During the lesson, you will be able to independently study the topic " Graphic solution equations, inequalities. The teacher in the lesson will analyze the graphical methods for solving equations and inequalities. It will teach you how to build graphs, analyze them and get solutions to equations and inequalities. The lesson will also discuss concrete examples on this topic.

Topic: Numeric functions

Lesson: Graphical solution of equations, inequalities

1. Lesson topic, introduction

We have looked at charts elementary functions, including graphics power functions c different indicators. We also considered the rules for shifting and transforming function graphs. All these skills must be applied when required. graphicdecision equations or graphic decisioninequalities.

2. Solving equations and inequalities graphically

Example 1. Graphically solve the equation:

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

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

The graph of the function is a straight line, we will build it according to the table.

Graphs intersect at a point There are no other intersection points, since the function is monotonically increasing, the function is decreasing monotonically, and, therefore, their intersection point is unique.

Example 2. Solve the 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 should be under the line. This is done when

Example 3. Solve the inequality

Let's build graphs of functions (Fig. 2).

Find the root of the equation When there are no solutions. There is one solution for .

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

Example 4. Solve graphically the inequality:

Domain:

Let's build graphs of functions for (Fig. 3).

a. The graph of the function should be located under the graph; this is done when

b. The graph of the function is located above the graph at But since we have a non-strict sign in the condition, it is important not to lose the isolated root

3. Conclusion

We have reviewed graphic method solving equations and inequalities; we considered specific examples, in the solution of which we used such properties of functions as monotonicity and evenness.

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

2. Mordkovich A. G. et al. Algebra grade 9: Task book for students educational institutions/ A. G. Mordkovich, T. N. Mishustina and others - 4th ed. — M.: Mnemosyne, 2002.-143 p.: ill.

3. Yu. N. Makarychev, Algebra. Grade 9: textbook. 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. Sh. A. Alimov, Yu. M. Kolyagin, and Yu. V. Sidorov, Algebra. Grade 9 16th ed. - M., 2011. - 287 p.

5. Mordkovich A. G. Algebra. Grade 9 At 2 pm Part 1. Textbook for students of educational institutions / A. G. Mordkovich, P. V. Semenov. - 12th ed., erased. — M.: 2010. — 224 p.: ill.

6. Algebra. Grade 9 At 2 hours. Part 2. Task book for students of educational 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 RESOLVE THE USE".

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

One of the most convenient methods for solving square inequalities is a graphic method. In this article, we will analyze how quadratic inequalities are solved graphically. First, let's discuss what the essence of this method is. And then we give the algorithm and consider examples of solving quadratic inequalities graphically.

Page navigation.

The essence of the graphic method

Generally graphical way to solve inequalities with one variable is used not only to solve square inequalities, but also inequalities of other types. The essence of the graphical method for solving inequalities next: consider the functions y=f(x) and y=g(x) that correspond to the left and right parts inequalities, build their graphs in one rectangular system coordinates and find out at what intervals the graph of one of them is located below or above 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 the function f below the graph of the function g are solutions to the inequality f(x)
• the graph of the function f not above the graph of the function g are solutions to the inequality f(x)≤g(x) .

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

Let us 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 (in this case f(x)=a x 2 +b x+c) corresponds to the left side of the quadratic inequality, the second y=0 (in this case g (x)=0 ) corresponds to the right side of the inequality. schedule quadratic function f is a parabola and the graph permanent function g is a straight line coinciding with the abscissa axis Ox .

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

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 it is possible not to depict the Oy axis, since its position does not affect the solution of the inequality):

In this drawing, we see a parabola whose branches are directed upwards and which intersects the axis Ox at two points, the abscissas of which are x 1 and x 2 . This drawing corresponds to the variant when the coefficient a is positive (it is responsible for the upward direction of the branches of the parabola), and when the value is positive discriminant of a square 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 draw in red the parts of the parabola located above the abscissa axis, and in blue - located below the abscissa axis.


