Geometric transformations of graphs of functions table. Transforming Graphs of Trigonometric Functions

Hypothesis: If you study the movement of the graph during the formation of the equation of functions, then you can see that all graphs obey general patterns therefore, it is possible to formulate general laws regardless of functions, which will make it possible not only to facilitate the construction of graphs of various functions, but also to use them in solving problems.

Purpose: To study the movement of graphs of functions:

1) The task of studying literature

2) Learn to build graphs of various functions

3) Learn how to convert charts linear functions

4) Consider the use of graphs in solving problems

Object of study: Graphs of functions

Subject of research: Movements of graphs of functions

Relevance: The construction of function graphs, as a rule, takes a lot of time and requires attention from the student, but knowing the rules for transforming function graphs and graphs of basic functions, you can quickly and easily build function graphs, which will allow you not only to complete tasks for plotting function graphs, but also solve problems related to it (to find the maximum (minimum height of time and meeting point))

This project is useful to all students of the school.

Literature review:

The literature discusses ways to construct a graph of various functions, as well as examples of the transformation of graphs of these functions. Graphs of almost all main functions are used in various technical processes, which makes it possible to more clearly present the course of the process and program the result

Permanent function. This function is given by the formula y = b, where b is some number. schedule permanent function is a straight line parallel to the x-axis and passing through the point (0; b) on the y-axis. The graph of the function y \u003d 0 is the abscissa axis.

Types of function 1Direct proportionality. This function is given by the formula y \u003d kx, where the coefficient of proportionality k ≠ 0. The direct proportionality graph is a straight line passing through the origin.

Linear function. Such a function is given by the formula y = kx + b, where k and b are real numbers. The graph of a linear function is a straight line.

Linear function graphs can intersect or be parallel.

So, the lines of the graphs of linear functions y \u003d k 1 x + b 1 and y \u003d k 2 x + b 2 intersect if k 1 ≠ k 2; if k 1 = k 2 , then the lines are parallel.

2 Inverse proportionality is a function that is given by the formula y \u003d k / x, where k ≠ 0. K is called the coefficient inverse proportionality. The inverse proportionality graph is a hyperbola.

The function y \u003d x 2 is represented by a graph called a parabola: on the interval [-~; 0] the function is decreasing, on the interval the function is increasing.

The function y \u003d x 3 increases along the entire number line and is graphically represented by a cubic parabola.

Power function with natural indicator. This function is given by the formula y \u003d x n, where n is natural number. Graphs power function with a natural exponent depend on n. For example, if n = 1, then the graph will be a straight line (y = x), if n = 2, then the graph will be a parabola, etc.

Power function with integer negative indicator represented by the formula y \u003d x -n, where n is a natural number. This function is defined for all x ≠ 0. The graph of the function also depends on the exponent n.

Power function with positive fractional indicator. This function is represented by the formula y \u003d x r, where r is a positive irreducible fraction. This function is also neither even nor odd.

Graph-line that displays the relationship of dependent and independent variables on the coordinate plane. The graph serves to visually display these elements.

An independent variable is a variable that can take on any value in the scope of a function (where given function makes sense (can't divide by zero)

To plot a function graph,

1) Find ODZ (range of acceptable values)

2) take some arbitrary values for the independent variable

3) Find the value of the dependent variable

4) Build coordinate plane mark these points on it

5) Connect their lines if necessary, investigate the resulting graph. Transformation of graphs of elementary functions.

Graph Conversion

AT pure form the basic elementary functions are encountered, unfortunately, not so often. Much more likely to deal with elementary functions obtained from the basic elementary ones by adding constants and coefficients. Graphs of such functions can be built by applying geometric transformations to the graphs of the corresponding basic elementary functions (or go to new system coordinates). For example, the quadratic function formula is quadratic parabola a formula compressed three times relative to the ordinate axis, symmetrically displayed relative to the abscissa axis, shifted against the direction of this axis by 2/3 units and shifted along the direction of the ordinate axis by 2 units.

