Amplitude and Period

Learning Objectives

Understand amplitude and period.
Graph the sine function with changes in amplitude and period.
Graph the cosine function with changes in amplitude and period.
Match a sine or cosine function to its graph and vice versa.

Introduction

You know how to graph the functions $y= sin x$ and $y=cosx$ . Now you’ll learn how to graph a whole “family” of sine and cosine functions. These functions have the form $y=a sin bx$ or $y=a cos bx$ , where $a$ and $b$ are constants.

Periodic Functions

We used the variable $theta$ previously to show an angle in standard position, and we also referred to the sine and cosine functions as $y=sin theta$ and $y=cos theta$ . Often the sine and cosine functions are used in applications that have nothing to do with triangles or angles, and the letter $x$ is used instead of $theta$ for the input (as well as to label the horizontal axis). So from this point forward, we’ll refer to these same functions as $y=sin x$ and $y=cos x$ . This change does not affect the graphs; they remain the same.

You know that the graphs of the sine and cosine functions have a pattern of hills and valleys that repeat. The length of this repeating pattern is $2pi$ . That is, the graph of $y=sinx$ (or $y=cosx$ ) on the interval $[0, 2pi]$ looks like the graph on the interval $[2pi, 4pi]$ or $[4pi, 6pi]$ or $[-2pi, 0]$ . This pattern continues in both directions forever.

The graph below shows four repetitions of a pattern of length $2pi$ . Each one contains exactly one complete copy of the “hill and valley” pattern.

If a function has a repeating pattern like sine or cosine, it is called a periodic functionA function whose graph has a pattern that repeats forever in both directions.. The periodThe length of the smallest interval that contains exactly one copy of the repeating pattern of a periodic function. is the length of the smallest interval that contains exactly one copy of the repeating pattern. So the period of $y=sinx$ or $y=cosx$ is $2pi$ . Any part of the graph that shows this pattern over one period is called a cycleAny part of a graph of a periodic function that is one period long.. For example, the graph of $y=cosx$ on the interval $[0, 2pi]$ is one cycle.

A graph shows the function Y equals cosine X. The curve has a repeating valley and hill pattern from negative 4 pi to 4 pi. It is highlighted from 0 to 2 pi to show 1 cycle.

You know from graphing quadratic functions of the form $y=ax^2$ that as you changed the value of $a$ , you changed the “width” of the graph. Now we’ll look at functions of the form $y=sinbx$ and see how changes to $b$ will affect the graph. For example, is $y=sin 2x$ periodic, and if so, what is the period? Here is a table with some inputs and outputs for this function.

$x$ (in radians)	$2x$ (in radians)	$sin2x$
$0$	$0$	$0$
$pi/6$	$pi/3$	$sqrt3/2$
$pi/4$	$pi/2$	$1$
$pi/3$	$(2pi)/3$	$sqrt3/2$
$pi/2$	$pi$	$0$
$(2pi)/3$	$(4pi)/3$	$-sqrt3/2$
$(3pi)/4$	$(3pi)/2$	$-1$
$(5pi)/6$	$(5pi)/3$	$-sqrt3/2$
$pi$	$2pi$	$0$

As the values of $x$ go from $0$ to $pi$ , the values of $2x$ go from $0$ to $2pi$ . We can see from the graph that the function $y=sin2x$ is a periodic function, and goes through one full cycle on the interval $[0, pi]$ , so its period is $pi$ . If you substituted values of $x$ from $pi$ to $2pi$ , the values of $2x$ would go from $2pi$ to $4pi$ , and $sin2x$ would go through another complete cycle of the sine function.

A graph shows the function Y equals sine 2 X. The curve has a repeating hill and valley pattern from negative 4 pi to 4 pi. One period of pi is highlighted from 0 to pi. There are 2 cycles on the interval 0 to 2 pi.

Notice that $y=sin2x$ has two cycles on the interval $[0, 2pi]$ which is the interval $y=sinx$ needs to complete one full cycle.

What is the smallest positive value for $x$ where $y=sin2x$ is at its minimum?

A) $-pi/4$

B) $pi/4$

C) $(3pi)/4$

D) $pi/2$

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What is the period of the function $y=sin3x$ ? Here is a table with some inputs and outputs for this function.

