Graphic solution of systems of equations and inequalities examples. “Advantages of the graphical method for solving equations and inequalities.” Topic: Numeric functions

January 1 was officially recognized as a holiday in the Labor Code adopted by the All-Russian Central Executive Committee on December 10, 1918. Clause 7 of Article 104 of the Appendix to the Code established rules on weekly rest and holidays. According to this regulatory document, January 1 became the first calendar day off of the year of six holiday dates established in honor of social and historical events.

In the same list of non-working holidays, January 1 was approved in 1922 by the Labor Code. Moreover, according to the same document, since 1922, calendar holidays became paid, and the duration of the working days preceding them was reduced to six hours. Shortened pre-holiday days were also paid in full.

The situation with the first day of the year continued until 1929, when, by the Decree of the Council of People's Commissars of the USSR of September 24, 1929 “On working time and rest time in enterprises and institutions switching to a continuous production week,” calendar holidays were replaced by “revolutionary days.” The document excluded January 1 from paid days off - now on the first day of the year it was necessary to work, just like on weekdays. So January 1 became a regular working day and remained so until 1947 inclusive.

This day again became a day off and holiday starting on December 23, 1947 after the decree of the Presidium Supreme Council THE USSR. It remained so until 1991 – that is, until the collapse of the USSR.

January 1, 1919 is the day of the formation of the Belarusian SSR.

On January 1, 1969, the first issue of the cartoon “Well, Just Wait!” was released.

World events

One of the most convenient methods for solving quadratic inequalities is the graphical method. In this article we will look at how quadratic inequalities are solved graphically. First, let's discuss what the essence of this method is. Next, we will present the algorithm and consider examples of solving quadratic inequalities graphically.

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The essence of the graphic method

At all graphical method for solving inequalities with one variable is used not only to solve quadratic inequalities, but also other types of inequalities. The essence graphic method solutions to inequalities next: consider the functions y=f(x) and y=g(x), which correspond to the left and right sides of the inequality, build their graphs in one rectangular coordinate system and find out at what intervals the graph of one of them is lower or higher than the other. Those intervals where

• the graph of the function f above the graph of the function g are solutions to the inequality f(x)>g(x) ;
• the graph of the function f not lower than the graph of the function g are solutions to the inequality f(x)≥g(x) ;
• the graph of f below the graph of g are solutions to the inequality f(x)
• the graph of a function f not higher than the graph of a function g are solutions to the inequality f(x)≤g(x) .

We will also say that the abscissas of the intersection points of the graphs of the functions f and g are solutions to the equation f(x)=g(x) .

Let's transfer these results to our case - to solve the quadratic inequality a x 2 +b x+c<0 (≤, >, ≥).

We introduce two functions: the first y=a x 2 +b x+c (with f(x)=a x 2 +b x+c) corresponding to the left side of the quadratic inequality, the second y=0 (with g (x)=0 ) corresponds to the right side of the inequality. Schedule quadratic function f is a parabola and the graph constant function g – straight line coinciding with the abscissa axis Ox.

Next, according to the graphical method of solving inequalities, it is necessary to analyze at what intervals the graph of one function is located above or below another, which will allow us to write down the desired solution to the quadratic inequality. In our case, we need to analyze the position of the parabola relative to the Ox axis.

Depending on the values ​​of the coefficients a, b and c, the following six options are possible (for our needs, a schematic representation is sufficient, and we do not need to depict the Oy axis, since its position does not affect the solutions to the inequality):

In this drawing we see a parabola, the branches of which are directed upward, and which intersects the Ox axis at two points, the abscissa of which are x 1 and x 2. This drawing corresponds to the option when the coefficient a is positive (it is responsible for the upward direction of the parabola branches), and when the value is positive discriminant of a quadratic trinomial a x 2 +b x+c (in this case, the trinomial has two roots, which we denoted as x 1 and x 2, and we assumed that x 1 0 , D=b 2 −4·a·c=(−1) 2 −4·1·(−6)=25>0, x 1 =−2 , x 2 =3 .

For clarity, let’s depict in red the parts of the parabola located above the x-axis, and in blue – those located below the x-axis.


