What is a graphical solution of equations. Graphical solution of mixed inequalities

Graphical solution of equations

Let's summarize our knowledge about charts functions. We have learned how to plot the following functions:

y \u003d b (straight line, parallel to the x axis);

y = kx (straight line passing through the origin);

y - kx + m (straight line);

y \u003d x 2 (parabola).

Knowledge of these graphs will allow us, if necessary, to replace the analytical model geometric (graphical), for example, instead of the model y \u003d x 2 (which is an equality with two variables x and y), consider a parabola in coordinate plane. In particular, it is sometimes useful for solving equations. Let's discuss how this is done with a few examples.

A. V. Pogorelov, Geometry for grades 7-11, Textbook for educational institutions

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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;
- practiced graphic ways solving quadratic equations using the selection method full square.

I wanted to give Special attention 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 learning activities, to the subject of mathematics;
- formation of tolerance (tolerance), the ability to work in a team.

DURING THE CLASSES

I. Organizing time

- Today in the lesson we will generalize and consolidate the graphical solution of quadratic equations different 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.

Decision:

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.

Decision quadratic equation are the abscissas of these points x 1 \u003d - 3, x 2 \u003d 2.

Answer: - 3; 2.

III. Frontal survey

What is a graph 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

Decision:

- 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

Decision:

- 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 the parameter graphic method using the third method using a parabola and a straight parallel 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. Homemade 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

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 of a 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.

AT ancient Babylon could solve some kinds of quadratic equations.

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= ax 2 + 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 2 n , y= x - 2 n , 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.

Function inverse proportionality y= k / x, where k¹ 0. The graph of this function is called a hyperbola.

Function ( x – a ) 2 + (y – b ) 2 = r 2 , 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 = ax 2 + bx + c where a, b , with- some numbers and a¹ 0. The graph of this function is a parabola.

The equation y 2 ( a – x ) = x 2 ( a + x ) . The graph of this equation will be a curve called a strophoid.

The equation ( x 2 + y 2 ) 2 = a ( x 2 – y 2 ) . The graph of this equation is called the Bernoulli lemniscate.

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

Curve (x 2 y 2 - 2 a x) 2 \u003d 4 a 2 (x 2 + y 2). 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 coordinate method, there was no 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 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:

We find the coordinates of the vertex of the parabola A (x 0; y 0): x 0 =- b /2 a ;

Y 0 \u003d ax about 2 + in 0 + c;

We find the axis of symmetry of the parabola (straight line x \u003d x 0);

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 = x 2 – 2 x – 3 . Abscissas of points of intersection with the axis x and are the roots of the quadratic equation x 2 – 2 x – 3 = 0.

There are five ways to graphically solve this equation.

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

3. Let's break the equation into two functions: y = x 2 –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 x 2 – 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 x 2 – 2 x – 3 = 0 on the 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 equations n

Example 1 solve the equation x 5 = 3 – 2 x .

y = x 5 , 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= ax 2 + 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.

On the next year I would also like 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.

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 the areas of land and earthworks of a military nature, as well as 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.

But the general rule for 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 linear function is 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.

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.

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);

PAGE_BREAK--

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 the 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.

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.

You have already met with quadratic equations in the 7th grade algebra course. Recall that a quadratic equation is an equation of the form ax 2 + bx + c \u003d 0, where a, b, c are any numbers (coefficients), and a. Using our knowledge of certain functions and their graphs, we are now able, without waiting for a systematic study of the topic "Quadricular Equations", to solve some quadratic equations, and in various ways; we will consider these methods using the example of one quadratic equation.

Example. Solve the equation x 2 - 2x - 3 = 0.
Decision.
I way . Let's build a graph of the function y \u003d x 2 - 2x - 3, using the algorithm from § 13:

1) We have: a \u003d 1, b \u003d -2, x 0 \u003d \u003d 1, y 0 \u003d f (1) = 1 2 - 2 - 3 \u003d -4. This means that the point (1; -4) is the vertex of the parabola, and the straight line x = 1 is the axis of the parabola.

2) Take two points on the x-axis that are symmetrical about the axis of the parabola, for example, the points x \u003d -1 and x \u003d 3.

We have f(-1) = f(3) = 0. Let's construct points (-1; 0) and (3; 0) on the coordinate plane.

