Slide 2

Mathematics is the science of the young. Otherwise it can not be. Classes in mathematics are such gymnastics of the mind, for which all the flexibility and all the endurance of youth is needed. Norbert Wiener (1894-1964), American scientist

Slide 3

the relation between the numbers a and b (mathematical expressions), connected by signs Inequality -

Slide 4

Historical reference The problems of proving equalities and inequalities arose in ancient times. To denote the signs of equality and inequality, special words or their abbreviations were used. IV century BC, Euclid, V Book of the Beginnings: if a, b, c, d - positive numbers and a - greatest number in the proportion a / b = c / d, then the inequality a + d = b + c is fulfilled. III century, the main work of Papp of Alexandria "Mathematical collection": if a, b, c, d are positive numbers and a / b> c / d, then the inequality ad> bc is fulfilled. More than 2000 BC the inequality was known Turns into true equality for a = b.

Slide 5

Modern special signs 1557. Introduced an equal sign = by the English mathematician R. Rikord. His motive: "No two objects can be more equal than two parallel segments." 1631 year. Introduced signs> and

Slide 6

Types of inequalities With variable (one or several) Strict Nonstrict With module With parameter Nonstandard Systems Collections Numeric Simple Double Multiple Algebraic integers: -linear -square -higher degrees Fractional-rational Irrational Trigonometric Exponential Logarithmic Mixed type

Slide 7

Methods for solving inequalities Graphic Basic Special Functional-graphic Use of properties of inequalities Transition to equivalent systems Transition to equivalent collections Variable change Interval method (including generalized) Algebraic Splitting method for nonstrict inequalities

Slide 8

is the value of a variable, which, when substituted, turns it into a true numerical inequality. Solve an inequality - find all its solutions or prove that there are none. Two inequalities are said to be equivalent if all solutions to each are solutions to the other inequality, or if both inequalities have no solutions. Inequalities Solving one variable inequality

Slide 9

Describe the inequalities. Solve orally 3) (x - 2) (x + 3)  0

Slide 10

Graphical method

Solve the inequality graphically 1) Build a graph 2) Build a graph in the same coordinate system. 3) Find the abscissas of the intersection points of the graphs (the values ​​are taken approximately, the accuracy is checked by substitution). 4) Determine the solution of this inequality according to the graph. 5) We write down the answer.

Slide 11

Functional-graphical method for solving the inequality f (x)

Slide 12

Functional-graphical method Solve the inequality: 3) The equation f (x) = g (x) has no more than one root. Solution. 4) By selection, we find that x = 2. II. Let us schematically depict on the numerical axis Ox the graphs of the functions f (x) and g (x) passing through the point x = 2. III. Let's define solutions and write down the answer. Answer. x -7 undefined 2

Slide 13

Solve inequalities:

Slide 14

Build the graphs of the USE-9 function, 2008

Slide 15

y x O 1 1 -1 -1 -2 -3 -4 2 3 4 -2 -3 -4 2 3 4 1) y = | x | 2) y = | x | -1 3) y = || x | -1 | 4) y = || x | -1 | -1 5) y = ||| x | -1 | -1 | 6) y = ||| x | -1 | -1 | -1 y = |||| x | -1 | -1 | -1 |

Slide 16

y x O 1 1 -1 -1 -2 -3 -4 2 3 4 -2 -3 -4 2 3 4 Determine the number of intervals of solutions of the inequality for each value of the parameter a

Slide 17

Build a graph of the function of the exam-9, 2008

Slide 18

Slide 19

The graph of a linear or square inequality is built in the same way as a graph of any function (equation) is built. The difference is that inequality implies multiple solutions, so the graph of inequality is not just a point on a number line or a line on coordinate plane... Using mathematical operations and the inequality sign, you can determine the set of solutions to the inequality.

Steps

Graphical representation of linear inequality on the number line

Solve inequality. To do this, isolate the variable using the same algebraic techniques that you use to solve any equation. Remember that when multiplying or dividing an inequality by a negative number (or term), reverse the sign of the inequality.

