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Solving quadratic inequalities using the graph of a parabola. Graphical solution of inequalities, systems of sets of inequalities with two variables

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 specified by this inequality. They call him 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 represents 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 represents yourself disjunction of these inequalities Set solution 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)

Lesson type:

Lesson type: Lecture, problem solving lesson.

Duration: 2 hours.

Goals:1) Learn the graphic method.

2) Show the use of the Maple program in solving systems of inequalities using the graphical method.

3) Develop perception and thinking on this topic.

Lesson plan:

Progress of the lesson.

Stage 1: The graphical method consists of constructing a set of feasible solutions to the PLP, and finding in this set the point corresponding to the max/min objective function.

Due to disabilities clear graphical representation this method applies only to systems linear inequalities with two unknowns and systems that can be reduced to a given form.

In order to clearly demonstrate the graphical method, let’s solve the following problem:

1. At the first stage, it is necessary to construct a region of feasible solutions. For this example, it is most convenient to select X2 as the abscissa, and X1 as the ordinate, and write the inequalities in the following form:

Since both the graphs and the area of feasible solutions are in the first quarter. In order to find the boundary points, we solve the equations (1)=(2), (1)=(3) and (2)=(3).

As can be seen from the illustration, the polyhedron ABCDE forms a region of feasible solutions.

If the region of feasible solutions is not closed, then either max(f)=+ ?, or min(f)= -?.

2. Now we can proceed to directly finding the maximum of the function f.

By alternately substituting the coordinates of the vertices of the polyhedron into the function f and comparing the values, we find that f(C)=f(4;1)=19 is the maximum of the function.

This approach is quite beneficial with a small number of vertices. But this procedure may take a long time if there are quite a lot of vertices.

In this case, it is more convenient to consider a level line of the form f=a. With a monotonic increase in the number a from -? to +? straight lines f=a are shifted along the normal vector. The normal vector has coordinates (C1;C2), where C1 and C2 are coefficients for the unknowns in the objective function f=C1?X1+C2?X2+C0.. If with such a movement of the level line there is a certain point X is the first common point of the domain of feasible solutions (polyhedron ABCDE) and the level line, then f(X) is the minimum of f on the set ABCDE. If X is the last point of intersection of the level line and the ABCDE set, then f(X) is the maximum on the set of feasible solutions. If for a>-? the straight line f=a intersects the set of feasible solutions, then min(f)= -?. If this happens for a>+?, then max(f)=+?.

In our example, the straight line f=a intersects the region ABCDE at point C(4;1). Since this is the last intersection point, max(f)=f(C)=f(4;1)=19.

Solve the system of inequalities graphically. Find corner solutions.

x1>= 0, x2>=0

> with(plots);

> with(plottools);

> S1:=solve((f1x = X6, f2x = X6), );

Answer: All points Si where i=1..10 for which x and y are positive.

Area limited by these points: (54/11.2/11) (5/7.60/7) (0.5) (10/3, 10/3)

Stage 3. Each student is given one of 20 options, in which the student is asked to independently solve the inequality using a graphical method, and the remaining examples are given as homework.

Lesson No. 4 Graphic solution linear programming problems

Lesson type: lesson of learning new material.

Lesson type: Lecture + problem solving lesson.

Duration: 2 hours.

Goals: 1) Study the graphical solution of the linear programming problem.

2) Learn to use the Maple program when solving a linear programming problem.

2) Develop perception and thinking.

Lesson plan: Stage 1: learning new material.

Stage 2: Working on new material in the Maple mathematical package.

Stage 3: checking the studied material and homework.

Progress of the lesson.

The graphical method is quite simple and intuitive for solving linear programming problems with two variables. It is based on geometric presentation of feasible solutions and TFs of the problem.

Each of the inequalities of the linear programming problem (1.2) defines a certain half-plane on the coordinate plane (Fig. 2.1), and the system of inequalities as a whole defines the intersection of the corresponding planes. The set of intersection points of these half-planes is called area of feasible solutions(ODR). ODR always represents convex figure, i.e. having the following property: if two points A and B belong to this figure, then the entire segment AB belongs to it. ODR can be graphically represented by a convex polygon, an unlimited convex polygonal area, a segment, a ray, or one point. If the system of constraints in problem (1.2) is inconsistent, the ODS is an empty set.

