how to find the period of a graph

Amplitude and Period

Learning Objective(s)

· Understand aamplitude and period.

· Graph the sine function with changes in amplitude and period.

· Graph the cosine function with changes in aamplitude and period.

· Lucifer a sine or cosine function to its graph and vice versa.

Introduction

You know how to graph the functions  and . Now you'll learn how to graph a whole "family" of sine and cosine functions. These functions have the grade  or , where a and b are constants.

Periodic Functions

We used the variable  previously to testify an angle in standard position, and we also referred to the sine and cosine functions equally  and . Often the sine and cosine functions are used in applications that have cypher to practice with triangles or angles, and the alphabetic character 10 is used instead of  for the input (as well every bit to label the horizontal axis). Then from this point frontwards, we'll refer to these same functions as  and . This change does non touch on the graphs; they remain the same.

You know that the graphs of the sine and cosine functions take a design of hills and valleys that echo. The length of this repeating blueprint is . That is, the graph of  (or ) on the interval  looks like the graph on the interval  or  or . This pattern continues in both directions forever.

The graph beneath shows four repetitions of a pattern of length . Each one contains exactly one consummate copy of the "hill and valley" pattern.

If a function has a repeating pattern like sine or cosine, it is chosen a periodic function. The period is the length of the smallest interval that contains exactly one copy of the repeating pattern. So the period of  or  is . Whatever function of the graph that shows this pattern over i menstruation is called a wheel. For case, the graph of  on the interval  is one cycle.

Y'all know from graphing quadratic functions of the form  that as yous inverse the value of a y'all changed the "width" of the graph. At present we'll look at functions of the form  and come across how changes to b volition affect the graph. For example, is  periodic, and if and then, what is the period? Here is a table with some inputs and outputs for this function.

ten (in radians)	2ten (in radians)	sin2x
0	0	0
		i
		0
		0

As the values of x become from 0 to , the values of  go from 0 to . We tin can run into from the graph that the function  is a periodic function, and goes through one total bike on the interval [0, ], then its menses is . If you substituted values of x from  to , the values of  would become from  to , and  would become through some other complete cycle of the sine function.

Discover that has ii cycles on the interval [0, 2 ], which is the interval needs to complete 1 full bike.

What is the smallest positive value for x where  is at its minimum?

A)

B)

C)

D)

Testify/Hide Reply

A)

Wrong. The function does attain its minimum value at this betoken, simply  is not a positive value. The right answer is .

B)

Incorrect. Perhaps you lot confused minimum and maximum. The right answer is .

C)

Right. The minimum value for the sine function is . Look at the graph of . It attains this minimum at the bottom of every valley. The bottom of the showtime valley where x is positive is at .

D)

Incorrect. You may have thought of 0 every bit the minimum value, but the sine function takes on negative values. The correct answer is .

What is the period of the function ? Here is a table with some inputs and outputs for this function.

10 (in radians)	iiix (in radians)	sin3x
0	0	0
		1
		0
		0

As the values of x go from 0 to, the values of  become from 0 to . We can see from the graph that  goes through ane full bike on the interval , and so its period is .

Notice that has three cycles on the interval [0, 2 ], which is the interval needs to complete ane full cycle.

What is the period of the part ? Here is a table with some inputs and outputs for this function.

10 (in radians)	x (in radians)	sin
0	0	0
		1
		0
		0

As the values of x go from 0 to , the values of  go from 0 to .

We can come across from the graph that  goes through i full cycle on the interval , and then its menstruation is .

Notice that has half of one total cycle on the interval [0, two ], which is the interval needs to complete one full cycle.

Let's put these results into a table. For the first three functions nosotros have rewritten their periods with the numerator and so that the design becomes clear. Tin can you run into a relationship between the part and the denominator in the periods?

In each instance, the catamenia could be found by dividing  by the coefficient of ten. In general, the period of  is , and the menses of  is . Since the period is the length of an interval, it must always be a positive number. Since it is possible for b to be a negative number, we must utilise  in the formula to be sure the menstruum, , is always a positive number.

You can remember of the different values of b as having an "squeeze box" (or a spring) event on the graphs of sine and cosine. A large value of b squeezes them in and a modest value of b stretches them out.