Let's understand these geometric transformations of a function graph step by step using specific examples.

With the help of geometric transformations of the graph of the function f (x), a graph of any function of the form formula can be constructed, where the formula is the compression or expansion coefficients along the oy and ox axes, respectively, the minus signs in front of the coefficients formula and formula indicate a symmetrical display of the graph with respect to coordinate axes, a and b define the shift relative to the abscissa and ordinate axes, respectively.

Thus, there are three types of geometric transformations of the function graph:

The first type is scaling (compression or expansion) along the abscissa and ordinate axes.

The need for scaling is indicated by formula coefficients other than one, if the number is less than 1, then the graph is compressed relative to oy and stretched relative to ox, if the number is greater than 1, then we stretch along the ordinate axis and shrink along the abscissa axis.

The second type is a symmetrical (mirror) display with respect to the coordinate axes.

The need for this transformation is indicated by the minus signs in front of the coefficients of the formula (in this case, we display the graph symmetrically with respect to the ox axis) and the formula (in this case, we display the graph symmetrically with respect to the y axis). If there are no minus signs, then this step is skipped.

Summary of the lesson of algebra and the beginning of analysis in grade 10

on the topic: "Chart conversion trigonometric functions»

The purpose of the lesson: to systematize knowledge on the topic "Properties and graphs of trigonometric functions y \u003d sin (x), y \u003d cos (x)".

Lesson objectives:

repeat the properties of trigonometric functions y \u003d sin (x), y \u003d cos (x);
repeat the reduction formulas;
conversion of graphs of trigonometric functions;
develop attention, memory, logical thinking; activate mental activity the ability to analyze, generalize and reason;
education of industriousness, diligence in achieving the goal, interest in the subject.

Lesson equipment:ict

Lesson type: learning new

During the classes

Before the lesson, 2 students on the board build graphs from their homework.

Organizing time:

Hello guys!

Today in the lesson we will convert the graphs of trigonometric functions y \u003d sin (x), y \u003d cos (x).

Oral work:

Checking homework.

solving puzzles.

Learning new material

All transformations of function graphs are universal - they are suitable for all functions, including trigonometric ones. Here we confine ourselves to a brief reminder of the main transformations of graphs.

Transformation of graphs of functions.

The function y \u003d f (x) is given. We start building all graphs from the graph of this function, then we perform actions with it.

Function

What to do with the schedule

y = f(x) + a

We raise all points of the first graph by a units up.

y = f(x) – a

All points of the first graph are lowered by a units down.

y = f(x + a)

We shift all points of the first graph by a units to the left.

y = f (x - a)

We shift all points of the first graph by a units to the right.

y = a*f(x),a>1

We fix the zeros in place, we shift the upper points higher by a times, the lower ones we lower lower by a times.

The graph will "stretch" up and down, the zeros remain in place.

y = a*f(x), a<1

We fix the zeros, the upper points will go down a times, the lower ones will rise a times. The graph will “shrink” to the x-axis.

y=-f(x)

Mirror the first graph about the x-axis.

y = f(ax), a<1

Fix a point on the y-axis. Each segment on the x-axis is increased by a times. The graph will stretch from the y-axis in different directions.

y = f(ax), a>1

Fix a point on the ordinate axis, each segment on the abscissa axis is reduced by a times. The graph will “shrink” to the y-axis on both sides.

y= | f(x)|

The parts of the graph located under the x-axis are mirrored. The entire graph will be located in the upper half-plane.

Solution schemes.

1)y = sin x + 2.

We build a graph y \u003d sin x. We raise each point of the graph up by 2 units (zeros too).

2)y \u003d cos x - 3.

We build a graph y \u003d cos x. We lower each point of the graph down by 3 units.

3)y = cos (x - /2)

We build a graph y \u003d cos x. We shift all points n/2 to the right.

4) y = 2 sin x .