All of the above also applies to the case when the system of restrictions (1.2) includes equalities, since any equality

can be represented as a system of two inequalities (see Fig. 2.1)

The digital filter at a fixed value defines a straight line on the plane. By changing the values of L, we get a family of parallel lines called level lines.

This is due to the fact that changing the value of L will entail a change only in the length of the segment cut off by the level line on the axis (initial ordinate), and the angular coefficient of the straight line will remain constant (see Fig. 2.1). Therefore, to solve it, it will be enough to construct one of the level lines, arbitrarily choosing the value of L.

The vector with coordinates from the CF coefficients at and is perpendicular to each of the level lines (see Fig. 2.1). The direction of the vector coincides with the direction increasing CF, which is important point to solve problems. Direction descending The CF is opposite to the direction of the vector.

The essence of the graphical method is as follows. In the direction (against the direction) of the vector in the ODR, the optimal point is searched. The optimal point is the point through which the level line passes, corresponding to the largest (smallest) value of the function. The optimal solution is always located on the boundary of the ODD, for example, at the last vertex of the ODD polygon through which the target line will pass, or on its entire side.

When searching for an optimal solution to linear programming problems, it is possible following situations: exists the only solution tasks; there is an infinite number of solutions (alternative); TF is not limited; the region of feasible solutions is a single point; the problem has no solutions.

Figure 2.1 Geometric interpretation of the constraints and CF of the problem.

Technique for solving LP problems using the graphical method

I. In the constraints of problem (1.2), replace the inequality signs with exact equality signs and construct the corresponding straight lines.

II. Find and shade the half-planes allowed by each of the inequality constraints of problem (1.2). To do this, you need to substitute the coordinates of a point [for example, (0;0)] into a specific inequality and check the truth of the resulting inequality.

If the inequality is true,

That it is necessary to shade the half-plane containing this point;

otherwise(the inequality is false) we must shade the half-plane that does not contain the given point.

Since and must be non-negative, their permissible values will always be above the axis and to the right of the axis, i.e. in the first quadrant.

Equality constraints allow only those points that lie on the corresponding line. Therefore, it is necessary to highlight such straight lines on the graph.

III. Define the ODR as a part of the plane that simultaneously belongs to all allowed areas and select it. In the absence of ODD, the problem has no solutions.

IV. If the ODR is not an empty set, then you need to construct the target line, i.e. any of the level lines (where L is an arbitrary number, for example, a multiple and, i.e., convenient for calculations). The construction method is similar to the construction of direct constraints.

V. Construct a vector that starts at the point (0;0) and ends at the point. If the target line and vector are constructed correctly, then they will perpendicular.

VI. When searching for the maximum CF, you need to move the target line in the direction vector, when searching for the minimum CF - against the direction vector. The last top of the ODR in the direction of movement will be the point of maximum or minimum of the CF. If such a point(s) does not exist, then we can conclude that unlimited TF on many plans from above (when searching for a maximum) or from below (when searching for a minimum).

VII. Determine the coordinates of the point max (min) of the digital filter and calculate the value of the digital filter. To calculate the coordinates of the optimal point, it is necessary to solve a system of equations of the lines at the intersection of which it is located.

Solve a linear programming problem

1. f(x)=2x1+x2 ->extr

x1>= 0, x2>=0

> plots((a+b<=3,a+3*b<=5,5*a-b<=5,a+b>=0,a>=0,b>=0), a=-2..5, b=-2..5, optionsfeasible=(color=red),

optionsopen=(color=blue, thickness=2),

optionsclosed=(color=green, thickness=3),

optionsexcluded=(color=yellow));

> with(simplex):

> C:=( x+y<=3, x+3*y <=5, 5*x-y <=5,x+y >=0};

> dp:=setup(( x+y<=3, x+3*y <=5, 5*x-y <=5,x+y >=0});

> n:=basis(dp);

Sh display(C,);

> L:=cterm(C);

Sh X:=dual(f,C,p);

Sh f_max:=subs(R,f);

Sh R1:=minimize(f,C ,NONNEGATIVE);

f_min:=subs(R1,f);

ANSWER: When x 1 =5/4 x 2 =5/4 f_max=15/4; At x 1 =0 x 2 =0 f_min=0;

Lesson No. 5. Solving matrix games using linear programming methods and the simplex method

Lesson type: lesson control + lesson learning new material. Lesson type: Lecture.