There is some other way to depict this upshot. In the interval ,  goes through one bike while  goes through two cycles. If you go dorsum and check all of the examples above, y'all will see that  has  cycles in the interval . Besides,  has  cycles in the interval .

Example
Problem	What are the periods of  and ?
		For , . Substitute this value into the formula.
		For , .
Answer	The menses of  is , and the period of  is .

Amplitude

Every bit y'all have seen, the graphs of all of these sine and cosine functions alternate between hills and valleys. The elevation of one hill (which equals the depth of one valley) is called the amplitude. You can see that for all the graphs we accept looked at so far, the amplitude equals 1.

The formal style to say this for whatever periodic function is:

Yous know that the maximum value of  or  is 1 and the minimum value of either is . So if you applied the above definition, you would go:

This result agrees with what was observed from the graph.

You have seen that changing the value of b in  or  either stretches or squeezes the graph similar an squeeze box or a spring, simply it does non modify the maximum or minimum values. For all of these functions, the maximum is 1 and the minimum is . So if y'all applied the definition of aamplitude, yous would be doing the exact same adding as we merely did above. The amplitude of any of these functions is i.

Let's look at a unlike kind of change to a function by graphing the office . Here'due south a table with some values of this function.

x (in radians)	sinten	2sinx
0	0	0
		1
	1	ii
		i
	0	0

We'll take the kickoff and third columns to make function of the graph and so extend that design to the left and to the right.

Now you can employ this graph in the following example.

Example
Problem	What is the amplitude of ?
		You can find the maximum and minimum values of the function from the graph. For instance, at  the value is 2, and at  the value is .
		Use the definition of amplitude.
	Notice that the tiptop of each hill is 2, and the depth of each valley is ii. This is equal to the aamplitude, as nosotros mentioned at the start.
	Observe likewise that the aamplitude is equal to the coefficient of the function: This is not a coincidence.
Answer	The amplitude is two.

Allow'southward compare the graph of this function to the graph of the sine function.

The issue of multiplying  by 2 is to stretch the graph vertically past a factor of 2. Considering it has been stretched vertically by this factor, the amplitude is twice as much, or ii. If we had looked at , the graph would have been stretched vertically by a gene of 3, and the amplitude of this office is iii. Once again, this is equal to the coefficient of the part. In general, we have the following rule.

The amplitude of  or  is .

Every bit the terminal instance, , shows, multiplying by a constant on the exterior affects the aamplitude. If yous multiply by a constant on the exterior and on the inside, as in , you will affect both the aamplitude and the period. Here is the graph of :

Example
Problem	Decide the amplitude and menstruation of .
		Employ the formula for aamplitude, with .
		Use the formula for menstruum, with .
Reply	The amplitude is three and the period is .

In this example, you lot could have establish the period past looking at the graph above. I complete cycle is shown, for case, on the interval , so the flow is .

In the functions  and , multiplying past the constant a merely affects the amplitude, not the period. As we said earlier, changing the value of b but affects the menstruation, non the amplitude. The general result is as follows.

The amplitude of  or  is .

The menses of  or  is .

To aid yous understand changes in amplitude and period for both the sine role and cosine function, try the following interactive practise:

This is a Java Applet created using GeoGebra from www.geogebra.org - information technology looks like y'all don't take Java installed, please get to www.java.com

What are the aamplitude and menstruation of ?

A) The amplitude is , and the menses is .

B) The amplitude is , and the menstruum is .

C) The amplitude is i, and the period is .

D) The amplitude is one, and the menstruum is .

Show/Hibernate Answer

A) The aamplitude is , and the period is .

Correct. In this function, , so this is the amplitude. The period equals .

B) The amplitude is , and the period is .

Wrong. The amplitude is right, but the menses is not. You probably multiplied  by 4 instead of dividing. The correct answer is A.

C) The amplitude is 1, and the period is .

Incorrect. The period is right, but the amplitude is not. You may accept idea the amplitude is the maximum minus the minimum, but information technology is half of this. The correct answer is A.

D) The amplitude is ane, and the period is .

Wrong. Yous may take idea the amplitude is the maximum minus the minimum, merely information technology is half of this. You probably multiplied  by 4, instead of dividing, to find the period. The correct respond is A.

