Mechanical transformations of the graph of a function. Converting graphs

Converting Function Graphs

In this article I will introduce you to linear transformations of function graphs and show you how to use these transformations to obtain a function graph from a function graph

A linear transformation of a function is a transformation of the function itself and/or its argument to the form , as well as a transformation containing an argument and/or function module.

The greatest difficulties when constructing graphs using linear transformations are caused by the following actions:

Isolation basic function, in fact, the graph of which we are transforming.
Definitions of the order of transformations.

AND It is on these points that we will dwell in more detail.

Let's take a closer look at the function

It is based on the function . Let's call her basic function.

When plotting a function we perform transformations on the graph of the base function.

If we were to perform function transformations in the same order in which its value was found when a certain value argument, then

Let's consider what types of linear transformations of argument and function exist, and how to perform them.

Argument transformations.

1. f(x) f(x+b)

1. Build a graph of the function

2. Shift the graph of the function along the OX axis by |b| units

left if b>0
right if b<0

Let's plot the function

1. Build a graph of the function

2. Shift it 2 units to the right:

2. f(x) f(kx)

1. Build a graph of the function

2. Divide the abscissas of the graph points by k, leaving the ordinates of the points unchanged.

Let's build a graph of the function.

1. Build a graph of the function

2. Divide all abscissas of the graph points by 2, leaving the ordinates unchanged:

3. f(x) f(-x)

1. Build a graph of the function

2. Display it symmetrically relative to the OY axis.

Let's build a graph of the function.

1. Build a graph of the function

2. Display it symmetrically relative to the OY axis:

4. f(x) f(|x|)

1. Build a graph of the function

2. The part of the graph located to the left of the OY axis is erased, the part of the graph located to the right of the OY axis is completed symmetrically relative to the OY axis:

The function graph looks like this:

Let's plot the function

1. We build a graph of the function (this is a graph of the function, shifted along the OX axis by 2 units to the left):

2. Part of the graph located to the left of the OY (x) axis<0) стираем:

3. We complete the part of the graph located to the right of the OY axis (x>0) symmetrically relative to the OY axis:

Important! Two main rules for transforming an argument.

1. All argument transformations are performed along the OX axis

2. All transformations of the argument are performed “vice versa” and “in reverse order”.

For example, in a function the sequence of argument transformations is as follows:

1. Take the modulus of x.

2. Add the number 2 to modulo x.

But we constructed the graph in reverse order:

First, transformation 2 was performed - the graph was shifted by 2 units to the left (that is, the abscissas of the points were reduced by 2, as if “in reverse”)

Then we performed the transformation f(x) f(|x|).

Briefly, the sequence of transformations is written as follows:

Now let's talk about function transformation . Transformations are taking place

1. Along the OY axis.

2. In the same sequence in which the actions are performed.

These are the transformations:

1. f(x)f(x)+D

2. Shift it along the OY axis by |D| units

up if D>0
down if D<0

Let's plot the function

1. Build a graph of the function

2. Shift it along the OY axis 2 units up:

2. f(x)Af(x)

1. Build a graph of the function y=f(x)

2. We multiply the ordinates of all points of the graph by A, leaving the abscissas unchanged.

Let's plot the function

1. Let's build a graph of the function

2. Multiply the ordinates of all points on the graph by 2:

3.f(x)-f(x)

1. Build a graph of the function y=f(x)

Let's build a graph of the function.

1. Build a graph of the function.

2. We display it symmetrically relative to the OX axis.

4. f(x)|f(x)|

1. Build a graph of the function y=f(x)

2. The part of the graph located above the OX axis is left unchanged, the part of the graph located below the OX axis is displayed symmetrically relative to this axis.

Let's plot the function

1. Build a graph of the function. It is obtained by shifting the function graph along the OY axis by 2 units down:

2. Now we will display the part of the graph located below the OX axis symmetrically relative to this axis:

And the last transformation, which, strictly speaking, cannot be called a function transformation, since the result of this transformation is no longer a function:

|y|=f(x)

1. Build a graph of the function y=f(x)

2. We erase the part of the graph located below the OX axis, then complete the part of the graph located above the OX axis symmetrically relative to this axis.

