Build a graph with module examples. Linear function plots with modules

Construction of graphs of functions containing the sign of the modulus.

I hope you have carefully studied point 23 and understand the difference between a view function and a . Now let's take a look at a couple more examples that should help you when plotting graphs.

Example 1. Graph a function

We have a function of the form , where .

1. We first construct a graph of a submodular function, i.e., a function. To do this, select the integer part of this fraction. I remind you that this can be done in two ways: by dividing the numerator by the denominator “in a column” or by painting the numerator so that an expression that is a multiple of the denominator appears in it. Let's select the whole part in the second way.

So the submodule function has the form . Hence, its graph is a hyperbola of the form shifted 1 unit to the right and 3 units up.

Let's build this chart.

2. To get the graph of the desired function , it is necessary to leave the part of the graph of the function that lies above the Ox axis unchanged, and the part of the graph lying below the Ox axis must be displayed symmetrically in the upper half-plane. Let's do these transformations.

The chart has been built.

The abscissa of the point of intersection of the graph with the x-axis can be calculated by solving the equation

y = 0, i.e. . We get that .

Now, according to the graph, you can determine all the properties of the function, find the smallest and largest values ​​of the function on the interval, solve problems with a parameter.

For example, you can answer this question. “For what values ​​of the parameter a equation has exactly one solution?

Let's draw straight y=a for different values ​​of the parameter a. (thin red lines in the following figure)

It is seen that if a<0 , then the graph of the constructed function and the straight line do not have common points, which means that the equation does not have a single solution.

If a 0< a<3 or a>3, then the straight line y=a and the constructed graph have two common points, i.e. the equation has two solutions.

If a = 0 or a = 3, then the equation has exactly one solution, because for these values a A line and a graph of a function have exactly one common point.

Example 2 Plot a function

Decision

Let us first construct a graph of the function for non-negative values ​​of x. If , then our function also takes the form , and the desired function is a function of the form .

The graph of the function is the branch of the parabola "directed" to the left, shifted by 4 units right. (Because we can imagine ).

Let's plot this function

and we will consider only that part of it, which is located to the right of the Oy axis. We'll erase the rest.

Please note that we have calculated the value of the ordinate of the graph point lying on the y-axis. To do this, it is enough to calculate the value of the function at x = 0. In our case, at x = 0 got y=2.

Now let's plot the function for X< 0 . To do this, we will build a line symmetrical to the one we have already built, relative to the Oy axis.

Thus, we have built a graph of the desired function.

Example 3. Graph a function

This is no longer an easy task. We see that both types of functions with a module are present here: and , and . Let's build in order:

First, let's plot the function graph without all modules: Then add the module for each argument. We get a function of the form , i.e. . To build such a graph, you need to apply symmetry about the Oy axis. Let's add an external module. Finally, we get the desired function . Since this function was obtained from the previous one using an external module, we have a function of the form , which means that it is necessary to apply symmetry with respect to Ox.

Now more.

This is a fractional linear function, to build a graph, you need to select the integer part, which we will do.

This means that the graph of this function is a hyperbola of the form shifted 2 to the right and 4 down.

Let's calculate the coordinates of the points of intersection with the coordinate axes.

y = 0 at x = 0, so the graph will pass through the origin.

2. Now let's plot the function .

To do this, in the original graph, first erase that part of it that is located to the left of the Oy axis:

, and then display it symmetrically about the Oy axis. Please note that the asymptotes are also displayed symmetrically!

Now let's build the final graph of the function: . To do this, we will leave the part of the previous graph lying above the Ox axis unchanged, and what is below the Ox axis we will symmetrically display in the upper half-plane. Again, don't forget that the asymptotes are displayed along with the graph!

The chart has been built.

Example 4: Using various graph transformations, plot a function

Something completely twisted and complex! Tons of modules! And the x-square has no modulus!!! It's impossible to build!

One way or something like this, an average student of the 8th grade, unfamiliar with the technique of plotting, can argue.

But not us! Because we know DIFFERENT ways to transform function graphs and we also know different properties of the module.

So, let's start in order.

The first problem is the lack of a module for x squared. No problem. We know that . Good. So our function can be written as . This is already better, because it looks like .

