.
Research method:
Theoretical (study and analysis), system-search, practical.
Main part.
In my research, I decided to try to give a graphical interpretation of the movement problems presented in our textbook, then answer the question posed in the problem according to the graph. For this method of solution, I took problems with a rectilinear uniform movement on one section of the route. It turned out that many problems can be solved in this way easier than in the usual way using the equation. The only drawback of this technique: to accurately obtain an answer to the question of the problem, you must be able to correctly select the scale of units of measurement on the coordinate axes. Big role in making the right choice The experience of solving plays on such a scale. Therefore, in order to master the art of solving problems using graphs, I had to look at them in large quantities.
define a coordinate system sOt with an abscissa axis Ot and a ordinate axis Os. To do this, according to the conditions of the problem, it is necessary to select the origin of reference: the beginning of the movement of the object or, from several objects, the one that began to move earlier or has traveled a greater distance is selected. On the abscissa axis, mark the time intervals in its units of measurement, and on the ordinate axis, mark the distance in the selected scale of its units of measurement.
Points on the coordinate plane must be marked in accordance with the scale according to the conditions of the problem, and the lines must be drawn carefully. The accuracy of solving the problem depends on this. Therefore, it is very important to successfully select the scale of divisions on the coordinate axes: it must be selected in such a way that the coordinates of the points are determined more accurately and, if possible, located at nodal points, i.e. at the intersections of the divisions of the coordinate axes. Sometimes it is useful to take as a unit segment on the x-axis the number of cells that is a multiple of the conditions of the problem with respect to time, and on the ordinate axis - the number of cells that is a multiple of the conditions of the problem with respect to distance. For example, 12 minutes in time requires choosing a number of cells that is a multiple of 5, because 12 minutes is a fifth of an hour.
Solving word problems on movement using graphs
Answer: 9 km.
Solution using equation:
x/12h. – time from A to B
x/18h. – time back
Answer: 9 km
Problem 2. (No. 156 in the textbook by Yu.N. Makarychev “Algebra 7.”)
Two cars are traveling along the highway at the same speed. If the first increases the speed by 10 km/h, and the second decreases by 10 km/h, then the first will cover the same distance in 2 hours as the second in 3 hours. How fast are the cars going?
Solution using equation:
Let x km/h the speed of the cars;
(x+10) and (x-10) respectively, the speed after increasing and decreasing;
2(x+10)=3(x-10)
Answer: 50km/h
Solving using a linear function graph:
1. Let’s set the coordinate plane sOt with the abscissa axis Ot, on which we mark the time intervals of movement, and the ordinate axis Os, on which we mark the distance traveled by the vehicles
2. Let’s plot divisions on a scale along the x-axis – one hour in 5 cells (in 1 cell – 12 minutes); We apply divisions along the ordinate axis, but do not indicate the scale.
3. Let's construct the line of movement of the first car I: the beginning of movement at point c
4. Construct the line of movement of the second car II: the beginning of movement at the point with coordinate (0;0). Next, we mark an arbitrary point (3;s 1) on the plane, because the car at the new speed was on the road for 3 hours.
4. Determine the speed of the cars v before it changes. Let us denote the difference in ordinates of points lying on straight lines with abscissa 1 by the symbol ∆s. According to the condition, this segment corresponds to a length of (10+10) km, because for one of them the speed decreased, and for the other the speed increased by 10 km/h. This means that the line of movement of the cars before changing speed must be equidistant from lines I and II and located on the coordinate plane between them. According to the graph, Δs = 2kl. corresponds to 20 km, v = 5 cells, which means we solve the proportion v = 50 km/h.
Answer: 50 km/h.
Problem 3
Solving using a linear function graph:
reference point is pier M
Let's mark point N (0; 162).
Answer: 2 hours 20 minutes.
Solution using equation:
162 -45(x +0.75)-36x =0
162-45x – 33.75 -36x =0
81x =128.25
2) 
Answer: 2 hours 20 minutes.
Task 4.
A cyclist left point A. At the same time, following him from point B, located at a distance of 20 km from A, a motorcyclist left at 16 km/h. The cyclist was traveling at a speed of 12 km/h. At what distance from point A will the motorcyclist catch up with the cyclist?
Solving using a linear function graph:
1. Let's set the coordinate plane sOt with the abscissa axis Ot, on which we will mark the time intervals of movement, and the ordinate axis Os, on which we will mark the distance covered by the motorcyclist and cyclist
2. Let’s draw divisions on a scale: along the ordinate axis – 8 km in 2 cells; along the abscissa axis – in 2 cells – 1 hour.