We build a graph y \u003d sin x. We leave the zeros in place, raise the upper points 2 times, lower the lower ones by the same amount.

PRACTICAL WORK Plotting trigonometric functions using the Advanced Grapher program.

Let's plot the function y = -cos 3x + 2.

Let's plot the function y \u003d cos x.
Reflect it about the x-axis.
This graph must be compressed three times along the x-axis.
Finally, such a graph must be lifted up by three units along the y-axis.

y = 0.5 sinx.

y=0.2 cos x-2

y = 5 cos 0 .5 x

y=-3sin(x+π).

2) Find the mistake and fix it.

V. Historical material. Euler's message.

Leonhard Euler is the greatest mathematician of the 18th century. Born in Switzerland. For many years he lived and worked in Russia, a member of the St. Petersburg Academy.

Why should we know and remember the name of this scientist?

By the beginning of the 18th century, trigonometry was still insufficiently developed: there were no symbols, formulas were written in words, it was difficult to assimilate them, the question of the signs of trigonometric functions in different quarters of the circle was also unclear, only angles or arcs were understood as an argument of a trigonometric function. It was only in the works of Euler that trigonometry received its modern form. It was he who began to consider the trigonometric function of a number, i.e. the argument came to be understood not only as arcs or degrees, but also as numbers. Euler deduced all trigonometric formulas from several basic ones, streamlined the question of the signs of the trigonometric function in different quarters of the circle. To designate trigonometric functions, he introduced symbols: sin x, cos x, tg x, ctg x.

On the threshold of the 18th century, a new direction appeared in the development of trigonometry - analytical. If before that the main goal of trigonometry was considered to be the solution of triangles, then Euler considered trigonometry as the science of trigonometric functions. The first part: the doctrine of function is part of the general doctrine of functions, which is studied in mathematical analysis. The second part: the solution of triangles - the chapter of geometry. Such innovations were made by Euler.

VI. Repetition

Independent work "Add the formula."

VII. Lesson summary:

1) What new did you learn at the lesson today?

2) What else do you want to know?

3) Grading.

The text of the work is placed without images and formulas.
The full version of the work is available in the "Job Files" tab in PDF format

Introduction

Transformation of graphs of a function is one of the basic mathematical concepts directly related to practical activities. The transformation of graphs of functions is first encountered in algebra grade 9 when studying the topic "Quadratic function". The quadratic function is introduced and studied in close connection with quadratic equations and inequalities. Also, many mathematical concepts are considered by graphical methods, for example, in grades 10-11, the study of a function makes it possible to find the domain of definition and the scope of the function, the areas of decrease or increase, asymptotes, intervals of constancy, etc. This important question is also submitted to the GIA. It follows that the construction and transformation of function graphs is one of the main tasks of teaching mathematics at school.

However, to plot many functions, a number of methods can be used to facilitate the construction. The above defines relevance research topics.

Object of study is the study of the transformation of graphs in school mathematics.

Subject of study - the process of constructing and transforming function graphs in a secondary school.

problem question: is it possible to build a graph of an unfamiliar function, having the skill of transforming graphs of elementary functions?

Target: plotting a function in an unfamiliar situation.

Tasks:

1. Analyze the educational material on the problem under study. 2. Identify schemes for transforming function graphs in a school mathematics course. 3. Select the most effective methods and tools for constructing and converting function graphs. 4. Be able to apply this theory in solving problems.

Necessary basic knowledge, skills, abilities:

Determine the value of the function by the value of the argument in various ways of specifying the function;

Build graphs of the studied functions;

Describe the behavior and properties of functions from the graph and, in the simplest cases, from the formula, find the largest and smallest values from the graph of the function;

Descriptions with the help of functions of various dependencies, their representation graphically, interpretation of graphs.

Main part

Theoretical part

As the initial graph of the function y = f(x), I will choose a quadratic function y=x 2 . I will consider cases of transformation of this graph associated with changes in the formula that defines this function and draw conclusions for any function.