Now let's find out what gaps correspond to these parts. The following drawing will help determine them (in the future, we will mentally make such selections in the form of rectangles):


So on the abscissa axis, two intervals (−∞, x 1) and (x 2, +∞) were highlighted in red, on them the parabola is higher than the Ox axis, they constitute the solution of the quadratic inequality a x 2 +b x+c>0 , and the interval (x 1 , x 2) is highlighted in blue, on it the parabola is below the axis Ox , it is a 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 way, x x2;
• the solution to the quadratic inequality a x 2 +b x+c≥0 is (−∞, x 1 ]∪ 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


Here we see a parabola, the branches of which are directed upwards, and which touches the abscissa axis, that is, it has one common point with it, let's denote the abscissa of this point as x 0. The presented case corresponds to a>0 (the branches are directed upwards) and D=0 ( square trinomial has one root x 0 ). For example, we 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 for the point of contact, that is, at the intervals (−∞, x 0) , (x 0 , ∞) . For clarity, we select 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 other 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 tangent point),

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 upwards, and it has no common points with the abscissa axis. Here we have the conditions a>0 (the branches are directed upwards) 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 where it is below the Ox axis, there is no point of contact).


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 are three options for the location of the parabola with branches directed downwards, and not upwards, relative to the axis Ox. In principle, they may not be considered, since multiplying both parts of the inequality by −1 allows us to pass to an equivalent inequality with a positive coefficient at x 2 . However, it does not hurt to get an idea about these cases. The reasoning here is similar, so we write down only the main results.

Solution algorithm

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

On the coordinate plane a schematic drawing is performed, which depicts the Ox axis (it is not necessary to depict the Oy axis) and a sketch of a parabola corresponding to a quadratic function y \u003d a x 2 +b x + c. To construct a sketch of a parabola, it is enough to find out two points:

• First, by the value of the coefficient a, it is found out where its branches are directed (for a>0 - upwards, for a<0 – вниз).
• And secondly, by the value of the discriminant of the square trinomial a x 2 + b x + c, it turns out whether the parabola intersects the x-axis at two points (for D> 0), touches it at one point (for D=0), or has no common points with the Ox axis (for 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, on it at the second step of the algorithm

  • when solving the quadratic inequality a·x 2 +b·x+c>0, the intervals at which the parabola is located above the abscissa axis are determined;
  • when solving the inequality a x 2 +b x+c≥0, the intervals are determined at which the parabola is located above the x-axis 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, there are intervals where the parabola is below the Ox axis and the abscissas of the intersection points (or the abscissa of the tangency point) are added to them;

  they constitute the desired solution of the quadratic inequality, and if there are no such intervals and no points of contact, then the original quadratic inequality has no solutions.

It remains only to solve a few quadratic inequalities using this algorithm.

Examples with Solutions

Example.

Solve the inequality .

Decision.

We need to solve a quadratic inequality, we will use the algorithm from the previous paragraph. In the first step, we need to draw a sketch of the graph of the quadratic function . The coefficient at x 2 is 2, it is positive, therefore, the branches of the parabola are directed upwards. Let us also find out whether the parabola with the abscissa axis has common points, for this we calculate the discriminant of the square 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 axis Ox 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. According to the clarified data, we obtain the following drawing (it fits the first template from the first paragraph of the article):

We pass 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 the abscissas of the intersection points to them.

It can be seen from the drawing that the parabola is below the abscissa in the interval (−3, 1/3) and we add the abscissas of the intersection points to it, that is, the numbers −3 and 1/3. As a result, we arrive at the numerical segment [−3, 1/3] . This is the desired solution. It can be written as a double inequality −3≤x≤1/3 .

Answer:

[−3, 1/3] or −3≤x≤1/3 .

Example.

Find a solution to the quadratic inequality −x 2 +16 x−63<0 .

Decision.

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 downwards. Let's calculate the discriminant, or better, its fourth part: D"=8 2 −(−1)(−63)=64−63=1. Its value is positive, we 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 initial 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 signed quadratic inequality<, 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 square inequalities, when the discriminant of a square trinomial on its left side is equal to zero, you need to be careful with the inclusion or exclusion of the abscissa of the tangent point from the answer. It depends on the sign of the inequality: if the inequality is strict, then it is not a solution to the inequality, and if it is non-strict, then it is.

Example.

Does the quadratic inequality 10 x 2 −14 x+4.9≤0 have at least one solution?

Decision.

