Individual project on the topic: “Graphical solution of equations and inequalities.” Solving inequalities graphically Solving equations and inequalities graphically

The graph of a linear or quadratic inequality is constructed in the same way as the graph of any function (equation). The difference is that inequality implies multiple solutions, so the graph of an 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 many solutions to the inequality.

Steps

Graphical representation of linear inequality on the number line

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

Draw a number line. On the number line, mark the value you 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 value found. If the variable is less than ( < {\displaystyle <} ) or more ( > (\displaystyle >)) of this value, the circle is not filled in because the solution set does 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 the solution set includes this value.

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

Graphic representation of linear inequality on the coordinate plane

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

   Draw a graph on the coordinate plane linear equation. To do this, convert the inequality into an equation and graph it as you would graph any linear equation. Plot the Y-intercept and then use the slope to plot the other points.

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

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

Graphical representation of the quadratic inequality on the coordinate plane

Determine that this inequality is quadratic. The quadratic 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 the inequality.

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 of the graphical method for solving inequalities next: consider the functions y=f(x) and y=g(x), which correspond to the left and right sides of the inequality, build their graphs in one rectangular coordinate system and find out at what intervals the graph of one of them is lower or higher than the other. Those intervals where

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

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

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

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

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

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

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

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


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


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

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

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

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


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

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


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

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

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


In this case, the branches of the parabola are directed upward, and it does not have common points with the abscissa axis. Here we have the conditions a>0 (branches are directed upward) and D<0 (квадратный трехчлен не имеет real roots). For example, you can plot the function y=2·x 2 +1, here a=2>0, D=0 2 −4·2·1=−8<0 .

Obviously, the parabola is located above the Ox axis along 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 also to them you need to add the abscissa of the points of intersection or the abscissa of the point of tangency. 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.

During the lesson, you will be able to independently study the topic “Graphical solution of equations and inequalities.” During the lesson, the teacher will examine graphical methods for solving equations and inequalities. Will teach you how to build graphs, analyze them and obtain solutions to equations and inequalities. The lesson will also cover specific examples on this topic.

Topic: Numeric functions

Lesson: Graphical solution of equations, inequalities

1. Lesson topic, introduction

We looked at graphs of elementary functions, including graphs of power functions with different exponents. We also looked at the rules for shifting and transforming function graphs. All these skills must be applied when required graphicsolution equations or graphical solutioninequalities.

2. Solving equations and inequalities graphically

Example 1: Solve the equation graphically:

Let's build graphs of functions (Fig. 1).

The graph of a function is a parabola passing through the points

The graph of the function is a straight line, let's build it using the table.

The graphs intersect at the point There are no other points of intersection, since the function increases monotonically, the function decreases monotonically, and, therefore, their point of intersection is the only one.

Answer:

Example 2: Solve inequality

a. For the inequality to hold, the graph of the function must be located above the straight line (Fig. 1). This is done when

b. In this case, on the contrary, the parabola must be under the straight line. This is done when

Example 3. Solve inequality

Let's build function graphs (Fig. 2).

Let's find the root of the equation When there are no solutions. There is one solution.

For the inequality to hold, the hyperbola must be located above the line. This is true when .

Answer:

Example 4. Solve graphically the inequality:

Domain:

Let's build function graphs for (Fig. 3).

a. The graph of the function must be located below the graph; this is done when

b. The graph of the function is located above the graph at But since the condition has a weak sign, it is important not to lose the isolated root

3. Conclusion

We looked at the graphical method for solving equations and inequalities; We looked at specific examples, the solution of which used such properties of functions as monotonicity and parity.

1. Mordkovich A.G. et al. Algebra 9th grade: Textbook. For general education Institutions.- 4th ed. - M.: Mnemosyne, 2002.-192 p.: ill.

2. Mordkovich A.G. et al. Algebra 9th grade: Problem book for students of general education institutions / A.G. Mordkovich, T.N. Mishustina et al. - 4th ed. - M.: Mnemosyne, 2002.-143 p.: ill.

3. Makarychev Yu. N. Algebra. 9th grade: educational. for general education students. institutions / Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, I. E. Feoktistov. — 7th ed., rev. and additional - M.: Mnemosyne, 2008.

4. Alimov Sh. A., Kolyagin Yu. M., Sidorov Yu. V. Algebra. 9th grade. 16th ed. - M., 2011. - 287 p.

5. Mordkovich A. G. Algebra. 9th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich, P. V. Semenov. — 12th ed., erased. - M.: 2010. - 224 p.: ill.

6. Algebra. 9th grade. In 2 parts. Part 2. Problem book for students of general education institutions / A. G. Mordkovich, L. A. Aleksandrova, T. N. Mishustina and others; Ed. A. G. Mordkovich. — 12th ed., rev. - M.: 2010.-223 p.: ill.

1. College section. ru in mathematics.

2. Internet project “Tasks”.

3. Educational portal “I WILL SOLVE the Unified State Exam”.

1. Mordkovich A.G. et al. Algebra 9th grade: Problem book for students of general education institutions / A.G. Mordkovich, T.N. Mishustina et al. - 4th ed. - M.: Mnemosyne, 2002.-143 p.: ill. No. 355, 356, 364.

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 can be a finite number or an infinite number of solutions. The area may be closed polygon or be unlimited.

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.

The graphical method is one of the main methods for solving quadratic inequalities. In the article we will present an algorithm for using the graphical method, and then consider special cases using examples.

The essence of the graphical method

The method is applicable to solving any inequalities, not only quadratic ones. Its essence is this: 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 look at which of the graphs is located above the other, and on which intervals. The intervals are estimated as follows:

Definition 1

• solutions to the inequality f (x) > g (x) are intervals where the graph of the function f is higher than the graph of the function g;
• solutions to 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 to the inequality f(x)< g (x) являются интервалы, где график функции f ниже графика функции g ;
• 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 to the equation f (x) = g (x).