Farther. The function has an external module, so it looks like you will have to use the rules for plotting a function. Let's see then what a submodule expression is. This is a function of the form . If not for -2, then the function would again contain an external module and we know how to graph the function using symmetries. Aha! But after all, if we build it, then, shifting it down by 2 units, we get what we are looking for!

So, something is starting to emerge. Let's try to create an algorithm for plotting a graph.

1.

5. And finally . Everything that lies below the Ox axis will be displayed symmetrically in the upper half-plane.

Hooray! The schedule is ready!

Good luck with your hard work of charting!

transcript

1 Regional scientific and practical conference of educational and research work of students in grades 6-11 "Applied and fundamental questions of mathematics" Methodological aspects of studying mathematics Construction of graphs of functions containing the module Gabova Anzhela Yurievna, grade 10, MOBU "Gymnasium 3" Kudymkar, Pikuleva Nadezhda Ivanovna, teacher of mathematics, MOBU "Gymnasium 3", Kudymkar, Perm, 2016

2 Contents: Introduction...3 p. I. Main part... 6 p. 1.1Historical background.. 6 p. 2.Basic definitions and properties of functions p. .8 p. 2.3 Fractional-rational function 8 p. 3. Algorithms for plotting graphs with modulus 9 p. 3.1 Determining the modulus .. 9 p. in the "nested modules" formula.10 p. 3.4 Algorithm for constructing graphs of functions of the form y = a 1 x x 1 + a 2 x x a n x x n + ax + b...13 p. 3.5 Algorithm for constructing a graph of a quadratic function with modulus.14 p. plotting a fractionally rational function with a modulo. 15p. 4. Changes in the graph of a quadratic function depending on the location of the sign of the absolute value ..17str. II. Conclusion ... 26 p. III. List of references and sources...27 p. IV. Application....28p. 2

3 Introduction Graphing functions is one of the most interesting topics in school mathematics. The greatest mathematician of our time, Israel Moiseevich Gelfand, wrote: “The process of plotting graphs is a way of turning formulas and descriptions into geometric images. This plotting is a means to see formulas and functions and see how these functions change. For example, if y \u003d x 2 is written, then you immediately see a parabola; if y = x 2-4, you see a parabola lowered by four units; if y \u003d - (x 2 4), then you see the previous parabola turned down. This ability to see the formula at once, and its geometric interpretation is important not only for the study of mathematics, but also for other subjects. It's a skill that stays with you for a lifetime, like learning to ride a bike, type, or drive a car." The basics of solving equations with modules were obtained in the 6th 7th grade. I chose this particular topic because I believe that it requires a deeper and more thorough study. I want to get more knowledge about the modulus of a number, different ways of plotting graphs containing the sign of the absolute value. When the “standard” equations of lines, parabolas, hyperbolas include the sign of the modulus, their graphs become unusual and even beautiful. To learn how to build such graphs, you need to master the techniques for constructing basic figures, as well as firmly know and understand the definition of the modulus of a number. In the school mathematics course, graphs with a module are not considered in depth enough, which is why I wanted to expand my knowledge on this topic, to conduct my own research. Without knowing the definition of the module, it is impossible to build even the simplest graph containing an absolute value. A characteristic feature of graphs of functions containing expressions with a module sign, 3

4 is the presence of kinks at those points at which the expression under the module sign changes sign. Purpose of the work: to consider the construction of a graph of linear, quadratic and fractionally rational functions containing a variable under the module sign. Tasks: 1) To study the literature on the properties of the absolute value of linear, quadratic and fractional-rational functions. 2) Investigate changes in the graphs of functions depending on the location of the sign of the absolute value. 3) Learn to plot equations graphs. Object of study: graphs of linear, quadratic and fractionally rational functions. Subject of study: changes in the graph of linear, quadratic and fractionally rational functions depending on the location of the sign of the absolute value. The practical significance of my work lies in: 1) using the acquired knowledge on this topic, as well as deepening it and applying it to other functions and equations; 2) in the use of research skills in further educational activities. Relevance: Graphing assignments are traditionally one of the most difficult topics in mathematics. Our graduates are faced with the problem of successfully passing the GIA and the Unified State Examination. Research problem: plotting functions containing the modulus sign from the second part of the GIA. Research hypothesis: the application of the methodology for solving tasks of the second part of the GIA, developed on the basis of general methods for constructing graphs of functions containing the sign of the module, will allow students to solve these tasks 4