Now let's find out which intervals correspond to these parts. The following drawing will help you identify them (in the future we will make similar selections in the form of rectangles mentally):


So on the abscissa axis two intervals (−∞, x 1) and (x 2 , +∞) were highlighted in red, on them the parabola is above the Ox axis, they constitute a solution to the quadratic inequality a x 2 +b x+c>0 , and the interval (x 1 , x 2) is highlighted in blue, there is a parabola below the Ox axis, it represents the solution to the inequality a x 2 +b x+c<0 . Решениями нестрогих квадратных неравенств a·x 2 +b·x+c≥0 и a·x 2 +b·x+c≤0 будут те же промежутки, но в них следует включить числа x 1 и x 2 , отвечающие равенству a·x 2 +b·x+c=0 .

And now briefly: for a>0 and D=b 2 −4 a c>0 (or D"=D/4>0 for an even coefficient b)

• the solution to the quadratic inequality a x 2 +b x+c>0 is (−∞, x 1)∪(x 2 , +∞) or in another notation x x2;
• the solution to the quadratic inequality a x 2 +b x+c≥0 is (−∞, x 1 ]∪ or in another notation x 1 ≤x≤x 2 ,

where x 1 and x 2 are the roots of the quadratic trinomial a x 2 +b x+c, and x 1


Here we see a parabola, the branches of which are directed upward, and which touches the abscissa axis, that is, it has one common point with it; we denote the abscissa of this point as x 0. The presented case corresponds to a>0 (the branches are directed upward) and D=0 (the square trinomial has one root x 0). For example, you can take the quadratic function y=x 2 −4·x+4, here a=1>0, D=(−4) 2 −4·1·4=0 and x 0 =2.

The drawing clearly shows that the parabola is located above the Ox axis everywhere except the point of contact, that is, on the intervals (−∞, x 0), (x 0, ∞). For clarity, let’s highlight areas in the drawing by analogy with the previous paragraph.


We draw conclusions: for a>0 and D=0

• the solution to the quadratic inequality a·x 2 +b·x+c>0 is (−∞, x 0)∪(x 0, +∞) or in another notation x≠x 0;
• the solution to the quadratic inequality a·x 2 +b·x+c≥0 is (−∞, +∞) or in another notation x∈R ;
• quadratic inequality a x 2 +b x+c<0 не имеет решений (нет интервалов, на которых парабола расположена ниже оси Ox );
• the quadratic inequality a x 2 +b x+c≤0 has a unique solution x=x 0 (it is given by the point of tangency),

where x 0 is the root of the square trinomial a x 2 + b x + c.


In this case, the branches of the parabola are directed upward, and it does not have common points with the abscissa axis. Here we have the conditions a>0 (branches are directed upward) and D<0 (квадратный трехчлен не имеет действительных корней). Для примера можно построить график функции y=2·x 2 +1 , здесь a=2>0 , D=0 2 −4·2·1=−8<0 .

Obviously, the parabola is located above the Ox axis throughout its entire length (there are no intervals at which it is below the Ox axis, there is no point of tangency).


Thus, for a>0 and D<0 решением квадратных неравенств a·x 2 +b·x+c>0 and a x 2 +b x+c≥0 is the set of all real numbers, and the inequalities a x 2 +b x+c<0 и a·x 2 +b·x+c≤0 не имеют решений.

And there remain three options for the location of the parabola with branches directed downward, not upward, relative to the Ox axis. In principle, they need not be considered, since multiplying both sides of the inequality by −1 allows us to go to an equivalent inequality with a positive coefficient for x 2. But it still doesn’t hurt to get an idea about these cases. The reasoning here is similar, so we will write down only the main results.

Solution algorithm

The result of all previous calculations is algorithm for solving quadratic inequalities graphically:

On coordinate plane a schematic drawing is made, which shows the Ox axis (it is not necessary to depict the Oy axis) and a sketch of a parabola corresponding to the quadratic function y=a·x 2 +b·x+c. To draw a sketch of a parabola, it is enough to clarify two points:

• Firstly, by the value of the coefficient a it is determined where its branches are directed (for a>0 - upward, for a<0 – вниз).
• And secondly, by the value of the discriminant of the square trinomial a x 2 + b x + c it is determined whether the parabola intersects the abscissa axis at two points (for D>0), touches it at one point (for D=0), or has no common points with the Ox axis (at D<0 ). Для удобства на чертеже указываются координаты точек пересечения или координата точки касания (при наличии этих точек), а сами точки изображаются выколотыми при решении строгих неравенств, или обычными при решении нестрогих неравенств.