$x$ (in radians)	$3x$ (in radians)	$sin3x$
$0$	$0$	$0$
$pi/9$	$pi/3$	$sqrt3/2$
$pi/6$	$pi/2$	$1$
$(2pi)/9$	$(2pi)/3$	$sqrt3/2$
$pi/3$	$pi$	$0$
$(4pi)/9$	$(4pi)/3$	$-sqrt3/2$
$pi/2$	$(3pi)/2$	$-1$
$(5pi)/9$	$(5pi)/3$	$-sqrt3/2$
$(2pi)/3$	$2pi$	$0$

As the values of $x$ go from $0$ to $(2pi)/3$ , the values of $3x$ go from $0$ to $2pi$ . We can see from the graph that $y=sin3x$ goes through one full cycle on the interval $[0,(2pi)/3]$ , so its period is $(2pi)/3$ .

A graph shows the function Y equals sine 3 X. The curve has a repeating hill and valley pattern from negative 2 pi to 2 pi. One period of 2 pi over 3 is highlighted from 0 to 2 pi over 3. There are 3 cycles on the interval 0 to 2 pi.

Notice that $y=sin3x$ has three cycles on the interval $[0, 2pi]$ , which is the interval $y=sin x$ needs to complete one full cycle.

What is the period of the function $y=sin(1/2)x$ ? Here is a table with some inputs and outputs for this function.

$x$ (in radians)	$1/2x$ (in radians)	$sin(1/2)x$
$0$	$0$	$0$
$(2pi)/3$	$pi/3$	$sqrt3/2$
$pi$	$pi/2$	$1$
$(4pi)/3$	$(2pi)/3$	$sqrt3/2$
$2pi$	$pi$	$0$
$(8pi)/3$	$(4pi)/3$	$-sqrt3/2$
$3pi$	$(3pi)/2$	$-1$
$(10pi)/3$	$(5pi)/3$	$-sqrt3/2$
$4pi$	$2pi$	$0$

As the values of $x$ go from $0$ to $4pi$ , the values of $1/2x$ go from $0$ to $2pi$ .

We can see from the graph that $y=sin(1/2x)$ goes through one full cycle on the interval $[0,4pi]$ , so its period is $4pi$ .

A graph shows the function Y equals sine one-half X. The curve has a repeating hill and valley pattern from negative 4 pi to 4 pi. One period of 4 pi is highlighted from 0 to 4 pi. There is half of 1 full cycle on the interval 0 to 2 pi.

Notice that $y=sin(1/2x)$ has half of one full cycle on the interval $[0, 2^pi]$ , which is the interval $y=sin x$ needs to complete one full cycle.

Let’s put these results into a table. For the first three functions, we have rewritten their periods with the numerator $2pi$ so that the pattern becomes clear. Can you see a relationship between the function and the denominator in the periods?

Function	Period
$y=sin(1/2x)$	$4pi=(2pi)/(1/2)$
$y=sin1x$	$2pi=(2pi)/1$
$y=sin2x$	$pi=(2pi)/2$
$y=sin(3x)$	$(2pi)/3=(2pi)/3$

In each case, the period could be found by dividing $2pi$ by the coefficient of $x$ . In general, the period of $y=sin bx$ is $(2pi)/|b|$ , and the period of $y=cosbx$ is $(2pi)/|b|$ . 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 use $|b|$ in the formula to be sure the period, $(2pi)/|b|$ , is always a positive number.

The period of $y=sinbx$ or $y=cosbx$ is $(2pi)/|b|$ .

You can think of the different values of $b$ as having an “accordion” (or a spring) effect on the graphs of sine and cosine. A large value of $b$ squeezes them in and a small value of $b$ stretches them out.

There is another way to describe this effect. In the interval $[0, 2pi]$ , $y=sinx$ goes through one cycle while $y=sin2x$ goes through two cycles. If you go back and check all of the examples above, you will see that $y=sinbx$ has $|b|$ cycles in the interval $[0, 2pi]$ . Likewise, $y=cosbx$ has $|b|$ cycles in the interval $[0, 2pi]$ .

