Image on a plane of a set of points defined by an inequality with two variables Performed the work: Ksenia Surova
Purpose: 1). Form: - the concept that the solution to an inequality with two variables is a set of points in the plane. - the ability to depict on a plane a set of points defined by an inequality with two variables. - learn to use the algorithm. 2). Develop: the ability to analyze the proposed situation; graphic skills. 3). Cultivate mindfulness.
Progress: 1. Preparation for the perception of new material: y-3x+4=0. . x 2 +6x-8=0 x 2 +y-16=0 -What is the solution to an equation with two variables? -Is it possible to depict the solution of an equation with two variables on the coordinate plane? What will be the solution to such an equation?
2. Studying new material. Each line divides the coordinate plane into two parts (half-planes). 0 x y y-3x+4=0 -4 2nd half-plane 1st half-plane What is the condition for points lying on a line? f (x;y)=0 (equation of a line) What condition do you think is satisfied by points not lying on a line? Consider the first figure: Take points A(-4;-1), B(-2;4). C(0;2). Which half-plane do these points belong to? Let's substitute the coordinates of the points into the equation of the straight line and compare the resulting values with zero. A(-4;-1) -1-3(-4)+4= -1+12+4=15, 15 0, B(-2;4) 4-3(-2)+4=4 +6+4=14, 14 0, C(0;2) 2-3 0+4=6, 6 0. The value of our polynomial f (x;y) at points A, B, C takes on a value greater 0.
How to write this condition using a mathematical model? y-3x+4 0 . Which half-plane do points D(6;0), E(0;-6), F (3;-3) belong to? Let's compare the values of the polynomial y-3x+4 at these points with zero. D(6;0) 0-36+4=-18+4=-14, -14 0, E(0;-6) -6-30+4= -2, -2 0, F (3 ;-3) -3-3 3+4= -3-9+4, -8 0. What condition do the points of the lower half-plane satisfy? y-3x+4 0 Conclusion: Points not lying on the line satisfy the inequality. f (x;y) 0 or f (x;y) 0.
3. Fill out the table. Which of the conditions are satisfied by the points of the coordinate plane: A(0;4), B(0;-4), O(0;0), C(-2;-2), D(5;0), E(4; 8), F (0;-6), K(4;1), M(-2;1), N (8;-2) F (x;y)=0 F (x;y) 0 F (x;y) 0
Define the set of points of the plane in the pictures as an inequality: x y 0 4 2 y=x 2 -6x+8 y x 0 4 -4 4 -4 x 2 +y 2 =16 Let's summarize: How to set the set of points of the plane with an inequality? I created an algorithm for my actions. 1. We build a graph of the function f (x;y) = 0 2. We take a control point. 3. Check the inequality f (x;y) 0 or f (x;y) 0
6 3 0 y x y+2x-6=0 6 3 0 y x y+2x-6=0 4. Set the inequality of points on the coordinate plane. What is the difference between these two cases? Conclusion: In the first case, the points of the line are included in the specified set, therefore these points define a set that satisfies the inequality f (x;y) 0, in the second case, the points of the line are not part of the set of the specified half-plane, therefore our set is defined by the inequality f (x; y) 0. And so, if the inequality sign is not strict, then we depict the graph of the equation as a solid line; if the inequality sign is strict, then the graph of the equation is depicted with a dotted line.
Independent work. Option 1 Option 2 Display on the plane a set of points specified by the inequality: A) y=2x-4 0 (2b) y-x -5 0 C) x 2 +4x+y 2 0 (3b) x 2 =y 2 -4у≤0 Define a set of points of the coordinate plane by the inequality: (2b) Graphically depict the solution to inequality (3b) How do you think this set can be defined: (The same drawing without shading is shown on the board.) What lines are shown? (straight circle) A straight line divides a plane into two half-planes. Which half-plane does the shaded part belong to and what condition does it satisfy? y+x-4≥0 The circle divides the plane into two parts: inside the circle and outside it. We are interested in the internal part. What condition does it satisfy? (x+y) 2 + (y-2) 2 -9
That is, this set is the result of the intersection of two sets. That is, by solving a system of inequalities: (x-2) 2 + (y-2) 2 -9 0 And so you and I have defined a certain set with a system of inequalities. Let's summarize: Let's create an algorithm for constructing a set of points on the plane, specified by a system of inequalities: We build a graph of the equation f 1 (x;y)=0 and f 2 (x;y)=0 We depict a set of points that satisfies the first inequality. We depict a set of points satisfying the second inequality. The result is the intersection of sets.