Draw a number line. On the number line, mark the value found (the variable can be less than, greater than or equal to this value). Draw a number line of the appropriate length (long or short).

Draw a circle to represent the found value. If the variable is less ( < {\displaystyle <} ) or more ( > (\ displaystyle>)) of this value, the circle is not filled, because many solutions do not include this value. If the variable is less than or equal to ( ≤ (\ displaystyle \ leq)) or greater than or equal to ( ≥ (\ displaystyle \ geq)) to this value, the circle is filled because many solutions include this value.

On the number line, shade the area that defines the set of solutions. If the variable is greater than the found value, shade the area to the right of it, because the solution set includes all values ​​that are greater than the found value. If the variable is less than the found value, shade the area to the left of it, because the solution set includes all values ​​that are less than the found value.

Graphical representation of linear inequality on the coordinate plane

1. Solve inequality (find the value y (\ displaystyle y) ). To get a linear equation, isolate the variable on the left side using the known algebraic methods... The variable should remain on the right side x (\ displaystyle x) and possibly some constant.

   Plot the linear equation on the coordinate plane. To do this, convert the inequality to an equation and plot the graph as you would any linear equation. Draw the y-intercept, and then use the slope to add more points.

   Draw a straight line. If the inequality is strict (includes the sign < {\displaystyle <} or > (\ displaystyle>)), draw the dashed line, because the set of solutions does not include values ​​that lie on the line. If the inequality is not strict (includes the sign ≤ (\ displaystyle \ leq) or ≥ (\ displaystyle \ geq)), draw a solid line, because many solutions include values ​​that lie on a line.

   Shade the appropriate area. If the inequality has the form y> m x + b (\ displaystyle y> mx + b), shade over the line. If the inequality has the form y< m x + b {\displaystyle y, shade the area under the line.

Plotting a square inequality on a coordinate plane

Determine that the given inequality is square. The square inequality has the form a x 2 + b x + c (\ displaystyle ax ^ (2) + bx + c)... Sometimes the inequality does not contain a first-order variable ( x (\ displaystyle x)) and / or a free term (constant), but necessarily includes a second-order variable ( x 2 (\ displaystyle x ^ (2))). Variables x (\ displaystyle x) and y (\ displaystyle y) must be isolated on different sides of inequality.

FEDERAL EDUCATION AGENCY

INSTITUTE FOR EDUCATION DEVELOPMENT

"Graphical methods for solving equations and inequalities with parameters"

Completed

mathematic teacher

MOU SOSH №62

Lipetsk 2008

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

NS;at) 4

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

1.2. Turn................................................. .................................................. nine

1.3. Homotetia. Compression to Straight ............................................... ................. 13

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

2. GRAPHIC TECHNIQUES. COORDINATE PLANE ( NS;a) 17

CONCLUSION................................................. .......................................... twenty

REFERENCES ................................................ ........ 22

INTRODUCTION

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

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

Other schoolchildren treat the parameter as an unknown quantity and, without embarrassment, can express the parameter in the answer through the variable NS.

In the final and entrance exams, there are mainly two types of problems with parameters. You will immediately recognize them by their wording. First: "For each value of the parameter, find all the solutions of some equation or inequality." Second: "Find all values ​​of the parameter, for each of which some conditions are satisfied for a given equation or inequality." Accordingly, the answers in these two types of problems differ essentially. In the answer to the problem of the first type, all possible values ​​of the parameter are listed and for each of these values ​​the solutions of the equation are written. In the answer to the problem of the second type, all parameter values ​​are indicated for 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 substituted into the equation, the latter turns into a true numerical equality. The solution to an inequality with a parameter is defined similarly. To solve an equation (inequality) with a parameter means for each admissible value of the parameter to find the set of all solutions of this equation (inequality).