Example
Problem	What are the periods of $y=sin4x$ and $y=cos(-1/2x)$ ?
	$"period"=(2pi)/\|b\|=(2pi)/4=pi/2$	For $y=sin4x$ , $b=4$ . Substitute this value into the formula.
	$"period"=(2pi)/\|b\|=(2pi)/(\|-1/2\|)=(2pi)/(1/2)$ $=(2pi)/1*2/1=4pi$	For $y=cos(-1/2x)$ , $b=-1/2$ .
Answer	The period of $y=sin4x$ is $pi/2$ , and the period of $y=cos(-1/2x)$ is $4pi$ .

Amplitude

As you have seen, the graphs of all of these sine and cosine functions alternate between hills and valleys. The height of one hill (which equals the depth of one valley) is called the amplitudeHalf the difference between the maximum and the minimum values of a periodic function.. You can see that for all the graphs we have looked at so far, the amplitude equals $1$ .

A graph shows the function Y equals sine X. The curve has a repeating hill and valley pattern from negative 2 pi to 2 pi. The height of the hills or the maximum is 1. The depth of the valleys or the minimum is negative 1.

The formal way to say this for any periodic function is:

$"amplitude"=("maximum"-"minimum")/2$

You know that the maximum value of $y=sinx$ or $y=cosx$ is $1$ and the minimum value of either is $-1$ . So if you applied the above definition, you would get:

$"amplitude"=(1-(-1))/2=(1+1)/2=2/2=1$

This result agrees with what was observed from the graph.

You have seen that changing the value of $b$ in $y=sinbx$ or $y=cosbx$ either stretches or squeezes the graph like an accordion or a spring, but it does not change the maximum or minimum values. For all of these functions, the maximum is $1$ and the minimum is $-1$ . So if you applied the definition of amplitude, you would be doing the exact same calculation as we just did above. The amplitude of any of these functions is $1$ .

Let’s look at a different kind of change to a function by graphing the function $y=2sinx$ . Here’s a table with some values of this function.

$x$ (in radians)	$sinx$	$2 sinx$
$0$	$0$	$0$
$pi/6$	$1/2$	$1$
$pi/4$	$sqrt2/2$	$sqrt2$
$pi/3$	$sqrt3/2$	$sqrt3$
$pi/2$	$1$	$2$
$(2pi)/3$	$sqrt3/2$	$sqrt3$
$(3pi)/4$	$sqrt2/2$	$sqrt2$
$(5pi)/6$	$1/2$	$1$
$pi$	$0$	$0$

We’ll take the first and third columns to make part of the graph and then extend that pattern to the left and to the right.

A graph shows the function Y equals 2 sine X. The curve has a repeating hill and valley pattern from negative 2 pi to 2 pi. The maximum is 2 and the minimum is negative 2. The curve completes 1 cycle on an interval from 0 to 2 pi.

Now you can use this graph in the following example.

Example
Problem	What is the amplitude of $y=2 sin x$ ?
	$"maximum"=2$ $"minimum"=-2$	You can find the maximum and minimum values of the function from the graph. For example, at $x=pi/2$ the value is $2$ , and at $x=-pi/2$ the value is $-2$ .
	$"amplitude"=(2-(-2))/2=(2+2)/2=4/2=2$	Use the definition of amplitude.
	Notice that the height of each hill is $2$ , and the depth of each valley is $2$ . This is equal to the amplitude, as we mentioned at the start.
	Notice also that the amplitude is equal to the coefficient of the function: $y=2sinx$ This is not a coincidence.
Answer	The amplitude is $2$ .

Let’s compare the graph of this function to the graph of the sine function.

On the left, a graph shows the function Y equals sine X. The curve has a repeating hill and valley pattern from negative 2 pi to 2 pi. The maximum is 1 and the minimum is negative 1. The curve completes 1 cycle on an interval from 0 to 2 pi. On the right, a graph shows the function Y equals 2 sine X. The curve has a repeating hill and valley pattern from negative 2 pi to 2 pi. The maximum is 2 and the minimum is negative 2. The curve completes 1 cycle on an interval from 0 to 2 pi.