5 on a conscious basis, choose the most rational solution method, apply different solution methods and pass the GIA more successfully. Research methods used in the work: 1. Analysis of mathematical literature and Internet resources on this topic. 2. Reproductive reproduction of the studied material. 3. Cognitive-search activity. 4. Analysis and comparison of data in search of a solution to problems. 5. Statement of hypotheses and their verification. 6. Comparison and generalization of mathematical facts. 7. Analysis of the obtained results. When writing this work, the following sources were used: Internet resources, OGE tests, mathematical literature. 5

6 I. Main part 1.1 Historical background. In the first half of the 17th century, the concept of a function began to take shape as a dependence of one variable on another. So, the French mathematicians Pierre Fermat () and Rene Descartes () imagined a function as a dependence of the ordinate of a curve point on its abscissa. And the English scientist Isaac Newton () understood the function as a coordinate of a moving point that changes depending on time. The term "function" (from the Latin function performance, commission) was first introduced by the German mathematician Gottfried Leibniz (). He associated a function with a geometric image (a graph of a function). Later, the Swiss mathematician Johann Bernoulli () and a member of the St. Petersburg Academy of Sciences, the famous mathematician of the 18th century Leonard Euler () considered the function as an analytical expression. Euler also has a general understanding of a function as the dependence of one variable on another. The word "module" comes from the Latin word "modulus", which means "measure" in translation. This is a multi-valued word (homonym), which has many meanings and is used not only in mathematics, but also in architecture, physics, engineering, programming and other exact sciences. In architecture, this is the initial unit of measurement established for a given architectural structure and used to express the multiple ratios of its constituent elements. In engineering, this is a term used in various fields of technology that does not have a universal meaning and serves to denote various coefficients and quantities, for example, the engagement modulus, the modulus of elasticity, etc. 6

7 Bulk modulus (in physics) is the ratio of the normal stress in the material to the relative elongation. 2.Basic definitions and properties of functions Function is one of the most important mathematical concepts. A function is such a dependence of the variable y on the variable x, in which each value of the variable x corresponds to a single value of the variable y. Ways of setting a function: 1) analytical method (the function is set using a mathematical formula); 2) tabular method (the function is specified using the table); 3) descriptive method (the function is given by a verbal description); 4) graphical method (the function is set using a graph). The graph of a function is the set of all points of the coordinate plane, the abscissas of which are equal to the value of the argument, and the ordinates to the corresponding values ​​of the function. 2.1 Quadratic function The function defined by the formula y=ax 2 +in+c, where x and y are variables, and the parameters a, b and c are any real numbers, and a = 0, is called quadratic. The graph of the function y \u003d ax 2 + in + c is a parabola; the axis of symmetry of the parabola y \u003d ax 2 + in + c is a straight line, for a> 0 the “branches” of the parabola are directed upwards, for a<0 вниз. Чтобы построить график квадратичной функции, нужно: 1) найти координаты вершины параболы и отметить её в координатной плоскости; 2) построить ещё несколько точек, принадлежащих параболе; 3) соединить отмеченные точки плавной линией.,. 2.2Линейная функция функция вида 7

8 (for functions of one variable). The main property of linear functions is that the increment of the function is proportional to the increment of the argument. That is, the function is a generalization of direct proportionality. The graph of a linear function is a straight line, hence its name. This concerns a real function of one real variable. 1) At, the straight line forms an acute angle with the positive direction of the x-axis. 2) When, the line forms an obtuse angle with the positive direction of the x-axis. 3) is an indicator of the ordinate of the point of intersection of the line with the y-axis. 4) When, the line passes through the origin. , 2.3 A fractional-rational function is a fraction whose numerator and denominator are polynomials. It has the form where, polynomials in any number of variables. Rational functions of one variable are a special case: where and are polynomials. 1) Any expression that can be obtained from variables using four arithmetic operations is a rational function. eight