Graphs of Sine Functions

Yous know the role  has amplitude  and period . You can employ these facts to draw the graph of any role in the form  by starting with the graph of  and modifying it.

For example, suppose you lot wanted the graph of . Since , this function has the aforementioned period as . Since , the aamplitude is 4. Therefore, y'all would take the graph of  and simply stretch it vertically by a gene of iv. Here is one bicycle for these two functions.

Notation that the points that were on the x-centrality "remain" on the ten-centrality. At these points (where ), the value of  is 0. If yous multiply 0 by 4 (or anything else), you will still have a value of 0. And then the points will even so be on the x-axis. On the other paw, the highest and lowest points accept moved away from the x-centrality. They had y-values of 1 and  for , and they have y-values of iv and  for .

Case
Problem	Graph  on the interval .
	The value of b is 1, then the graph has a period of , every bit does .
	The value of a is , and so the graph has an aamplitude of 1, as does .
	Though the aamplitude and the menstruum are the same as the function , the graph is not exactly the same. The effect of multiplying by  is to replace y-values by their opposites. So the graph of  gets reflected over the 10-centrality.
Answer

If you want to check these graphs with a graphing computer, make certain that the graphing window has the correct settings. For this final example, you would apply  and . In full general, you would probably want to suit the ten-values to evidence one full bike and the y-values using the amplitude.

If the values of a and b are both dissimilar from 1, then yous need to combine the effects of the two changes.

Example
Trouble	Graph  on the interval .
	The value of a is 3, so the graph has an amplitude of 3. This has the effect of taking the graph of  and stretching it vertically by a gene of iii.
	The value of b is , and so the graph has a period of . This is twice the flow of . This has the effect of taking the graph of  and stretching it horizontally past a factor of 2. (The culling manner to say this is that  has  of a wheel on the interval .)
	To make the graph of , you must combine the two effects described to a higher place.
Answer

Sometimes you need to stretch the graph of the sine function, and sometimes y'all need to shrink it.

Example
Problem	Graph two cycles of a sine function whose amplitude is  and whose menses is.
	At that place are different functions of the grade  that fit this description considering a and b could be positive or negative. We will draw the graph bold these are positive.
	Considering the amplitude is , this has the outcome of taking the graph of  and shrinking it vertically by a factor of ii.
	The flow is , which is  the period of . This has the effect of taking the graph of  and shrinking it horizontally by a cistron of 3.
	To make the graph, you must combine the two effects described higher up.
Respond

Even without knowing the specific value of a constant, y'all can sometimes still narrow down the possibilities for the shape of a graph.

The graph of a office , where a is a abiding, is drawn on the interval .

Which of the following options could be this graph?

A)                                                               B)

C)                                                               D)

Show/Hide Respond

A)

Incorrect. This has the correct shape and menses, but information technology is in the incorrect position. Regardless of the value of a, the graph must pass through the ten-axis at , which it does non. The correct answer is D.

B)

Wrong. This is the graph of a cosine function. Regardless of the value of a, the graph must laissez passer through the x-axis at . The right answer is D.

C)

Incorrect. Considering the coefficient of 10 is 1, the graph should take a flow of , but this graph has a flow of . The correct respond is D.

D)

Correct. Because the coefficient of ten is one, the graph has a period of , which this selection has. The cistron a could stretch or shrink the graph, but it must notwithstanding pass through the x-axis at the points , which it does.

Graphs of Cosine Functions

You lot know the function  has amplitude  and period . Merely as you did with sine functions, you can employ these facts to describe the graph of any function in the form  by starting with the graph of  and modifying information technology.

For case, suppose you wanted the graph of . Since , it has the same amplitude as . And, because , the period is given by:

Since this is twice the menses of , yous would take the graph of  and stretch it horizontally by a gene of ii. Here is a side-by-side comparing of these two graphs.

Note that in the interval , the graph of  has one total bicycle. Since , the graph of  has  of a cycle in that interval.

If yous are using a graphing reckoner, you need to adjust the settings for each graph to get a graphing window that shows all the features of the graph. For the last example, you would utilize  and . In general, you would probably desire to adjust the x-values to show one full cycle and the y-values to testify the highest and everyman points. In the adjacent example, you would use  and  for the graphing window considering you are specifically asked to graph it over the domain  and the graph will have an amplitude of 2, going as depression as -2 and as high as +2.

