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Graphical solution of inequalities. Solving systems of linear inequalities graphically

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

The essence of the graphical method

The method is applicable to solve any inequalities, not just square ones. Its essence is this: the right and left parts of the inequality are considered as two separate functions y \u003d f (x) and y \u003d g (x), their graphs are built in a rectangular coordinate system and they look at which of the graphs is located above the other, and on which intervals. The intervals are evaluated as follows:

Definition 1

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

Consider the above algorithm with an example. To do this, take the quadratic inequality a x 2 + b x + c< 0 (≤ , >, ≥) and derive two functions from it. Left side 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 y = 0 (in this case g (x) = 0) .

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

The branches of the parabola are directed upwards. It intersects the x-axis at points x 1 And x2. Coefficient a to this case positive, since it is he who is responsible for the direction of the branches of the parabola. The discriminant is positive, which indicates the presence of two roots in square trinomiala x 2 + b x + c. We denote the roots of the trinomial as x 1 And x2, and it was accepted that x 1< x 2 , since on the O x axis they depicted a point with an abscissa x 1 to the left of the point with the abscissa x2.

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

Let's select the gaps 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 . In blue, we marked 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 short note 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 way 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 other notation x ≤ x 1 , x ≥ x 2 ;
• solution of 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 other notation x 1 ≤ x ≤ x 2 ,

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

In this figure, the parabola touches the O x axis at only one point, which is indicated as x0 a > 0. D=0, therefore, the square trinomial has one root x0.

The parabola is located completely above the O x axis, except for the point of contact of the coordinate axis. Color the gaps (− ∞ , x 0) , (x 0 , ∞) .

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

• solution of the quadratic inequality a x 2 + b x + c > 0 is (− ∞ , x 0) ∪ (x 0 , + ∞) or in other notation x ≠ x0;
• solution of the quadratic inequality a x 2 + b x + c ≥ 0 is (− ∞ , + ∞) or in another notation x ∈ R ;
• square inequality a x 2 + b x + c< 0 has no solutions (there are no intervals on which the parabola is located below the axis O x);
• square inequality a x 2 + b x + c ≤ 0 It has only decision x = x0(it is given by the point of contact),

Where x0- the root of a square trinomial a x 2 + b x + c.

Consider the third case, when the branches of the parabola are directed upwards and do not touch the axis O x. The branches of the parabola point 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 solution of square 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 do not have solutions.

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

Consideration of the previous section of the article prepared us for the perception of the algorithm for solving inequalities using a graphical method. To carry out calculations, we will need to use a drawing every time, which will show the coordinate line O x and the parabola that corresponds to 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 intersection points of the parabola and the abscissa axis, which are determined by the value of the discriminant of the square trinomial a · x 2 + b · x + c.

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

Having a finished 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 gaps and intersection points are the solution to the quadratic inequality. If there are no intersection or tangency points and no intervals, then it is considered that the inequality specified in the conditions of the problem has no solutions.

Now let's solve some 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 x2 positive, because 2 . This means that the branches of the parabola will be directed upwards.

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

D \u003d 5 1 3 2 - 4 2 (- 2) \u003d 400 9

As you can see, D Above zero, therefore, we have two intersection points: x 1 \u003d - 5 1 3 - 400 9 2 2 and x 2 \u003d - 5 1 3 + 400 9 2 2, that is, x 1 = − 3 And x 2 = 1 3.

We are solving a non-strict inequality, therefore we put ordinary points on the graph. We draw a parabola. As you can see, the drawing has the same appearance as in the first template we reviewed.

Our inequality has the sign ≤ . Therefore, we need to select the gaps 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 . Adding intersection points to it, we get number line− 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 graphic method.

Solution

The square of the variable has a negative numerical coefficient, so the branches of the parabola will point downwards. 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 \u003d - 8 + 1 - 1 and x 2 \u003d - 8 - 1 - 1, x 1 \u003d 7 and x2 = 9.

It turns out that the parabola intersects the x-axis at points 7 And 9 . We mark these points on the graph as empty, since we are working with strict inequality. After that, we 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. Mark these intervals in blue.