Duration: 2 hours.

Goals:1) Check and consolidate knowledge of past material in previous lessons.

2) Learn a new method for solving matrix games.

3) develop memory, mathematical thinking and attention.

Stage 1: check your homework as independent work.

Stage 2: give a brief description of the zigzag method

Stage 3: consolidate new material and assign homework.

Progress of the lesson.

Linear programming methods are numerical methods for solving optimization problems that can be reduced to formal linear programming models.

As is known, any linear programming problem can be reduced to a canonical model for minimizing a linear objective function with linear equality-type constraints. Since the number of variables in a linear programming problem is greater than the number of constraints (n > m), it is possible to obtain a solution by setting (n - m) variables, called free. The remaining m variables, called basic, can be easily determined from a system of equality constraints using the usual methods of linear algebra. If a solution exists, then it is called basic. If the basic solution is admissible, then it is called basic permissible. Geometrically, the basic feasible solutions correspond to the vertices (extreme points) of a convex polyhedron, which bounds the set of feasible solutions. If a linear programming problem has optimal solutions, then at least one of them is basic.

The above considerations mean that when searching for an optimal solution to a linear programming problem, it is sufficient to limit ourselves to enumerating basic feasible solutions. The number of basic solutions is equal to the number of combinations of n variables in m:

C = mn! / n m! * (n - m)!

and can be large enough to enumerate them by direct search real time. The fact that not all basic solutions are admissible does not change the essence of the problem, since in order to assess the admissibility of a basic solution, it must be obtained.

The problem of rational enumeration of basis solutions to a linear programming problem was first solved by J. Danzig. The simplex method he proposed is still the most widespread general method linear programming. The simplex method implements a directed search of admissible basic solutions along the corresponding extreme points of a convex polyhedron of admissible solutions in the form of an iterative process, where at each step the values of the objective function strictly decrease. The transition between extreme points is carried out along the edges of a convex polyhedron of admissible solutions in accordance with simple linear algebraic transformations of the system of restrictions. Since the number of extreme points is finite, and the objective function is linear, then by searching through the extreme points in the direction of decreasing the objective function, the simplex method converges to a global minimum in a finite number of steps.

Practice has shown that for most applied linear programming problems, the simplex method allows one to find the optimal solution in a relatively small number of steps compared to the total number of extreme points of the admissible polyhedron. At the same time, it is known that for some linear programming problems with a specially selected form of the admissible region, the use of the simplex method leads to a complete enumeration of the extreme points. This fact to a certain extent stimulated the search for new effective methods solutions to linear programming problems, based on ideas other than the simplex method, which allow solving any linear programming problem in a finite number of steps, significantly less than the number of extreme points.

Among polynomial linear programming methods invariant to the domain configuration acceptable values, the most common is the L.G. method. Khachiyan. However, although this method has a polynomial complexity estimate depending on the dimension of the problem, it nevertheless turns out to be non-competitive compared to the simplex method. The reason for this is that the dependence of the number of iterations of the simplex method on the dimension of the problem is expressed by a 3rd order polynomial for most practical problems, while in Khachiyan’s method, this dependence always has an order of at least four. This fact has crucial for practice, where applied problems difficult for the simplex method are extremely rare.

It should also be noted that for applied linear programming problems that are important in a practical sense, special methods, taking into account the specific nature of the constraints of the problem. In particular, for a homogeneous transport problem, special algorithms for selecting the initial basis are used, the most famous of which are the northwest corner method and the approximate Vogel method, and the algorithmic implementation of the simplex method itself is close to the specifics of the problem. To solve the linear assignment problem (selection problem), instead of the simplex method, either the Hungarian algorithm is usually used, based on the interpretation of the problem in terms of graph theory as the problem of finding the maximum weight perfect matching in a bipartite graph, or Mack's method.