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

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

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

On the plane (x; y) the function y =f (NS; a) defines a family of curves depending on the parameter a. It is clear that each family f has certain properties. First of all, we will be interested in what kind of transformation of the plane (parallel translation, rotation, etc.) it is possible to pass from one curve of the family to any other. A separate paragraph will be devoted to each of these transformations. It seems to us that such a classification makes it easier for the decisive to find the necessary graphic image. Note that with this approach, the conceptual part of the solution does not depend on which figure (line, circle, parabola, etc.) is a member of the family of curves.

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

Speaking of graphic methods, it is impossible to get around one problem, "born" of the practice of the competitive exam. We are referring to the question of severity, and hence 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 rigorously. However, by whom, when and where is 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 by common sense. We understand the degree of subjectivity of this point of view. Moreover, the graphical method is just one of the means of visualization. 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 ">

Plotting a function with the domain of definition and (Fig. 1). The resulting graph is a family of lines y = a should only cross at one point. It can be seen from the figure that this requirement is fulfilled only for 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 straight line is 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 new series tasks..gif "width =" 107 "height =" 27 src = ">.

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

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

Further, shifting the "semi-parabola" to the left, we fix the last moment when the graphs at = NS+ 1 and have two common points (position III). This arrangement is provided by the requirement a= 1.

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

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

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

Consider the functions and . Among them, only one defines a family of curves. Now we see that the replacement made is of undoubted benefit. In parallel, we note that in the previous problem, by a similar replacement, one can make a straight line rather than a “semi-parabola” move. Let's turn to fig. 4. Obviously, if the abscissa of the "semi-parabola" vertex is greater than one, that is, –3 a > 1, , then the equation of the roots has no..gif "width =" 89 "height =" 29 "> and have a different character of monotony.

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

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

Answer. .

Let's make one remark. In some examples of this section, we will have to solve a standard problem: for a straight family, find its slope corresponding to the moment of tangency with the curve. Let's show how to do this in general view using the derivative.

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

Desired slope k is equal.

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, also at one point they intersect the arc AB OB and OM (tangent) .. gif "width =" 16 "height =" 48 src = ">. The slope of the tangent is equal. Easily found out of 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. Point - is the maximum point.

The 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. between straight lines OA and OV, satisfy the condition of the problem..gif "width =" 17 "height =" 47 src = ">.

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

1.3. Homotetia. 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. 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 the 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 a circle, which means that the system will have eight solutions. . Obviously for, 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 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 larger and smaller, that is. Note that there is.

Answer. or .

1.4. Two straight lines on a plane

In essence, the idea of ​​solving the problems of this paragraph is based on the question of studying the mutual arrangement of two straight lines: and ... It is not difficult to show the general solution of this problem. We will turn directly to specific typical examples, which, in our opinion, will not harm the general side of the issue.

Example 10. For what a and b is 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 system inequality defines a half-plane with a boundary at= 2x- 1 (fig. 10). It is easy to understand that the resulting system has a solution if the straight line ah +by = 5 intersects the boundary of the 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, the equation Oh+ by = 5 defines a vertical line that obviously intersects the line y = 2NS - 1. However, this statement is true only if ..gif "width =" 43 "height =" 20 src = "> the system has solutions..gif" width = "99" height = "48">. In this case, the condition of intersection of straight lines is achieved at, ie ..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 plot the function.

- Consider the straight lines and select those intervals of the Oa axis, on 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 through a for each of the found intervals of the value of a separately.

Looking at 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 these methods is limited by the difficulties that can be encountered when constructing a graphical image. Is it really that bad? Apparently not. Indeed, with this approach in to a large extent the main didactic value of tasks with parameters as a model of miniature research is 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 competition problems with parameters follow the path from a picture to a condition.

In these problems, the possibilities of solving problems with a parameter were discussed, which open up to us when the graphs of functions included in the left and right sides of equations or inequalities are displayed 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 surface. We can say that we get a whole family of graphs corresponding to different values ​​of the parameter.

