Graphical equations. Graphical solution of mixed equations

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 average increased by six times arithmetic roots equations 3 – 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 graphical representation functions allows you to solve 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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Graphs of Quadratic Functions

In the last lesson we learned how to build a graph of any quadratic function. With the help of such functions we can solve the so-called quadratic equations, which in general view are written as follows: $ax^2+bx+c=0$,
$a, b, c$ are any numbers, but $a≠0$.
Guys, compare the equation written above and this: $y=ax^2+bx+c$.
They are almost identical. The difference is that instead of $y$ we wrote $0$, i.e. $y=0$. How then to solve quadratic equations? The first thing that comes to mind is to construct a graph of the parabola $ax^2+bx+c$ and find the points of intersection of this graph with the straight line $y=0$. There are other solutions. Let's look at them using a specific example.

Methods for solving quadratic functions

Example.
Solve the equation: $x^2+2x-8=0$.

Solution.
Method 1. Let's plot the function $y=x^2+2x-8$ and find the points of intersection with the straight line $y=0$. The coefficient of the highest degree is positive, which means the branches of the parabola point upward. Let's find the coordinates of the vertex:
$x_(c)=-\frac(b)(2a)=\frac(-2)(2)=-1$.
$y_(в)=(-1)^2+2*(-1)-8=1-2-8=-9$.

We take the point with coordinates $(-1;-9)$ as the beginning new system coordinates and construct a graph of the parabola $y=x^2$ in it.

We see two points of intersection. They are marked with black dots on the graph. We are solving the equation for x, so we need to choose the abscissas of these points. They are equal to $-4$ and $2$.
Thus, the solution to the quadratic equation $x^2+2x-8=0$ is two roots: $ x_1=-4$ and $x_2=2$.

Method 2. Transform the original equation to the form: $x^2=8-2x$.
Thus, we can solve this equation in the usual graphical way by finding the abscissa of the intersection points of the two graphs $y=x^2$ and $y=8-2x$.
We obtained two intersection points, the abscissas of which coincide with the solutions obtained in the first method, namely: $x_1=-4$ and $x_2=2$.

Method 3.
Let's transform the original equation to this form: $x^2-8=-2x$.
Let's build two graphs $y=x^2-8$ and $y=-2x$ and find their intersection points.
The graph of $y=x^2-8$ is a parabola shifted down 8 units.
We obtained two intersection points, and the abscissas of these points are the same as in the two previous methods, namely: $x_1=-4$ and $x_2=2$.

Method 4.
Let's select the perfect square in the original equation: $x^2+2x-8=x^2+2x+1-9=(x+1)^2-9$.
Let's construct two graphs of the functions $y=(x+1)^2$ and $y=9$. The graph of the first function is a parabola shifted one unit to the left. The graph of the second function is a straight line parallel to the abscissa axis and passing through the ordinate equal to $9$.
Once again we obtained two points of intersection of the graphs, and the abscissas of these points coincide with those obtained in the previous methods $x_1=-4$ and $x_2=2$.

Method 5.
Divide the original equation by x: $\frac(x^2)(x)+\frac(2x)(x)-\frac(8)(x)=\frac(0)(x)$.
$x+2-\frac(8)(x)=0$.
$x+2=\frac(8)(x)$.
Let's solve this equation graphically, construct two graphs $y=x+2$ and $y=\frac(8)(x)$.
Again we got two points of intersection, and the abscissas of these points coincide with those obtained above $x_1=-4$ and $x_2=2$.

Algorithm for graphical solution of quadratic functions

Guys, we looked at five ways to solve this graphically quadratic equations. In each of these methods, the roots of the equations turned out to be the same, which means the solution was obtained correctly.

Basic methods for graphically solving quadratic equations $ax^2+bx+c=0$, $a, b, c$ - any numbers, but $a≠0$:
1. Construct a graph of the function $y=ax^2+bx+c$, find the points of intersection with the abscissa axis, which will be the solution to the equation.
2. Construct two graphs $y=ax^2$ and $y=-bx-c$, find the abscissa of the intersection points of these graphs.
3. Construct two graphs $y=ax^2+c$ and $y=-bx$, find the abscissa of the intersection points of these graphs. The graph of the first function will be a parabola, shifted either down or up, depending on the sign of the number c. The second graph is a straight line passing through the origin.
4. Select a complete square, that is, bring the original equation to the form: $a(x+l)^2+m=0$.
Construct two graphs of the function $y=a(x+l)^2$ and $y=-m$, find their intersection points. The graph of the first function will be a parabola shifted either to the left or to the right, depending on the sign of the number $l$. The graph of the second function will be a straight line parallel to the abscissa axis and intersecting the ordinate axis at a point equal to $-m$.
5. Divide the original equation by x: $ax+b+\frac(c)(x)=0$.
Convert to the form: $\frac(c)(x)=-ax-b$.
Construct two graphs again and find their intersection points. The first graph is a hyperbola, the second graph is a straight line. Unfortunately, graphic method solving quadratic equations is not always in a good way solutions. The intersection points of different graphs are not always integers or may have very large numbers in the abscissa (ordinate). big numbers, which cannot be built on a regular sheet of paper.

Let us demonstrate all these methods more clearly with an example.