Let's plot the equation

1. We build a graph of the function:

2. We erase the part of the graph located below the OX axis:

3. We complete the part of the graph located above the OX axis symmetrically relative to this axis.

And finally, I suggest you watch a VIDEO TUTORIAL in which I show a step-by-step algorithm for constructing a graph of a function

The graph of this function looks like this:

Depending on the conditions of physical processes, some quantities take on constant values and are called constants, others change under certain conditions and are called variables.

A careful study of the environment shows that physical quantities are dependent on each other, that is, a change in some quantities entails a change in others.

Mathematical analysis deals with the study of quantitative relationships between mutually varying quantities, abstracting from the specific physical meaning. One of the basic concepts of mathematical analysis is the concept of function.

Consider the elements of the set and the elements of the set
(Fig. 3.1).

If some correspondence is established between the elements of the sets
And in the form of a rule , then they note that the function is defined
.

Definition 3.1. Correspondence , which associates with each element not empty set
some well-defined element not empty set ,called a function or mapping
V .

Symbolically display
V is recorded in the following way:

At the same time, many
is called the domain of definition of the function and is denoted
.

In turn, many is called the range of values of the function and is denoted
.

In addition, it should be noted that the elements of the set
are called independent variables, the elements of the set are called dependent variables.

Methods for specifying a function

The function can be specified in the following main ways: tabular, graphical, analytical.

If, based on experimental data, tables are compiled that contain the values of the function and the corresponding argument values, then this method of specifying the function is called tabular.

At the same time, if some studies of the experimental result are displayed on a recorder (oscilloscope, recorder, etc.), then it is noted that the function is specified graphically.

The most common is the analytical way of specifying a function, i.e. a method in which an independent and dependent variable is linked using a formula. In this case, the domain of definition of the function plays a significant role:

different, although they are given by the same analytical relations.

If you only specify the function formula
, then we consider that the domain of definition of this function coincides with the set of those values of the variable , for which the expression
has the meaning. In this regard, the problem of finding the domain of definition of a function plays a special role.

Task 3.1. Find the domain of a function

Solution

The first term takes real values when
, and the second at. Thus, to find the domain of definition for this function it is necessary to solve the system of inequalities:

As a result, the solution to such a system is obtained. Therefore, the domain of definition of the function is the segment
.

The simplest transformations of function graphs

The construction of function graphs can be significantly simplified if you use the well-known graphs of basic elementary functions. Main elementary functions The following functions are called:

1) power function
Where
;

2)exponential function
Where
And
;

3) logarithmic function
, Where - any positive number other than one:
And
;

4) trigonometric functions

;
.

5) inverse trigonometric functions
;
;
;
.

Elementary functions are functions that are obtained from basic elementary functions using four arithmetic operations and superpositions applied a finite number of times.

Simple geometric transformations also allow you to simplify the process of constructing a graph of functions. These transformations are based on the following statements:

The graph of the function y=f(x+a) is the graph y=f(x), shifted (for a >0 to the left, for a< 0 вправо) на |a| единиц параллельно осиOx.

The graph of the function y=f(x) +b is the graph of y=f(x), shifted (at b>0 up, at b< 0 вниз) на |b| единиц параллельно осиOy.

The graph of the function y = mf(x) (m0) is the graph of y = f(x), stretched (at m>1) m times or compressed (at 0

The graph of the function y = f(kx) is the graph of y = f(x), compressed (for k >1) k times or stretched (for 0< k < 1) вдоль оси Ox. При –< k < 0 график функции y = f(kx) есть зеркальное отображение графика y = f(–kx) от оси Oy.