9 2) The set of rational functions is closed under arithmetic operations and the operation of composition. 3) Any rational function can be represented as a sum of simple fractions - this is used in analytical integration .., 3. Graphing algorithms with a module if a is negative. a = 3.2 Algorithm for constructing a graph of a linear function with a modulus To plot the graphs of functions y= x, you need to know that for positive x we ​​have x = x. This means that for positive values ​​of the argument, the graph y=x coincides with the graph y=x, that is, this part of the graph is a ray emerging from the origin at an angle of 45 degrees to the abscissa axis. For x< 0 имеем x = -x; значит, для отрицательных x график y= x совпадает с биссектрисой второго координатного угла. Впрочем, вторую половину графика (для отрицательных X) легко получить из первой, если заметить, что функция y= x чётная, так как -a = a. Значит, график функции y= x симметричен относительно оси Oy, и вторую половину графика можно приобрести, отразив относительно оси ординат часть, начерченную для положительных x. Получается график:y= x 9

10 For construction, we take points (-2; 2) (-1; 1) (0; 0) (1; 1) (2; 2). Now let's build a graph y= x-1. If A is the graph point y= x with coordinates (a; a), then the graph point y= x-1 with the same value of the Y ordinate will be the point A1 (a+1; a). This point of the second graph can be obtained from point A(a; a) of the first graph by shifting parallel to the Ox axis to the right. This means that the entire graph of the function y= x-1 is obtained from the graph of the function y= x by shifting parallel to the Ox axis to the right by 1. Let's build graphs: y= x-1 To build, we take points (-2; 3) (-1; 2) (0; 1) (1; 0) (2; 1). 3.3 Construction of graphs of functions containing "nested modules" in the formula Let's consider the construction algorithm using a specific example.

11 y \u003d i-2-ix + 5ii 1. We build a graph of the function. 2. We display the graph of the lower half-plane upwards symmetrically with respect to the OX axis and get the graph of the function. eleven

12 3. We display the graph of the function down symmetrically about the OX axis and get the graph of the function. 4. We display the graph of the function down symmetrically with respect to the OX axis and get the graph of the function 5. Display the graph of the function with respect to the OX axis and get the graph. 12

13 6. As a result, the graph of the function looks like this 3.4. An algorithm for constructing graphs of functions of the form y = a 1 x x 1 + a 2 x x a n x x n + ax + b. In the previous example, it was easy enough to expand the module signs. If there are more sums of modules, then it is problematic to consider all possible combinations of signs of submodule expressions. How can we graph the function in this case? Note that the graph is a polyline, with vertices at points having abscissas -1 and 2. For x = -1 and x = 2, the submodule expressions are equal to zero. In a practical way, we approached the rule for constructing such graphs: The graph of a function of the form y \u003d a 1 x x 1 + a 2 x x a n x x n + ax + b is a polyline with infinite extreme links. To construct such a polyline, it is enough to know all its vertices (the abscissas of the vertices are zeros of submodule expressions) and one control point each on the left and right infinite links. thirteen

14 Task. Plot the function y = x + x 1 + x + 1 and find its smallest value. Solution: 1. Zeros of submodule expressions: 0; -one; Polyline vertices (0; 2); (-thirteen); (1; 3). (zeros of submodule expressions are substituted into the equation) We build a graph (Fig. 7), the smallest value of the function is Algorithm for plotting a graph of a quadratic function with the module Drawing up algorithms for converting graphs of functions. 1.Construction of a graph of the function y= f(x). According to the definition of the module, this function is decomposed into a set of two functions. Therefore, the graph of the function y= f(x) consists of two graphs: y= f(x) in the right half-plane, y= f(-x) in the left half-plane. Based on this, we can formulate a rule (algorithm). The graph of the function y= f(x) is obtained from the graph of the function y= f(x) as follows: at x 0 the graph is preserved, and at x< 0полученная часть графика отображается симметрично относительно оси ОУ. 2.Построение графика функции y= f(x). а). Строим график функции y= f(x). б). Часть графика y= f(x), лежащая над осью ОХ, сохраняется, часть его, лежащая под осью ОХ, отображается симметрично относительно оси ОХ. 14

15 3. To build a graph of the function y= f(x), you must first graph the function y= f(x) for x> 0, then for x< 0 построить изображение, симметричное ему относительно оси ОУ, а затем на интервалах, где f(x) <0,построить изображение, симметричное графику y= f(x) относительно оси ОХ. 4.Для построения графиков вида y = f(x)достаточно построить график функции y= f(x) для тех х из области определения, при которых f(х) 0, и отобразить полученную часть графика симметрично относительно оси абсцисс. Пример Построим график функции у = х 2 6х +5. Сначала построим параболу у= х 2 6х +5. Чтобы получить из неё график функции у = х 2-6х + 5, нужно каждую точку параболы с отрицательной ординатой заменить точкой с той же абсциссой, но с противоположной (положительной) ординатой. Иными словами, часть параболы, расположенную ниже оси Ох, нужно заменить линией, ей симметричной относительно оси Ох (Рис.1). Рис Алгоритм построения графика дробно рациональной функции с модулем 1. Начнем с построения графика В основе его лежит график функции и все мы знаем, как он выглядит: Теперь построим график 15