Thank you for your attention!!!
5 -4 -3 -2 -1 0 1 2 3 4 5 6
What is the name of the straight line shown in the figure?
Name the coordinates of the points
A, B, C, D, O.
A(4), B(-4), C(5.5), D(-1.5), O(0)
Оx – abscissa axis
Oy - ordinate axis
Point 0 – origin
3 – abscissa of point M
4 - ordinate of point M
A plane with a coordinate system indicated on it is called a coordinate plane.
Numbers used to indicate where an object is located, call it coordinates.
( from the Latin words co - “together”
ordinatus - “definite”)
The rectangular coordinate system, consisting of two mutually perpendicular axes with a common origin, was invented in the 16th century. The famous French mathematician Rene Descartes.
The Cartesian coordinate system made it possible to combine the numerical and geometric lines of mathematics.
Name the coordinates of the points
A, B, C, D, E, F
- A(3;1)
- B(2;-2)
- C (-2;4)
- D (-4;-2)
- E(0;2)
- F(-4;0)
This you need to know:
- If a point lies on the ordinate axis, its abscissa is zero.
2. If a point lies on the x-axis, its ordinate is zero.
Draw coordinate axes in your notebook, taking a unit segment of 1 cm.
Plot the points:
A (4;1), B (-1;4), C (3;-2),
D(-3;-1); K (0;3), N (-2;1)
F (-2.5;-4.5), S (0.5;-2.5)
Let's check ourselves
Write down the coordinates of points B, A, R, S, I, K
- B(3;1)
- A(2:-5)
- R(0;-9)
- S (-3;-5)
- I (-2;3)
- K(-1;9)
Construct a figure by sequentially connecting points with coordinates with segments And
(3; 7), (1; 5), (2; 4), (4; 3), (5; 2), (6; 2),
(8; 4), (8;-1), (6; 0), (0;-3), (2;-6), (-2;-3), (-4;-2), (-5;-1), (-6; 1), (-6; 2), (-3; 5), (3; 7) Separately: (-3; 3) Separately: (-6; 1), (-4; 1) Separately: (-3; 5), (-2; 2), (-2; 0), (-4;-2) (take 1 cell of a notebook as a unit segment)
3. Draw a set of points x≤2 on the coordinate axis. Draw a set of points 2 ≤ y ≤5 on the coordinate plane." width="640" - Draw a set of points y on the coordinate axis
- Draw a set of points x 3 on the coordinate axis.
- Draw a set of points x≤2 on the coordinate axis.
- Draw a set of points 2 ≤ y ≤5 on the coordinate plane.
y 
3" width="640"
Let it be given equation with two variables F(x; y). You have already become familiar with ways to solve such equations analytically. Many solutions of such equations can be represented in graph form.
The graph of the equation F(x; y) is the set of points in the coordinate plane xOy whose coordinates satisfy the equation.
To graph equations in two variables, first express the y variable in the equation in terms of the x variable.
Surely you already know how to build various graphs of equations with two variables: ax + b = c – straight line, yx = k – hyperbola, (x – a) 2 + (y – b) 2 = R 2 – circle whose radius is equal to R, and the center is at point O(a; b).
Example 1.
Graph the equation x 2 – 9y 2 = 0.
Solution.
Let's factorize the left side of the equation.
(x – 3y)(x+ 3y) = 0, that is, y = x/3 or y = -x/3.
Answer: Figure 1.