The effect of multiplying $sinx$ by $2$ is to stretch the graph vertically by a factor of $2$ . Because it has been stretched vertically by this factor, the amplitude is twice as much, or $2$ . If we had looked at $y=3sinx$ , the graph would have been stretched vertically by a factor of $3$ , and the amplitude of this function is $3$ . Again, this is equal to the coefficient of the function. In general, we have the following rule.

The amplitude of $y=asinbx$ or $y=acosbx$ is $|a|$ .

As the last example, $y=2sinx$ , shows, multiplying by a constant on the outside affects the amplitude. If you multiply by a constant on the outside and on the inside, as in $y=-3cos2x$ , you will affect both the amplitude and the period. Here is the graph of $y=-3cos2x$ :

A graph shows Y equals negative 3 cosine 2 X. The curve has a repeating hill and valley pattern from negative 2 pi to 2 pi. The maximum is 3 and the minimum is negative 3. The curve completes 1 cycle on an interval from 0 to pi.

Example
Problem	Determine the amplitude and period of $y=-3cos2x$ .
	$"amplitude"=\|a\|=\|-3\|=3$	Use the formula for amplitude, with $a=-3$ .
	$"period"=(2pi)/\|b\|=(2pi)/2=pi$	Use the formula for period, with $b=2$ .
Answer	The amplitude is $3$ and the period is $pi$ .

In this example, you could have found the period by looking at the graph above. One complete cycle is shown, for example, on the interval $[0, pi]$ , so the period is $pi$ .

In the functions $y=asinbx$ and $y=acosbx$ , multiplying by the constant $a$ only affects the amplitude, not the period. As we said earlier, changing the value of $b$ only affects the period, not the amplitude. The general result is as follows.

The amplitude of $y=asinbx$ or $y=acosbx$ is $|a|$ .

The period of $y=asinbx$ or $y=acosbx$ is $(2pi)/|b|$ .

Supplemental Interactive Activity

To help you understand changes in amplitude and period for both the sine function and cosine function, try the following interactive exercise.

What are the amplitude and period of $y=1/2sin4x$ ?

A) The amplitude is $1/2$ , and the period is $(pi)/2$ .

B) The amplitude is $1/2$ , and the period is $8pi$ .

C) The amplitude is $1$ , and the period is $pi/2$ .

D) The amplitude is $1$ , and the period is $8pi$ .

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Graphs of Sine Functions

You know the function $y=asinbx$ has amplitude $|a|$ and period $(2pi)/|b|$ . You can use these facts to draw the graph of any function in the form $y=asinbx$ by starting with the graph of $y=sinx$ and modifying it.

For example, suppose you wanted the graph of $y=4sinx$ . Since $b=1$ , this function has the same period as $y=sinx$ . Since $a=4$ , the amplitude is $4$ . Therefore, you would take the graph of $y=sinx$ and simply stretch it vertically by a factor of $4$ . Here is one cycle for these two functions.

A graph shows Y equals sine X. A curve with a hill and valley is plotted with the following coordinates: (0, 0), (pi over 2, 1), (pi, 0), (3 pi over 2, negative 1), and (2 pi, 0). A graph shows Y equals 4 sine X. A curve with a hill and valley is plotted with the following coordinates: (0, 0), (pi over 2, 4), (pi, 0), (3 pi over 2, negative 4), and (2 pi, 0).

Note that the points that were on the $x$ -axis “remain” on the $x$ -axis. At these points (where $x=0, pi, 2pi$ ), the value of $sinx$ is $0$ . If you multiply $0$ by $4$ (or anything else), you will still have a value of $0$ . So the points will still be on the $x$ -axis. On the other hand, the highest and lowest points have moved away from the $x$ -axis. They had $y$ -values of $1$ and $-1$ for $sinx$ , and they have $y$ -values of $4$ and $-4$ for $4sinx$ .

Example
Problem	Graph $y=-sinx$ on the interval $[0, 2pi]$ .
	The value of $b$ is $1$ , so the graph has a period of $2pi$ , as does $y=sinx$ .
	The value of $a$ is $-1$ , so the graph has an amplitude of $1$ , as does $y=sinx$ .
	Though the amplitude and the period are the same as the function $y=sinx$ , the graph is not exactly the same. The effect of multiplying by $-1$ is to replace $y$ -values by their opposites. So the graph of $y=sinx$ gets reflected over the $x$ -axis.
Answer

If you want to check these graphs with a graphing calculator, make sure that the graphing window has the correct settings. For this last example, you would use $0<=x<=2pi$ and $-1<=y<=1$ . In general, you would probably want to adjust the $x$ -values to show one full cycle and the $y$ -values using the amplitude.