16 To get this graph, it is enough just to shift the previously obtained graph by three units to the right. Note that if the denominator of the fraction were x + 3, then we would shift the graph to the left: Now we need to multiply by two all the ordinates to get the graph of the function Finally, we shift the graph up by two units: The last thing left for us to do , is to plot the given function if it is enclosed under the sign of the modulus. To do this, we reflect symmetrically upwards the entire part of the graph, the ordinates of which are negative (the part that lies below the x-axis): Fig.4 16

17 4. Changes in the graph of a quadratic function depending on the location of the sign of the absolute value. Plot the function y \u003d x 2 - x -3 1) Since x \u003d x at x 0, the required graph coincides with the parabola y \u003d 0.25 x 2 - x - 3. If x<0, то поскольку х 2 = х 2, х =-х и требуемый график совпадает с параболой у=0,25 х 2 + х) Если рассмотрим график у=0,25 х 2 - х - 3 при х 0 и отобразить его относительно оси ОУ мы получим тот же самый график. (0; - 3) координаты точки пересечения графика функции с осью ОУ. у =0, х 2 -х -3 = 0 х 2-4х -12 = 0 Имеем, х 1 = - 2; х 2 = 6. (-2; 0) и (6; 0) - координаты точки пересечения графика функции с осью ОХ. Если х<0, ордината точки требуемого графика такая же, как и у точки параболы, но с положительной абсциссой, равной х. Такие точки симметричны относительно оси ОУ(например, вершины (2; -4) и -(2; -4). Значит, часть требуемого графика, соответствующая значениям х<0, симметрична относительно оси ОУ его же части, соответствующей значениям х>0. b) Therefore, I complete for x<0 часть графика, симметричную построенной относительно оси ОУ. 17

18 Fig. 4 The graph of the function y \u003d f (x) coincides with the graph of the function y \u003d f (x) on the set of non-negative values ​​of the argument and is symmetrical to it with respect to the y-axis on the set of negative values ​​of the argument. Proof: If x 0, then f (x) = f (x), i.e. on the set of non-negative values ​​of the argument, the graphs of the functions y = f (x) and y = f (x) coincide. Since y \u003d f (x) is an even function, then its graph is symmetrical with respect to the OS. Thus, the graph of the function y \u003d f (x) can be obtained from the graph of the function y \u003d f (x) as follows: 1. plot the function y \u003d f (x) for x>0; 2. For x<0, симметрично отразить построенную часть относительно оси ОУ. Вывод: Для построения графика функции у = f (х) 1. построить график функции у = f(х) для х>0; 2. For x<0, симметрично отразить построенную часть относительно оси ОУ. Построить график функции у = х 2-2х Освободимся от знака модуля по определению Если х 2-2х 0, т.е. если х 0 и х 2, то х 2-2х = х 2-2х Если х 2-2х<0, т.е. если 0<х< 2, то х 2-2х =- х 2 + 2х Видим, что на множестве х 0 и х 2 графики функции у = х 2-2х и у = х 2-2х совпадают, а на множестве (0;2) графики функции у = -х 2 + 2х и у = х 2-2х совпадают. Построим их. График функции у = f (х) состоит из части графика функции у = f(х) при у?0 и симметрично отражённой части у = f(х) при у <0 относительно оси ОХ. Построить график функции у = х 2 - х -6 1) Если х 2 - х -6 0, т.е. если х -2 и х 3, то х 2 - х -6 = х 2 - х