3. Let’s construct the line of movement of motorcyclist II: mark the beginning of his movement at the origin of coordinates B(0;0). The motorcyclist was traveling at a speed of 16 km/h, which means that straight line II must pass through the point with coordinates (1; 16).
4. Let’s construct a line of movement for cyclist I: its beginning will be at point A(0;20), because point B is located 20 km from point A, and he left at the same time as the motorcyclist. The cyclist was traveling at a speed of 12 km/h, which means that straight line I must pass through the point with coordinates (1;32).
5. Let’s find P (5; 80) – the point of intersection of lines I and II, reflecting the movement of the motorcyclist and the cyclist: its ordinate will show the distance from point B, at which the motorcyclist will catch up with the cyclist.
Р(5; 80) |=s = 80, |=80 – 20 = 60(km) – the distance from point A at which the motorcyclist will catch up with the cyclist..
Answer: 60 km.
Solution using equation:
Let x km be the distance from point A to the meeting place
x /12 cyclist time
(x +20)/16 motorcyclist time
x /12=(x +20)/16
16x =12x +240
4x =240
x =60
Answer: 60 km
Task 5.
The motorcyclist covered the distance between the cities in 2 hours, and the cyclist in 5 hours. The speed of the cyclist is 18 km/h less than the speed of the motorcyclist. Find the speeds of the cyclist and motorcyclist and the distance between the cities.
Solving using a linear function graph:
1. Let's define the coordinate plane sOt with the abscissa axis Ot, on which we mark the time intervals of movement, and the ordinate axis Os, on which we mark the distance.
2. Let's mark the division along the abscissa axis in 2 cells 1 hour. Along the ordinate axis we will leave the distance without divisions.
3. Let us draw the line of movement of cyclist I in 5 hours and the line of movement of motorcyclist II in 2 hours. The end of both lines must have the same ordinate.
4. Let's draw a segment with abscissa 1 between lines I and II. The length of this segment reflects a distance of 18 km. From the drawing we get that 3 cells are equal to 18 km, which means there are 6 km in 1 cell.
5. Then, according to the graph, we determine the speed of the cyclist is 12 km/h, the speed of the motorcyclist is 30 km/h, the distance between cities is 60 km.
Solution using equation:
Let x km/h the speed of the cyclist, then (x +18) km/h the speed of the motorcyclist
2(x +18)=5x
2x +36=5x
x =12
2) 12+18=30(km/h) motorcyclist speed
3) (km) distance between cities
Answer: 12 km/h; 30 km/h; 60 km
Answer: 60 km.
Task 6.
Along the river flow, a boat covers a distance of 30 km in 3 hours and 20 minutes, and against the current in 4 hours, a distance of 28 km. How far will the boat travel across the lake in 1.5 hours?
Solving using a linear function graph:
1. Let’s set the coordinate plane sOt with the abscissa axis Ot, on which we mark the time intervals of movement, and the ordinate axis Os, on which we mark the distance traveled by the boat
2. Let's draw divisions on a scale: along the ordinate axis - in two cells 4 km; along the abscissa axis – in 6 cells – 1 hour (in 1 cell – 10 minutes), because According to the conditions of the problem, time in minutes is given.
3. Let's build a line of movement of the boat along the river I: the beginning of the line will be at the point with coordinate (0;0). The boat floats 30 km in 3 hours 20 minutes, which means the line must pass through the point with coordinate (;30), because 3h 20min = h.
4. Let’s construct a line of movement of the boat against the flow of the river II: let’s take the beginning of the movement at the point with coordinate (0;0). The boat floats 28 km in 4 hours, which means that the straight line of motion must pass through the point with coordinate (4;28).
5. Let’s construct a line of movement of the boat on the lake: let’s take the beginning of the movement at the point with coordinate (0; 0). The line of the boat's own movement will have to be located equidistantly between the lines of the boat's movement along the river. This means that we must divide the segment consisting of all points with abscissa 1 between the lines of movement along the river in half and mark its middle. From (0; 0) through this marked point we draw a ray, which will be the line of movement along the lake.
6. According to the conditions of the problem, we need to find the distance covered by the boat on the lake in 1.5 hours, which means we must determine on this line the ordinate of the point with the abscissa t = 1.5, |=s = 12, |= 12 km the boat will travel along the lake in 1.5 hours.
Answer: 12 km.
Solution using a system of equations:
Let x km/h be the speed of the lake, and y km/h be the speed of the river
Answer: 12 km.