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

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

Yandex.RTB R-A-339285-1

Geometric transformations of the graph of a function

Applying geometric transformations of a given graph, we obtain that the graph is depicted by a function of the form ± k 1 · f (± k 2 · (x + a)) + b, when k 1 > 0, k 2 > 0 are compression coefficients 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 a symmetrical display of the graph relative to the axes, a and b shift it along O x and along O y.

Definition 1

There are 3 types geometric transformations of the graph:

Scaling along O x and O y. This is influenced by the coefficients k 1 and k 2 provided they are not equal to 1 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.
Symmetrical display relative to coordinate axes. If there is a “-” sign in front of k 1, the symmetry is relative to O x, and in front of k 2 it is relative to O y. If “-” is missing, then the item is skipped when solving;
Parallel transfer (shift) along O x and O y. The transformation is carried out if there are coefficients a and b unequal to 0. If a is positive, the graph is shifted to the left by | a | units, if a is negative, then to the right at the same distance. The b value determines the movement along the O y axis, which means that when b is positive, the function moves up, and when b is negative, it moves down.

Let's look at solutions using examples, starting with power function.

Example 1

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

Solution

Let's represent the functions this way:

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 = 2, it is worth paying attention to the presence of “-”, a = - 1 2, b = 3. From here we get that geometric transformations are carried out by stretching along O y twice, displayed symmetrically relative to O x, shifted to the right by 1 2 and upward by 3 units.

If we depict the original power function, we get that

when stretched twice along O y we have that

The mapping, symmetric with respect to O x, has the form

and move to the right by 1 2

a movement of 3 units up looks like

Let's look at transformations of exponential functions using examples.

Example 2

Construct a graph of the exponential function y = - 1 2 1 2 (2 - x) + 8.

Solution.

Let's transform the function based on the properties of a 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

From this we can see 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 find that the original exponential function has the form

Squeezing twice along O y gives

Stretching along O x

Symmetrical mapping with respect to O x

The mapping is symmetrical with respect to O y

Move up 8 units

Let's look at the solution using 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) .

Solution

To solve it is necessary 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 a 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

Let's plot the original logarithmic function

We compress the system according to O y

We stretch along O x

We perform a mapping with respect to O y

We shift up by 2 units, we get

To convert graphs trigonometric function it is necessary to fit a solution scheme of the form ± k 1 · f (± k 2 · (x + a)) + b. It is necessary that k 2 be equal to T k 2 . From here 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.

Let's look at examples of solving problems with transformations y = sin x.

Example 4

Construct a graph of y = - 3 sin 1 2 x - 3 2 - 2 using transformations of the function y=sinx.

Solution

It is necessary to reduce 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 = 3, k 2 = 1 2, a = - 3, b = - 2. Since there is a “-” 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 transformation. When plotting the original sinusoid y = sin (x), we find that the smallest positive period is considered to be T = 2 π. Finding the maximum at points π 2 + 2 π · k; 1, and the minimum - - π 2 + 2 π · k; - 1, k ∈ Z.

The O y is stretched threefold, which means 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 by half, we find that the smallest positive period increases by 2 times and is equal to T = 2 π k 2 = 4 π. The maxima go to π + 4 π · k; 3, k ∈ Z, minimums – in - π + 4 π · k; - 3, k ∈ Z.

The image is produced symmetrically with respect to O x. The smallest positive period in in this case does not change and is equal to 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. The minimum common period does not change. Finding maxima with transition to points - π + 3 + 4 π · k; 1, k ∈ Z, minimums - π + 3 + 4 π · k; - 5 , k ∈ Z .

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

Let's consider detailed conversion functions y = cos x.

Example 5

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

Solution

According to the algorithm it is necessary 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

From the condition it is clear that k 1 = 3 2, k 2 = 2, a = - 1, b = 1, where k 2 has “-”, but before k 1 it is absent.

From this we see 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 graphical illustration.

At given schedule y = cos (x) it is clear that the smallest general period equals T = 2 π. Finding maxima in 2 π · k ; 1, k ∈ Z, and there are π + 2 π · k minima; - 1, k ∈ Z.