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

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

When the discriminant of a square trinomial zero, it is necessary to carefully approach the question of whether it is worth including the abscissas of the touch point in the answer. In order to make the right decision, it is necessary to take into account the inequality sign. In strict inequalities, the point of contact of the abscissa axis is not a solution to the inequality, in non-strict ones it is.

Example 3

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

Solution

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

Let's plot the function y = 10 x 2 − 14 x + 4, 9. Its branches are directed upwards, since the coefficient at x2 positive, and it touches the x-axis at the point with the x-axis 0 , 7 , because D" = (− 7) 2 − 10 4 , 9 = 0, whence x 0 = 7 10 or 0 , 7 .

Put a point and draw a parabola.

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

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

Example 4

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

Solution

The branches of the parabola point downwards. The discriminant is zero. Intersection point x0 = 4.

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

We are dealing with a strict inequality. Therefore, we are interested in the intervals on 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 is not located below the O x axis at it. Therefore, we get two intervals (− ∞ , 4) , (4 , + ∞) .

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

Not always with a negative value of the discriminant, the inequality will not have solutions. There are cases when the solution will be the set of all real numbers.

Example 5

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

Solution

The coefficient a is positive. The discriminant is negative. The branches of the parabola will be directed upwards. There are no points of intersection of the parabola with the O x axis. Let's turn to 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 graphic way.

Solution

The branches of the parabola point downwards. The discriminant is negative, so common points there is no parabola and x-axis. Let's turn to the drawing.

We work with a non-strict inequality with the sign ≥ , therefore, we are interested in the intervals on which the parabola is located above the x-axis. Judging by the schedule, there are no such gaps. This means that the inequality given in the condition of the problem has no solutions.

Answer: There are no solutions.

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

Steps

Graphical representation of a linear inequality on a number line

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

   • For example, given the inequality 3y + 9 > 12 (\displaystyle 3y+9>12). To isolate the variable, subtract 9 from both sides of the inequality, and then divide both sides by 3:
     3y + 9 > 12 (\displaystyle 3y+9>12)
     3 y + 9 − 9 > 12 − 9 (\displaystyle 3y+9-9>12-9)
     3 y > 3 (\displaystyle 3y>3)
     3 y 3 > 3 3 (\displaystyle (\frac (3y)(3))>(\frac (3)(3)))
     y > 1 (\displaystyle y>1)
   • An inequality must have only one variable. If the inequality has two variables, it is better to plot the graph on the coordinate plane.

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

   • For example, if you calculated that y > 1 (\displaystyle y>1), mark the value 1 on the number line.

3. Draw a circle to represent the found value. If the variable is less than ( < {\displaystyle <} ) or more ( > (\displaystyle >)) of this value, the circle is not filled 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.

   • y > 1 (\displaystyle y>1), on the number line, draw an open circle at point 1 because 1 is not in the solution set.

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

   • For example, given the inequality y > 1 (\displaystyle y>1), on the number line, shade the area to the right of 1 because the solution set includes all values ​​greater than 1.

   Graphical representation of a 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. The variable should remain on the right side x (\displaystyle x) and possibly some constant.

      • For example, given the inequality 3y + 9 > 9x (\displaystyle 3y+9>9x). To isolate a variable y (\displaystyle y), subtract 9 from both sides of the inequality, and then divide both sides by 3:
        3y + 9 > 9x (\displaystyle 3y+9>9x)
        3 y + 9 − 9 > 9 x − 9 (\displaystyle 3y+9-9>9x-9)
        3 y > 9 x − 9 (\displaystyle 3y>9x-9)
        3 y 3 > 9 x − 9 3 (\displaystyle (\frac (3y)(3))>(\frac (9x-9)(3)))
        y > 3 x − 3 (\displaystyle y>3x-3)

   2. Plot on the coordinate plane linear equation. plot the graph as you plot any linear equation. Plot the point of intersection with the Y-axis, and then plot other points using the slope.

      • y > 3 x − 3 (\displaystyle y>3x-3) plot the equation y = 3 x − 3 (\displaystyle y=3x-3). The point of intersection with the Y-axis has coordinates , and slope is 3 (or 3 1 (\displaystyle (\frac (3)(1)))). So first plot a point with coordinates (0 , − 3) (\displaystyle (0,-3)); the point above the point of intersection with the y-axis has coordinates (1 , 0) (\displaystyle (1,0)); the point below the point of intersection with the y-axis has coordinates (− 1 , − 6) (\displaystyle (-1,-6))

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

      • For example, in case of inequality y > 3 x − 3 (\displaystyle y>3x-3) draw the dotted line, because the set of solutions does not include values ​​lying on the line.