Solve a 3x3 matrix game

f(x)=x 1 +x 2 +x 3

x1>= 0, x2>=0, x3>=0

> with(simplex):

> C:=( 0*x+3*y+2*z<=1, 2*x+0*y+1*z <=1, 3*x+0*y+0*z <=1};

Sh display(C,);

> feasible(C, NONNEGATIVE , "NewC", "Transform");

> S:=dual(f,C,p);

Sh R:=maximize(f,C ,NONNEGATIVE);

Sh f_max:=subs(R,f);

Sh R1:=minimize(S ,NONNEGATIVE);

> G:=p1+p2+p3;

> f_min:=subs(R1,G);

Let's find the price of the game

> V:=1/f_max;

Let's find the optimal strategy of the first player > X:=V*R1;

Let's find the optimal strategy of the second player

ANSWER: When X=(3/7, 3/7,1/7) V=9/7; When Y=(3/7.1/7.3/7) V=9/7;

Each student is given one of 20 options, in which the student is asked to independently solve a 2x2 matrix game, and the remaining examples as homework.

One of the most convenient solution methods quadratic inequalities- This is a 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.

Page navigation.

The essence of the graphic method

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

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

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

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

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

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

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

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

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

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

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

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

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

where x 1 and x 2 are the roots of the square 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 quadratic function y=x 2 −4·x+4, here a=1>0, D=(−4) 2 −4·1·4=0 and x 0 =2.

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

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

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

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

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

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

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

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

Solution algorithm

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

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

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

When the drawing is ready, use it in the second step of the algorithm
- when solving the quadratic inequality a·x 2 +b·x+c>0, the intervals are determined at which the parabola is located above the abscissa;
- when solving the inequality a·x 2 +b·x+c≥0, the intervals at which the parabola is located above the abscissa axis are determined and the abscissas of the intersection points (or the abscissa of the tangent point) are added to them;
- when solving the inequality a x 2 +b x+c<0 находятся промежутки, на которых парабола ниже оси Ox ;
- finally, when solving a quadratic inequality of the form a·x 2 +b·x+c≤0, intervals are found in which the parabola is below the Ox axis and the abscissa of the intersection points (or the abscissa of the tangent point) is added to them;
they constitute the desired solution to the quadratic inequality, and if there are no such intervals and no points of tangency, then the original quadratic inequality has no solutions.

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

Examples with solutions

Example.

Solve the inequality .

Solution.

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

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

Let's move on to the second step of the algorithm. Since we are solving a non-strict quadratic inequality with the sign ≤, we need to determine the intervals at which the parabola is located below the abscissa axis 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 the drawing clearly shows that there are no such gaps (since the parabola is everywhere below the abscissa axis), just as there are no points of intersection, and there are no points of tangency. Therefore, the original quadratic inequality has no solutions.

Answer:

no solutions or in another entry ∅.

References.

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 the beginnings of mathematical analysis. 11th grade. 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.

Goals:

1. Review knowledge about the quadratic function.

2. Get acquainted with the method of solving a quadratic inequality based on the properties of a quadratic function.

Equipment: multimedia, presentation “Solving quadratic inequalities”, cards for independent work, table “Algorithm for solving quadratic inequalities”, test sheets with carbon paper.

PROGRESS OF THE LESSON

I. Organizational moment (1 min).

II. Updating of reference knowledge(10 min).

1. Plotting a graph of the quadratic function y=x 2 -6x+8<Рисунок 1. Приложение >

determining the direction of the parabola branches;
determining the coordinates of the vertex of a parabola;
determination of the axis of symmetry;
determination of intersection points with coordinate axes;
finding additional points.

2. Determine from the drawing the sign of the coefficient a and the number of roots of the equation ax 2 + in + c = 0.<Рисунок 2. Приложение >

3. Using the graph of the function y=x 2 -4x+3, determine:

What are the zeros of the function?
Find the intervals at which the function takes positive values;
Find the intervals at which the function takes negative values;
At what values of x does the function increase and at what values does it decrease?<Рисунок 3>

4. Learning new knowledge (12 min.)

Problem 1: Solve the inequality: x 2 +4x-5 > 0.

The inequality is satisfied by the values of x for which the values of the function y = x 2 + 4x-5 are equal to zero or positive, that is, those values of x for which the points of the parabola lie on the x-axis or above this axis.

Let's build a graph of the function y=x 2 +4x-5.

With the oh axis: X 2 +4x-5=0. According to Vieta's theorem: x 1 = 1, x 2 = -5. Points(1;0),(-5;0).