• When the drawing is ready, use it in the second step of the algorithm

  • when solving the quadratic inequality a·x 2 +b·x+c>0, the intervals are determined at which the parabola is located above the abscissa;
  • when solving the inequality a·x 2 +b·x+c≥0, the intervals at which the parabola is located above the abscissa axis are determined and the abscissas of the intersection points (or the abscissa of the tangent point) are added to them;
  • when solving the inequality a x 2 +b x+c<0 находятся промежутки, на которых парабола ниже оси Ox ;
  • finally, when solving a quadratic inequality of the form a·x 2 +b·x+c≤0, intervals are found in which the parabola is below the Ox axis and the abscissa of the intersection points (or the abscissa of the tangent point) is added to them;

  they constitute the desired solution to the quadratic inequality, and if there are no such intervals and no points of tangency, then the original quadratic inequality has no solutions.

All that remains is to solve a few quadratic inequalities using this algorithm.

Examples with solutions

Example.

Solve the inequality .

Solution.

We need to solve a quadratic inequality, let's use the algorithm from the previous paragraph. In the first step we need to sketch the graph of the quadratic function . The coefficient of x 2 is equal to 2, it is positive, therefore, the branches of the parabola are directed upward. Let’s also find out whether the parabola has common points with the x-axis; to do this, we’ll calculate the discriminant of the quadratic trinomial . We have . The discriminant turned out to be greater than zero, therefore, the trinomial has two real roots: And , that is, x 1 =−3 and x 2 =1/3.

From this it is clear that the parabola intersects the Ox axis at two points with abscissas −3 and 1/3. We will depict these points in the drawing as ordinary points, since we are solving a non-strict inequality. Based on the clarified data, we obtain the following drawing (it fits the first template from the first paragraph of the article):

Let's move on to the second step of the algorithm. Since we are solving a non-strict quadratic inequality with the sign ≤, we need to determine the intervals at which the parabola is located below the abscissa and add to them the abscissas of the intersection points.

From the drawing it is clear that the parabola is below the x-axis on the interval (−3, 1/3) and to it we add the abscissas of the intersection points, that is, the numbers −3 and 1/3. As a result, we arrive at the numerical interval [−3, 1/3] . This is the solution we are looking for. It can be written as a double inequality −3≤x≤1/3.

Answer:

[−3, 1/3] or −3≤x≤1/3 .

Example.

Find the solution to the quadratic inequality −x 2 +16 x−63<0 .

Solution.

As usual, we start with a drawing. The numerical coefficient for the square of the variable is negative, −1, therefore, the branches of the parabola are directed downward. Let's calculate the discriminant, or better yet, its fourth part: D"=8 2 −(−1)·(−63)=64−63=1. Its value is positive, let's calculate the roots of the square trinomial: And , x 1 =7 and x 2 =9. So the parabola intersects the Ox axis at two points with abscissas 7 and 9 (the original inequality is strict, so we will depict these points with an empty center). Now we can make a schematic drawing:

Since we are solving a strict quadratic inequality with a sign<, то нас интересуют промежутки, на которых парабола расположена ниже оси абсцисс:

The drawing shows that the solutions to the original quadratic inequality are two intervals (−∞, 7) , (9, +∞) .

Answer:

(−∞, 7)∪(9, +∞) or in another notation x<7 , x>9 .

When solving quadratic inequalities, when the discriminant of a quadratic trinomial on its left side is zero, you need to be careful about including or excluding the abscissa of the tangent point from the answer. This depends on the sign of the inequality: if the inequality is strict, then it is not a solution to the inequality, but if it is not strict, then it is.

Example.

Does the quadratic inequality 10 x 2 −14 x+4.9≤0 have at least one solution?

Solution.

Let's plot the function y=10 x 2 −14 x+4.9. Its branches are directed upward, since the coefficient of x 2 is positive, and it touches the abscissa axis at the point with the abscissa 0.7, since D"=(−7) 2 −10 4.9=0, whence or 0.7 in the form of a decimal fraction. Schematically it looks like this:

Since we are solving a quadratic inequality with the ≤ sign, its solution will be the intervals on which the parabola is below the Ox axis, as well as the abscissa of the tangent point. From the drawing it is clear that there is not a single gap where the parabola would be below the Ox axis, so its solution will be only the abscissa of the tangent point, that is, 0.7.

Answer:

this inequality has a unique solution 0.7.

Example.

Solve the quadratic inequality –x 2 +8 x−16<0 .

Solution.

We follow the algorithm for solving quadratic inequalities and start by constructing a graph. The branches of the parabola are directed downward, since the coefficient of x 2 is negative, −1. Let us find the discriminant of the square trinomial –x 2 +8 x−16, we have D’=4 2 −(−1)·(−16)=16−16=0 and then x 0 =−4/(−1) , x 0 =4 . So, the parabola touches the Ox axis at the abscissa point 4. Let's make the drawing:

We look at the sign of the original inequality, it is there<. Согласно алгоритму, решение неравенства в этом случае составляют все промежутки, на которых парабола расположена строго ниже оси абсцисс.