1. Function y = f(x) + a

In the new formula, the function values (the coordinates of the graph points) are changed by the number a, compared to the "old" function value. This leads to a parallel translation of the graph of the function along the OY axis:

up if a > 0; down if a< 0.

CONCLUSION

Thus, the graph of the function y=f(x)+a is obtained from the graph of the function y=f(x) by means of a parallel translation along the ordinate axis by a units up if a > 0, and by a units down if a< 0.

2. Function y = f(x-a),

In the new formula, the argument values (the abscissas of the graph points) are changed by the number a, compared to the "old" argument value. This leads to a parallel transfer of the graph of the function along the OX axis: to the right if a< 0, влево, если a >0.

CONCLUSION

So the graph of the function y= f(x - a) is obtained from the graph of the function y=f(x) by parallel translation along the abscissa axis by a units to the left if a > 0, and by a units to the right if a< 0.

3. Function y = k f(x), where k > 0 and k ≠ 1

In the new formula, the function values (the coordinates of the graph points) change k times compared to the "old" function value. This leads to: 1) "stretching" from the point (0; 0) along the OY axis by k times, if k > 1, 2) "compression" to the point (0; 0) along the OY axis by a factor of 0, if 0< k < 1.

CONCLUSION

Therefore: to build a graph of the function y = kf(x), where k > 0 and k ≠ 1, you need to multiply the ordinates of the points of the given graph of the function y = f(x) by k. Such a transformation is called stretching from the point (0; 0) along the OY axis by k times if k > 1; contraction to the point (0; 0) along the OY axis by a factor if 0< k < 1.

4. Function y = f(kx), where k > 0 and k ≠ 1

In the new formula, the values of the argument (the abscissas of the graph points) change k times compared to the "old" value of the argument. This leads to: 1) “stretching” from the point (0; 0) along the OX axis by 1/k times if 0< k < 1; 2) «сжатию» к точке (0; 0) вдоль оси OX. в k раз, если k > 1.

CONCLUSION

And so: to build a graph of the function y = f(kx), where k > 0 and k ≠ 1, you need to multiply the abscissas of the points of the given graph of the function y=f(x) by k. Such a transformation is called stretching from the point (0; 0) along the OX axis by 1/k times if 0< k < 1, сжатием к точке (0; 0) вдоль оси OX. в k раз, если k > 1.

5. Function y = - f (x).

In this formula, the values of the function (the coordinates of the graph points) are reversed. This change results in a symmetrical display of the original graph of the function about the x-axis.

CONCLUSION

To build a graph of the function y = - f (x), you need a graph of the function y = f (x)

reflect symmetrically about the OX axis. Such a transformation is called a symmetry transformation about the OX axis.

6. Function y = f (-x).

In this formula, the values of the argument (the abscissas of the graph points) are reversed. This change results in a symmetrical display of the original function graph with respect to the OY axis.

An example for the function y \u003d - x² this transformation is not noticeable, since this function is even and the graph does not change after the transformation. This transformation is visible when the function is odd and when neither even nor odd.

7. Function y = |f(x)|.

In the new formula, the function values (the coordinates of the graph points) are under the module sign. This leads to the disappearance of parts of the graph of the original function with negative ordinates (that is, those located in the lower half-plane relative to the Ox axis) and a symmetrical display of these parts relative to the Ox axis.

8. Function y= f (|x|).

In the new formula, the argument values (the abscissas of the graph points) are under the module sign. This leads to the disappearance of parts of the graph of the original function with negative abscissas (that is, those located in the left half-plane relative to the OY axis) and their replacement with parts of the original graph that are symmetrical about the OY axis.

Practical part

Consider a few examples of the application of the above theory.

EXAMPLE 1.

Decision. Let's transform this formula:

1) Let's build a graph of the function

EXAMPLE 2.

