Graphic solution equations

Heyday, 2009

Introduction

The need to solve quadratic equations in ancient times was caused by the need to solve problems related to finding areas land plots and with earthworks of a military nature, as well as with the development of astronomy and mathematics itself. The Babylonians were able to solve quadratic equations around 2000 BC. The rule for solving these equations, set out in the Babylonian texts, essentially coincides with modern ones, but it is not known how the Babylonians arrived at this rule.

Formulas for solving quadratic equations in Europe were first set forth in the Book of Abacus, written in 1202 by the Italian mathematician Leonardo Fibonacci. His book contributed to the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries.

But general rule solutions to quadratic equations for all possible combinations of coefficients b and c were formulated in Europe only in 1544 by M. Stiefel.

In 1591 Francois Viet introduced formulas for solving quadratic equations.

In ancient Babylon they could solve some types of quadratic equations.

Diophantus of Alexandria And Euclid, Al-Khwarizmi And Omar Khayyam solved equations using geometric and graphical methods.

In 7th grade we studied functions y = C, y =kx, y =kx+ m, y =x 2,y = –x 2, in 8th grade – y = √x, y =|x|, y =ax2 + bx+ c, y =k/ x. In the 9th grade algebra textbook, I saw functions that were not yet known to me: y =x 3, y =x 4,y =x 2n, y =x- 2n, y = 3√x, (x– a) 2 + (y –b) 2 = r 2 and others. There are rules for constructing graphs of these functions. I wondered if there were other functions that obey these rules.

My job is to study function graphs and solve equations graphically.

1. What are the functions?

The graph of a function is the set of all points coordinate plane, the abscissas of which are equal to the values ​​of the arguments, and the ordinates are equal to the corresponding values ​​of the function.

Linear function given by the equation y =kx+ b, Where k And b- some numbers. The graph of this function is a straight line.

Function inverse proportionality y =k/ x, where k ¹ 0. The graph of this function is called a hyperbola.

Function (x– a) 2 + (y –b) 2 = r2 , Where A, b And r- some numbers. The graph of this function is a circle of radius r with center at point A ( A, b).

Quadratic function y= ax2 + bx+ c Where A,b, With– some numbers and A¹ 0. The graph of this function is a parabola.

The equation at2 (a– x) = x2 (a+ x) . The graph of this equation will be a curve called a strophoid.

/>Equation (x2 + y2 ) 2 = a(x2 – y2 ) . The graph of this equation is called Bernoulli's lemniscate.

The equation. The graph of this equation is called an astroid.

Curve (x2 y2 – 2 ax)2 =4a2 (x2 + y2 ) . This curve is called a cardioid.

Functions: y =x 3 – cubic parabola, y =x 4, y = 1/x 2.

2. The concept of an equation and its graphical solution

The equation– an expression containing a variable.

Solve the equation- this means finding all its roots, or proving that they do not exist.

Root of the equation is a number that, when substituted into an equation, produces a correct numerical equality.

Solving equations graphically allows you to find the exact or approximate value of the roots, allows you to find the number of roots of the equation.

When constructing graphs and solving equations, the properties of a function are used, which is why the method is often called functional-graphical.

To solve the equation, we “divide” it into two parts, introduce two functions, build their graphs, and find the coordinates of the points of intersection of the graphs. The abscissas of these points are the roots of the equation.

3. Algorithm for plotting a function graph

Knowing the graph of a function y =f(x) , you can build graphs of functions y =f(x+ m) ,y =f(x)+ l And y =f(x+ m)+ l. All these graphs are obtained from the graph of the function y =f(x) using parallel carry transformation: to │ m│ units of scale to the right or left along the x-axis and on │ l│ units of scale up or down along an axis y.

4. Graphical solution of the quadratic equation

Using a quadratic function as an example, we will consider the graphical solution of a quadratic equation. The graph of a quadratic function is a parabola.

What did the ancient Greeks know about the parabola?

Modern mathematical symbolism originated in the 16th century.

The ancient Greek mathematicians did not coordinate method, there was no concept of function. Nevertheless, the properties of the parabola were studied in detail by them. The ingenuity of ancient mathematicians is simply amazing - after all, they could only use drawings and verbal descriptions dependencies.

Most fully explored the parabola, hyperbola and ellipse Apollonius of Perga, who lived in the 3rd century BC. He gave these curves names and indicated what conditions the points lying on this or that curve satisfy (after all, there were no formulas!).

There is an algorithm for constructing a parabola:

Find the coordinates of the vertex of the parabola A (x0; y0): X=- b/2 a;

y0=axo2+in0+s;

Find the axis of symmetry of the parabola (straight line x=x0);

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We compile a table of values ​​for constructing control points;

We construct the resulting points and construct points that are symmetrical to them relative to the axis of symmetry.

