How to solve an equation graphically. Graphical methods for solving equations

Let there be a complete quadratic equation: A*x2+B*x+C=0, where A, B and C are any numbers, and A is not equal to zero. This is a general case of a quadratic equation. There is also a reduced form in which A=1. To solve any equation graphically, you need to move the term with the highest degree to another part and equate both parts to some variable.

After this, A*x2 will remain on the left side of the equation, and B*x-C on the right side (we can assume that B is a negative number, this does not change the essence). The resulting equation is A*x2=B*x-C=y. For clarity, in this case both parts are equated to the variable y.

Plotting graphs and processing results

Now we can write two equations: y=A*x2 and y=B*x-C. Next, you need to plot a graph of each of these functions. The graph y=A*x2 is a parabola with a vertex at the origin, the branches of which are directed upward or downward, depending on the sign of the number A. If it is negative, the branches are directed downward, if positive, the branches are directed upward.

The graph y=B*x-C is a regular straight line. If C=0, the line passes through the origin. In the general case, it cuts off a segment equal to C from the ordinate axis. The angle of inclination of this line relative to the abscissa axis is determined by the coefficient B. It is equal to the tangent of the inclination of this angle.

After the graphs are plotted, it will be seen that they intersect at two points. The coordinates of these points along the x-axis determine the roots of the quadratic equation. To accurately determine them, you need to clearly build graphs and choose the right scale.

Another graphical solution

There is another way to solve a quadratic equation graphically. It is not necessary to move B*x+C to the other side of the equation. You can immediately plot the function y=A*x2+B*x+C. Such a graph is a parabola with a vertex at an arbitrary point. This method is more complicated than the previous one, but you can build only one graph to...

First you need to determine the vertex of the parabola with coordinates x0 and y0. Its abscissa is calculated using the formula x0=-B/2*a. To determine the ordinate, you need to substitute the resulting abscissa value into the original function. Mathematically, this statement is written as follows: y0=y(x0).

Then you need to find two points symmetrical to the axis of the parabola. In them, the original function must vanish. After this, you can build a parabola. The points of its intersection with the X axis will give two roots of the quadratic equation.

Sometimes equations are solved graphically. To do this, you need to transform the equation so (if it is not already presented in a transformed form) that to the left and right of the equal sign there are expressions for which you can easily draw function graphs. For example, given the following equation:
x² – 2x – 1 = 0

If we have not yet studied solving quadratic equations algebraically, we can try to do this either by factoring or graphically. To solve such an equation graphically, we present it in this form:
x² = 2x + 1

From this representation of the equation it follows that it is necessary to find such values ​​of x for which the left side will be equal to the right.

As you know, the graph of the function y = x² is a parabola, and y = 2x + 1 is a straight line. The x coordinate of the points of the coordinate plane lying both on the first graph and on the second (that is, the points of intersection of the graphs) are precisely those x values ​​at which the left side of the equation will be equal to the right. In other words, the x-coordinates of the intersection points of the graphs are the roots of the equation.

Graphs can intersect at several points, at one point, or not intersect at all. It follows that an equation can have several roots, or one root, or none at all.

Let's look at a simpler example:
x² – 2x = 0 or x² = 2x

Let's draw graphs of the functions y = x² and y = 2x:

As can be seen from the drawing, the parabola and the straight line intersect at points (0; 0) and (2; 4). The x coordinates of these points are respectively equal to 0 and 2. This means that the equation x² – 2x = 0 has two roots - x 1 = 0, x 2 = 2.

Let's check this by solving the equation by taking the common factor out of brackets:
x² – 2x = 0
x(x – 2) = 0

The zero on the right side can occur either when x is 0 or 2.

The reason why we did not graphically solve the equation x² – 2x – 1 = 0 is that in most equations the roots are real (fractional) numbers, and it is difficult to accurately determine the value of x on a graph. Therefore, for most equations, the graphical solution is not the best. However, knowledge of this method provides a deeper understanding of the relationship between equations and functions.

In this lesson we will look at solving systems of two equations in two variables. First, let's look at the graphical solution of a system of two linear equations and the specifics of the set of their graphs. Next, we will solve several systems using the graphical method.

Topic: Systems of equations

Lesson: Graphical method for solving a system of equations

Consider the system

A pair of numbers that is simultaneously a solution to both the first and second equations of the system is called solving a system of equations.