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

Example
Problem	Graph $y=3sin(1/2x)$ on the interval $[0, 4pi]$ .
	The value of $a$ is $3$ , so the graph has an amplitude of $3$ . This has the effect of taking the graph of $y=sinx$ and stretching it vertically by a factor of $3$ .
	The value of $b$ is $1/2$ , so the graph has a period of $(2pi)/(1/2)=(2pi)/1*2/1=4pi$ . This is twice the period of $y=sinx$ . This has the effect of taking the graph of $y=sinx$ and stretching it horizontally by a factor of $2$ . (The alternative way to say this is that $y=3sin(1/2x)$ has $1/2$ of a cycle on the interval $[0, 2pi]$ .)
	To make the graph of $y=3sin(1/2x)$ , you must combine the two effects described above.

Answer

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

Example
Problem	Graph two cycles of a sine function whose amplitude is $1/2$ and whose period is $(2pi)/3$ .
	There are different functions of the form $y=asinbx$ that fit this description because $a$ and $b$ could be positive or negative. We will draw the graph assuming these are positive.
	Because the amplitude is $1/2$ , this has the effect of taking the graph of $y=sinx$ and shrinking it vertically by a factor of $2$ .
	The period is $(2pi)/3=1/3*2pi$ , which is $1/3$ the period of $y=sinx$ . This has the effect of taking the graph of $y=sinx$ and shrinking it horizontally by a factor of $3$ .
	To make the graph, you must combine the two effects described above.
Answer

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

The graph of a function $y=asinx$ , where $a$ is a constant, is drawn on the interval $0<=x<=2pi$ .

Which of the following options could be this graph?

The graph of a curve starts at (0, negative 1). The maximum is 1 and the minimum is negative 3. The curve completes 1 cycle on an interval from 0 to 2 pi.

The graph of a curve starts at (0, 2). The maximum is 2 and the minimum is negative 2. The curve completes 1 cycle on an interval from 0 to 2 pi.

The graph of a curve starts at (0, 0). The maximum is 1 and the minimum is negative 1. The curve completes 2 cycles on an interval from 0 to 2 pi.

The graph of a curve starts at (0, 0). The maximum is 2 and the minimum is negative 2. The curve completes 1 cycle on an interval from 0 to 2 pi.

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Graphs of Cosine Functions

You know the function $y=acosbx$ has amplitude $|a|$ and period $(2pi)/|b|$ . Just as you did with sine functions, you can use these facts to draw the graph of any function in the form $y=acosbx$ by starting with the graph of $y=cosx$ and modifying it.

For example, suppose you wanted the graph of $y=cos(1/2x)$ . Since $a=1$ , it has the same amplitude as $y=cosx$ . And, because $b=1/2$ , the period is given by:

$(2pi)/|b|=(2pi)/(1/2)=(2pi)/1*2/1=4pi$

Since this is twice the period of $y=cosx$ , you would take the graph of $y=cosx$ and stretch it horizontally by a factor of $2$ . Here is a side-by-side comparison of these two graphs.

On the left, a graph shows the function Y equals cosine X. The curve has a repeating valley and hill pattern from 0 to 4 pi. The maximum is 1 and the minimum is negative 1. The curve completes 1 cycle on an interval from 0 and 2 pi. On the right, a graph shows the function Y equals cosine one-half X. The curve has a valley and hill pattern from 0 to 4 pi. The maximum is 1 and the minimum is negative 1. The curve completes 1 full cycle between 0 and 4 pi, so only half of 1 full cycle occurs between 0 and 2 pi.

Note that in the interval $[0, 2pi]$ , the graph of $y=cosx$ has one full cycle. Since $b=1/2$ , the graph of $y=cos(1/2x)$ has $1/2$ of a cycle in that interval.