When stretched along Oy by 3 2 times, the amplitude of oscillations increases by 3 2 times. 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 by half, we find that the smallest positive period is the number T = 2 π k 2 = π. The transition of maxima to π · k occurs; 3 2 , k ∈ Z , minimums - π 2 + π · k ; - 3 2 , k ∈ Z .

Symmetrical mapping with respect to Oy. Since the graph is odd, it will not change.

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

When shifted by 1, the smallest positive period is equal to 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 cosine function transformation is complete.

Let's consider transformations using the example y = t g x.

Example 6

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

Solution

To begin with, it is necessary to reduce 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 visible that k 1 = 1 2, k 2 = 2 3, a = - π 2, b = π 3, and in front of the coefficients k 1 and k 2 there is a “-”. This means that after transforming the tangentsoids 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 tangents with graphical representation.

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

We compress it 2 times along Oy. T = π is considered the smallest positive period, where the domain of definition has the form - 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 it was equal to T = π k 2 = 3 2 π . And the domain of definition of the function with coordinates is 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 O x side. The period will not change at this point.

It is necessary to display coordinate axes symmetrically. The domain of definition in this case is unchanged. The schedule coincides with the previous one. This suggests that the tangent function is odd. If to odd function set a symmetric mapping of O x and O y, then transform to the original function.

Back forward

Attention! Slide previews are for informational purposes only and may not represent all the features of the presentation. If you are interested this work, please download the full version.

The purpose of the lesson: Determine the patterns of transformation of function graphs.

Tasks:

Educational:

Teach students to construct graphs of functions by transforming the graph of a given function, using parallel transfer, compression (tension), different kinds symmetry.

Educational:

Bring up personal qualities students (ability to listen), goodwill towards others, attentiveness, accuracy, discipline, ability to work in a group.
Cultivate interest in the subject and the need to acquire knowledge.

Developmental:

Develop spatial imagination And logical thinking students, the ability to quickly navigate the environment; develop intelligence, resourcefulness, and train memory.

Equipment:

Multimedia installation: computer, projector.

Literature:

Bashmakov, M. I. Mathematics [Text]: textbook for institutions beginning. and Wednesday prof. education / M.I. Bashmakov. - 5th ed., revised. – M.: Publishing center“Academy”, 2012. – 256 p.
Bashmakov, M. I. Mathematics. Problem book [Text]: textbook. allowance for education institutions early and Wednesday prof. education / M. I. Bashmakov. – M.: Publishing Center “Academy”, 2012. – 416 p.

Lesson plan:

Organizational moment (3 min).
Updating knowledge (7 min).
Explanation of new material (20 min).
Consolidation of new material (10 min).
Lesson summary (3 min).
Homework(2 minutes).

During the classes

1. Org. moment (3 min).

Checking those present.

Communicate the purpose of the lesson.

The basic properties of functions as dependencies between variable quantities should not change significantly when changing the method of measuring these quantities, i.e., when changing the measurement scale and reference point. However, due to more rational choice measurement method variables It is usually possible to simplify the recording of the dependency between them and bring this recording to some standard form. In geometric language, changing the way values are measured means some simple transformations of graphs, which we will study today.

2. Updating knowledge (7 min).

Before we talk about graph transformations, let's review the material we covered.

Oral work. (Slide 2).

Functions given:

3. Describe the graphs of functions: , , , .

3. Explanation of new material (20 min).

The simplest transformations of graphs are their parallel transfer, compression (stretching) and some types of symmetry. Some transformations are presented in the table (Annex 1), (Slide 3).

Work in groups.

Each group constructs graphs of given functions and presents the result for discussion.

Function	Transforming the graph of a function	Function examples	Slide
	OU on A units up if A>0, and on \|A\| units down if A<0.	,	(Slide 4)
	Parallel transfer along the axis Oh on A units to the right if A>0, and on - A units to the left if A<0.	,	(Slide 5)