Plot the function given by the formula

Decision. Let us transform this formula by highlighting in this square trinomial binomial square:

1) Let's build a graph of the function

2) Perform a parallel transfer of the constructed graph to the vector

EXAMPLE 3.

TASK FROM THE USE Plotting a piecewise function

Function graph Function graph y=|2(x-3)2-2|; one

Parallel transfer.

TRANSFER ALONG THE Y-AXIS

f(x) => f(x) - b
Let it be required to plot the function y \u003d f (x) - b. It is easy to see that the ordinates of this graph for all values of x on |b| units less than the corresponding ordinates of the graph of functions y = f(x) for b>0 and |b| units more - at b 0 or up at b To plot the function y + b = f(x), plot the function y = f(x) and move the x-axis to |b| units up for b>0 or by |b| units down at b

TRANSFER ALONG THE X-AXIS

f(x) => f(x + a)
Let it be required to plot the function y = f(x + a). Consider a function y = f(x), which at some point x = x1 takes the value y1 = f(x1). Obviously, the function y = f(x + a) will take the same value at the point x2, the coordinate of which is determined from the equality x2 + a = x1, i.e. x2 = x1 - a, and the equality under consideration is valid for the totality of all values from the domain of the function. Therefore, the graph of the function y = f(x + a) can be obtained by parallel displacement of the graph of the function y = f(x) along the x-axis to the left by |a| ones for a > 0 or to the right by |a| units for a To plot the function y = f(x + a), plot the function y = f(x) and move the y-axis to |a| units to the right for a>0 or |a| units to the left for a

Examples:

1.y=f(x+a)

2.y=f(x)+b

Reflection.

GRAPHING OF A FUNCTION OF THE VIEW Y = F(-X)

f(x) => f(-x)
Obviously, the functions y = f(-x) and y = f(x) take equal values at points whose abscissas are equal in absolute value, but opposite in sign. In other words, the ordinates of the graph of the function y = f(-x) in the region of positive (negative) values of x will be equal to the ordinates of the graph of the function y = f(x) with negative (positive) x values corresponding in absolute value. Thus, we get the following rule.
To plot the function y = f(-x), you should plot the function y = f(x) and reflect it along the y-axis. The resulting graph is the graph of the function y = f(-x)

GRAPHING OF A FUNCTION OF THE VIEW Y = - F(X)

f(x) => - f(x)
The ordinates of the graph of the function y = - f(x) for all values of the argument are equal in absolute value, but opposite in sign to the ordinates of the graph of the function y = f(x) for the same values of the argument. Thus, we get the following rule.
To plot the function y = - f(x), you should plot the function y = f(x) and reflect it about the x-axis.

Examples:

1.y=-f(x)

2.y=f(-x)

3.y=-f(-x)

Deformation.

DEFORMATION OF THE GRAPH ALONG THE Y-AXIS

f(x) => kf(x)
Consider a function of the form y = k f(x), where k > 0. It is easy to see that for equal values of the argument, the ordinates of the graph of this function will be k times greater than the ordinates of the graph of the function y \u003d f (x) for k > 1 or 1/k times less than the ordinates of the graph of the function y = f (x) for k To plot the graph of the function y = k f (x) you should plot the function y = f(x) and increase its ordinates by k times for k > 1 (stretch the graph along the ordinate axis) or decrease its ordinates by 1/k times for k
k > 1- stretching from the Ox axis
0 - compression to the OX axis

GRAPH DEFORMATION ALONG THE X-AXIS

f(x) => f(kx)
Let it be required to plot the function y = f(kx), where k>0. Consider a function y = f(x), which in arbitrary point x = x1 takes the value y1 = f(x1). Obviously, the function y = f(kx) takes the same value at the point x = x2, the coordinate of which is determined by the equality x1 = kx2, and this equality is valid for the totality of all values of x from the domain of the function. Consequently, the graph of the function y = f(kx) is compressed (for k 1) along the abscissa axis relative to the graph of the function y = f(x). Thus, we get the rule.
To plot the function y = f(kx), plot the function y = f(x) and reduce its abscissa by k times for k>1 (shrink the graph along the abscissa) or increase its abscissa by 1/k times for k
k > 1- compression to the Oy axis
0 - stretching from the OY axis