Task 7.
A boat travels 34 km downstream of a river in the same time as it travels 26 km upstream. The boat's own speed is 15 km/h. Find the speed of the river flow.
Solving using a linear function graph:
1. Let's set the coordinate plane sOt with the abscissa axis Ot, on which we mark the time intervals of movement, and the ordinate axis Os, on which we mark the distance traveled by the boat.
2. Let’s draw divisions on a scale: along the ordinate axis – 1 km in 1 cell; On the abscissa axis we will leave time without divisions.
3. Let’s construct line I of the boat’s movement along the river from 0 km to a point at 34 km: the beginning of the line will be at the point with coordinate (0; 0). The second coordinate will be (x; 34).
4. Let’s construct line II of the boat’s movement against the flow of the river from 0 km to a point at 26 km: the beginning of the line will be at the point with coordinate (0; 0). The second coordinate will be (x; 26).
5. Let us draw ray III from the origin (0; 0) through the middle of an arbitrary segment consisting of all points with the same abscissa between two lines of motion I and II. This beam will reflect the boat's own movement, because The boat's own speed is the arithmetic average of 2 speeds along the current and against the current of the river. On the resulting ray we will find a point with ordinate 15, because The boat's own speed is 15 km/h. The abscissa of the found point will correspond to the division of 1 hour.
6. To find the speed of the river flow, it is enough to find the length of the segment with abscissa 1 from line III to line II. The speed of the river is 2 km/h.
Answer: 2 km/h.
Solution using equation:
River flow speed x km/h
34/(15+x)=26/(15-x) Solving the proportion, we get:
Answer: 2 km/h.
Conclusion.
Advantages:
You can write down the tasks briefly;
Flaws:
LITERATURE.
1. Makarychev Yu. N., Mindyuk N. G., Neshkov K. I., Suvorova S. B., Algebra: Textbook for the 7th grade of educational institutions, “Prosveshchenie”, M., 2000.
2. Bulynin V., Application of graphic methods in solving text problems, educational and methodological newspaper “Mathematics”, No. 14, 2005.
3. Zvavich L.I. Didactic materials on algebra for grade 7.
In algebra lessons in 7th grade, I learned about the topic “Linear function. Mutual arrangement of graphs of linear functions.” I learned how to build graphs of a linear function, learned its properties, learned how to given formulas determine relative position graphs. I noticed that in the textbook by Yu.N. Makarychev
"Algebra.7 grade" examines problems in which, according to a given schedule, it is necessary to answer a number of questions. An example of such a task is presented on the slide.
Based on the given schedule, it can be determined that
And I had a question: is it possible to solve movement problems not by actions or using equations, but by using linear function graphs for this?
The hypothesis, goals and objectives are presented on the slide
In my research, I decided to try to give a graphical interpretation of the movement problems presented in our textbook, then answer the question posed in the problem according to the graph. For this method of solving, I took problems with rectilinear uniform motion on one section of the path.
It turned out that many problems are solved in this way. The only drawback of this technique: to accurately obtain an answer to the question of the problem, you must be able to correctly select the scale of units of measurement on the coordinate axes. Decision experience plays a big role in making the right choice of this scale. Therefore, to master the art of solving problems using graphs, I had to look at a lot of them.
Methods for solving word problems using linear function graphs.
In order to solve a word problem using linear function graphs, you need to:
set a coordinate system To do this, according to the conditions of the problem, you need to select the origin: the beginning of the movement of the object or, from several objects, select the one that began to move earlier or has traveled a greater distance. On the abscissa axis, mark the time intervals in its units of measurement, and on the ordinate axis, mark the distance in the selected scale of its units of measurement.
Draw the lines of movement of each of the objects specified in the problem statement through the coordinates of at least two straight points. Typically, the speed of an object provides information about the distance traveled in one unit of time from the beginning of its movement. If the object begins to move later, then the point at which its movement begins is shifted by a given number of units to the right from the origin along the abscissa axis. If an object begins to move from a place distant from the origin by a certain distance, then the point of origin of its motion is shifted upward along the ordinate axis.
The meeting place of several objects on the coordinate plane is indicated by the point of intersection of the lines depicting their movement, which means that the coordinates of this point provide information about the time of the meeting and the distance of the meeting place from the origin.
The difference in the speed of movement of two objects is determined by the length of the segment consisting of all points with an abscissa of 1 located between the lines of movement of these objects.
