At the lesson, students demonstrated the knowledge and skills of the program:
- recognize the types of functions, build their graphs;
– practiced the skills of constructing a quadratic function;
– worked out graphical methods for solving quadratic equations using the full square selection method.
I wanted to pay special attention to solving problems with a parameter, since the USE in mathematics offers a lot of tasks of this type.
The opportunity to apply this type of work in the classroom was given to me by the students themselves, as they have a sufficient knowledge base that can be deepened and expanded.
Pre-prepared templates by students allowed to save lesson time. During the lesson, I managed to implement the tasks at the beginning of the lesson and get the expected result.
The use of a physical education minute helped to avoid overwork of students, to maintain a productive motivation for obtaining knowledge.
In general, I am satisfied with the result of the lesson, but I think that there are still reserve opportunities: modern innovative technological tools, which, unfortunately, we do not have the opportunity to use.
Lesson type: consolidation of the studied material.
Lesson Objectives:
- General education and didactic:
- develop a variety of ways of mental activity of students;
- to form the ability to independently solve problems;
- educate the mathematical culture of students;
- develop students' intuition and the ability to use the knowledge gained.
- learning goals:
- summarize previously studied information on the topic "Graphical solution of quadratic equations";
- repeat plotting quadratic functions;
- to form the skills of using algorithms for solving quadratic equations by a graphical method.
- Educational:
- instilling interest in educational activities, in the subject of mathematics;
- formation of tolerance (tolerance), the ability to work in a team.
DURING THE CLASSES
I. Organizational moment
- Today in the lesson we will generalize and consolidate the graphical solution of quadratic equations in various ways.
In the future, we will need these skills in high school in mathematics lessons when solving trigonometric and logarithmic equations, finding the area of a curvilinear trapezoid, as well as in physics lessons.
II. Checking homework
Let's analyze on the board No. 23.5 (g).
Solve this equation using a parabola and a straight line.
Solution:
x 2 + x - 6 = 0
Let's transform the equation: x 2 \u003d 6 - x
Let's introduce functions:
y \u003d x 2; quadratic function y \u003d 6 - x linear,
chart yavl. parabola, graph yavl. straight,
We build graphs of functions in one coordinate system (according to a template)
We got two points of intersection.
The solution to the quadratic equation is the abscissas of these points x 1 = - 3, x 2 = 2.
Answer: - 3; 2.
III. Frontal survey
- What is the graph of a quadratic function?
- Can you tell me the algorithm for plotting a graph of a quadratic function?
- What is a quadratic equation?
- Give examples of quadratic equations?
- Write on the board your example of a quadratic equation. What are the coefficients?
- What does it mean to solve an equation?
- How many ways do you know of graphical solution of quadratic equations?
- What is the graphical methods for solving quadratic equations:
IV. Fixing the material
On the board, students decide in the first, second, third ways.
Class decides fourth
- x 2 + 6x - 5 \u003d 0
I will transform the quadratic equation, highlighting the full square of the binomial:
- x 2 + 6x - 5 \u003d - (x 2 - 6x + 5) \u003d - (x 2 - 6x + 32 - 9 + 5) \u003d - ((x - 3) 2 - 4) \u003d - (x - 3) 2+4
We got a quadratic equation:
- (x - 3) 2 + 4 \u003d 0
Let's introduce a function:
y \u003d - (x 2 - 3) 2 + 4
Quadratic function of the form y \u003d a (x + L) 2 + m
Graph yavl. parabola, branches directed downwards, shift of the main parabola along the Ox axis to the right by 3 units, upwards by 4 units along the Oy axis, top (3; 4).
We build according to the template.
Found the points of intersection of the parabola with the x-axis. Abscissas of these points yavl. solution of this equation. x=1, x=5.
Let's see other graphic solutions at the board. Comment on your way of solving quadratic equations.