In our case, these are open rays (−∞, 4) , (4, +∞) . Separately, we note that 4 - the abscissa of the point of contact - is not a solution, since at the point of contact the parabola is not lower than the Ox axis.

Answer:

(−∞, 4)∪(4, +∞) or in another notation x≠4 .

Pay special attention to cases where the discriminant of the quadratic trinomial on the left side of the quadratic inequality is less than zero. There is no need to rush here and say that the inequality has no solutions (we are used to making such a conclusion for quadratic equations with a negative discriminant). The point is that the quadratic inequality for D<0 может иметь решение, которым является множество всех действительных чисел.

Example.

Find the solution to the quadratic inequality 3 x 2 +1>0.

Solution.

As usual, we start with a drawing. The coefficient a is 3, it is positive, therefore, the branches of the parabola are directed upward. We calculate the discriminant: D=0 2 −4·3·1=−12 . Since the discriminant is negative, the parabola has no common points with the Ox axis. The information obtained is sufficient for a schematic graph:

We solve a strict quadratic inequality with a > sign. Its solution will be all intervals in which the parabola is above the Ox axis. In our case, the parabola is above the x-axis along its entire length, so the desired solution will be the set of all real numbers.

Ox , and you also need to add the abscissa of the points of intersection or the abscissa of the tangency to them. But from the drawing it is clearly visible that there are no such intervals (since the parabola is everywhere below the abscissa axis), just as there are no points of intersection, just as there are no points of tangency. Therefore, the original quadratic inequality has no solutions.

Answer:

no solutions or in another entry ∅.

Bibliography.

• Algebra: textbook for 8th grade. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
• Algebra: 9th grade: educational. for general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2009. - 271 p. : ill. - ISBN 978-5-09-021134-5.
• Mordkovich A. G. Algebra. 8th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich. - 11th ed., erased. - M.: Mnemosyne, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.
• Mordkovich A. G. Algebra. 9th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich, P. V. Semenov. - 13th ed., erased. - M.: Mnemosyne, 2011. - 222 p.: ill. ISBN 978-5-346-01752-3.
• Mordkovich A. G. Algebra and beginning of mathematical analysis. Grade 11. In 2 hours. Part 1. Textbook for students of general education institutions (profile level) / A. G. Mordkovich, P. V. Semenov. - 2nd ed., erased. - M.: Mnemosyne, 2008. - 287 p.: ill. ISBN 978-5-346-01027-2.

L.A. Kustova

mathematic teacher

Voronezh, MBOU Lyceum No. 5

Project

“Advantages of the graphical method for solving equations and inequalities.”

Class:

7-11

Item:

Mathematics

Research objective:

To figure outadvantages of the graphical method of solving equations and inequalities.

Hypothesis:

Some equations and inequalities are easier and more aesthetically pleasing to solve graphically.

Research stages:

Compare analytical and graphical solution methodsequations and inequalities.

Find out in what cases the graphical method has advantages.

Consider solving equations with modulus and parameter.

Research results:

1.The beauty of mathematics is a philosophical problem.

2.When solving some equations and inequalities, a graphical solutionmost practical and attractive.

3. You can apply the attractiveness of mathematics at school using a graphical solutionequations and inequalities.

“The mathematical sciences have attracted special attention since ancient times,

Currently, they have received even more interest in their influence on art and industry.”

Pafnutiy Lvovich Chebyshev.

Starting from grade 7, various methods of solving equations and inequalities are considered, including graphical ones. Those who think that mathematics is a dry science, I think, change their opinions when they see how beautifully some types can be solvedequations and inequalities. Let me give you a few examples:

1).Solve the equation: = .

You can solve it analytically, that is, raise both sides of the equation to the third power and so on.

The graphical method is convenient for given equation, if you just need to indicate the number of solutions.

Similar tasks are often encountered when solving the “geometry” block of the 9th grade OGE.

2).Solve the equation with the parameter:

││ x│- 4│= a

Not the most complex example, but if you solve it analytically, you will have to open the module brackets twice, and for each case consider the possible values ​​of the parameter. Graphically everything is very simple. We draw function graphs and see that:

Sources:

Computer programAdvanced Graph .

see also Solving a linear programming problem graphically, Canonical form of linear programming problems

The system of constraints for such a problem consists of inequalities in two variables:
and the objective function has the form F = C 1 x + C 2 y which needs to be maximized.