The work was carried out by Alexander Chichkanov, Dmitry Leonov under the supervision of Tkach T.V., Vyazovov S.M., Ostroverkhova I.V.
©2014

Basic elementary functions in their pure form without transformation are rare, so most often you have to work with elementary functions that are obtained from the basic ones by adding constants and coefficients. Such graphs are built using geometric transformations of given elementary functions.

Let's look at an example quadratic function of the form y \u003d - 1 3 x + 2 3 2 + 2, the graph of which is a parabola y \u003d x 2, which is compressed three times with respect to O y and symmetrical with respect to O x, moreover, shifted by 2 3 by O x to the right, by 2 units by O at up. On the coordinate line, it looks like this:

Yandex.RTB R-A-339285-1

Geometric transformations of a graph of a function

Applying geometric transformations of the given graph, we obtain that the graph is represented by a function of the form ± k 1 f (± k 2 (x + a)) + b when k 1 > 0, k 2 > 0 are the compression ratios at 0< k 1 < 1 , 0 < k 2 < 1 или растяжения при k 1 >1 , k 2 > 1 along O y and O x. The sign in front of the coefficients k 1 and k 2 indicates the symmetrical display of the graph relative to the axes, a and b shift it along O x and O y.

Definition 1

There are 3 types geometric transformation graphics:

Scaling along O x and O y. This is affected by the coefficients k 1 and k 2, provided that 1 is not equal, when 0< k 1 < 1 , 0 < k 2 < 1 , то график сжимается по О у, а растягивается по О х, когда k 1 >1, k 2 > 1, then the graph is stretched along O y and compressed along O x.
Symmetric display about coordinate axes. If there is a “-” sign in front of k 1, the symmetry goes with respect to O x, before k 2 it goes with respect to O y. If "-" is missing, then the decision point is skipped;
Parallel translation (shift) along O x and O y. The transformation is performed when the coefficients a and b are not equal to 0 . If the value of a is positive, then the graph is shifted to the left by | a | units, if negative a , then to the right by the same distance. The value of b determines the movement along the O y axis, which means that if b is positive, the function moves up, and if b is negative, it moves down.

Consider solutions using examples, starting with a power function.

Example 1

Transform y = x 2 3 and plot the function y = - 1 2 · 8 x - 4 2 3 + 3 .

Decision

Let's represent the functions like this:

y = - 1 2 8 x - 4 2 3 + 3 = - 1 2 8 x - 1 2 2 3 + 3 = - 2 x - 1 2 2 3 + 3

Where k 1 \u003d 2, you should pay attention to the presence of "-", a \u003d - 1 2, b \u003d 3. From here we get that geometric transformations are made from stretching along O y twice, displayed symmetrically with respect to O x, shifted to the right by 1 2 and up by 3 units.

If we represent the original power function, we get that

when stretched twice along O y, we have that

A mapping symmetric with respect to O x has the form

and move to the right by 1 2

moving 3 units up has the form

We will consider the transformations of the exponential function using examples.

Example 2

Graph the exponential function y = - 1 2 1 2 (2 - x) + 8 .

Decision.

We transform the function based on the properties of the power function. Then we get that

y = - 1 2 1 2 (2 - x) + 8 = - 1 2 - 1 2 x + 1 + 8 = - 1 2 1 2 - 1 2 x + 8

This shows that we get a chain of transformations y = 1 2 x:

y = 1 2 x → y = 1 2 1 2 x → y = 1 2 1 2 1 2 x → → y = - 1 2 1 2 1 2 x → y = - 1 2 1 2 - 1 2 x → → y = - 1 2 1 2 - 1 2 x + 8

We get that the original exponential function has the form

Squeezing twice along O y gives

Stretching along O x

Symmetric mapping with respect to O x

The mapping is symmetrical with respect to O y

Shift up 8 units

Consider the solution with an example logarithmic function y = log(x) .