With the y axis: y(0)=-5. Point (0;-5).

Additional points: y(-1)=-8, y(2)=7.<Рисунок 4>

Result: The function values are positive and equal to zero (non-negative) at

Is it necessary to graph the quadratic function in detail each time to solve an inequality?
Do you need to find the coordinates of the vertex of a parabola?
What's important? (a, x 1, x 2)

Conclusion: To solve a quadratic inequality, it is enough to determine the zeros of the function, the direction of the branches of the parabola and draw a sketch of the graph.

Problem 2: Solve the inequality: x 2 -6x+8 < 0.

Solution: Let's determine the roots of the equation x 2 -6x+8=0.

According to Vieta's theorem: x 1 =2, x 2 =4.

a>0 – the branches of the parabola are directed upward.

Let's build a sketch of the graph.<Рисунок 5>

Let us mark with “+” and “–” signs the intervals at which the function takes on positive and negative values. Let's choose the interval we need.

Answer: X€.

5. Consolidation of new material (7 min).

No. 660 (3). The student decides on the board.

Solve the inequality x 2 -3x-2<0.

X 2 -3x-2=0; x 2 +3x+2=0;

roots of the equation: x 1 = -1, x 2 = -2.

A<0 – ветви вниз. <Рисунок 6>

No. 660 (1) - Working with a hidden board.

Solve the inequality x 2 -3x+2 < 0.

Solution: x 2 -3x+2=0.

Let's find the roots: ; x 1 =1, x 2 =2.

a>0 – branches upward. We build a sketch of the graph of the function.<Рисунок 7>

Algorithm:

Find the roots of the equation ax 2 + in + c = 0.
Mark them on the coordinate plane.
Determine the direction of the branches of the parabola.
Draw a sketch of the graph.
Mark with “+” and “-” signs the intervals at which the function takes on positive and negative values.
Select the required interval.

6. Independent work (10 min.).

(Reception - carbon paper).

The control sheet is signed and handed over to the teacher for checking and determining correction.

Self-test on the board.

Additional task:

No. 670. Find the values of x at which the function takes values not greater than zero: y=x 2 +6x-9.

7. Homework (2 min).

№ 660 (2, 4), № 661 (2, 4).

Fill out the table:

D	Inequality	a	Drawing	Solution
D>0	ah 2 +in+s > 0	a>0
D>0	ah 2 +in+s > 0	a<0
D>0	ah 2 +in+s < 0	a>0
D>0	ah 2 +in+s < 0	a<0

8. Lesson summary (3 min).

Reproduce the algorithm for solving inequalities.
Who did a great job?
What did you find difficult?

The graphical method consists of constructing a set of admissible solutions to the PLP, and finding in this set the point corresponding to the max/min objective function.

Due to the limited possibilities of visual graphical representation, this method is used only for systems of linear inequalities with two unknowns and systems that can be reduced to this form.

In order to clearly demonstrate the graphical method, let’s solve the following problem:

Since both the graphs and the area of feasible solutions are in the first quarter. In order to find the boundary points, we solve the equations (1)=(2), (1)=(3) and (2)=(3).

As can be seen from the illustration, the polyhedron ABCDE forms a region of feasible solutions.

If the region of feasible solutions is not closed, then either max(f)=+ ?, or min(f)= -?.

2. Now we can proceed to directly finding the maximum of the function f.

By alternately substituting the coordinates of the vertices of the polyhedron into the function f and comparing the values, we find that f(C)=f (4; 1)=19 is the maximum of the function.

This approach is quite beneficial with a small number of vertices. But this procedure can take a long time if there are quite a lot of vertices.

In this case, it is more convenient to consider a level line of the form f=a. With a monotonic increase in the number a from -? to +? straight lines f=a are shifted along the normal vector. If, with such a movement of the level line, there is a certain point X - the first common point of the region of feasible solutions (polyhedron ABCDE) and the level line, then f(X) is the minimum of f on the set ABCDE. If X is the last point of intersection of the level line and the ABCDE set, then f(X) is the maximum on the set of feasible solutions. If for a>-? the straight line f=a intersects the set of feasible solutions, then min(f)= -?. If this happens for a>+?, then max(f)=+?.