Example
Trouble	Graph  on the interval .
	The amplitude equals . This has the outcome of taking the graph of  and stretching it vertically by a factor of 2. The negative sign on the "outside" has an boosted effect: the y-values are replaced past their opposites, and so the graph is besides flipped over the 10-axis.
	The value of b is 1, and then the graph has a period of , equally does .
	When the but change is a vertical stretch, compression, or flip, the x-intercepts remain the same. So the graph will laissez passer through the x-axis at  and .
Answer

Again, if the values of a and b are both different from 1, you lot need to combine the furnishings of the two changes. In the next case, y'all will see a variation that yous accept not seen before. Upward to this point, all of the values of b have been rational numbers, but hither nosotros are using the irrational number . This situation does not really change the procedure, simply you volition come across that information technology changes the scale on the ten-axis in a new manner.

Example
Problem	Graph  on the interval .
	The value of a is iv, then the graph has an amplitude of 4. This has the effect of taking the graph of  and stretching it vertically by a factor of 4.
	The value of b is , so the graph has a period of . This has the effect of shrinking the graph of  horizontally by a factor of , causing information technology to consummate one complete cycle on the interval [0, 2].
	Annotation the effect on the x-values of the intercepts, the high points, and the depression points. Considering the period is two, the start bicycle of the graph volition have loftier points at  and 2. The depression bespeak volition still be midway between these, then it is at . The x-intercepts are still midway between the loftier and the depression points, so they will exist at  and . The second bicycle of the graph has all of these points shifted to the right 2 units.
	To brand the graph of , you must combine the furnishings described above.
Answer

In two previous examples ( and ) yous saw that a negative sign on the exterior (a negative value of a) has the issue of flipping the graph effectually the x-axis. In the next instance, you will see the effect of a negative sign on the "inside" (a negative value of b).

I last hint: as well trying to figure out the overall issue of the value of a or b on the graph, you might want to check specific points. For example, simply substitute  into the function and meet where that indicate will terminate upwards.

Which of the following options is the graph of  on the interval ?

A)                                                            B)

C)                                                            D)

Prove/Hide Answer

A)

Incorrect. This is the graph of . Remember to check specific points similar . At that point, . Then the betoken  should be on the graph. The correct answer is C.

B)

Incorrect. You confused the effects of a and b. This is the graph of . The right respond is C.

C)

Correct. The value of a is , which will stretch the graph vertically by a cistron of . The period of the graph is , as is the catamenia of . The effect of the negative sign on the inside is to supercede ten-values by their opposites. This will flip the graph around the y-axis. However, because the graph of cosine is symmetric about the y-axis, this has no effect at all. Then the but change to the graph of  is the vertical stretch.

D)

Incorrect. This graph has the correct menstruation and amplitude. Still, you have confused the upshot of a minus sign on the inside with a minus sign on the outside. The correct answer is C.

Matching Graphs and Functions

Given whatsoever function of the form  or , you know how to find the aamplitude and menstruum and how to utilize this data to graph the functions.

Given a graph of a sine or cosine function, y'all also can determine the amplitude and period of the part. From this information, you can find values of a and b, and so a function that matches the graph.

Recollect that along with finding the aamplitude and period, it'southward a good idea to await at what is happening at . If a and b are any nonzero constants, the functions  and  will have the following values at :

This tells you lot that the graph of  passes through  regardless of the values of a and b, and the graph of  never passes through  regardless of the values of a and b. So, for example, if you lot are given a graph passing through the origin and are asked to decide which function it represents, y'all know correct away that it is not in the grade .

You need to be conscientious almost the sign of a. You might determine that a role has an amplitude of 4, for example. Though it is possible that , it is also possible that . Hither is an example of each of these two possibilities.

You will demand to compare the graph to that of  or  to see if, in improver to whatsoever stretching or shrinking, there has been a reflection over the x-axis. The graph to a higher place on the right can be thought of as the result of stretching and reflecting the graph of across the x-axis. If at that place has been a reflection, and then the value of a volition exist negative. Once you have determined if a is positive or negative, you can always choose a positive value of b.