1. Using the algorithm, we will construct a parabola y= x2 – 2 x– 3 . Abscissas of points of intersection with the axis x and there are roots of the quadratic equation x2 – 2 x– 3 = 0.

There are five ways to solve this equation graphically.

2. Let's split the equation into two functions: y= x2 And y= 2 x+ 3

3. Let's split the equation into two functions: y= x2 –3 And y=2 x. The roots of the equation are the abscissas of the points of intersection of the parabola and the line.

4. Transform the equation x2 – 2 x– 3 = 0 by isolating a complete square into functions: y= (x–1) 2 And y=4. The roots of the equation are the abscissas of the points of intersection of the parabola and the line.

5. Divide both sides of the equation term by term x2 – 2 x– 3 = 0 on x, we get x– 2 – 3/ x= 0 , let's split this equation into two functions: y= x– 2, y= 3/ x. The roots of the equation are the abscissas of the points of intersection of the line and the hyperbola.

5. Graphical solution of degree equationsn

Example 1. Solve the equation x5 = 3 – 2 x.

y= x5 , y= 3 – 2 x.

Answer: x = 1.

Example 2. Solve the equation 3 √ x= 10 – x.

Roots given equation is the abscissa of the point of intersection of the graphs of two functions: y= 3 √ x, y= 10 – x.

Answer: x = 8.

Conclusion

Having looked at the graphs of the functions: y =ax2 + bx+ c, y =k/ x, у = √x, y =|x|, y =x 3, y =x 4,y = 3√x, I noticed that all these graphs are built according to the rule of parallel translation relative to the axes x And y.

Using the example of solving a quadratic equation, we can conclude that the graphical method is also applicable for equations of degree n.

Graphical methods for solving equations are beautiful and understandable, but do not provide a 100% guarantee of solving any equation. The abscissas of the intersection points of the graphs can be approximate.

In 9th grade and in high school, I will continue to get acquainted with other functions. I'm interested to know whether those functions obey the rules of parallel transfer when constructing their graphs.

Next year I would also like to consider the issues of graphically solving systems of equations and inequalities.

Literature

1. Algebra. 7th grade. Part 1. Textbook for educational institutions/ A.G. Mordkovich. M.: Mnemosyne, 2007.

2. Algebra. 8th grade. Part 1. Textbook for educational institutions / A.G. Mordkovich. M.: Mnemosyne, 2007.

3. Algebra. 9th grade. Part 1. Textbook for educational institutions / A.G. Mordkovich. M.: Mnemosyne, 2007.

4. Glazer G.I. History of mathematics at school. VII–VIII grades. – M.: Education, 1982.

5. Journal Mathematics No. 5 2009; No. 8 2007; No. 23 2008.

6. Graphical solution of equations websites on the Internet: Tol VIKI; stimul.biz/ru; wiki.iot.ru/images; berdsk.edu; pege 3–6.htm.

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) that correspond to the left and right side inequalities, 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 x 2 ;
• 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 (branches directed upward) and D=0 ( quadratic trinomial has one root x 0 ). For example, you can take 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:

A schematic drawing is made on the coordinate plane, which depicts 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, based on 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 this equation if you simply 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's 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.
This implies the following method for graphically solving 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 may be solutions final number And infinite set. 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.

Many tasks that we are used to calculating purely algebraically can be solved much easier and faster; using function graphs will help us with this. You say “how so?” draw something, and what to draw? Believe me, sometimes it is more convenient and easier. Shall we get started? Let's start with the equations!

Graphical solution of equations

Graphical solution of linear equations

As you already know, the graph of a linear equation is a straight line, hence the name of this type. Linear equations are quite easy to solve algebraically - we transfer all the unknowns to one side of the equation, everything we know to the other, and voila! We found the root. Now I'll show you how to do it graphically.

So you have the equation:

How to solve it?
Option 1, and the most common one is to move the unknowns to one side and the knowns to the other, we get:

Now let's build. What did you get?

What do you think is the root of our equation? That's right, the coordinate of the intersection point of the graphs is:

Our answer is

That's the whole wisdom of the graphic solution. As you can easily check, the root of our equation is a number!

As I said above, this is the most common option, close to algebraic solution, but you can solve it differently. To consider an alternative solution, let's return to our equation:

This time we will not move anything from side to side, but will construct the graphs directly, as they are now:

Built? Let's see!

What is the solution this time? That's right. The same thing - the coordinate of the intersection point of the graphs:

And, again, our answer is.

As you can see, with linear equations everything is extremely simple. It's time to look at something more complex... For example, graphical solution of quadratic equations.

Graphical solution of quadratic equations

So, now let's start solving the quadratic equation. Let's say you need to find the roots of this equation:

Of course, you can now start counting through the discriminant, or according to Vieta’s theorem, but many people, out of nerves, make mistakes when multiplying or squaring, especially if the example is with large numbers, and, as you know, you won’t have a calculator for the exam... Therefore, let’s try to relax a little and draw while solving this equation.