Let's plot the function y=10 x 2 −14 x+4.9 . Its branches are directed upwards, since the coefficient at x 2 is positive, and it touches the abscissa at the point with the abscissa 0.7, since D "=(−7) 2 −10 4.9=0, whence or 0.7 as a decimal.Schematically, it looks like this:

Since we are solving a quadratic inequality with the sign ≤, then its solution will be the intervals on which the parabola is below the Ox axis, as well as the abscissa of the tangent point. It can be seen from the drawing that there is not a single gap where the parabola would be below the axis Ox, therefore, its solution will be only the abscissa of the point of contact, that is, 0.7.

Answer:

this inequality has a unique solution 0.7 .

Example.

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

Decision.

We act according to the algorithm for solving quadratic inequalities and start by plotting. The branches of the parabola are directed downwards, since the coefficient at x 2 is negative, −1. Find the discriminant of the square trinomial –x 2 +8 x−16 , we have D'=4 2 −(−1)(−16)=16−16=0 and further x 0 =−4/(−1) , x 0 =4 . So, the parabola touches the Ox axis at the point with the abscissa 4 . Let's make a drawing:

We look at the sign of the original inequality, it is<. 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 tangent point - is not a solution, since at the tangent point the parabola is not lower than the Ox axis.

Answer:

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

Pay special attention to cases where the discriminant of the square trinomial on the left side of the square 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 .

Decision.

As usual, we start with a drawing. The coefficient a is 3, it is positive, therefore, the branches of the parabola are directed upwards. Calculate the discriminant: D=0 2 −4 3 1=−12 . Since the discriminant is negative, the parabola has no common points with the x-axis. The information obtained is sufficient for a schematic diagram:

We are solving a strict quadratic inequality with > sign. Its solution will be all the intervals where 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 you need to add the abscissa of the intersection points or the abscissa of the touch point to them. But the drawing clearly shows that there are no such gaps (since the parabola is everywhere below the abscissa axis), as well as there are no intersection points, just as there are no points of contact. Therefore, the original quadratic inequality has no solutions.

Answer:

there are no solutions or in another notation ∅.

Bibliography.

• Algebra: textbook for 8 cells. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 16th ed. - M. : Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
• Algebra: Grade 9: textbook. for general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 16th ed. - M. : Education, 2009. - 271 p. : ill. - ISBN 978-5-09-021134-5.
• Mordkovich A. G. Algebra. 8th grade. At 2 pm Part 1. A textbook for students of educational institutions / A. G. Mordkovich. - 11th ed., erased. - M.: Mnemozina, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.
• Mordkovich A. G. Algebra. Grade 9 At 2 pm Part 1. Textbook for students of educational institutions / A. G. Mordkovich, P. V. Semenov. - 13th ed., Sr. - M.: Mnemosyne, 2011. - 222 p.: ill. ISBN 978-5-346-01752-3.
• Mordkovich A. G. Algebra and beginning of mathematical analysis. Grade 11. At 2 pm Part 1. A textbook for students of educational institutions (profile level) / A. G. Mordkovich, P. V. Semenov. - 2nd ed., erased. - M.: Mnemosyne, 2008. - 287 p.: ill. ISBN 978-5-346-01027-2.

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 get 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 the linear equation on the coordinate plane. 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. 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 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)).

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

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

In order to visually demonstrate the graphical method, we will solve the following problem:

1. At the first stage, it is necessary to construct the area of ​​feasible solutions. For this example, it is most convenient to choose X2 for the abscissa, and X1 for the ordinate, and write the inequalities in the following form:

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

As can be seen from the illustration, the polyhedron ABCDE forms an area of ​​feasible solutions.

If the domain of admissible 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.

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 - the maximum of the function.

This approach is quite beneficial for a small number of vertices. But this procedure can be delayed 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 monotonous increase in the number a from -? to +? straight lines f=a are displaced along the normal vector. If, with such a displacement of the level line, there exists some 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 set ABCDE, then f(X) is the maximum on the set of feasible solutions. If for a>-? the line f=a intersects the set of admissible solutions, then min(f)= -?. If this happens when a>+?, then max(f)=+?.

First level

Solving equations, inequalities, systems using function graphs. visual guide (2019)

Many tasks that we are used to calculating purely algebraically can be solved much easier and faster, using function graphs will help us with this. You say "how so?" to draw something, and what to draw? Trust me, sometimes it's more convenient and easier. Shall we start? Let's start with equations!