Two details are emphatically emphasized.

Firstly, we are not talking about a "graphical" solution. All values, coordinates, roots are calculated strictly, analytically, as solutions of the corresponding equations, systems. The same applies to the cases of touching or crossing the charts. They are not determined by eye, but using 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 significantly increase. Precisely imagining what happens in the problem for different values ​​of the parameter, you may find the correct solution algorithm.

Therefore, we will end these words with an insistent sentence: if in even the slightest bit difficult task there are functions whose graphs you can draw, be sure to do it, you will not regret it.

BIBLIOGRAPHIC LIST

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

2. Gorshtein, with parameters [Text]: 3rd edition, supplemented and revised /,. - M .: Ileksa, Kharkov: Gymnasium, 1999 .-- 336 p.

Many tasks that we are used to calculating purely algebraically can be solved much easier and faster, using function graphs will help us in this. You say "how so?" to draw something, and what to draw? Trust me, sometimes it's more convenient and easier. Let's 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 species. Linear equations are easy enough to solve algebraically - we transfer all unknowns to one side of the equation, everything that we know - to the other and voila! We found the root. Now I will show you how to do it graphically.

So you have an equation:

How to solve it?
Option 1, and the most common is to transfer unknowns in one direction, and known in the other, we get:

Now we are building. What did you do?

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

Our answer is

That's all the 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 in another way. To consider an alternative solution, let's return to our equation:

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

Have you built it? We look!

What is the solution this time? Everything is correct. The same is 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 consider something more difficult ... For example, graphical solution of quadratic equations.

Graphical solution of quadratic equations

So now let's get down to 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 are on their nerves making mistakes when multiplying or squaring, especially if the example is with large numbers, and, as you know, you will not have a calculator on the exam ... So let's try to relax a bit and draw while solving this equation.

You can graphically find solutions to this equation different ways... Consider various options, and you yourself will choose which one you like best.

Method 1. Directly

We just build a parabola according to this equation:

To do this quickly, I will give you one little tip: it is convenient to start construction by defining the vertex of the parabola. The following formulas will help to determine the coordinates of the vertex of the 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" direct "construction of the parabola in order to find its roots. Nevertheless, let's count to the end, and then I'll show you how to make it much (much!) Easier!

Have you counted? What are the coordinates of 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, and to build a parabola, we need more ... points. How many 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 to the opposite side:

We return to our parabola. For our case, the point. We need two more points, respectively, can we take positive ones, or can we take negative ones? What points are more convenient for you? It is more convenient for me to work with positive ones, so I will calculate at and.

Now we have three points, and we can safely build our parabola, reflecting the last two points relative to its vertex:

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

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

Just? We finished solving the equation with you in a complex graphical way, or else it will be!

Of course, you can check our answer algebraically - count the roots using Vieta's theorem or the Discriminant. What did you do? Same? You see! Now let's see a very simple graphic solution, I'm sure you will like it very much!

Method 2.Divided into several functions

Let's take all our equation too:, but write it down a little differently, namely:

Can we write it down like that? We can, because the transformation is equivalent. We look further.

Let's construct two functions separately:

1. - a graph is a simple parabola that you can easily build even without defining a vertex using formulas and compiling a table to determine other points.
2. - a graph is a straight line, which you can just as easily plot, having estimated the values ​​and in your head without even resorting to a calculator.

Have you built it? Compare with what came out for me:

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

Accordingly, the solution to this equation is:

What do you say? You must admit that this solution is much easier than the previous one and even easier than looking for roots through the discriminant! If so, try to solve the following equation in this way:

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

The graphs show that the answers are:

Did you manage? Well done! Now let's look at the chuuuut 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, you can bring everything to common denominator, find the roots of the resulting equation, not forgetting to take into account the ODV, 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. - a graph is a straight line that you can easily plot, having estimated the values ​​and in your head without even resorting to a calculator.