   4. Shade the corresponding area. If the inequality has the form y > m x + b (\displaystyle y>mx+b), fill in the area above the line. If the inequality has the form y< m x + b {\displaystyle y, fill in the area under the line.

      • For example, in case of inequality y > 3 x − 3 (\displaystyle y>3x-3) shade the area above the line.

   Graphical representation of a quadratic inequality on the coordinate plane

   1. Determine that this inequality is square. Square inequality has the form a x 2 + b x + c (\displaystyle ax^(2)+bx+c). Sometimes the inequality does not contain a first order variable ( x (\displaystyle x)) and/or free term (constant), but must include a second-order variable ( x 2 (\displaystyle x^(2))). Variables x (\displaystyle x) And y (\displaystyle y) should be isolated to different sides inequalities.

      • For example, you need to plot the inequality y< x 2 − 10 x + 16 {\displaystyle y.

   2. Draw a graph on the coordinate plane. To do this, convert the inequality into an equation and build a graph, as you build a graph of any quadratic equation. Remember that the graph of a quadratic equation is a parabola.

      • For example, in case of inequality y< x 2 − 10 x + 16 {\displaystyle y plot quadratic equation y = x 2 − 10 x + 16 (\displaystyle y=x^(2)-10x+16). The apex of the parabola is at the point (5 , − 9) (\displaystyle (5,-9)), and the parabola intersects the x-axis at points (2 , 0) (\displaystyle (2,0)) And (8 , 0) (\displaystyle (8,0)).

Let f(x,y) And g(x, y)- two expressions with variables X And at and domain of definition 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, under which the inequality turns into a true numerical inequality, is called its decision and denoted (x, y). Solve the inequality is to find a set of such pairs.

If each pair of numbers (x, y) from the set of solutions to the inequality, put in correspondence a point M(x, y), we obtain the set of points on the plane given by this inequality. He is called graph of this inequality . An inequality plot 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 equals sign and find a line that has the equation f(x,y) = g(x,y). This line divides the plane into several parts. After that, it suffices to take one point in each part and check whether the inequality holds at this point f(x, y) > g(x, y). If it is executed at this point, then it will also be executed in the entire part where this point lies. Combining such parts, we obtain a set of solutions.

Task. y > x.

Solution. First, we replace the inequality sign with an equals 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 that, we take one point in each part and check whether the inequality holds at this point y > x.

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

Rice. 18.

Solution. First, replace the inequality sign with an equals 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 validity of the inequality X 2 + at 2 £ 25 in each part, we get that the graph is the set of points of the circle and part of the 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 any value (x, y), which turns each of the inequalities into a true numerical inequality. Lots of solutions systems inequalities is the intersection of the solution sets of inequalities that form the given system.

Set of inequalities is yourself disjunction of these inequalities. Set decision is any value (x, y), which turns into a true numerical inequality at least one of the inequalities in the set. Lots of solutions aggregates is the union of sets of solutions to inequalities forming a set.

Task. Solve graphically a 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 a set of points in the plane that are the intersection (double hatching) of the solution sets of the first and second inequalities.

Task. Solve graphically a set of inequalities

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

Exercises for independent work

1. Solve graphically 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) c)

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 is to be maximized.

Let's answer the question: what pairs of numbers ( x; y) are solutions to the system of inequalities, i.e., do they 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 of one linear inequality with two unknowns.
To solve a linear inequality with two unknowns means to determine all pairs of values ​​of the unknowns for which the inequality is satisfied.
For example, inequality 3 x – 5y≥ 42 satisfy the pairs ( x , y) : (100, 2); (3, –10), etc. The problem is to find all such pairs.
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, take a point with coordinate x = x 0; then a point lying on a straight line and having an abscissa x 0 , has an ordinate

Let for definiteness a<0, b>0, c>0. All points with abscissa x 0 above P(e.g. dot M), have yM>y 0 , and all points below the point P, with abscissa x 0 , have yN<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 hand, 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 of graphical solution of systems linear inequalities from two variables. To solve the system, you need:

1. For each inequality, write down the equation corresponding to the given inequality.
2. Construct lines that are graphs of functions given by equations.
3. For each straight line, determine the half-plane, which is given by the inequality. To do this, take an arbitrary point that does not lie on a straight 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 in the system.