Let us answer the question: what pairs of numbers ( x; y) are solutions to the system of inequalities, i.e., satisfy each of the inequalities simultaneously? In other words, what does it mean to solve a system graphically?
First you need to understand what is the solution to one linear inequality with two unknowns.
Solving a linear inequality with two unknowns means determining all pairs of unknown values ​​for which the inequality holds.
For example, inequality 3 x – 5y≥ 42 satisfy pairs ( x , y) : (100, 2); (3, –10), etc. The task is to find all such pairs.
Let's consider two inequalities: ax + by≤ c, ax + by≥ c. Straight ax + by = c divides the plane into two half-planes so that the coordinates of the points of one of them satisfy the inequality ax + by >c, and the other inequality ax + +by <c.
Indeed, let us take a point with coordinate x = x 0 ; then a point lying on a line and having an abscissa x 0, has an ordinate

Let for certainty a< 0, b>0, c>0. All points with abscissa x 0 lying above P(for example, dot M), have y M>y 0 , and all points below the point P, with abscissa x 0 , have y N<y 0 . Because the x 0 is an arbitrary point, then there will always be points on one side of the line for which ax+ by > c, forming a half-plane, and on the other side - points for which ax + by< c.

Picture 1

The inequality sign in the half-plane depends on the numbers a, b , c.
It follows from this next way graphical solution of systems of linear inequalities in two variables. To solve the system you need:

1. For each inequality, write the equation corresponding to this inequality.
2. Construct straight lines that are graphs of functions specified by equations.
3. For each line, determine the half-plane, which is given by the inequality. To do this, take an arbitrary point that does not lie on a line and substitute its coordinates into the inequality. if the inequality is true, then the half-plane containing the chosen point is the solution to the original inequality. If the inequality is false, then the half-plane on the other side of the line is the set of solutions to this inequality.
4. To solve a system of inequalities, it is necessary to find the area of ​​intersection of all half-planes that are the solution to each inequality of the system.

This area may turn out to be empty, then the system of inequalities has no solutions and is inconsistent. Otherwise, the system is said to be consistent.
There can be a finite number or an infinite number of solutions. The area can be a closed polygon or unbounded.

Let's look at three relevant examples.

Example 1. Solve the system graphically:
x + y – 1 ≤ 0;
–2x – 2y + 5 ≤ 0.

• consider the equations x+y–1=0 and –2x–2y+5=0 corresponding to the inequalities;
• Let's construct straight lines given by these equations.

Figure 2

Let us define the half-planes defined by the inequalities. Let's take an arbitrary point, let (0; 0). Let's consider x+ y– 1 0, substitute the point (0; 0): 0 + 0 – 1 ≤ 0. This means that in the half-plane where the point (0; 0) lies, x + y – 1 ≤ 0, i.e. the half-plane lying below the line is a solution to the first inequality. Substituting this point (0; 0) into the second, we get: –2 ∙ 0 – 2 ∙ 0 + 5 ≤ 0, i.e. in the half-plane where the point (0; 0) lies, –2 x – 2y+ 5≥ 0, and we were asked where –2 x – 2y+ 5 ≤ 0, therefore, in the other half-plane - in the one above the straight line.
Let's find the intersection of these two half-planes. The lines are parallel, so the planes do not intersect anywhere, which means that the system of these inequalities has no solutions and is inconsistent.

Example 2. Find graphically solutions to the system of inequalities:

Figure 3
1. Let's write out the equations corresponding to the inequalities and construct straight lines.
x + 2y– 2 = 0

x	2	0
y	0	1

y – x – 1 = 0

x	0	2
y	1	3

y + 2 = 0;
y = –2.
2. Having chosen the point (0; 0), we determine the signs of inequalities in the half-planes:
0 + 2 ∙ 0 – 2 ≤ 0, i.e. x + 2y– 2 ≤ 0 in the half-plane below the straight line;
0 – 0 – 1 ≤ 0, i.e. y –x– 1 ≤ 0 in the half-plane below the straight line;
0 + 2 =2 ≥ 0, i.e. y+ 2 ≥ 0 in the half-plane above the straight line.
3. The intersection of these three half-planes will be an area that is a triangle. It is not difficult to find the vertices of the region as the intersection points of the corresponding lines


Thus, A(–3; –2), IN(0; 1), WITH(6; –2).