Example.
Solve the equation: $x^2+3x-12=0$,

Solution.
Let's plot the parabola and find the coordinates of the vertices: $x_(c)=-\frac(b)(2a)=\frac(-3)(2)=-1.5$.
$y_(в)=(-1.5)^2+2*(-1.5)-8=2.25-3-8=-8.75$.
When constructing such a parabola, problems immediately arise, for example, in correctly marking the vertex of the parabola. In order to accurately mark the ordinate of the vertex, you need to select one cell equal to 0.25 scale units. At this scale, you need to go down 35 units, which is inconvenient. Anyway, let's build our schedule.
The second problem we encounter is that the graph of our function intersects the x-axis at a point with coordinates that cannot be accurately determined. An approximate solution is possible, but mathematics is an exact science.
Thus, the graphical method is not the most convenient. Therefore, solving quadratic equations requires more universal method, which we will study in the next lessons.

Problems to solve independently

1. Solve the equation graphically (in all five ways): $x^2+4x-12=0$.
2. Solve the equation using any graphical method: $-x^2+6x+16=0$.

Graphic solution 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 were able to solve quadratic equations around 2000 BC. The rule for solving these equations, set out in the Babylonian texts, essentially coincides with modern ones, but it is not known how the Babylonians arrived at this rule.

Formulas for solving quadratic equations in Europe were first set forth in the Book of 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 solutions to quadratic equations for all possible combinations of coefficients b and c were formulated in Europe only in 1544 by M. Stiefel.

In 1591 Francois Viet introduced formulas for solving quadratic equations.

IN ancient Babylon could solve some types of quadratic equations.

Diophantus of Alexandria And Euclid, Al-Khwarizmi And Omar Khayyam solved equations using geometric and graphical methods.

In 7th grade we studied functions y = C, y =kx, y =kx+ m, y =x 2,y = –x 2, in 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 wondered if there were other functions that obey these rules.

My job is to study function graphs and solve equations graphically.

1. What are the functions?

The graph of a function is the set of all points 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 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 with center 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 Bernoulli's lemniscate.

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

Curve (x2 y2 – 2 ax)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 and its graphical solution

The equation– 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 an equation, produces a 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 constructing graphs and solving equations, the properties of a function are used, which is why the method is often called functional-graphical.

To solve the equation, we “divide” it into two parts, introduce two functions, build their graphs, and find the coordinates of the points of intersection of the graphs. The abscissas of these points are the roots of the equation.

3. Algorithm for plotting a function graph

Knowing the graph of a function y =f(x) , you can build graphs of 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 parallel carry transformation: to │ m│ units of scale to the right or left along the x-axis and on │ l│ units of scale up or down along an axis y.

4. Graphical solution of the quadratic equation

Using a quadratic function as an example, we will consider the 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 did not coordinate method, there was no concept of function. Nevertheless, the properties of the parabola were studied in detail by them. The ingenuity of ancient mathematicians is simply amazing - after all, 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 gave these curves names and indicated what conditions the points lying on this or that curve satisfy (after all, there were no formulas!).

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=axo2+in0+s;

Find the axis of symmetry of the parabola (straight line x=x0);

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We compile a table of values ​​for constructing control points;

We construct the resulting points and construct points that are symmetrical to them relative to the axis of symmetry.

1. Using the algorithm, we will construct a parabola y= x2 – 2 x– 3 . Abscissas of points of intersection with the axis x and there are roots of the quadratic equation x2 – 2 x– 3 = 0.

There are five ways to solve this equation graphically.

2. Let's split the equation into two functions: y= x2 And y= 2 x+ 3

3. Let's split 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 and the line.

4. Transform the equation x2 – 2 x– 3 = 0 using selection full square to functions: y= (x–1) 2 And y=4. The roots of the equation are the abscissas of the points of intersection of the parabola and the line.

5. Divide both sides of the equation term by term x2 – 2 x– 3 = 0 on 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.

Roots given equation is the abscissa of the point of intersection of the graphs of two functions: y= 3 √ x, y= 10 – x.

Answer: x = 8.

Conclusion

Having looked at the graphs of the functions: y =ax2 + bx+ c, y =k/ x, у = √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 for equations of degree n.

Graphic methods solutions to equations are beautiful and understandable, but do not provide a 100% guarantee of solving any equation. The abscissas of the intersection points of the graphs can be approximate.

In 9th grade and in high school, I will continue to get acquainted with other functions. I'm interested to know whether those functions obey the rules of parallel transfer when constructing their graphs.

On next year I would also like to consider the issues of graphically solving systems of equations and inequalities.

Literature

1. Algebra. 7th grade. Part 1. Textbook for educational institutions/ A.G. Mordkovich. M.: Mnemosyne, 2007.

2. Algebra. 8th grade. Part 1. Textbook for educational institutions / A.G. Mordkovich. M.: Mnemosyne, 2007.

3. Algebra. 9th grade. Part 1. Textbook for educational institutions / A.G. Mordkovich. M.: Mnemosyne, 2007.

4. Glazer G.I. History of mathematics at school. VII–VIII grades. – M.: Education, 1982.

5. Journal Mathematics No. 5 2009; No. 8 2007; No. 23 2008.

6. Graphical solution of equations websites on the Internet: Tol VIKI; stimul.biz/ru; wiki.iot.ru/images; berdsk.edu; page 3–6.htm.

>>Mathematics: Graphical solution of equations

Graphical solution of equations

Let's summarize our knowledge about graphs functions. We have learned how to build graphs of the following functions:

y =b (straight line parallel to the x axis);

y = kx (line passing through the origin);

y - kx + m (straight line);

y = 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 = x 2 (which represents an equality with two variables x and y), consider a parabola in the coordinate plane. In particular, it is sometimes useful for solving equations. Let's discuss how this is done using several examples.

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

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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 (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 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 equal to tangent 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. For their precise definition 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. This graph is a parabola with a vertex at 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.