19 If x 2 - x -6<0, т.е. если -2<х< 3, то х 2 - х -6 = -х 2 + х +6. Построим их. 2) Построим у = х 2 - х -6. Нижнюю часть графика симметрично отбражаем относительно ОХ. Сравнивая 1) и 2), видим что графики одинаковые. Работа на тетрадях. Докажем, что график функции у = f (х) совпадает с графиком функции у = f (х) для f(х) >0 and symmetrically reflected part y \u003d f (x) at y<0 относительно оси ОХ. Действительно, по определению абсолютной величины, можно данную функцию рассмотреть как совокупность двух линий: у = f(х), если f(х) 0; у = - f(х), если f(х) <0 Для любой функции у = f(х), если f(х) >0, then f (x) \u003d f (x), which means that in this part the graph of the function y \u003d f (x) coincides with the graph of the function itself y \u003d f (x). If f(x)<0, то f (х) = - f(х),т.е. точка (х; - f(х)) симметрична точке (х; f (х)) относительно оси ОХ. Поэтому для получения требуемого графика отражаем симметрично относительно оси ОХ "отрицательную" часть графика у = f(х). Вывод: действительно для построения графика функции у = f(х) достаточно: 1.Построить график функции у = f(х) ; 2. На участках, где график расположен в нижней полуплоскости, т.е., где f(х) <0, симметрично отражаем относительно оси абсцисс. (Рис.5) 19

20 Fig.5 Conclusion: To plot the function y= f(x) 1. Plot the function y=f(x) ; 2. In areas where the graph is located in the lower half-plane, i.e., where f (x)<0, строим кривые, симметричные построенным графикам относительно оси абсцисс. (Рис.6, 7.) 20

21 Research work on plotting function graphs y \u003d f (x) Applying the definition of the absolute value and the previously considered examples, we will plot the function graphs: y \u003d 2 x - 3 y \u003d x 2-5 x y \u003d x 2-2 and made conclusions. In order to build a graph of the function y = f (x) it is necessary: ​​1. Build a graph of the function y = f (x) for x>0. 2. Build the second part of the graph, i.e. reflect the constructed graph symmetrically with respect to the OS, because this function is even. 3. The sections of the resulting graph located in the lower half-plane should be converted to the upper half-plane symmetrically to the OX axis. Construct a graph of the function y \u003d 2 x - 3 (1st method for determining the module) X< -1,5 и х>1.5 a) y = 2x - 3, for x>0 b) for x<0, симметрично отражаем построенную часть относительно оси ОУ. 2. Строим у = -2 х + 3, для 2 х - 3 < 0. т.е. -1,5<х<1,5 а) у = -2х + 3, для х>0 b) for x<0, симметрично отражаем построенную часть относительно оси ОУ. У = 2 х - 3 1) Строим у = 2х-3, для х>0. 2) We build a straight line symmetrical to the one built with respect to the OS axis. 3) The sections of the graph located in the lower half-plane are displayed symmetrically about the OX axis. Comparing both graphs, we see that they are the same. 21

22 Examples of problems Example 1. Consider the graph of the function y = x 2 6x +5. Since x is squared, then regardless of the sign of the number x after squaring it will be positive. It follows from this that the graph of the function y \u003d x 2-6x +5 will be identical to the graph of the function y \u003d x 2-6x +5, i.e. graph of a function that does not contain an absolute value sign (Fig. 2). Fig.2 Example 2. Consider the graph of the function y \u003d x 2 6 x +5. Using the definition of the modulus of a number, we replace the formula y \u003d x 2 6 x +5 Now we are dealing with a piecewise assignment of dependence that is well known to us. We will build a graph like this: 1) build a parabola y \u003d x 2-6x +5 and circle that part of it, which is 22

23 corresponds to non-negative x values, i.e. the part to the right of the y-axis. 2) in the same coordinate plane, we construct a parabola y \u003d x 2 +6x +5 and circle that part of it that corresponds to negative values ​​of x, i.e. the part to the left of the y-axis. The circled parts of the parabolas together form a graph of the function y \u003d x 2-6 x +5 (Fig. 3). Fig.3 Example 3. Consider the graph of the function y \u003d x 2-6 x +5. Because the graph of the equation y \u003d x 2 6x +5 is the same as the graph of the function without the modulus sign (considered in example 2), it follows that the graph of the function y \u003d x 2 6 x +5 is identical to the graph of the function y \u003d x 2 6 x +5 , considered in example 2 (Fig. 3). Example 4. Let's build a graph of the function y \u003d x 2 6x +5. To do this, we construct a graph of the function y \u003d x 2-6x. To get from it the graph of the function y \u003d x 2-6x, you need to replace each point of the parabola with a negative ordinate with a point with the same abscissa, but with the opposite (positive) ordinate. In other words, the part of the parabola located below the x-axis must be replaced by a line symmetrical about the x-axis. Because we need to build a graph of the function y \u003d x 2-6x +5, then the graph of the function we have considered y \u003d x 2-6x just needs to be raised along the y axis by 5 units up (Fig. 4). 23