1 student
Solution:
- x 2 + 6x - 5 \u003d 0
We introduce the function y \u003d - x + 6x - 5, a quadratic function, the graph is a parabola, the branches are directed downwards, the top
x 0 \u003d - in / 2a
x 0 \u003d - 6 / - 2 \u003d 3
y 0 \u003d - 3 2 + 18 \u003d 9; dot (3; 9)
axis of symmetry x = 3
We build according to the template
We got points of intersection with the Ox axis, the abscissas of these points are the solution of a quadratic equation. Two roots x 1 = 1, x 2 = 5
2 student
Solution:
- x 2 + 6x - 5 \u003d 0
Let's transform: - x 2 + 6x \u003d 5
We introduce the functions: y1 \u003d - x 2 + 6x, y2 \u003d 5, linear function, quadratic function, graph graph yavl. line y || Oh yavl. parabola, branches directed downwards, vertex x 0 \u003d - in / 2a
x 0 \u003d - 6 / - 2 \u003d 3
y 0 \u003d - 3 2 + 18 \u003d 9;
(3; 9).
axis of symmetry x = 3
We build according to the template
Got intersection points
parabolas and a straight line, their abscissas are the solution of a quadratic equation. Two roots x 1 = 1, x 2 = 5
So, the same equation can be solved in different ways, and the answer should be the same.
V. Physical education
VI. Solving a problem with a parameter
At what values R equation x 2 + 6x + 8 = p:
- Has no roots?
- Has one root?
Does it have two roots?
How is this equation different from the previous one?
That's right, letter!
We will refer to this letter as parameter, R.
As long as she doesn't tell you anything. But we will continue to solve various problems with a parameter.
Today we will solve a quadratic equation with a parameter using a graphical method using the third method using a parabola and a straight line parallel to the x-axis.
The student helps the teacher to solve at the blackboard.
Where do we start to decide?
Let's set the functions:
y 1 \u003d x 2 + 6x + 8 y 2 \u003d p linear function,
quadratic function, the graph is a straight line
chart yavl. parabola,
branches pointing down
x 0 \u003d - in / 2a,
x 0 = - 6/2 = - 3
y 0 \u003d (- 3) 2 + 6 (- 3) + 8 \u003d - 1
(– 3; – 1)
The axis of symmetry x = 3, I will not build a table, but I will take the template y = x 2 and attach it to the top of the parabola.
The parabola is built! Now we need to draw a line y = p.
Where should a line be drawn? R to get two roots?
Where should a line be drawn? R to get one root?
Where should a line be drawn? R without roots?
– So, how many roots can our equation have?
Did you like the task? Thanks for the help! Grade 5.
VII. Independent work by options (5 min.)
y \u003d x 2 - 5x + 6 y \u003d - x 2 + x - 6
Solve a quadratic equation in a graphical way, choosing a convenient way for you. If someone completes the task earlier, check your solution in another way. This will be subject to additional marks.
VIII. Lesson summary
- What did you learn in today's lesson?
- Today in the lesson we solved quadratic equations using a graphical method, using various methods of solving, and considered a graphical method for solving a quadratic equation with a parameter!
- Let's move on to homework.
IX. Homework
1. Home test on page 147, from Mordkovich's problem book for options I and II.
2. On the circle, on Wednesday, we will solve the V-th method, (hyperbola and straight line).
X. Literature:
1. A.G. Mordkovich. Algebra-8. Part 1. Textbook for students of educational institutions. Moscow: Mnemosyne, 2008
2. A.G. Mordkovich, L.A. Aleksandrova, T.N. Mishustin, E.E. Tulchinskaya. Algebra - 8. Part 2. Task book for students of educational institutions. Moscow: Mnemosyne, 2008
3. A.G. Mordkovich. Algebra 7-9. Methodological guide for a teacher. M .: Mnemosyne, 2004
4. L.A. Alexandrova. Algebra-8. Independent work for students of educational institutions./Ed. A.G. Mordkovich. Moscow: Mnemosyne, 2009
If you want to learn how to swim, then boldly enter the water, and if you want to learn how to solve problems, solve them.