3) Through the points (-1; 0), (1; -4), (3; 0) we draw a parabola (Fig. 68).

The roots of the equation x 2 - 2x - 3 \u003d 0 are the abscissas of the points of intersection of the parabola with the x-axis; so the roots of the equation are: x 1 \u003d - 1, x 2 - 3.

II way. We transform the equation to the form x 2 \u003d 2x + 3. Let us construct graphs of the functions y - x 2 and y \u003d 2x + 3 in one coordinate system (Fig. 69). They intersect at two points A(-1; 1) and B(3; 9). The roots of the equation are the abscissas of points A and B, which means that x 1 \u003d - 1, x 2 - 3.

III method . Let's transform the equation to the form x 2 - 3 = 2x. Let us construct graphs of the functions y \u003d x 2 - 3 and y \u003d 2x in one coordinate system (Fig. 70). They intersect at two points A (-1; - 2) and B (3; 6). The roots of the equation are the abscissas of points A and B, therefore x 1 \u003d - 1, x 2 \u003d 3.

IV way. Let's transform the equation to the form x 2 -2x 4-1-4 \u003d 0
and beyond
x 2 - 2x + 1 = 4, i.e. (x - IJ = 4.
Let's build a parabola y \u003d (x - 1) 2 and a straight line y \u003d 4 in one coordinate system (Fig. 71). They intersect at two points A(-1; 4) and B(3; 4). The roots of the equation are the abscissas of points A and B, therefore x 1 \u003d -1, x 2 \u003d 3.

V way. Dividing term by term both sides of the equation by x, we get

We construct a hyperbola and a straight line y \u003d x - 2 in the same coordinate system (Fig. 72).

They intersect at two points A (-1; -3) and B(3; 1). The roots of the equation are the abscissas of points A and B, therefore, x 1 \u003d - 1, x 2 \u003d 3.

So, we solved the quadratic equation x 2 - 2x - 3 \u003d 0 graphically in five ways. Let's analyze the essence of these methods.

I way. Build a graph of the function at the point of its intersection with the x-axis.

II way. They transform the equation to the form ax 2 \u003d -bx - c, build a parabola y \u003d ax 2 and a straight line y \u003d -bx - c, find their intersection points (the roots of the equation are the abscissas of the intersection points, if, of course, there are any).

III way. Transform the equation to the form ax 2 + c \u003d - bx, build a parabola y - ax 2 + c and a straight line y \u003d -bx (it passes through the origin); find their points of intersection.

IV way. Applying the full square selection method, the equation is converted to the form

Build a parabola y \u003d a (x + I) 2 and a straight line y \u003d - m, parallel to the x axis; find the points of intersection of the parabola and the line.

V way. Transform the equation to the form

They build a hyperbola (this is a hyperbola provided that ) and a straight line y \u003d - ax - b; find their points of intersection.

Note that the first four methods are applicable to any equations of the form ax 2 + bx + c \u003d 0, and the fifth - only to those with c. In practice, you can choose the method that you think is most adapted to this equation or whichever you like better (or more understandable).

Comment . Despite the abundance of ways to graphically solve quadratic equations, the confidence that any quadratic equation we
we can solve graphically, no. Let, for example, you need to solve the equation x 2 - x - 3 \u003d 0 (we will specially take an equation similar to what was in
considered example). Let's try to solve it, for example, in the second way: we transform the equation to the form x 2 \u003d x + 3, we construct a parabola y \u003d x 2 and
straight line y \u003d x + 3, they intersect at points A and B (Fig. 73), which means that the equation has two roots. But what are these roots, we use the drawing
we can’t say - points A and B do not have such “good” coordinates as in the above example. Now consider the equation
x 2 - 16x - 95 = 0. Let's try to solve it, say, in the third way. Let's transform the equation to the form x 2 - 95 = 16x. Here you need to build a parabola
y \u003d x 2 - 95 and a straight line y \u003d 16x. But the limited size of the notebook sheet does not allow this, because the parabola y \u003d x 2 must be lowered 95 cells down.

So, graphical methods for solving a quadratic equation are beautiful and pleasant, but they do not give a 100% guarantee of solving any quadratic equation. We will take this into account later.