24 Fig.4 Example 5. Let's build a graph of the function y \u003d x 2-6x + 5. To do this, we use the well-known piecewise function. Find the zeros of the function y \u003d 6x +5 6x + 5 \u003d 0 at. Consider two cases: 1) If, then the equation takes the form y = x 2 6x -5. Let's build this parabola and circle that part of it where. 2) If, then the equation takes the form y \u003d x 2 + 6x +5. Let's build this parabola and circle that part of it, which is located to the left of the point with coordinates (Fig. 5). 24

25 Fig.5 Example6. Let's plot the function y \u003d x 2 6 x +5. To do this, we will plot the function y \u003d x 2-6 x +5. We plotted this graph in Example 3. Since our function is completely under the module sign, in order to plot the function graph y \u003d x 2 6 x +5, you need each point of the function graph y \u003d x 2 6 x + 5 with a negative ordinate, replace with a point with the same abscissa, but with the opposite (positive) ordinate, i.e. the part of the parabola located below the Ox axis must be replaced by a line that is symmetrical with respect to the Ox axis (Fig. 6). Fig.6 25

26 II. Conclusion "Mathematical information can be used skillfully and profitably only if it is mastered creatively, so that the student sees for himself how it would be possible to arrive at it independently." A.N. Kolmogorov. These tasks are of great interest to ninth grade students, as they are very common in the OGE tests. The ability to build these graphs of functions will allow you to pass the exam more successfully. French mathematicians Pierre Fermat () and Rene Descartes () imagined a function as a dependence of the ordinate of a curve point on its abscissa. And the English scientist Isaac Newton () understood the function as a coordinate of a moving point that changes depending on time. 26

27 III. List of references and sources 1. Galitsky M. L., Goldman A. M., Zvavich L. I. Collection of problems in algebra for grades 8 9: Proc. allowance for school students. and classes with deepening. study Mathematics 2nd ed. M .: Enlightenment, Dorofeev G.V. Mathematics. Algebra. Functions. Data analysis. Grade 9: m34 Proc. for general education studies. manager 2nd ed., stereotype. M .: Bustard, Solomonik V.S. Collection of questions and problems in mathematics M .: "Higher school", Yashchenko I.V. GIA. Mathematics: typical exam options: About options.m .: "National Education", p. 5. Yashchenko I.V. OGE. Mathematics: typical exam options: About options.m .: "National Education", p. 6. Yashchenko I.V. OGE. Mathematics: typical exam options: About options.m .: "National Education", p.

28 Appendix 28

29 Example 1. Plot the function y = x 2 8 x Solution. Let us define the parity of the function. The value for y(-x) is the same as the value for y(x), so this function is even. Then its graph is symmetrical with respect to the Oy axis. We build a graph of the function y \u003d x 2 8x + 12 for x 0 and display the graph symmetrically relative to Oy for negative x (Fig. 1). Example 2. The following graph of the form y \u003d x 2 8x This means that the graph of the function is obtained as follows: they build a graph of the function y \u003d x 2 8x + 12, leave the part of the graph that lies above the Ox axis unchanged, and the part of the graph that lies under the abscissa axis, is displayed symmetrically with respect to the Ox axis (Fig. 2). Example 3. To plot the function y \u003d x 2 8 x + 12, a combination of transformations is carried out: y \u003d x 2 8x + 12 y \u003d x 2 8 x + 12 y \u003d x 2 8 x Answer: Figure 3. Example 4 The expression standing under the module sign, changes sign at the point x=2/3. At x<2/3 функция запишется так: 29

30 For x>2/3, the function will be written as follows: That is, the point x=2/3 divides our coordinate plane into two regions, in one of which (to the right) we build the function and in the other (to the left) the graph of the function We build: Example 5 Next the graph is also broken, but has two breakpoints, since it contains two expressions under the module signs:

31 Expand the modules on the first interval: On the second interval: On the third interval: Thus, on the interval (- ; 1.5] we have the graph written by the first equation, on the interval the graph written by the second equation, and on the interval )