Solutions to this equation can be found graphically in various ways. Let's consider various options, and you can choose which one you like best.

Method 1. Directly

We simply build a parabola using this equation:

To do this quickly, I'll give you one little hint: It is convenient to start the construction by determining the vertex of the parabola. The following formulas will help determine the coordinates of the vertex of a parabola:

You will say “Stop! The formula for is very similar to the formula for finding the discriminant,” yes, it is, and this is a huge disadvantage of “directly” constructing a parabola to find its roots. However, let's count to the end, and then I'll show you how to do it much (much!) easier!

Did you count? What coordinates did you get for the vertex of the parabola? Let's figure it out together:

Exactly the same answer? Well done! And now we already know the coordinates of the vertex, but to construct a parabola we need more... points. How many minimum points do you think we need? Right, .

You know that a parabola is symmetrical about its vertex, for example:

Accordingly, we need two more points on the left or right branch of the parabola, and in the future we will symmetrically reflect these points on the opposite side:

Let's return to our parabola. For our case, period. We need two more points, so we can take positive ones, or we can take negative ones? Which points are more convenient for you? It’s more convenient for me to work with positive ones, so I’ll calculate at and.

Now we have three points, we can easily construct our parabola by reflecting the last two points relative to its vertex:

What do you think is the solution to the equation? That's right, points at which, that is, and. Because.

And if we say that, it means that it must also be equal, or.

Just? We have finished solving the equation with you in a complex graphical way, or there will be more!

Of course, you can check our answer algebraically - you can calculate the roots using Vieta's theorem or Discriminant. What did you get? The same? Here you see! Now let's look at a very simple graphic solution, I'm sure you'll really like it!

Method 2. Divided into several functions

Let’s take our same equation: , but we’ll write it a little differently, namely:

Can we write it like this? We can, since the transformation is equivalent. Let's look further.

Let's construct two functions separately:

1. - the graph is a simple parabola, which you can easily construct even without defining the vertex using formulas and drawing up a table to determine other points.
2. - the graph is a straight line, which you can just as easily construct by estimating the values ​​in your head without even resorting to a calculator.

Built? Let's compare with what I got:

What do you think are the roots of the equation in this case? Right! The coordinates obtained by the intersection of two graphs and, that is:

Accordingly, the solution to this equation is:

What do you say? Agree, this method of solution is much easier than the previous one and even easier than looking for roots through a discriminant! If so, try solving the following equation using this method:

What did you get? Let's compare our graphs:

The graphs show that the answers are:

Did you manage? Well done! Now let's look at the equations a little more complicated, namely, the solution of mixed equations, that is, equations containing functions of different types.

Graphical solution of mixed equations

Now let's try to solve the following:

Of course, we can bring everything to common denominator, find the roots of the resulting equation, not forgetting to take into account the ODZ, but again, we will try to solve it graphically, as we did in all previous cases.

This time let's build the following 2 graphs:

1. - the graph is a hyperbola
2. - the graph is a straight line, which you can easily construct by estimating the values ​​in your head without even resorting to a calculator.

Realized it? Now start building.

Here's what I got:

Looking at this picture, tell me what are the roots of our equation?

That's right, and. Here's the confirmation:

Try plugging our roots into the equation. Happened?

That's right! Agree, solving such equations graphically is a pleasure!

Try to solve the equation graphically yourself:

I’ll give you a hint: move part of the equation to the right side so that the simplest functions to construct are on both sides. Did you get the hint? Take action!

Now let's see what you got:

Respectively:

1. - cubic parabola.
2. - ordinary straight line.

Well, let's build:

As you wrote down long ago, the root of this equation is - .

Having decided this a large number of examples, I'm sure you realized how easily and quickly you can solve equations graphically. It's time to figure out how to solve systems in this way.

Graphic solution of systems

Graphically solving systems is essentially no different from graphically solving equations. We will also build two graphs, and their intersection points will be the roots of this system. One graph is one equation, the second graph is another equation. Everything is extremely simple!

Let's start with the simplest thing - solving systems of linear equations.

Solving systems of linear equations

Let's say we have the following system:

First, let's transform it so that on the left there is everything that is connected with, and on the right - everything that is connected with. In other words, let’s write these equations as a function in our usual form:

Now we just build two straight lines. What is the solution in our case? Right! The point of their intersection! And here you need to be very, very careful! Think about it, why? Let me give you a hint: we are dealing with a system: in the system there is both, and... Got the hint?

That's right! When solving a system, we must look at both coordinates, and not just as when solving equations! Another important point is to write them down correctly and not confuse where we have the meaning and where the meaning is! Did you write it down? Now let's compare everything in order:

And the answers: and. Do a check - substitute the found roots into the system and make sure whether we solved it correctly graphically?

