Parallel transfer.
TRANSLATION ALONG THE Y-AXIS
f(x) => f(x) - b
Suppose you want to build a graph of the function y = 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 function graph y = f(x) for b>0 and |b| units more - at b 0 or up at b To plot the graph of the function y + b = f(x), you should construct a graph of the function y = f(x) and move the x-axis to |b| units up at b>0 or by |b| units down at b
TRANSFER ALONG THE ABSCISS AXIS
f(x) => f(x + a)
Suppose you want to plot the function y = f(x + a). Consider the 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 definition of the function. Therefore, the graph of the function y = f(x + a) can be obtained by parallel moving the graph of the function y = f(x) along the x-axis to the left by |a| units for a > 0 or to the right by |a| units for a To construct a graph of the function y = f(x + a), you should construct a graph of the function y = f(x) and move the ordinate axis to |a| units to the right when a>0 or by |a| units to the left at a
Examples:
1.y=f(x+a)
2.y=f(x)+b
Reflection.
CONSTRUCTION OF A GRAPH OF A FUNCTION OF THE FORM Y = F(-X)
f(x) => f(-x)
It is obvious that 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) for the corresponding negative (positive) values of x 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 relative to the ordinate. The resulting graph is the graph of the function y = f(-x)
CONSTRUCTION OF A GRAPH OF A FUNCTION OF THE FORM 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 a graph of the function y = - f(x), you should plot a graph of the function y = f(x) and reflect it relative to the x-axis.
Examples:
1.y=-f(x)
2.y=f(-x)
3.y=-f(-x)
Deformation.
GRAPH DEFORMATION ALONG THE Y-AXIS
f(x) => k f(x)
Consider a function of the form y = k f(x), where k > 0. It is easy to see that with 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 = 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 construct a graph of the function y = k f(x), you should construct a graph of the function y = f(x) and increase its ordinates by k times for k > 1 (stretch the graph along the ordinate axis ) or reduce its ordinates by 1/k times at k
k > 1- stretching from the Ox axis
0 - compression to the OX axis
GRAPH DEFORMATION ALONG THE ABSCISS AXIS
f(x) => f(k x)
Let it be necessary to construct a graph of the function y = f(kx), where k>0. Consider the function y = f(x), which at an arbitrary point x = x1 takes the value y1 = f(x1). It is obvious that 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 definition of the function. Consequently, the graph of the function y = f(kx) turns out to be compressed (for k 1) along the abscissa axis relative to the graph of the function y = f(x). Thus, we get the rule.
To construct a graph of the function y = f(kx), you should construct a graph of the function y = f(x) and reduce its abscissas by k times for k>1 (compress the graph along the abscissa axis) or increase its abscissas 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 guidance of T.V. Tkach, S.M. Vyazov, I.V. Ostroverkhova.
©2014
Basic elementary functions in their 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 us consider the example of a 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 with respect to Oy and symmetrical with respect to Ox, and shifted by 2 3 along Ox to the right, up 2 units along Oy. 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 a 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 consider the solution using the example of a logarithmic function y = ln (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 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 . 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 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 .
At this stage, the graph of the trigonometric function is considered transformed.
Let's consider a detailed transformation of the function 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 to reduce the given function 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.
Given the graph y = cos(x), it is clear that the shortest total period is T = 2π. 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 we assign a symmetric mapping of O x and O y to an odd function, then we transform it to the original function.
Summary of algebra lesson and beginning of analysis in 10th grade
on the topic: “Transformation of graphs of trigonometric functions”
The purpose of the lesson: to systematize knowledge on the topic “Properties and graphs of trigonometric functions y=sin (x), y=cos (x)”.
Lesson objectives:
- repeat the properties of trigonometric functions y=sin (x), y=cos (x);
- repeat reduction formulas;
- converting graphs of trigonometric functions;
- develop attention, memory, logical thinking; intensify mental activity, the ability to analyze, generalize and reason;
- fostering hard work, diligence in achieving goals, interest in the subject.
Lesson equipment: ICT
Lesson type: learning new things
During the classes
Before the lesson, 2 students draw graphs from their homework on the board.
Organizing time:
Hello guys!
Today in the lesson we will transform the graphs of trigonometric functions y=sin (x), y=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 will limit ourselves to a brief reminder of the main transformations of graphs.
Transformation of function graphs.
The function y = 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 the points of the first graph by a units up.
y = f(x) – a
We lower all the points of the first graph down a units.
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, move the upper points higher by a times, and lower the lower ones 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” towards the x-axis.
y = -f(x)
Mirror the first graph about the x-axis.
y = f (ax), a<1
Fix a point on the ordinate axis. Each segment on the abscissa axis is increased by a times. The graph will stretch from the ordinate axis in different directions.
y = f (ax), a >1
Fix a point on the ordinate axis, reduce each segment on the abscissa axis by a factor. The graph will “shrink” towards the y-axis on both sides.
y = | f(x)|
The parts of the graph located under the abscissa 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 = sin x. We raise each point of the graph upward by 2 units (zeros too).
2)y = cos x – 3.
We build a graph y = cos x. We lower each point of the graph down by 3 units.
3)y = cos (x - /2)
We build a graph y = cos x. We shift all points by p/2 to the right.
4)y = 2 sinx.
We build a graph y = sin x. We leave the zeros in place, raise the upper points by 2 times, and lower the lower ones by the same amount.
PRACTICAL WORK Plotting graphs of trigonometric functions using the Advanced Grapher program.
Let's plot the function y = -cos 3x + 2.
- Let's plot the function y = cos x.
- Let's reflect it relative to the abscissa axis.
- This graph must be compressed three times along the x-axis.
- Finally, such a graph must be raised up by three units along the y-axis.
y = 0.5 sin x.
y = 0.2 cos x-2
y = 5cos 0 .5 x
y= -3sin(x+π).
2) Find the mistake and fix it.
V. Historical material. A message about Euler.
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 not sufficiently developed: there were no conventional notations, formulas were written in words, it was difficult to learn them, the question of the signs of trigonometric functions in different quarters of a circle was unclear, and the argument of a trigonometric function meant only angles or arcs. Only in the works of Euler did trigonometry receive its modern form. It was he who began to consider the trigonometric function of a number, i.e. Argument began to be understood not only as arcs or degrees, but also as numbers. Euler derived all trigonometric formulas from several basic ones and streamlined the question of the signs of the trigonometric function in different quarters of the circle. To denote trigonometric functions, he introduced the symbolism: sin x, cos x, tan x, ctg x.
At the threshold of the 18th century, a new direction appeared in the development of trigonometry - analytical. If before this 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 functions is part of the general doctrine of functions, which is studied in mathematical analysis. Part two: solving triangles - geometry chapter. Such innovations were made by Euler.
VI. Repetition
Independent work “Add the formula.”
VII. Lesson summary:
1) What new did you learn in class today?
2) What else do you want to know?
3) Grading.