Let's consider another example in which the resulting solution domain of the system is not limited.

FEDERAL AGENCY FOR EDUCATION

INSTITUTE FOR EDUCATIONAL DEVELOPMENT

“Graphical methods for solving equations and inequalities with parameters”

Completed

mathematic teacher

Municipal educational institution secondary school No. 62

Lipetsk 2008

INTRODUCTION........................................................ ........................................................ .3

X;at) 4

1.1. Parallel transfer................................................... ........................... 5

1.2. Turn................................................. ........................................................ 9

1.3. Homothety. Compression to straight line................................................... ................. 13

1.4. Two straight lines on a plane................................................... ....................... 15

2. GRAPHIC TECHNIQUES. COORDINATE PLANE ( X;A) 17

CONCLUSION................................................. .......................................... 20

BIBLIOGRAPHICAL LIST.................................................................... ........ 22

INTRODUCTION

The problems that schoolchildren have when solving non-standard equations and inequalities are caused both by the relative complexity of these problems and by the fact that school, as a rule, focuses on solving standard problems.

Many schoolchildren perceive the parameter as a “regular” number. Indeed, in some problems a parameter can be considered a constant value, but this constant value takes on unknown values! Therefore, it is necessary to consider the problem for all possible values ​​of this constant. In other problems, it may be convenient to artificially declare one of the unknowns as a parameter.

Other schoolchildren treat a parameter as an unknown quantity and, without embarrassment, can express the parameter in terms of a variable in their answer X.

At graduations and entrance exams There are mainly two types of problems with parameters. You can immediately distinguish them by their wording. First: “For each parameter value, find all solutions to some equation or inequality.” Second: “Find all values ​​of the parameter, for each of which certain conditions are satisfied for a given equation or inequality.” Accordingly, the answers in problems of these two types differ in essence. The answer to a problem of the first type lists all possible values ​​of the parameter and for each of these values ​​the solutions to the equation are written. The answer to a problem of the second type indicates all parameter values ​​under which the conditions specified in the problem are met.

The solution of an equation with a parameter for a given fixed value of the parameter is such a value of the unknown, when substituting it into the equation, the latter turns into a correct numerical equality. The solution to an inequality with a parameter is determined similarly. Solving an equation (inequality) with a parameter means, for each admissible value of the parameter, finding the set of all solutions to a given equation (inequality).

1. GRAPHIC TECHNIQUES. COORDINATE PLANE ( X;at)

Along with the basic analytical techniques and methods for solving problems with parameters, there are ways to use visual and graphical interpretations.

Depending on what role the parameter is assigned to in the problem (unequal or equal to the variable), two main graphical techniques can be distinguished accordingly: the first is the construction of a graphical image on the coordinate plane (X;y), the second - on (X; A).

On the plane (x; y) the function y =f (X; A) defines a family of curves depending on the parameter A. It is clear that every family f has certain properties. We will be primarily interested in what kind of plane transformation (parallel translation, rotation, etc.) can be used to move from one curve of the family to another. A separate paragraph will be devoted to each of these transformations. It seems to us that such a classification makes it easier for the decider to find the necessary graphic image. Note that with this approach, the ideological part of the solution does not depend on which figure (straight line, circle, parabola, etc.) will be a member of the family of curves.

Of course, the graphic image of the family is not always y =f (X;A) described by a simple transformation. Therefore, in such situations, it is useful to focus not on how the curves of the same family are related, but on the curves themselves. In other words, we can distinguish another type of problem in which the idea of ​​a solution is primarily based on the properties of specific geometric shapes, and not the family as a whole. What figures (more precisely, families of these figures) will interest us first of all? These are straight lines and parabolas. This choice is due to the special (basic) position of the linear and quadratic functions in school mathematics.

Speaking about graphical methods, it is impossible to avoid one problem “born” from the practice of competitive exams. We are referring to the question of the rigor, and therefore the legality, of a decision based on graphic considerations. Undoubtedly, from a formal point of view, the result taken from the “picture”, not supported analytically, was not obtained strictly. However, who, when and where determines the level of rigor that a high school student should adhere to? In our opinion, the requirements for the level of mathematical rigor for a student should be determined common sense. We understand the degree of subjectivity of such a point of view. Moreover, the graphical method is just one of the means of clarity. And visibility can be deceiving..gif" width="232" height="28"> has only one solution.