Solving a system of equations means finding all its solutions, or establishing that there are no solutions. We have looked at the graphs of the basic equations, let's move on to considering systems.

Example 1. Solve the system

Solution:

These are linear equations, the graph of each of them is a straight line. The graph of the first equation passes through the points (0; 1) and (-1; 0). The graph of the second equation passes through the points (0; -1) and (-1; 0). The lines intersect at the point (-1; 0), this is the solution to the system of equations ( Rice. 1).

The solution to the system is a pair of numbers. Substituting this pair of numbers into each equation, we obtain the correct equality.

We have obtained a unique solution to the linear system.

Recall that when solving a linear system, the following cases are possible:

the system has a unique solution - the lines intersect,

the system has no solutions - the lines are parallel,

the system has an infinite number of solutions - the straight lines coincide.

We considered a special case of the system when p(x; y) and q(x; y) are linear expressions of x and y.

Example 2. Solve a system of equations

Solution:

The graph of the first equation is a straight line, the graph of the second equation is a circle. Let's build the first graph by points (Fig. 2).

The center of the circle is at point O(0; 0), the radius is 1.

The graphs intersect at point A(0; 1) and point B(-1; 0).

Example 3. Solve the system graphically

Solution: Let's build a graph of the first equation - it is a circle with a center at t.O(0; 0) and radius 2. The graph of the second equation is a parabola. It is shifted upward by 2 relative to the origin, i.e. its vertex is point (0; 2) (Fig. 3).

The graphs have one common point - i.e. A(0; 2). It is the solution to the system. Let's plug a couple of numbers into the equation to check if it's correct.

Example 4. Solve the system

Solution: Let's construct a graph of the first equation - this is a circle with a center at t.O(0; 0) and radius 1 (Fig. 4).

Let's plot the function This is a broken line (Fig. 5).

Now let's move it 1 down along the oy axis. This will be the graph of the function

Let's place both graphs in the same coordinate system (Fig. 6).

We get three intersection points - point A(1; 0), point B(-1; 0), point C(0; -1).

We looked at the graphical method for solving systems. If you can plot a graph of each equation and find the coordinates of the intersection points, then this method is quite sufficient.

But often the graphical method makes it possible to find only an approximate solution of the system or answer the question about the number of solutions. Therefore, other methods are needed, more accurate, and we will deal with them in the following lessons.

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

2. Mordkovich A.G. and others. Algebra 9th grade: Problem book for students of general education institutions / A. G. Mordkovich, T. N. Mishustina, etc. - 4th ed. - M.: Mnemosyne, 2002.-143 p.: ill.

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

5. Mordkovich A. G. Algebra. 9th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich, P. V. Semenov. — 12th ed., erased. - M.: 2010. - 224 p.: ill.

6. Algebra. 9th grade. In 2 parts. Part 2. Problem book for students of general education 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.ru section on mathematics ().

2. Internet project “Tasks” ().

3. Educational portal “I WILL SOLVE the Unified State Exam” ().

1. Mordkovich A.G. and others. Algebra 9th grade: Problem book for students of general education institutions / A. G. Mordkovich, T. N. Mishustina, etc. - 4th ed. - M.: Mnemosyne, 2002.-143 p.: ill. No. 105, 107, 114, 115.

If you want to learn to swim, then boldly enter the water, and if you want to learn how to solve problems, solve them.

D. Polya

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 at which it turns into a correct numerical equality, or establishing that there are no such values.

Range of acceptable values equations (O.D.Z.) is the set of all those values ​​of the variable (variables) at which all expressions included in the equation are defined.

Many equations presented in the Unified State Examination are solved using standard methods. But no one forbids using something unusual, even in the simplest cases.

So, for example, consider the equation 3 – x 2 = 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 – x 2 And y = 6 / (2 – x) and build their graphs.

The function y = 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 downward (since a = -1< 0).

The vertex of the parabola will be shifted along the ordinate axis by 3 units upward. Thus, the coordinate of the vertex 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 abscissa axis (Fig. 1).

The graph of the function y = 6 / (2 – x) is a hyperbola.

The graph of this function can be plotted using the following transformations:

1) y = 6 / x – inverse proportionality. The graph of a function is a hyperbola. It can be built point by point; to do this, let’s create 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 step 1 is symmetrically displayed relative to the ordinate axis (Fig. 3).