Realized? Now start building.

Here's what happened to me:

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

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

Try to plug our roots into the equation. Happened?

That's right! Agree, it is a pleasure to graphically solve such equations!

Try to solve the equation yourself in a graphical way:

Let me give you a hint: move part of the equation to the right side, so that the simplest functions to build are on both sides. Got the hint? Take action!

Now let's see what happened:

Respectively:

1. is a cubic parabola.
2. - an ordinary straight line.

Well, we build:

As you wrote down for a long time, the root of this equation is -.

Having solved such a large number of examples, I'm sure you realized how easy and quick you can solve equations graphically. It's time to figure out how to solve the system in a similar way.

Graphic solution of systems

Graphic solution systems are essentially no different from the graphical solution of 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 - 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 everything that is connected with, and on the right - that is connected with. In other words, we 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 why? Let me give you a hint: we are dealing with a system: the system has both and, and ... Understood the hint?

That's right! When solving the 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 to confuse where we have the meaning and where the meaning! Did you write it down? Now let's compare everything in order:

And the answers are: and. Make a check - substitute the found roots into the system and make sure if we solved it correctly in a graphical way?

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 to solve the following system:

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

And now, in general, the matter is small - I built it quickly and here's a solution for you! We build:

Are the graphs the same? Now mark the system solutions in the figure and write down the identified answers correctly!

I've done everything? Compare with my posts:

Is that correct? Well done! You already click similar tasks like nuts! And if so, we will give you a more complex system:

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

I will give you a little hint, as the system looks well, not very simple! 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!

How is it? Beautiful? How many intersection points did you get? I have three! Let's compare our charts:

Same way? Now, carefully write down all the decisions of our system:

Now take another look at the system:

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

Graphical solution of inequalities

Graphical solution of linear inequalities

After the last example, you can do everything! Exhale now - compared to the previous sections, this one will be very, very light!

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

To begin with, we will carry out the simplest transformations - we will open the brackets of perfect squares and give similar terms:

The inequality is not strict, therefore - is not included in the gap, 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 two-variable inequality:

Let's draw a function in the coordinate system.

Have you got such a schedule? And now we are carefully looking at what we have there in inequality? Smaller? So, we paint over everything that is to the left of our straight line. And if there were more? That's right, then they would paint over everything that is to the right of our straight line. It's simple.

All solutions to this inequality are shaded in orange. That's it, the two-variable inequality is solved. This means that the coordinates of any point from the shaded area are the solutions.

Graphical solution of square inequalities

Now we will deal with how to graphically solve square inequalities.

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

And 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 read exactly the theory of quadratic functions).

Anyway, here's a little reminder sign:

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

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 (in the same way as when solving quadratic equations):

Have you counted? What did you do?

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

We start building one branch of the parabola:

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

Now, back to our inequality.

We need it to be less than zero, respectively:

Since in our inequality the sign is strictly less, then we exclude the end points - "gouge out".

Answer:

Long way, right? Now I will show you a simpler version of a 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's write down the answer now:

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

Let's multiply the left and right sides by:

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

Did you manage?

See how the graph turned out for me:

Answer: .

Graphical solution of mixed inequalities

Now let's move on to more complex inequalities!

How do you like this:

Creepy, right? To be honest, I have no idea how to solve this algebraically ... But, it is not necessary. Graphically, there is nothing complicated about it! The eyes are afraid, but the hands are doing!

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

I will not paint a table for each one - I am sure you can handle it perfectly on your own (still, there are so many examples to solve!).

Have you painted it? Now build two graphs.

Let's compare our drawings?

Is it the same for you? Fine! Now we will place the intersection points and determine by color which graph we have, in theory, should be larger, that is. See what happened in the end:

And now we just look, where is the selected chart higher than the chart? Feel free to take a pencil and paint over this area! She will be the solution to our complex inequality!