D. Poya
The equation is an equality containing one or more unknowns, provided that the task is to find those values of the unknowns for which it is true.
solve the equation- this means finding all the values of the unknowns for which it turns into the correct numerical equality, or establishing that there are no such values.
Valid Range equations (O.D.Z.) is the set of all those values of the variable (variables) for which all expressions included in the equation are defined.
Many equations presented in the exam are solved by standard methods. But no one forbids using something unusual, even in the simplest cases.
So, for example, consider the equation 3 – x 2 \u003d 6 / (2 - x).
Let's solve it graphically, and then find the arithmetic mean of its roots increased by six times.
To do this, consider the functions y=3 – x2 and y = 6 / (2 - x) and plot their graphs.
The function y \u003d 3 - x 2 is quadratic.
Let's rewrite this function in the form y = -x 2 + 3. Its graph is a parabola, the branches of which are directed downwards (because a = -1< 0).
The top of the parabola will be shifted along the y-axis by 3 units up. So the vertex coordinate is (0; 3).
To find the coordinates of the points of intersection of the parabola with the abscissa axis, we equate this function to zero and solve the resulting equation:
Thus, at points with coordinates (√3; 0) and (-√3; 0) the parabola intersects the x-axis (Fig. 1).
The graph of the function y = 6 / (2 - x) is a hyperbola.
This function can be graphed using the following transformations:
1) y = 6 / x - inverse proportionality. The function graph is a hyperbola. It can be built by points, for this we will compile a table of values for x and y:
x | -6 | -3 | -2 | -1 | 1 | 2 | 3 | 6 |
y | -1 | -2 | -3 | -6 | 6 | 3 | 2 | 1 |
2) y = 6 / (-x) - the graph of the function obtained in paragraph 1 is displayed symmetrically with respect to the y-axis (Fig. 3).
3) y = 6 / (-x + 2) - we shift the graph obtained in paragraph 2 along the x-axis by two units to the right (Fig. 4). 
Now let's draw the graphs of the functions y = 3 –
x 2 and y = 6 / (2 - x) in the same coordinate system (Fig. 5). 
The figure shows that the graphs intersect at three points.
It is important to understand that the graphical method of solving does not allow you to find the exact value of the root. So the numbers -1; 0; 3 (the abscissas of the intersection points of the graphs of functions) are so far only the supposed roots of the equation.
By means of check we will be convinced that numbers -1; 0; 3 - really the roots of the original equation:
Root -1:
3 – 1 = 6 / (2 – (-1));
3 – 0 = 6 / (2 – 0);
3 – 9 = 6 / (2 – 3);
Their arithmetic mean:
(-1 + 0 + 3) / 3 = 2/3.
Let's increase it six times: 6 2/3 = 4.
This equation, of course, can be solved in a more familiar way. – algebraic.
So, find the arithmetic mean of the roots of equation 3 increased by six times – x 2 \u003d 6 / (2 - x).
Let's start the solution of the equation with the search for O.D.Z. The denominator of a fraction should not be zero, therefore:
To solve the equation, we use the basic property of proportion, this will get rid of the fraction.
(3 – x 2)(2 - x) = 6.
Let's open the brackets and give like terms:
6-3x – 2x2 + x3 = 6;
x 3 – 2x 2 - 3x = 0.
Let's take the common factor out of brackets:
x(x2 – 2x - 3) = 0.
We use the fact that the product is equal to zero only when at least one of the factors is equal to zero, so we have:
x = 0 or x2 – 2x - 3 = 0.
Let's solve the second equation.
x2 – 2x - 3 = 0. It's square, so let's use the discriminant.
D=4 – 4 (-3) = 16;
x 1 \u003d (2 + 4) / 2 \u003d 3;
x 2 = (2 – 4) / 2 = -1.
All three obtained roots satisfy O.D.Z.