Solution. For convenience, we denote lg b = a. Let's write an equation equivalent to the original one: https://pandia.ru/text/78/074/images/image004_56.gif" width="125" height="92">

Building a graph of a function with the domain of definition and (Fig. 1). The resulting graph is a family of straight lines y = a must intersect at only one point. The figure shows that this requirement is met only when a > 2, i.e. lg b> 2, b> 100.

Answer. https://pandia.ru/text/78/074/images/image010_28.gif" width="15 height=16" height="16"> determine the number of solutions to the equation .

Solution. Let's plot the function 102" height="37" style="vertical-align:top">

Let's consider. This is a straight line parallel to the OX axis.

Answer..gif" width="41" height="20">, then 3 solutions;

if , then 2 solutions;

if , 4 solutions.

Let's move on to a new series of tasks..gif" width="107" height="27 src=">.

Solution. Let's build a straight line at= X+1 (Fig. 3)..gif" width="92" height="57">

have one solution, which is equivalent for the equation ( X+1)2 = x + A have one root..gif" width="44 height=47" height="47"> the original inequality has no solutions. Note that someone who is familiar with the derivative can obtain this result differently.

Next, shifting the “semi-parabola” to the left, we will fix the last moment when the graphs at = X+ 1 and have two common points (position III). This arrangement is ensured by the requirement A= 1.

It is clear that for the segment [ X 1; X 2], where X 1 and X 2 – abscissas of the points of intersection of the graphs, will be the solution to the original inequality..gif" width="68 height=47" height="47">, then

When a "semi-parabola" and a straight line intersect at only one point (this corresponds to the case a > 1), then the solution will be the segment [- A; X 2"], where X 2" – the largest of the roots X 1 and X 2 (position IV).

Example 4..gif" width="85" height="29 src=">.gif" width="75" height="20 src="> . From here we get .

Let's look at the functions and . Among them, only one defines a family of curves. Now we see that the replacement brought undoubted benefits. In parallel, we note that in the previous problem, using a similar replacement, you can make not a “semi-parabola” move, but a straight line. Let's turn to Fig. 4. Obviously, if the abscissa of the vertex is a “semi-parabola” more than one, i.e. –3 A > 1, , then the equation has no roots..gif" width="89" height="29"> and have a different monotonicity.

Answer. If then the equation has one root; if https://pandia.ru/text/78/074/images/image039_10.gif" width="141" height="81 src=">

has solutions.

Solution. It is clear that direct families https://pandia.ru/text/78/074/images/image041_12.gif" width="61" height="52">..jpg" width="259" height="155" >

Meaning k1 we will find by substituting the pair (0;0) into the first equation of the system. From here k1 =-1/4. Meaning k 2 we get by demanding from the system

https://pandia.ru/text/78/074/images/image045_12.gif" width="151" height="47"> when k> 0 have one root. From here k2= 1/4.

Answer. .

Let's make one remark. In some examples of this point, we will have to solve a standard problem: for a line family, find its angular coefficient corresponding to the moment of tangency with the curve. We will show how to do this in general form using the derivative.

If (x0; y 0) = center of rotation, then the coordinates (X 1; at 1) points of tangency with the curve y =f(x) can be found by solving the system

The required slope k equal to .

Example 6. For what values ​​of the parameter does the equation have a unique solution?

Solution..gif" width="160" height="29 src=">..gif" width="237" height="33">, arc AB.

All rays passing between OA and OB intersect the arc AB at one point, and also intersect the arc AB OB and OM (tangent) at one point..gif" width="16" height="48 src=">. The angular coefficient of the tangent is equal to . Easily found from the system

So, direct families https://pandia.ru/text/78/074/images/image059_7.gif" width="139" height="52">.

Answer. .

Example 7..gif" width="160" height="25 src="> has a solution?

Solution..gif" width="61" height="24 src="> and decreases by . The point is the maximum point.

A function is a family of straight lines passing through the point https://pandia.ru/text/78/074/images/image062_7.gif" width="153" height="28"> is the arc AB. The straight lines that will be located between straight lines OA and OB, satisfy the conditions of the problem..gif" width="17" height="47 src=">.

Answer..gif" width="15" height="20">no solutions.

1.3. Homothety. Compression to a straight line.

Example 8. How many solutions does the system have?

https://pandia.ru/text/78/074/images/image073_1.gif" width="41" height="20 src="> the system has no solutions. For a fixed a > 0 the graph of the first equation is a square with vertices ( A; 0), (0;-A), (-a;0), (0;A). Thus, the members of the family are homothetic squares (the center of homothety is the point O(0; 0)).