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

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

BRIEFLY ABOUT THE MAIN

Algorithm for solving equations using graphs of functions:

1. Let us express in terms of
2. Define the type of the function
3. Let's build the graphs of the resulting functions
4. Find the intersection points of the graphs
5. Correctly write down the answer (taking into account the ODZ and inequality signs)
6. Check the answer (substitute the roots in the equation or system)

For more details about plotting functions, see the topic "".

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The graphical method is one of the main methods for solving square inequalities. In the article we will give an algorithm for applying the graphical method, and then we will consider special cases with examples.

The essence of the graphical method

The method is applicable for solving any inequalities, not only square ones. Its essence is as follows: the right and left sides of the inequality are considered as two separate functions y = f (x) and y = g (x), their graphs are plotted in a rectangular coordinate system and they see which of the graphs is located above the other, and on which intervals. The intervals are assessed as follows:

Definition 1

• the solutions of the inequality f (x)> g (x) are intervals, where the graph of the function f is higher than the graph of the function g;
• the solutions of the inequality f (x) ≥ g (x) are intervals where the graph of the function f is not lower than the graph of the function g;
• solutions of the inequality f (x)< g (x) являются интервалы, где график функции f ниже графика функции g ;
• the solutions to the inequality f (x) ≤ g (x) are intervals where the graph of the function f is not higher than the graph of the function g;
• the abscissas of the intersection points of the graphs of the functions f and g are solutions of the equation f (x) = g (x).

Let's consider the above algorithm with an example. To do this, take the square inequality a x 2 + b x + c< 0 (≤ , >, ≥) and derive two functions from it. The left-hand side of the inequality will correspond to y = ax2 + bx + c (in this case f (x) = a 0).

The graph of the first function is a parabola, the second is a straight line that coincides with the abscissa axis O x. Let us analyze the position of the parabola relative to the O x axis. To do this, we will perform a schematic drawing.

The branches of the parabola are directed upwards. It crosses the O x axis at points x 1 and x 2... The coefficient a in this case is positive, since it is he who is responsible for the direction of the branches of the parabola. The discriminant is positive, which indicates the presence of two roots in square trinomiala x 2 + b x + c... We designated the roots of the trinomial as x 1 and x 2, and assumed that x 1< x 2 , since a point with an abscissa is depicted on the O x axis x 1 to the left of the point with the abscissa x 2.

The parts of the parabola located above the O x axis will be denoted in red, below - in blue. This will allow us to make the drawing more descriptive.

Select the gaps that correspond to these parts and mark them in the figure with fields of a certain color.

In red we marked the intervals (- ∞, x 1) and (x 2, + ∞), on them the parabola is above the O x axis. They are a x 2 + b x + c> 0. We marked in blue the interval (x 1, x 2), which is a solution to the inequality a x 2 + b x + c< 0 . Числа x 1 и x 2 будут отвечать равенству a · x 2 + b · x + c = 0 .

Let's make a short record of the solution. For a> 0 and D = b 2 - 4 · a · c> 0 (or D "= D 4> 0 for an even coefficient b) we get:

• the solution to the square inequality a x 2 + b x + c> 0 is (- ∞, x 1) ∪ (x 2, + ∞) or in another notation x< x 1 , x >x 2;
• the solution to the square inequality a · x 2 + b · x + c ≥ 0 is (- ∞, x 1] ∪ [x 2, + ∞) or in another notation x ≤ x 1, x ≥ x 2;
• solution of the square inequality a x 2 + b x + c< 0 является (x 1 , x 2) или в другой записи x 1 < x < x 2 ;
• the solution to the square inequality a · x 2 + b · x + c ≤ 0 is [x 1, x 2] or in another notation x 1 ≤ x ≤ x 2,

where x 1 and x 2 are the roots of the square trinomial a x 2 + b x + c, and x 1< x 2 .

In this figure, the parabola touches the O x axis only at one point, which is designated as x 0 a> 0. D = 0, therefore, the square trinomial has one root x 0.