Graphical solution of equations

Graphical solution of linear equations

As you already know, the graph of a linear equation is a straight line, hence the name of this type. Linear equations are quite easy to solve algebraically - we transfer all unknowns to one side of the equation, everything that we know - to the other, and voila! We have found the root. Now I'll show you how to do it graphic way.

So you have an equation:

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

And now we are building. What did you get?

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

Our answer is

That's the whole wisdom of the graphic solution. As you can easily check, the root of our equation is a number!

As I said above, this is the most common option, close to algebraic solution, but it can also be done in a different way. To consider an alternative solution, let's return to our equation:

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

Built? Look!

What is the solution this time? All right. The same is the coordinate of the point of intersection of the graphs:

And, again, our answer is .

As you can see, with linear equations everything is extremely simple. It's time to consider something more complicated... For example, graphic solution of quadratic equations.

Graphical solution of quadratic equations

So, now let's start solving the quadratic equation. Let's say you need to find the roots of this equation:

Of course, you can now start counting through the discriminant, or according to the Vieta theorem, but many nerves make mistakes when multiplying or squaring, especially if the example is with big numbers, and, as you know, you won’t have a calculator at the exam ... Therefore, let's try to relax a bit and draw while solving this equation.

Graphically find solutions given equation can different ways. Consider various options and you can choose which one you like best.

Method 1. Directly

We just build a parabola according to this equation:

To make it quick, I'll give you one little hint: it is convenient to start the construction by determining the vertex of the parabola. The following formulas will help determine the coordinates of the vertex of the parabola:

You say "Stop! The formula for is very similar to the formula for finding the discriminant "yes, it is, and it is a huge minus"direct" construction of a parabola to find its roots. However, let's count to the end, and then I'll show you how to make it much (much!) easier!

Did you count? What are the coordinates of the vertex of the parabola? Let's figure it out together:

Exactly the same answer? Well done! And now we already know the coordinates of the vertex, and to build a parabola, we need more ... points. What do you think, how many minimum points do we need? Correctly, .

You know that a parabola is symmetrical about its vertex, for example:

Accordingly, we need two more points along the left or right branch of the parabola, and in the future we will symmetrically reflect these points on the opposite side:

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

Now we have three points, and we can easily build our parabola by reflecting the last two points about its top:

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

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

Just? We have finished solving the equation with you in a complex graphical way, or there will be more!

Of course, you can check our answer algebraically - you can calculate the roots through the Vieta theorem or the Discriminant. What did you get? The same? You see! Now let's see a very simple graphical solution, I'm sure you'll like it very much!

Method 2. Split into several functions

Let's take everything, too, our equation: , but we write it in a slightly different way, namely:

Can we write it like this? We can, since the transformation is equivalent. Let's look further.

Let's build two functions separately:

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

Built? Compare with what I got:

Do you think that in this case are the roots of the equation? Correctly! Coordinates by, which are obtained by crossing two graphs and, that is:

Accordingly, the solution to this equation is:

What do you say? Agree, this solution method is much easier than the previous one and even easier than looking for roots through the discriminant! If so, try this method to solve the following equation:

What did you get? Let's compare our charts:

The graphs show that the answers are:

Did you manage? Well done! Now let's look at the equations a little bit more complicated, namely, the solution of mixed equations, that is, equations containing functions of different types.

Graphical solution of mixed equations

Now let's try to solve the following:

Of course, everything can be brought to common denominator, find the roots of the resulting equation, not forgetting to take into account the ODZ, but again, we will try to solve graphically, as we did in all previous cases.

This time let's plot the following 2 graphs:

1. - the graph is a hyperbola
2. - a graph is a straight line that you can easily build by estimating the values ​​and in your head without even resorting to a calculator.

Realized? Now start building.

Here's what happened to me:

Looking at this picture, what are the roots of our equation?

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

Try plugging our roots into the equation. Happened?

All right! Agree, graphically solving such equations is a pleasure!

Try to solve the equation yourself graphically:

I give you a hint: move part of the equation to right side so that both sides have the simplest functions to build. Got the hint? Take action!

Now let's see what you got:

Respectively:

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

Well, we are building:

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

Having solved this a large number of examples, I'm sure you realized how you can easily and quickly solve equations graphically. It's time to figure out how to decide in a similar way systems.