Solving systems of nonlinear equations

What if, instead of one straight line, we have quadratic equation? It's okay! You just build a parabola instead of a straight line! Do not believe? Try solving the following system:

What's our next step? That’s right, write it down so that it’s convenient for us to build graphs:

And now it’s all a matter of small things - build it quickly and here’s your solution! We are building:

Did the graphs turn out the same? Now mark the solutions of the system in the figure and correctly write down the identified answers!

I've done everything? Compare with my notes:

Is everything right? Well done! You are already cracking these types of tasks like nuts! If so, let’s give you a more complicated system:

What are we doing? Right! We write the system so that it is convenient to build:

I’ll give you a little hint, since the system looks very complicated! When building graphs, build them “more”, and most importantly, do not be surprised by the number of intersection points.

So, let's go! Exhaled? Now start building!

So how? Beautiful? How many intersection points did you get? I have three! Let's compare our graphs:

Also? Now carefully write down all the solutions of our system:

Now look at the system again:

Can you imagine that you solved this in just 15 minutes? Agree, mathematics is still simple, especially when looking at an expression you are not afraid to make a mistake, but just take it and solve it! You're a big lad!

Graphical solution of inequalities

Graphical solution of linear inequalities

After the last example, you can do anything! Now breathe out - compared to the previous sections, this one will be very, very easy!

We will start, as usual, with a graphical solution to a linear inequality. For example, this one:

First, let's carry out the simplest transformations - open the brackets full squares and give similar terms:

The inequality is not strict, therefore it is not included in the interval, and the solution will be all points that are to the right, since more, more, and so on:

Answer:

That's all! Easily? Let's solve a simple inequality with two variables:

Let's draw a function in the coordinate system.

Did you get such a schedule? Now let’s look carefully at what inequality we have there? Less? This means we paint over everything that is to the left of our straight line. What if there were more? That's right, then we would paint over everything that is to the right of our straight line. It's simple.

All solutions to this inequality are “shaded out” orange. That's it, the inequality with two variables is solved. This means that the coordinates of any point from the shaded area are the solutions.

Graphical solution of quadratic inequalities

Now we will understand how to graphically solve quadratic inequalities.

But before we get down to business, let's review some material regarding the quadratic function.

What is the discriminant responsible for? That’s right, for the position of the graph relative to the axis (if you don’t remember this, then definitely read the theory about quadratic functions).

In any case, here's a little reminder for you:

Now that we have refreshed all the material in our memory, let's get down to business - solve the inequality graphically.

I’ll tell you right away that there are two options for solving it.

Option 1

We write our parabola as a function:

Using the formulas, we determine the coordinates of the vertex of the parabola (exactly the same as when solving quadratic equations):

Did you count? What did you get?

Now let's take two more different points and calculate for them:

Let's start building one branch of the parabola:

We symmetrically reflect our points onto another branch of the parabola:

Now let's return to our inequality.

We need it to be less than zero, respectively:

Since in our inequality the sign is strictly less than, we exclude the end points - “puncture out”.

Answer:

Long way, right? Now I will show you a simpler version of the graphical solution using the example of the same inequality:

Option 2

We return to our inequality and mark the intervals we need:

Agree, it's much faster.

Let us now write down the answer:

Let's consider another solution that simplifies the algebraic part, but the main thing is not to get confused.

Multiply the left and right sides by:

Try to solve the following quadratic inequality yourself in any way you like: .

Did you manage?

Look how my graph turned out:

Answer: .

Graphical solution of mixed inequalities

Now let's move on to more complex inequalities!

How do you like this:

It's creepy, isn't it? Honestly, I have no idea how to solve this algebraically... But it’s not necessary. Graphically there is nothing complicated about this! The eyes are afraid, but the hands are doing!

The first thing we will start with is by constructing two graphs:

I won’t write out a table for each one - I’m sure you can do it perfectly on your own (wow, there are so many examples to solve!).

Did you paint it? Now build two graphs.

Let's compare our drawings?

Is it the same with you? Great! Now let’s arrange the intersection points and use color to determine which graph we should have larger in theory, that is. Look what happened in the end:

Now let’s just look at where our selected graph is higher than the graph? Feel free to take a pencil and paint over it this area! She will be the solution to our complex inequality!

At what intervals along the axis is we located higher than? Right, . This is the answer!

Well, now you can handle any equation, any system, and even more so any inequality!

BRIEFLY ABOUT THE MAIN THINGS

Algorithm for solving equations using function graphs:

1. Let's express it through
2. Let's define the function type
3. Let's build graphs of the resulting functions
4. Let's find the intersection points of the graphs
5. Let’s write the answer correctly (taking into account the ODZ and inequality signs)
6. Let's check the answer (substitute the roots into the equation or system)

For more information about constructing function graphs, see the topic “”.

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