The parabola is located above the O x axis completely, except for the point of tangency of the coordinate axis. Let's color the intervals (- ∞, x 0), (x 0, ∞).

Let's write down the results. At a> 0 and D = 0:

• by solving the square inequality a x 2 + b x + c> 0 is (- ∞, x 0) ∪ (x 0, + ∞) or in another notation x ≠ x 0;
• by solving the square inequality a x 2 + b x + c ≥ 0 is an (− ∞ , + ∞) or in another notation x ∈ R;
• square inequality a x 2 + b x + c< 0 has no solutions (there are no intervals on which the parabola is located below the axis O x);
• square inequality a x 2 + b x + c ≤ 0 has the only solution x = x 0(it is given by the point of contact),

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

Consider the third case, when the branches of the parabola are directed upwards and do not touch the axis O x... The branches of the parabola are directed upwards, which means that a> 0... A square trinomial has no real roots, since D< 0 .

There are no intervals on the graph where the parabola would be below the abscissa axis. We will take this into account when choosing a color for our drawing.

It turns out that for a> 0 and D< 0 solution of square inequalities a x 2 + b x + c> 0 and a x 2 + b x + c ≥ 0 is a multitude of all real numbers, and inequalities a x 2 + b x + c< 0 and a x 2 + b x + c ≤ 0 have no solutions.

It remains for us to consider three options, when the branches of the parabola are directed downward. We do not need to dwell on these three options in detail, since when multiplying both sides of the inequality by - 1, we get an equivalent inequality with a positive coefficient at x 2.

Consideration of the previous section of the article prepared us for the perception of an algorithm for solving inequalities using a graphical method. To carry out the calculations, we will need each time to use the drawing, which will depict the coordinate line O x and the parabola, which corresponds to quadratic function y = a x 2 + b x + c... In most cases, we will not depict the O axis for y, since it is not needed for calculations and will only overload the drawing.

To build a parabola, we need to know two things:

Definition 2

• the direction of the branches, which is determined by the value of the coefficient a;
• the presence of intersection points of the parabola and the abscissa axis, which are determined by the value of the discriminant of the square trinomial a x 2 + b x + c.

We will denote the points of intersection and tangency in the usual way when solving nonstrict inequalities and empty when solving strict ones.

Having a ready-made drawing allows you to proceed to the next step of the solution. It involves determining the intervals at which the parabola is located above or below the O x axis. Gaps and intersection points are the solution to a square inequality. If there are no intersection or tangency points and there are no intervals, then it is considered that the inequality specified in the conditions of the problem has no solutions.

Now let's solve some of the square inequalities using the above algorithm.

Example 1

It is necessary to solve the inequality 2 · x 2 + 5 1 3 · x - 2 graphically.

Solution

Let's draw a graph of the quadratic function y = 2 x 2 + 5 1 3 x - 2. Coefficient at x 2 positive, since it is 2 ... This means that the branches of the parabola will be directed upwards.

Let us calculate the discriminant of the square trinomial 2 · x 2 + 5 1 3 · x - 2 in order to find out whether the parabola has common points with the abscissa axis. We get:

D = 5 1 3 2 - 4 2 (- 2) = 400 9

As you can see, D is greater than zero, therefore, we have two intersection points: x 1 = - 5 1 3 - 400 9 2 2 and x 2 = - 5 1 3 + 400 9 2 2, that is, x 1 = - 3 and x 2 = 1 3.

We solve a non-strict inequality, therefore we put down ordinary points on the graph. We draw a parabola. As you can see, the picture looks the same as in the first template we reviewed.

Our inequality has the sign ≤. Therefore, we need to select the intervals on the graph where the parabola is located below the O x axis and add intersection points to them.

The interval we need is 3, 1 3. Add intersection points to it and get a numerical segment - 3, 1 3. This is the solution to our problem. The answer can be written as a double inequality: - 3 ≤ x ≤ 1 3.