Graphic solution of systems

The graphical solution of systems is essentially no different from the graphical solution of equations. We will also build two graphs, and their intersection points will be the roots of this system. One graph is one equation, the second graph is another equation. Everything is extremely simple!

Let's start with the simplest - solving systems of linear equations.

Solving systems of linear equations

Let's say we have the following system:

To begin with, we will transform it in such a way that on the left there is everything that is connected with, and on the right - what is connected with. In other words, we write these equations as a function in the usual form for us:

And now we just build two straight lines. What is the solution in our case? Correctly! The point of their intersection! And here you need to be very, very careful! Think why? I'll give you a hint: we're dealing with a system: the system has both, and... Got the hint?

All right! When solving the system, we must look at both coordinates, and not only, as when solving equations! Another important point- write them down correctly and not confuse where we have the value, and where the value is! Recorded? Now let's compare everything in order:

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

Solving systems of nonlinear equations

But what if instead of one straight line, we will have quadratic equation? It's okay! You just build a parabola instead of a straight line! Do not believe? Try to solve the following system:

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

And now it’s all about the small thing - I built it quickly and here’s the solution for you! Building:

Are the graphics the same? Now mark the solutions of the system in the picture and correctly write down the revealed answers!

I've done everything? Compare with my notes:

All right? Well done! You already click on such tasks like nuts! And if so, let's give you a more complicated system:

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

I’ll give you a little hint, since the system looks very complicated! When building graphs, build them "more", and most importantly, do not be surprised at the number of intersection points.

So let's go! Exhaled? Now start building!

Well, how? Beautiful? How many intersection points did you get? I have three! Let's compare our graphs:

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

Now look at the system again:

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

Graphical solution of inequalities

Graphical solution of linear inequalities

After last example you have everything on your shoulder! Now exhale - compared to the previous sections, this one will be very, very easy!

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

To begin with, we will carry out the simplest transformations - we will open the brackets full squares and add like terms:

The inequality is not strict, therefore - is not included in the interval, and the solution will be all points that are to the right, since more, more, and so on:

Answer:

That's all! Easily? Let's solve a simple inequality with two variables:

Let's draw a function in the coordinate system.

Do you have such a chart? And now we carefully look at what we have in inequality? Smaller? So, we paint over everything that is to the left of our straight line. What if there were more? That's right, then they would paint over everything that is to the right of our straight line. Everything is simple.

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

Graphical solution of quadratic inequalities

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

But before we get straight to the point, let's recap some stuff about the square function.

What is the discriminant responsible for? That's right, for the position of the graph relative to the axis (if you don't remember this, then read the theory about quadratic functions for sure).

In any case, here's a little reminder for you:

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

I will tell you right away that there are two options for solving it.

Option 1

We write our parabola as a function:

Using the formulas, we determine the coordinates of the vertex of the parabola (in the same way as when solving quadratic equations):

Did you count? What did you get?

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

We begin to build one branch of the parabola:

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

Now back to our inequality.

We need it to be less than zero, respectively:

Since in our inequality there is a sign strictly less, we exclude the end points - we “poke out”.

Answer:

Long way, right? Now I will show you a simpler version of the graphical solution using the same inequality as an example:

Option 2

We return to our inequality and mark the intervals we need:

Agree, it's much faster.

Let's write down the answer now:

Consider another solution that simplifies and algebraic part, but the main thing is not to get confused.

Multiply the left and right sides by:

Try to solve the following quadratic inequality on your own in any way you like: .

Did you manage?

See how my chart turned out:

Answer: .

Graphical solution of mixed inequalities

Now let's move on to more complex inequalities!

How do you like this:

Horrible, right? Honestly, I have no idea how to solve this algebraically ... But, it is not necessary. Graphically, there is nothing complicated in this! The eyes are afraid, but the hands are doing!

The first thing we start with is by building two graphs:

I will not write a table for everyone - I'm sure you can do it perfectly on your own (of course, there are so many examples to solve!).

Painted? Now build two graphs.

Let's compare our drawings?

Do you have the same? Fine! Now let's place the intersection points and determine with a color which graph we should have, in theory, should be larger, that is. Look what happened in the end:

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

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

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

BRIEFLY ABOUT THE MAIN

Algorithm for solving equations using function graphs:

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

For more information about plotting function graphs, see the topic "".