Solving equations and inequalities graphically. Graphical solution of equations, inequalities

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 program Advanced Graph .

Let f(x,y) And g(x, y)- two expressions with variables X And at and scope X. Then inequalities of the form f(x, y) > g(x, y) or f(x, y) < g(x, y) called inequality with two variables .

Meaning of Variables x, y from many X, at which the inequality becomes true numerical inequality, it's called decision and is designated (x, y). Solve inequality - this means finding many such pairs.

If each pair of numbers (x, y) from the set of solutions to the inequality, match the point M(x, y), we obtain the set of points on the plane defined by this inequality. He is called graph of this inequality . The graph of an inequality is usually an area on a plane.

To depict the set of solutions to the inequality f(x, y) > g(x, y), proceed as follows. First, replace the inequality sign with an equal sign and find a line that has the equation f(x,y) = g(x,y). This line divides the plane into several parts. After this, it is enough to take one point in each part and check whether the inequality is satisfied at this point f(x, y) > g(x, y). If it is executed at this point, then it will be executed in the entire part where this point lies. Combining such parts, we obtain many solutions.

Task. y > x.

Solution. First, we replace the inequality sign with an equal sign and construct a line in a rectangular coordinate system that has the equation y = x.

This line divides the plane into two parts. After this, take one point in each part and check whether the inequality is satisfied at this point y > x.

Task. Solve graphically the inequality
X 2 + at 2 £25.

Rice. 18.

Solution. First, replace the inequality sign with an equal sign and draw a line X 2 + at 2 = 25. This is a circle with a center at the origin and a radius of 5. The resulting circle divides the plane into two parts. Checking the satisfiability of the inequality X 2 + at 2 £ 25 in each part, we find that the graph is a set of points on a circle and parts of a plane inside the circle.

Let two inequalities be given f 1(x, y) > g 1(x, y) And f 2(x, y) > g 2(x, y).

Systems of sets of inequalities with two variables

System of inequalities is yourself conjunction of these inequalities. System solution is every meaning (x, y), which turns each of the inequalities into a true numerical inequality. Many solutions systems inequalities is the intersection of sets of solutions to inequalities that form a given system.

Set of inequalities is yourself disjunction of these inequalities By the solution of the totality is every meaning (x, y), which converts at least one of the set of inequalities into a true numerical inequality. Many solutions totality is a union of sets of solutions to inequalities that form a set.

Task. Solve graphically the system of inequalities

Solution. y = x And X 2 + at 2 = 25. We solve each inequality of the system.

The graph of the system will be the set of points on the plane that are the intersection (double hatching) of the sets of solutions to the first and second inequalities.

Task. Solve graphically a set of inequalities

Solution. First, we replace the inequality sign with an equal sign and draw lines in one coordinate system y = x+ 4 and X 2 + at 2 = 16. Solve each inequality in the population. The graph of the population will be a set of points on the plane, which are the union of the sets of solutions to the first and second inequalities.

Exercises for independent work

1. Solve graphically the inequalities: a) at> 2x; b) at< 2x + 3;

V) x 2+ y 2 > 9; G) x 2+ y 2 £4.

2. Solve graphically systems of inequalities:

a) b)

Slide 2

Mathematics is the science of the young. Otherwise it can not be. Mathematics is a form of mental gymnastics that requires all the flexibility and endurance of youth. Norbert Wiener (1894-1964), American scientist

Slide 3

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

Slide 4

Historical background Problems of proving equalities and inequalities arose in ancient times. Special words or their abbreviations were used to denote equality and inequality signs. IV century BC, Euclid, V book of “Beginnings”: if a, b, c, d are positive numbers and a is greatest number in the proportion a/b=c/d, then the inequality a+d=b+c holds. III century, the main work of Pappus of Alexandria “Mathematical collection”: if a, b, c, d are positive numbers and a/b>c/d, then the inequality ad>bc is satisfied. More than 2000 BC the inequality was known turns into a true equality when a=b.

Slide 5

Modern special signs 1557. The equal sign = was introduced by the English mathematician R. Ricord. His motive: “No two objects can be more equal than two parallel segments.” 1631 Signs > and

Slide 6

Types of inequalities With a variable (one or more) Strict Non-strict With a modulus With a parameter Non-standard Systems Collections Numerical Simple Double Multiples Algebraic integers: -linear -quadratic -higher powers Fractional-rational Irrational Trigonometric Exponential Logarithmic Mixed type

Slide 7

Methods for solving inequalities Graphic Basic Special Functional-graphical Using the properties of inequalities Transition to equivalent systems Transition to equivalent collections Replacing a variable Interval method (including generalized) Algebraic Splitting method for non-strict inequalities

Slide 8

is the value of a variable that, 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 both inequalities have no solutions. Inequalities Solving inequalities in one variable

Slide 9

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

Slide 10

Graphical method

Solve graphically the inequality 1) Construct a graph 2) Construct a graph in the same coordinate system. 3) Find the abscissa of the intersection points of the graphs (the values ​​are taken approximately, we check the accuracy by substitution). 4) We determine from the graph the solution to this inequality. 5) 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 at most 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 determine the solutions and write down the answer. Answer. x -7 undefined 2

Slide 13

Solve the inequalities:

Slide 14

Build graphs of the Unified State Examination-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 to the inequality for each value of parameter a

Slide 17

Build a graph of the Unified State Examination-9 function, 2008

Slide 18

Slide 19

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 an 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 move anything from side to side, but will construct the graphs directly, since they now exist:

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, graphic solution 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 look at the different 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, you can bring everything to a common denominator, find the roots of the resulting equation, without 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 worked through such a large number of examples, I’m sure you realized how easy and quick it is to 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 a 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 of perfect squares and present 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 in 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 yourself quadratic inequality 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 “”.

THE REMAINING 2/3 ARTICLES ARE AVAILABLE ONLY TO YOUCLEVER STUDENTS!

Become a YouClever student,

Prepare for the Unified State Exam or Unified State Exam in mathematics for the price of “a cup of coffee per month”,

And also get unlimited access to the textbook "YouClever", the Preparation Program (workbook) "100gia", unlimited trial Unified State Examination and OGE, 6000 problems with analysis of solutions and other services YouClever and 100gia.

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 arbitrary point, not lying on a line, substitute its coordinates into the inequality. if the inequality is true, then the half-plane containing the chosen point is the solution to the original inequality. If the inequality is false, then the half-plane on the other side of the line is the set of solutions to this inequality.
4. To solve a system of inequalities, it is necessary to find the area of ​​intersection of all half-planes that are the solution to each inequality of the system.

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

Let's look at three relevant examples.

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

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

Figure 2

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

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

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

x	2	0
y	0	1

y – x – 1 = 0

x	0	2
y	1	3

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


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

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