Let's look at the above algorithm using an example. To do this, take the quadratic inequality a x 2 + b x + c< 0 (≤ , >, ≥) and derive two functions from it. The left side of the inequality will correspond to y = a · x 2 + b · x + c (in this case f (x) = a · x 2 + b · x + c), and the right side y = 0 (in this case g (x) = 0).

The graph of the first function is a parabola, the second is a straight line, which coincides with the x-axis Ox. Let's analyze the position of the parabola relative to the O x axis. To do this, let's make a schematic drawing.

The branches of the parabola are directed upward. It intersects the O x axis at points x 1 And x 2. Coefficient a in this case is positive, since it is it that is responsible for the direction of the branches of the parabola. The discriminant is positive, indicating that the quadratic trinomial has two roots a x 2 + b x + c. We denote the roots of the trinomial as x 1 And x 2, and it was accepted 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 abscissa point 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 visual.

Let's select the spaces that correspond to these parts and mark them in the figure with fields of a certain color.

We marked in red 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 the 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 brief summary 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 quadratic inequality a x 2 + b x + c > 0 is (− ∞ , x 1) ∪ (x 2 , + ∞) or in another notation x< x 1 , x >x2;
• the solution to the quadratic inequality a · x 2 + b · x + c ≥ 0 is (− ∞ , x 1 ] ∪ [ x 2 , + ∞) or in another notation x ≤ x 1 , x ≥ x 2 ;
• solving the quadratic 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 quadratic 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 quadratic 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 contact coordinate axis. Let's color the intervals (− ∞ , x 0) , (x 0 , ∞) .

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

• solving the quadratic inequality a x 2 + b x + c > 0 is (− ∞ , x 0) ∪ (x 0 , + ∞) or in another notation x ≠ x 0;
• solving 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 has no solutions (there are no intervals at which the parabola is located below the axis O x);
• quadratic inequality a x 2 + b x + c ≤ 0 has a unique solution x = x 0(it is given by the point of contact),

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

Let's consider the third case, when the branches of the parabola are directed upward and do not touch the axis O x. The branches of the parabola are directed upward, which means that a > 0. The square trinomial has no real roots because D< 0 .

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

It turns out that when a > 0 And D< 0 solving quadratic inequalities 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 And a x 2 + b x + c ≤ 0 have no solutions.

We have three options left to consider when the branches of the parabola are directed downward. There is no need to dwell on these three options in detail, since when we multiply both sides of the inequality by − 1, we obtain an equivalent inequality with a positive coefficient for 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 calculations, we will need to use a drawing each time, which will depict the coordinate line O x and a parabola that corresponds to the quadratic function y = a x 2 + b x + c. In most cases, we will not depict the O y axis, since it is not needed for calculations and will only overload the drawing.

To construct a parabola, we will 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 points of intersection of the parabola and the abscissa axis, which are determined by the value of the discriminant of the quadratic trinomial a · x 2 + b · x + c .

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

Having a completed drawing allows you to move on 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. The intervals and points of intersection are the solution to the quadratic inequality. If there are no points of intersection or tangency 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 several quadratic 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 because it is equal 2 . This means that the branches of the parabola will be directed upward.

Let us calculate the discriminant of the quadratic 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 we 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 ordinary points on the graph. Let's draw a parabola. As you can see, the drawing has the same appearance as in the first template we considered.

Our inequality has the sign ≤. Therefore, we need to highlight 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. We 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. Let's calculate the fourth part of the discriminant D " = 8 2 − (− 1) · (− 63) = 64 − 63 = 1. This result tells us that there will be two points of intersection.

Let's 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 x-axis at the points 7 And 9 . Let's mark these points on the graph as empty, since we are working with strict inequality. After this, 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. Let's 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 a quadratic trinomial is zero, it is necessary to carefully consider whether to include the abscissa of the tangent points in the answer. In order to accept correct solution, it is necessary to take into account the inequality sign. In strict inequalities, the point of tangency of the x-axis is not a solution to the inequality, but in non-strict ones it is.

Example 3

Solve quadratic inequality 10 x 2 − 14 x + 4, 9 ≤ 0 graphical method.

Solution

The branches of the parabola in this case will be directed upward. It will touch the O x axis at 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 x-axis at the x-axis point 0 , 7 , because D " = (− 7) 2 − 10 4, 9 = 0, from where x 0 = 7 10 or 0 , 7 .

Let's put a point and draw a parabola.

We solve a non-strict inequality with a sign ≤. Hence. We will be interested in the intervals at which the parabola is located below the x-axis and the point of tangency. There are no intervals in the figure that would satisfy our conditions. There is only a point of contact 0, 7. This is the solution we are looking for.

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

Example 4

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

Solution

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

We mark the point of tangency on the x-axis and draw a parabola.

We are dealing with severe inequality. Consequently, we are interested in the intervals at which the parabola is located below the O x axis. Let's mark them in blue.

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

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

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

Example 5

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

Solution

Coefficient a is positive. The discriminant is negative. The branches of the parabola will be directed upward. There are no points of intersection of the parabola with the O x axis. Let's look at the drawing.

We work with strict inequality, which has a > sign. This means that we are interested in the intervals at which the parabola is located above the x-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 downwards. The discriminant is negative, therefore, there are no common points between the parabola and the x-axis. Let's look at the drawing.

We are working with a non-strict inequality with the sign ≥, therefore, the intervals in which the parabola is located above the x-axis are of interest to us. Judging by the graph, there are no such gaps. This means that the inequality given in the problem conditions has no solutions.

Answer: No solutions.

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