A special place is occupied by defining figures on a plane by equations containing the sign of the absolute value, which we will dwell on in detail. Let's consider the stages of constructing graphs of equations of the form |y| = f(x) and |y| = |f(x)|.
The first equation is equivalent to the system
(f(x) ≥ 0,
(y = f(x) or y = -f(x).
That is, its graph consists of graphs of two functions: y = f(x) and y = -f(x), where f(x) ≥ 0.
To plot the second equation, plot two functions: y = f(x) and y = -f(x).
Example 2.
Graph the equation |y| = 2 + x.
Solution.
The given equation is equivalent to the system
(x + 2 ≥ 0,
(y = x + 2 or y = -x – 2.
We build many points.
Answer: Figure 2.
Example 3.
Plot the equation |y – x| = 1.
Solution.
If y ≥ x, then y = x + 1, if y ≤ x, then y = x – 1.
Answer: Figure 3.
When constructing graphs of equations containing a variable under the modulus sign, it is convenient and rational to use area method, based on dividing the coordinate plane into parts in which each submodular expression retains its sign.
Example 4.
Graph the equation x + |x| + y + |y| = 2.
Solution.
In this example, the sign of each submodular expression depends on the coordinate quadrant.
1) In the first coordinate quarter x ≥ 0 and y ≥ 0. After expanding the module, the given equation will look like:
2x + 2y = 2, and after simplification x + y = 1.
2) In the second quarter, where x< 0, а y ≥ 0, уравнение будет иметь вид: 0 + 2y = 2 или y = 1.
3) In the third quarter x< 0, y < 0 будем иметь: x – x + y – y = 2. Перепишем этот результат в виде уравнения 0 · x + 0 · y = 2.
4) In the fourth quarter, when x ≥ 0, and y< 0 получим, что x = 1.
We will plot this equation by quarters.
Answer: Figure 4.
Example 5.
Draw a set of points whose coordinates satisfy the equality |x – 1| + |y – 1| = 1.
Solution.
The zeros of the submodular expressions x = 1 and y = 1 divide the coordinate plane into four regions. Let's break down the modules by region. Let's arrange this in the form of a table.
| Region |
Submodular expression sign |
The resulting equation after expanding the module |
| I | x ≥ 1 and y ≥ 1 | x + y = 3 |
| II | x< 1 и y ≥ 1 | -x + y = 1 |
| III | x< 1 и y < 1 | x + y = 1 |
| IV | x ≥ 1 and y< 1 | x – y = 1 |
Answer: Figure 5.
On the coordinate plane, figures can be specified and inequalities.
Inequality graph with two variables is the set of all points of the coordinate plane whose coordinates are solutions to this inequality.
Let's consider algorithm for constructing a model for solving inequalities with two variables:
- Write down the equation corresponding to the inequality.
- Graph the equation from step 1.
- Select an arbitrary point in one of the half-planes. Check whether the coordinates of the selected point satisfy this inequality.
- Draw graphically the set of all solutions to the inequality.
Let us first consider the inequality ax + bx + c > 0. The equation ax + bx + c = 0 defines a straight line dividing the plane into two half-planes. In each of them, the function f(x) = ax + bx + c retains its sign. To determine this sign, it is enough to take any point belonging to the half-plane and calculate the value of the function at this point. If the sign of the function coincides with the sign of the inequality, then this half-plane will be the solution to the inequality.
Let's look at examples of graphical solutions to the most common inequalities with two variables.
1) ax + bx + c ≥ 0. Figure 6.
2)
|x| ≤ a, a > 0. Figure 7.
3) x 2 + y 2 ≤ a, a > 0. Figure 8.
4) y ≥ x 2 . Figure 9.
5)
xy ≤ 1. Figure 10. 
If you have questions or want to practice drawing on a plane model the sets of all solutions to inequalities in two variables using mathematical modeling, you can conduct free 25-minute lesson with an online tutor after . To further work with a teacher, you will have the opportunity to choose the one that suits you
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