3) y = 6 / (-x + 2) – shift the graph obtained in step 2 along the x-axis by two units to the right (Fig. 4).

Now let's plot 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 solution does not allow you to find the exact value of the root. So the numbers are -1; 0; 3 (abscissas of the intersection points of the function graphs) are so far only the assumed roots of the equation.

Using a check, we will make sure that the numbers are -1; 0; 3 are indeed 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 average:

(-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 = 6 / (2 – x).

Let's start solving the equation by searching for O.D.Z. The denominator of the fraction should not be zero, therefore:

To solve the equation, we use the basic property of proportion, this will allow us to get rid of the fraction.

(3 – x 2)(2 – x) = 6.

Let's open the brackets and present similar terms:

6 – 3x – 2x 2 + x 3 = 6;

x 3 – 2x 2 – 3x = 0.

Let's take the common factor out of brackets:

x(x 2 – 2x – 3) = 0.

Let's take advantage of 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 x 2 – 2x – 3 = 0.

Let's solve the second equation.

x 2 – 2x – 3 = 0. It’s square, so we’ll use the discriminant.

D=4 – 4 · (-3) = 16;

x 1 = (2 + 4) / 2 = 3;

x 2 = (2 – 4) / 2 = -1.

All three obtained roots satisfy O.D.Z.

Therefore, let’s find their arithmetic mean and increase it six times:

6 · (-1 + 3 + 0) / 3 = 4.

In fact, the graphical method of solving equations is used quite rarely. This is due to the fact that the graphical representation of functions allows solving equations only approximately. This method is mainly used in those problems where it is important to search not for the roots of the equation themselves - their numerical values, but only for their quantity.

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In this video lesson, the topic “Function y=x 2” is offered for study. Graphic solution of equations." During this lesson, students will be able to get acquainted with a new way of solving equations - graphically, which is based on knowledge of the properties of graphs of functions. The teacher will show how to solve the function y=x 2 graphically.

Subject:Function

Lesson:Function. Graphical solution of equations

Graphical solution of equations is based on knowledge of function graphs and their properties. Let us list the functions whose graphs we know:

1), the graph is a straight line parallel to the abscissa axis, passing through a point on the ordinate axis. Let's look at an example: y=1:

For different values, we get a family of straight lines parallel to the x-axis.

2) Function of direct proportionality, the graph of this function is a straight line passing through the origin of coordinates. Let's look at an example:

We have already constructed these graphs in previous lessons; recall that to construct each line, you need to select a point that satisfies it, and take the origin of coordinates as the second point.

Let us 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, the following relationship exists between two parameters k of the same sign: for positive k, the larger it is, the faster the function increases, and for negative ones, the function decreases faster for large values ​​of k in absolute value.

3) Linear function. When - we obtain the point of intersection with the ordinate axis and all straight lines of this type pass through the point (0; m). In addition, 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. And of course the value of k affects the rate of change of the function value.

4). The graph of this function is a parabola.

Let's look at examples.

Example 1 - Solve the equation graphically:

We don’t know functions of this type, so we need to transform the given equation to work with known functions:

We get familiar functions on both sides of the equation:

Let's build graphs of functions:

The graphs have two intersection points: (-1; 1); (2; 4)

Let's check whether the solution is found correctly and substitute the coordinates into the equation:

The first point was found correctly.

, , , , , ,

The second point was also found correctly.

So, the solutions to the equation are and

We proceed similarly to the previous example: we transform the given equation to functions known to us, construct their graphs, find the intersection currents and from here indicate the solutions.

We get two functions:

Let's build graphs:

These graphs do not have intersection points, which means the given equation has no solutions

Conclusion: in this lesson we reviewed the functions and their graphs known to us, remembered their properties and looked at the graphical method of solving equations.

1. Dorofeev G.V., Suvorova S.B., Bunimovich E.A. and others. 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.: Enlightenment. 2006

Task 1: Makarychev Yu.N., Mindyuk N.G., Neshkov K.I. and others. Algebra 7, No. 494, Art. 110;

Task 2: Makarychev Yu.N., Mindyuk N.G., Neshkov K.I. and others. Algebra 7, No. 495, Art. 110;

Task 3: Makarychev Yu.N., Mindyuk N.G., Neshkov K.I. and others. Algebra 7, No. 496, Art. 110;