Let's turn to Fig. 8..gif" width="80" height="25"> each side of the square has two common points with the circle, which means the system will have eight solutions. When the circle turns out to be inscribed in the square, i.e. there will be four solutions again . Obviously, the system has no solutions.

Answer. If A< 1 или https://pandia.ru/text/78/074/images/image077_1.gif" width="56" height="25 src=">, then there are four solutions; if , then there are eight solutions.

Example 9. Find all values ​​of the parameter, for each of which the equation is https://pandia.ru/text/78/074/images/image081_0.gif" width="181" height="29 src=">. Consider the function ..jpg" width="195" height="162">

The number of roots will correspond to the number 8 when the radius of the semicircle is greater and less than , that is. Note that there is .

Answer. or .

1.4. Two straight lines on a plane

Essentially, the idea of ​​solving the problems of this paragraph is based on the question of research relative position two straight lines: And . It is easy to show the solution to this problem in general form. We will turn directly to specific typical examples, which, in our opinion, will not cause damage common side question.

Example 10. For what a and b does the system

https://pandia.ru/text/78/074/images/image094_0.gif" width="160" height="25 src=">..gif" width="67" height="24 src="> , t..gif" width="116" height="55">

The inequality of the system defines a half-plane with boundary at= 2x– 1 (Fig. 10). It is easy to realize that the resulting system has a solution if the straight line ah +by = 5 intersects the boundary of a half-plane or, being parallel to it, lies in the half-plane at– 2x + 1 < 0.

Let's start with the case b = 0. Then it would seem that the equation Oh+ by = 5 defines a vertical line that obviously intersects the line y = 2X - 1. However, this statement is true only when ..gif" width="43" height="20 src="> the system has solutions ..gif" width="99" height="48">. In this case, the condition for the intersection of lines is achieved at , i.e. ..gif" width="52" height="48">.gif" width="41" height="20"> and , or and , or and https ://pandia.ru/text/78/074/images/image109_0.gif" width="69" height="24 src=">.

− In the coordinate plane xOa we build a graph of the function.

− Consider the straight lines and select those intervals of the Oa axis at which these straight lines satisfy the following conditions: a) does not intersect the graph of the function https://pandia.ru/text/78/074/images/image109_0.gif" width="69" height ="24"> at one point, c) at two points, d) at three points and so on.

− If the task is to find the values ​​of x, then we express x in terms of a for each of the found intervals of the value of a separately.

The view of a parameter as an equal variable is reflected in graphical methods..jpg" width="242" height="182">

Answer. a = 0 or a = 1.

CONCLUSION

We hope that the analyzed problems convincingly demonstrate the effectiveness of the proposed methods. However, unfortunately, the scope of application of these methods is limited by the difficulties that can be encountered when constructing a graphic image. Is it really that bad? Apparently not. Indeed, with this approach, the main didactic value of problems with parameters as a model of miniature research is largely lost. However, the above considerations are addressed to teachers, and for applicants the formula is quite acceptable: the end justifies the means. Moreover, let us take the liberty of saying that in a considerable number of universities, compilers of competitive problems with parameters follow the path from the picture to the condition.

In these problems, we discussed the possibilities for solving problems with a parameter that open up to us when we draw graphs of functions included in the left and right sides of equations or inequalities on a sheet of paper. Due to the fact that the parameter can take arbitrary values, one or both of the displayed graphs move in a certain way on the plane. We can say that a whole family of graphs is obtained corresponding to different values ​​of the parameter.

Let us strongly emphasize two details.

Firstly, we are not talking about a “graphical” solution. All values, coordinates, roots are calculated strictly, analytically, as solutions to the corresponding equations and systems. The same applies to cases of touching or crossing graphs. They are determined not by eye, but with the help of discriminants, derivatives and other tools available to you. The picture only gives a solution.

Secondly, even if you do not find any way to solve the problem associated with the graphs shown, your understanding of the problem will expand significantly, you will receive information for self-testing and the chances of success will increase significantly. By accurately imagining what happens in a problem when different meanings parameter, you may find the correct solution algorithm.

Therefore, we will conclude these words with an urgent sentence: if in the slightest degree difficult task There are functions for which you know how to draw graphs, be sure to do it, you won’t regret it.

BIBLIOGRAPHICAL LIST

1. Cherkasov,: Handbook for high school students and applicants to universities [Text] /, . – M.: AST-PRESS, 2001. – 576 p.

2. Gorshtein, with parameters [Text]: 3rd edition, expanded and revised / , . – M.: Ilexa, Kharkov: Gymnasium, 1999. – 336 p.