Example 3

Construct the function y = ln e 2 · - 1 2 x 3 using the transformation y = ln (x) .

Decision

To solve it, you need to use the properties of the logarithm, then we get:

y = ln e 2 - 1 2 x 3 = ln (e 2) + ln - 1 2 x 1 3 = 1 3 ln - 1 2 x + 2

The transformations of the logarithmic function look like this:

y = ln (x) → y = 1 3 ln (x) → y = 1 3 ln 1 2 x → → y = 1 3 ln - 1 2 x → y = 1 3 ln - 1 2 x + 2

Draw a graph of the original logarithmic function

We compress the system according to O y

We stretch along O x

We make a mapping with respect to O y

We make a shift up by 2 units, we get

To transform the graphs of a trigonometric function, it is necessary to fit solutions of the form ± k 1 · f (± k 2 · (x + a)) + b to the scheme. It is necessary that k 2 be equal to T k 2 . Hence we get that 0< k 2 < 1 дает понять, что график функции увеличивает период по О х, при k 1 уменьшает его. От коэффициента k 1 зависит амплитуда колебаний синусоиды и косинусоиды.

Consider examples of solving tasks with transformations y = sin x .

Example 4

Plot y = - 3 sin 1 2 x - 3 2 - 2 using the transformations of the y=sinx function.

Decision

It is necessary to bring the function to the form ± k 1 · f ± k 2 · x + a + b . For this:

y = - 3 sin 1 2 x - 3 2 - 2 = - 3 sin 1 2 (x - 3) - 2

It can be seen that k 1 \u003d 3, k 2 \u003d 1 2, a \u003d - 3, b \u003d - 2. Since there is “-” before k 1, but not before k 2, then we get a chain of transformations of the form:

y = sin (x) → y = 3 sin (x) → y = 3 sin 1 2 x → y = - 3 sin 1 2 x → → y = - 3 sin 1 2 x - 3 → y = - 3 sin 1 2 (x - 3) - 2

Detailed sine wave conversion. When plotting the original sinusoid y \u003d sin (x), we find that T \u003d 2 π is considered the smallest positive period. Finding the maximum at points π 2 + 2 π · k ; 1 , and the minimum - π 2 + 2 π · k ; - 1 , k ∈ Z .

Stretching along O y is performed three times, which means that the increase in the amplitude of oscillations will increase by 3 times. T = 2 π is the smallest positive period. The maxima go to π 2 + 2 π · k ; 3 , k ∈ Z , minima - - π 2 + 2 π · k ; - 3 , k ∈ Z .

When stretching along O x twice, we obtain that the smallest positive period increases by 2 times and equals T \u003d 2 π k 2 \u003d 4 π. The maxima go to π + 4 π · k ; 3 , k ∈ Z , minima - in - π + 4 π · k ; - 3 , k ∈ Z .

The image is produced symmetrically with respect to O x. The smallest positive period in this case does not change and equals T = 2 π k 2 = 4 π . The maximum transition looks like - π + 4 π · k ; 3 , k ∈ Z , and the minimum is π + 4 π · k ; - 3 , k ∈ Z .

The graph is shifted down by 2 units. There is no change in the smallest common period. Finding maxima with transition to points - π + 3 + 4 π · k ; 1 , k ∈ Z , minima - π + 3 + 4 π · k ; - 5 , k ∈ Z .

On the this stage the graph of a trigonometric function is considered transformed.

Consider detailed conversion functions y = cos x .

Example 5

Plot the function y = 3 2 cos 2 - 2 x + 1 using a function transformation of the form y = cos x .