Therefore, we find their arithmetic mean and increase it six times:
6 (-1 + 3 + 0) / 3 = 4.
In fact, the graphical method of solving equations is rarely used. This is due to the fact that the graphical representation of functions allows solving equations only approximately. Basically, this method is used in those tasks where it is important to search not for the roots of the equation themselves - their numerical values, but only their number.
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One way to solve equations is a graphical method. It is based on plotting functions and determining their intersection points. Consider a graphical way to solve the quadratic equation a*x^2+b*x+c=0.
First way to solve
Let's transform the equation a*x^2+b*x+c=0 to the form a*x^2 =-b*x-c. We build graphs of two functions y= a*x^2 (parabola) and y=-b*x-c (straight line). Looking for intersection points. The abscissas of the intersection points will be the solution of the equation.
Let's show with an example: solve the equation x^2-2*x-3=0.
Let's transform it into x^2 =2*x+3. We build graphs of functions y= x^2 and y=2*x+3 in one coordinate system.
Graphs intersect at two points. Their abscissas will be the roots of our equation.
Formula solution
To be convincing, we check this solution analytically. We solve the quadratic equation by the formula:
D = 4-4*1*(-3) = 16.
X1= (2+4)/2*1 = 3.
X2 = (2-4)/2*1 = -1.
Means, solutions match.
The graphical method of solving equations also has its drawback, with the help of it it is not always possible to obtain an exact solution of the equation. Let's try to solve the equation x^2=3+x.
Let's build a parabola y=x^2 and a straight line y=3+x in the same coordinate system.
Again got a similar pattern. A line and a parabola intersect at two points. But we cannot say the exact values of the abscissas of these points, only approximate ones: x≈-1.3 x≈2.3.
If we are satisfied with the answers of such accuracy, then we can use this method, but this rarely happens. Usually exact solutions are needed. Therefore, the graphical method is rarely used, and mainly to check existing solutions.
Need help with your studies?
Previous topic:
In this video lesson, the topic “Function y \u003d x 2. Graphical solution of equations. During this lesson, students will be able to get acquainted with a new way of solving equations - graphical, which is based on knowledge of the properties of function graphs. The teacher will show you how to graphically solve the function y=x 2 .
Topic:Function
Lesson:Function. Graphical solution of equations
The graphical solution of equations is based on the knowledge of function graphs and their properties. We list the functions whose graphs we know:
1), the graph is a straight line parallel to the x-axis, passing through a point on the y-axis. Consider an example: y=1:
For different values, we get a family of straight lines parallel to the x-axis.
2) Direct proportionality function the graph of this function is a straight line passing through the origin. Consider an example:
We have already built these graphs in previous lessons, recall that to build each line, you need to select a point that satisfies it, and take the origin as the second point.
Recall the role of the coefficient k: as the function increases, the angle between the straight line and the positive direction of the x-axis is acute; when the function decreases, the angle between the straight line and the positive direction of the x-axis is obtuse. In addition, there is the following relationship between two parameters k of the same sign: for positive k, the larger it is, the faster the function increases, and for negative, the function decreases faster for large values of k modulo.
3) Linear function. When - we get the point of intersection with the y-axis and all lines of this kind pass through the point (0; m). In addition, as the function increases, the angle between the line and the positive direction of the x-axis is acute; when the function decreases, the angle between the straight line and the positive direction of the x-axis is obtuse. And of course, the value of k affects the rate of change of the value of the function.
four). The graph of this function is a parabola.
Consider examples.
Example 1 - graphically solve the equation:
We do not know functions of this type, so we need to transform the given equation in order to work with known functions:
We got familiar functions in both parts of the equation:
Let's build graphs of functions:
Graphs have two intersection points: (-1; 1); (2; 4)
Let's check if the solution is found correctly, substitute the coordinates into the equation:
The first point is found correctly.
, 
, , , , ,
The second point is also found correctly.