Here are some examples of this process.

Example
Problem	Determine a function of the grade  or  whose graph is shown beneath.
	First, observe that the graph passes through the origin, so you lot are looking for a function of the form .
	Adjacent, observe that the maximum value of the office is two and the minimum is , so the amplitude is 2. The graph has the same "orientation" as . (It starts with a colina to the right of the y-axis.) This implies that a is positive, and in particular, .
	Finally, find that the graph shows 2 cycles and that one unabridged cycle is independent in the interval . Therefore, the period is . Since , . According to our process, once y'all accept determined if a is positive or negative, you can e'er choose a positive value of b . And then .
	Combine these three pieces of information.
Answer

Case
Problem	Determine a office of the grade  or  whose graph is shown below.
	First, detect that the graph does not pass through the origin, but rather crests, reaching a maximum when x = 0, then you lot are looking for a office of the form .
	Next, observe that the maximum value of the part is  and the minimum is , so the amplitude is . The graph has the same "orientation" equally . (It has a hill with the y-axis running through the heart.) This implies that a is positive, and in particular, .
	Finally, observe that the graph shows two cycles and that one entire cycle is contained in the interval . Therefore, the period is . Because , . According to our process, once you accept adamant if a is positive or negative, yous can e'er choose a positive value of b . So .
	Combine these 3 pieces of data.
Answer

Brand sure that you recognize where a cycle starts and ends. The menses is the length of the interval over which the one cycle runs.

Which of the following functions is represented by the graph below?

A)

B)

C)

D)

Prove/Hide Respond

A)

Wrong. You correctly recognized the graph as a reflected sine office, but the catamenia is incorrect. Peradventure you saw the  on the right and used that equally the length of one cycle. Nonetheless, the entire graph is one cycle, and the period equals . The correct answer is D.

B)

Wrong. You lot correctly found the amplitude and menses of this sine function. However, you also need to cheque the orientation of the graph. Find that to the right of the y-centrality you lot take a valley instead of a hill. The right reply is D.

C)

Incorrect. You lot correctly found the aamplitude and the orientation of this sine function. Nevertheless, the menstruation is wrong. Perhaps you recognized that the period of the graph is twice the catamenia of , and thought that the value of b would be ii. But to find the value of b, you must gear up the period equal to . The correct reply is D.

D)

Correct. The graph passes through the origin, so the function could have the course , just not . The amplitude is 1. The graph has a valley on the right, which could be the result of a reflection of  over the x-axis. Therefore, . The graph shows one cycle, and so the menstruation is . Since , the value of b could be . So this could be the graph of .

Remember that when writing a office you can use the notation  in place of the variable y.

Which of the following graphs represents ?

A)                                                                B)

C)                                                                D)

Bear witness/Hide Answer

A)

Wrong. This is the graph of a function of the form . Call up to bank check the value of the function at . Since , the role  passes through , not the origin as shown in this graph. The correct answer is B.

B)

Correct. First, this graph has the shape of a cosine function. Second, considering  in the equation, the amplitude is three. Finally, because , the period of this function is . The graph in this answer completes 1 total cycle between and  and then its flow is as needed.

C)

Incorrect. This graph does have the shape of a cosine part, and the amplitude is 3, which is correct. However, the period is wrong. You probably multiplied  by  instead of dividing. The correct answer is B.

D)

Incorrect. This graph does have the shape of a cosine function. However, in determining the graph, it appears that y'all switched the values of a and b. The correct reply is B.

Summary

The functions  and  are periodic functions: their graphs have a repeating pattern of hills and valleys that continues in both directions forever. The height of the loma or the depth of the valley is chosen the amplitude, and is equal to . Any one total pattern in the graph is called a bicycle, and the length of an interval over which a cycle occurs is called the period. The menstruation is equal to the value .

Yous can use this data to graph any of these functions by starting with the basic graph of  or  and then doing a combination of stretching or shrinking the graph vertically based on the value of a, stretching or shrinking the graph horizontally based on the value of b, or reflecting information technology based on the signs of a and b.

You tin also get-go with a graph, determine the values of a and b, so make up one's mind a function that information technology represents.

Source: http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L2_T3_text_final.html

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