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

Let's look at three relevant examples.

Example 1. Graphically solve the system:
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 us construct the straight lines given by these equations.

Figure 2

Let us define the half-planes given by the inequalities. Take an arbitrary point, let (0; 0). Consider x+ y– 1 0, we substitute the point (0; 0): 0 + 0 – 1 ≤ 0. hence, in the half-plane where the point (0; 0) lies, x + y – 1 ≤ 0, i.e. the half-plane lying below the straight line is the solution to the first inequality. Substituting this point (0; 0) into the second one, 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 another half-plane - in the one above the straight line.
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, it is inconsistent.

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

Figure 3
1. Write down 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 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 points of intersection of the corresponding lines


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

Let us consider one more example, in which the resulting domain of the solution of the system is not limited.

Graphical solution of equations

Heyday, 2009

Introduction

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

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

But general rule solving quadratic equations, with all possible combinations of coefficients b and c, was formulated in Europe only in 1544 by M. Stiefel.

In 1591 François Viet introduced formulas for solving quadratic equations.

Some kinds of quadratic equations could be solved in ancient Babylon.

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

In 7th grade we studied functions y \u003d C, y=kx, y =kx+ m, y =x 2,y = -x 2, in the 8th grade - y = √x, y =|x|, y=ax2 + bx+ c, y =k/ x. In the 9th grade algebra textbook, I saw functions that were not yet known to me: y=x 3, y=x 4,y=x 2n, y=x- 2n, y= 3√x, (x– a) 2 + (y -b) 2 = r 2 and others. There are rules for constructing graphs of these functions. I was wondering if there are other functions that obey these rules.

My job is to study graphs of functions and solve equations graphically.

1. What are the functions

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

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

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

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

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

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

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

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

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

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

2. The concept of an equation, its graphical solution

The equation is an expression containing a variable.

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

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

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

When plotting graphs and solving equations, the properties of a function are used, so the method is often called functional-graphic.

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

3. Algorithm for constructing a graph of a function

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

4. Graphical solution of the quadratic equation

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

What did the ancient Greeks know about the parabola?

Modern mathematical symbolism originated in the 16th century.

The ancient Greek mathematicians had neither the coordinate method nor the concept of a function. However, the properties of the parabola were studied by them in detail. The inventiveness of ancient mathematicians is simply amazing, because they could only use drawings and verbal descriptions of dependencies.

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

There is an algorithm for constructing a parabola:

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

y0=aho2+in0+s;

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

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Compiling a table of values ​​for building control points;

We construct the obtained points and construct points symmetrical to them with respect to the axis of symmetry.

1. Let's build a parabola according to the algorithm y= x2 – 2 x– 3 . Abscissas of points of intersection with the axis x and are the roots of the quadratic equation x2 – 2 x– 3 = 0.

There are five ways to graphically solve this equation.

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

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

4. Transform the equation x2 – 2 x– 3 = 0 by selecting the full square on the function: y= (x–1) 2 And y=4. The roots of the equation are the abscissas of the points of intersection of the parabola with the line.

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

5. Graphical solution of degree equationsn

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

y= x5 , y= 3 – 2 x.

Answer: x = 1.

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

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

Answer: x=8.

Conclusion

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

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

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

In the 9th grade and in the senior classes, I will still get acquainted with other functions. I'm interested to know if those functions obey the rules of parallel translation when plotting their graphs.

Next year I also want to consider the issues of graphical solution of systems of equations and inequalities.

Literature

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

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

3. Algebra. Grade 9 Part 1. Textbook for educational institutions / A.G. Mordkovich. Moscow: Mnemosyne, 2007.

4. Glazer G.I. History of mathematics at school. VII-VIII classes. – M.: Enlightenment, 1982.

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

6. Graphic solution of equations Internet sites: Tol WIKI; stimul.biz/en; wiki.iot.ru/images; berdsk.edu; pege 3–6.htm.