Points on the coordinate plane must be marked in accordance with the scale according to the conditions of the problem, and the lines must be drawn carefully. The accuracy of solving the problem depends on this.
Problem 1. (No. 673 in the textbook by Yu.N. Makarychev “Algebra 7.”)
A cyclist travels along path AB at a speed of 12 km/h. Returning, he reached a speed of 18 km/h and spent 15 minutes less on the return journey than on the way from A to B. How many kilometers from A to B.
Solution using equation:
Let x km be the distance from A to B.
x/12h. – time from A to B
x/18h. – time back
Since he spent 15 minutes less on the way back, we will create the equation
Answer: 9 km
Solving using a linear function graph:
1. Let us define the coordinate plane sOtc by the abscissa axis Оt, on which we will mark the time intervals of movement, and by the ordinate axis Os, on which we will mark the distance.
2. Let’s draw divisions on a scale: along the ordinate axis – 3 km in one cell; on the abscissa - one hour in 4 cells (in 1 cell - 15 minutes).
3. Let's build a line of movement there: mark the beginning of the movement with a dot (0;0). The cyclist was traveling at a speed of 12 km/h, which means the straight line must pass through the point (1;12).
4. Let’s build a line of movement back: mark the end of the line with a dot (; 0), because The cyclist spent 15 minutes less on the return trip. He was driving at a speed of 18 km/h, which means that the next point on the straight line has coordinates (;18).
5. Note (; 9) - the point of intersection of the lines: its ordinate will show the distance: s = 9
Answer: 9 km.
Problem 2 (No. 757 in the textbook by Yu.N. Makarychev “Algebra 7”)
The distance between piers M and N is 162 km. The motor ship departed from pier M at a speed of 45 km/h. After 45 minutes, another motor ship departed from pier N to meet him, the speed of which was 36 km/h. How many hours after the departure of the first ship will they meet?
Solution using equation:
Let there be a meeting in x hours
162 -45(x +0.75)-36x =0
162-45x – 33.75 -36x =0
81x =128.25
2) 
Answer: 2 hours 20 minutes.
Solving using a linear function graph:
1. Let us define the coordinate plane sOt with the abscissa axis Ot, on which we mark the time intervals of movement, and the ordinate axis Os, on which
Let us note the distance from pier M to pier N, equal to 162 km. The beginning
reference point is pier M
2. Let’s draw divisions on a scale: along the ordinate axis – in two cells 18 km; The abscissa is one hour in 6 cells (1 cell is 10 minutes), because The task conditions indicate the time in minutes.
Let's mark point N (0; 162).
3. Let's construct the line of movement of the first motor ship I: the beginning of its movement will be at the point with coordinates (0;0). The first motor ship was sailing at a speed of 45 km/h, which means the straight line must pass through the point with coordinates (1;45).
4. Let's construct the line of movement of the second motor ship II: the beginning of movement will be at point c
coordinates (; 162), since he left point N, 162 km away from M, for 45 minutes. later than first, and 45 min. = h. The second motor ship sailed at a speed of 36 km/h, which means that the straight line must pass through the point (; 126), since the second motor ship left in the direction of point M: 162 – 36 = 126 (km).
5. The point of intersection of lines I and II is point A (;108). The abscissa of the point shows the time after which they met after the departure of the first ship: t =, |=h = 2h20min. – the time of meeting of two ships after the departure of the first ship.
Answer: 2 hours 20 minutes.
Conclusion.
At the end of the study, I was able to identify the advantages and disadvantages of solving problems graphically.
Advantages:
You can write down the tasks briefly;
It's quite easy to work with small numbers.
Flaws:
Difficult to work with large numbers.
OSR. "Solving Equations Using Graphs."
Exercise:
1) Basic summary.
A graph is a set of points on a coordinate plane that have x and y values.
are connected by some dependence and each value x corresponds to a single value y.
The graphical method is one of the most convenient and visual ways of presentation and analysis.
information.
In practice, the graphical method for solving equations often turns out to be useful. He
is as follows: to solve the equations f(x)=0, plot the function y=f(x) and find
abscissas of the points of intersection of the graph with the Ox axis: these abscissas are the roots of the equation.
Algorithm for solving equations graphically
To solve graphically an equation of the form f(x) = g(x), you need:
1. Construct function graphs in one coordinate plane:
y = f(x) and y = g(x).
2. Find the intersection points of these graphs.
3. Indicate the abscissa of each of these intersections.
4. Write down the answer.
It is quite easy to solve a system of equations graphically, since each
the equation of the system on the coordinate plane represents some kind of
line.