Answer:- 3, 1 3 or - 3 ≤ x ≤ 1 3.

Example 2

- x 2 + 16 x - 63< 0 graphical method.

Solution

The square of the variable has a negative numerical coefficient, so the branches of the parabola will be directed downward. We calculate the fourth part of the discriminant D "= 8 2 - (- 1) · (- 63) = 64 - 63 = 1... This result tells us that there will be two intersection points.

We calculate the roots of the square trinomial: x 1 = - 8 + 1 - 1 and x 2 = - 8 - 1 - 1, x 1 = 7 and x 2 = 9.

It turns out that the parabola intersects the abscissa axis at points 7 and 9 ... Let's mark these points on the graph as empty, since we are working with strict inequality. After that, draw a parabola that intersects the O x axis at the marked points.

We will be interested in the intervals at which the parabola is located below the O x axis. We mark these intervals in blue.

We get the answer: the solution to the inequality is the intervals (- ∞, 7), (9, + ∞).

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

In cases where the discriminant of the square trinomial is zero, it is necessary to carefully consider the question of whether it is worth including the abscissas in the answer. To accept correct solution, the inequality sign must be taken into account. In strict inequalities, the point of tangency of the abscissa axis is not a solution to the inequality; in non-strict inequalities, it is.

Example 3

Solve square inequality 10 x 2 - 14 x + 4, 9 ≤ 0 graphical method.

Solution

The branches of the parabola in this case will be directed upwards. It will touch the O x axis at the point 0, 7, since

Let's plot the function y = 10 x 2 - 14 x + 4, 9... Its branches are directed upward, since the coefficient at x 2 positive, and it touches the abscissa at the point with the abscissa 0 , 7 , because D "= (- 7) 2 - 10 4, 9 = 0, whence x 0 = 7 10 or 0 , 7 .

Let's put a point and draw a parabola.

We solve the weak inequality with the ≤ sign. Hence. We will be interested in the intervals at which the parabola is located below the abscissa axis and the point of tangency. There are no intervals in the figure that would satisfy our conditions. There is only a touch point 0, 7. This is the desired solution.

Answer: The inequality has only one solution, 0, 7.

Example 4

Solve square inequality - x 2 + 8 x - 16< 0 .

Solution

The branches of the parabola are directed downward. The discriminant is zero. Intersection point x 0 = 4.

Mark the point of tangency on the abscissa axis and draw a parabola.

We are dealing with severe inequality. Therefore, we are interested in the intervals at which the parabola is located below the O x axis. Mark them in blue.

The point with abscissa 4 is not a solution, since the parabola in it is not located below the O x axis. Therefore, we get two intervals (- ∞, 4), (4, + ∞).

Answer: (- ∞, 4) ∪ (4, + ∞) or in another notation x ≠ 4.

Not always with negative value the discriminant inequality will have no solutions. There are cases when the solution will be the set of all real numbers.

Example 5

Solve the square inequality 3 x 2 + 1> 0 graphically.

Solution

Coefficient a is positive. The discriminant is negative. The branches of the parabola will point upward. There are no points of intersection of the parabola with the O x axis. Let's refer to the figure.

We are working with a strict inequality that has a> sign. This means that we are interested in the intervals at which the parabola is located above the abscissa axis. This is exactly the case when the answer is the set of all real numbers.

Answer:(- ∞, + ∞) or so x ∈ R.

Example 6

It is necessary to find a solution to the inequality - 2 x 2 - 7 x - 12 ≥ 0 graphically.

Solution

The branches of the parabola are directed downward. The discriminant is negative, therefore, there are no common points of the parabola and the abscissa axis. Let's refer to the figure.

We are working with a nonstrict inequality with a ≥ sign; therefore, we are interested in the intervals at which the parabola is located above the abscissa axis. Judging by the schedule, there are no such gaps. This means that the inequality given for the condition of the problem has no solutions.

Answer: There are no solutions.

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