Decision

According to the algorithm, given function reduce to the form ± k 1 · f ± k 2 · x + a + b . Then we get that

y = 3 2 cos 2 - 2 x + 1 = 3 2 cos (- 2 (x - 1)) + 1

It can be seen from the condition that k 1 \u003d 3 2, k 2 \u003d 2, a \u003d - 1, b \u003d 1, where k 2 has "-", and it is absent before k 1.

From here we get that we get a graph of a trigonometric function of the form:

y = cos (x) → y = 3 2 cos (x) → y = 3 2 cos (2 x) → y = 3 2 cos (- 2 x) → → y = 3 2 cos (- 2 (x - 1 )) → y = 3 2 cos - 2 (x - 1) + 1

Step by step cosine transformation with graphic illustration.

At given schedule y = cos (x) it can be seen that the smallest common period is equal to T = 2 π . Finding maxima in 2 π · k ; 1 , k ∈ Z , and minima π + 2 π · k ; - 1 , k ∈ Z .

When stretched along O y by a factor of 32, the oscillation amplitude increases by a factor of 32. T = 2 π is the smallest positive period. Finding maxima in 2 π · k ; 3 2 , k ∈ Z , minima in π + 2 π · k ; - 3 2 , k ∈ Z .

When compressed along O x twice, we obtain that the smallest positive period is the number T = 2 π k 2 = π . The maxima are transferred to π · k ; 3 2 , k ∈ Z , minima - π 2 + π · k ; - 3 2 , k ∈ Z .

Symmetric mapping with respect to O y. Since the graph is odd, it will not change.

When shifting the graph by 1 . There are no changes in the smallest positive period T = π . Finding maxima in π · k + 1 ; 3 2 , k ∈ Z , minima - π 2 + 1 + π · k ; - 3 2 , k ∈ Z .

When shifted by 1, the smallest positive period is T = π and is not changed. Finding maxima in π · k + 1 ; 5 2 , k ∈ Z , minima in π 2 + 1 + π · k ; - 1 2 , k ∈ Z .

The transformation of the cosine function is complete.

Consider transformations using the example y = t g x .

Example 6

Plot the function y = - 1 2 t g π 3 - 2 3 x + π 3 using the transformations of the function y = t g (x) .

Decision

To begin with, it is necessary to bring the given function to the form ± k 1 f ± k 2 x + a + b, after which we obtain that

y = - 1 2 t g π 3 - 2 3 x + π 3 = - 1 2 t g - 2 3 x - π 2 + π 3

It is clearly seen that k 1 \u003d 1 2, k 2 \u003d 2 3, a \u003d - π 2, b \u003d π 3, and before the coefficients k 1 and k 2 there is a "-". So, after transforming the tangentoids, we get

y = t g (x) → y = 1 2 t g (x) → y = 1 2 t g 2 3 x → y = - 1 2 t g 2 3 x → → y = - 1 2 t g - 2 3 x → y = - 1 2 t g - 2 3 x - π 2 → → y = - 1 2 t g - 2 3 x - π 2 + π 3

Step-by-step transformation of a tangentoid with a graphic image.

We have that the original graph is y = t g (x) . The positive period change is T = π . The domain of definition is - π 2 + π · k ; π 2 + π · k , k ∈ Z .

We squeeze 2 times along O y. T \u003d π is considered the smallest positive period, where the domain of definition is - π 2 + π · k ; π 2 + π · k , k ∈ Z .

Stretch along O x 3 2 times. Let's calculate the smallest positive period, and was equal to T = π k 2 = 3 2 π . And the domain of the function with coordinates - 3 π 4 + 3 2 π · k ; 3 π 4 + 3 2 π · k , k ∈ Z , only the domain of definition changes.

Symmetry goes on the side of O x. The period will not change at this point.

It is necessary to display the coordinate axes symmetrically. The domain of definition in this case is unchanged. The chart is the same as before. This suggests that the tangent function is odd. If to odd function set a symmetric mapping O x and O y, then we will transform to the original function.