So, the solutions of the equation are and
We act similarly to the previous example: we transform the given equation to the functions known to us, plot their graphs, find the intersection currents, and from here we indicate the solutions.
We get two functions:
Let's build graphs:
These graphs do not have intersection points, which means that the given equation has no solutions
Conclusion: in this lesson, we reviewed the functions known to us and their graphs, remembered their properties and considered a graphical way to solve equations.
1. Dorofeev G.V., Suvorova S.B., Bunimovich E.A. et al. Algebra 7. 6th edition. M.: Enlightenment. 2010
2. Merzlyak A.G., Polonsky V.B., Yakir M.S. Algebra 7. M.: VENTANA-GRAF
3. Kolyagin Yu.M., Tkacheva M.V., Fedorova N.E. and others. Algebra 7 .M .: Education. 2006
Task 1: Makarychev Yu.N., Mindyuk N.G., Neshkov K.I. et al. Algebra 7, no. 494, p. 110;
Task 2: Makarychev Yu.N., Mindyuk N.G., Neshkov K.I. and others. Algebra 7, No. 495, item 110;
Task 3: Makarychev Yu.N., Mindyuk N.G., Neshkov K.I. et al. Algebra 7, no. 496, p. 110;
In linear programming, a graphical method is used to determine convex sets (solution polyhedron). If the main linear programming problem has an optimal plan, then the objective function takes a value at one of the vertices of the decision polyhedron (see figure).
Service assignment. Using this service, you can solve the problem of linear programming using the geometric method online, as well as get a solution to the dual problem (estimate the optimal use of resources). Additionally, a solution template is created in Excel.
Instruction. Select the number of rows (number of limits).
If the number of variables is more than two, it is necessary to bring the system to SZLP (see example and example No. 2). If the constraint is double, for example, 1 ≤ x 1 ≤ 4 , then it is split into two: x 1 ≥ 1 , x 1 ≤ 4 (that is, the number of rows increases by 1).You can also build a feasible solution area (DDR) using this service.
The following are also used with this calculator:
Simplex method for solving LLP
Solution of the transport problem
Matrix game solution
Using the service online, you can determine the price of a matrix game (lower and upper bounds), check for a saddle point, find a solution to a mixed strategy using the following methods: minimax, simplex method, graphical (geometric) method, Brown's method.
Extremum of a function of two variables
Limit Calculation
Solving a linear programming problem by a graphical method includes the following steps:
- Lines are built on the plane X 1 0X 2.
- Half planes are defined.
- Define a decision polygon;
- Build a vector N(c 1 ,c 2), which indicates the direction of the objective function;
- Move the direct objective function c 1 x 2 + c 2 x 2= 0 in the direction of the vector N to the extreme point of the solution polygon.
- Calculate the coordinates of the point and the value of the objective function at this point.
Example. The company manufactures two types of products - P1 and P2. For the production of products, two types of raw materials are used - C1 and C2. The wholesale price of a unit of production is equal to: CU 5 for P1 and 4 c.u. for P2. The consumption of raw materials per unit of production of type P1 and type P2 is given in the table.
Table - Consumption of raw materials for production
It is required to determine:
How many products of each type should the company produce in order to maximize the income from the sale of products?
- Formulate a mathematical model of a linear programming problem.
- Solve a linear programming problem graphically (for two variables).
Let us formulate a mathematical model of a linear programming problem.
x 1 - production P1, units.
x 2 - production of P2 products, units.
x 1 , x 2 ≥ 0
Resource limits
6x1 + 4x2 ≤ 24
x1 + 2x2 ≤ 6
Demand limits
x 1 +1 ≥ x 2
x2 ≤ 2
objective function
5x1 + 4x2 → max
Then we get the following LLP:
6x1 + 4x2 ≤ 24
x1 + 2x2 ≤ 6
x 2 - x 1 ≤ 1
x2 ≤ 2
x 1 , x 2 ≥ 0
5x1 + 4x2 → max