By constructing graphs of these equations and finding the coordinates of their points
intersections (if they exist), we obtain the desired solution.
Graphic solution inequalities, comes down to finding such points x,
in which one graph lies above or below another.
Examples:
#1: Solve the equation
x
4
5
x
points
cross
I
graphs
functions
№
2.
Decide
is
drawing
abscissa
1
.
equations
5
cm.
:
X
X
4
By decision
at
ui
Examination
1
4
15
4
4
right
Answer
.1:
equation
x
3
3
x
By decision
equations
is
at
3
X
ui
3
X
cm.
drawing
abscissa
.
2
points
cross
I
graphs
functions
No. 3. Re
1
3
Examination
:
3
1
right
1:
33
Answer
.
sew equation
Solution: Let's build graphs of functions
and y = x
The graphs of the functions do not intersect, and, therefore, the equation has no roots (see figure).
Answer: no roots.
No. 4. Find the value of the expression x + y, if (x
;y
is a solution to the system of equations.
Solution:
to the left.
parallel transfer by 1 unit
parallel translation 2 units to the left.
= 1, y
=1
+ y
=0.
X
X
Answer: 0.
No. 5. Solve the inequality
Answer: x>2.
>12 1.5x. No. 6. Solve the inequality
. Answer: x>0.
No. 7. Solve the equation sinx + cosx=1. Let's plot the functions y=sinx u y=1cosx. (Figure 5) From
The graph shows that the equation has 2 solutions: x = 2 n, where nЄZ and x = /2+2 k, where kЄZ.
π
π
π
2
sin x(
1
cos x(
6
4
2
1
2
2
1
1
0
x
2
4
6
2
No. 8. Solve the equation: 3x = (x1) 2 + 3
Solution: apply functional method solutions to equations:
because this system has a unique solution, then using the selection method we find x = 1
Answer: 1.
No. 9.Solve inequality: cos x 1 + 3x
Solution:
Answer: (
;
).
No. 10. Solve the equation
In our case, the function
increases when x>0, and the function y = 3 – x decreases when
all values of x, including for x>0, which means
equation
root Note that for x = 2 the equation becomes
V true equality, because
has at most one
.
Answer: 2.
2)Solve the task:
1) Does the equation have a root and if so, is it positive or negative?
A)
; b)
, c) 6x =1/6, d)
.
2) Solve graphical method equation
.
1
3
X
3
X
3) Solve the equation graphically:
A)
b)
.
3
х
3
X
5
1
2
X
4) The figure shows a graph of the function y=f(x).
1) 1 2) 6 3) 7 4) 8
5) Which of the figures shows the graph of the function
?
at
log
x
1
2
1) at 2) at 3) at 4)
at
1 1 1
6) The graph of which function is shown in the figure?
1) y = 2x1.5; 2) y = 2x – 2;
3) y = 2x – 3; 4) y = 2x – 2.
7)Which function is graphed in the figure?
1) y = sinx; 2)
at
sin
x
6
; 3)
at
sin
x
3
; 4)
.
at
sin
x
6
8) The figure shows a graph of functions
y = f (x) and y = g (x), given on the interval
[5;6]. Specify those values of x for which
the inequality g(x) holds
y
y
)(xg
f(x)1
1) [5; 0] 2) [5; 2]
0 1 x
3) [2; 2] 4)
9) The figure shows a graph of the function y=f(x).
Find the number of integer roots of the equation f(x)= 0.
1) 3 2) 4 3) 2 4) 1
)(xf
y
10) The figure shows a graph of the function y=f(x).
Find the number of integer roots of the equation f(x)+2= 0.
1) 3 2) 5 3) 4 4) 1
Graphic solution of a quadratic equation Strengthen the ability to build graphs of various functions; Develop decision-making skills quadratic equations graphically. Brdsk 2009 Municipal educational institution - Economic Lyceum General lesson on the topic “Quadratic function”, algebra 8th grade teacher Fedoseeva T.M.
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Second method: a). Let's break the equation x 2 -2x-3=0 into parts x 2 = 2x+3 Let's write two functions y= x 2; y=2x+3 We build graphs of these functions in one coordinate system. The abscissas of the intersection points are the roots of the equation. 0 1 x y Solve the equation x 2 +2x-3=0
Third method: x 2 -3 = 2x y = x 2 -3; y=2x We build graphs of these functions in one coordinate system. The abscissas of the intersection points are the roots of the equation. 0 1 x y Solve the equation x 2 +2x-3=0