If you are using a graphing calculator, 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 use $0<=x<=4pi$ and $-1<=y<=1$ . In general, you would probably want to adjust the $x$ -values to show one full cycle and the $y$ -values to show the highest and lowest points. In the next example, you would use $-pi<=x<=pi$ and $-2<=y<=2$ for the graphing window because you are specifically asked to graph it over the domain $[-pi,pi]$ and the graph will have an amplitude of $2$ , going as low as $-2$ and as high as $+2$ .

Example
Problem	Graph $y=-2cosx$ on the interval $[-pi,pi]$ .
	The amplitude equals $\|a\|=\|-2\|=2$ . This has the effect of taking the graph of $y=cosx$ and stretching it vertically by a factor of $2$ . The negative sign on the “outside” has an additional effect: the $y$ -values are replaced by their opposites, so the graph is also flipped over the $x$ -axis.
	The value of $b$ is $1$ , so the graph has a period of $2pi$ , as does $y=cosx$ .
	When the only change is a vertical stretch, compression, or flip, the $x$ -intercepts remain the same. So the graph will pass through the $x$ -axis at $-pi/2$ and $pi/2$ .
Answer

Again, if the values of $a$ and $b$ are both different from $1$ , you need to combine the effects of the two changes. In the next example, you will see a variation that you have not seen before. Up to this point, all of the values of $b$ have been rational numbers, but here we are using the irrational number $pi$ . This situation does not really change the procedure, but you will see that it changes the scale on the $x$ -axis in a new way.

Example
Problem	Graph $y=4cospi x$ on the interval $[0,4]$ .
	The value of $a$ is $4$ , so the graph has an amplitude of $4$ . This has the effect of taking the graph of $y=cosx$ and stretching it vertically by a factor of $4$ .
	The value of $b$ is $pi$ , so the graph has a period of $(2pi)/pi=2$ . This has the effect of shrinking the graph of $y=cosx$ horizontally by a factor of $pi$ , causing it to complete one complete cycle on the interval $[0, 2]$ .
	Note the effect on the $x$ -values of the intercepts, the high points, and the low points. Because the period is $2$ , the first cycle of the graph will have high points at $x=0$ and $2$ . The low point will still be midway between these, so it is at $x=1$ . The $x$ -intercepts are still midway between the high and the low points, so they will be at $x=1/2$ and $3/2$ . The second cycle of the graph has all of these points shifted to the right $2$ units.
	To make the graph of $y=4cospix$ , you must combine the effects described above.
Answer

In two previous examples ( $y=-sinx$ and $y=-2cosx$ ), you saw that a negative sign on the outside (a negative value of $a$ ) has the effect of flipping the graph around the $x$ -axis. In the next example, you will see the effect of a negative sign on the “inside” (a negative value of $b$ ).

One last hint: besides trying to figure out the overall effect of the value of $a$ or $b$ on the graph, you might want to check specific points. For example, just substitute $x=0$ into the function and see where that point will end up.

Which of the following options is the graph of $y=3/2 cos(-x)$ on the interval $-pi<=x<=pi$ ?

A graph shows a curve that has a maximum of 3 over 2, a minimum of negative 3 over 2, and it crosses the y-axis at the origin. It completes 1 cycle on the interval from negative pi to pi.

A graph shows a curve that has a maximum of 1, a minimum of negative 1, and it crosses the y-axis at (0, negative 1). It completes one and one-half cycles on the interval from negative pi to pi.

A graph shows a curve that has a maximum of 3 over 2, a minimum of negative 3 over 2, and it crosses the y-axis at (0, 3 over 2). It completes 1 cycle on the interval from negative pi to pi.

A graph shows a curve that has a maximum of 3 over 2, a minimum of negative 3 over 2, and it crosses the y-axis at (0, negative 3 over 2). It completes 1 cycle on the interval from negative pi to pi.

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Matching Graphs and Functions

Given any function of the form $y=asinbx$ or $y=acosbx$ , you know how to find the amplitude and period and how to use this information to graph the functions.

Given a graph of a sine or cosine function, you also can determine the amplitude and period of the function. From this information, you can find values of $a$ and $b$ , and then a function that matches the graph.

Remember that along with finding the amplitude and period, it’s a good idea to look at what is happening at $x=0$ . If $a$ and $b$ are any nonzero constants, the functions $y=asinbx$ and $y=acosbx$ will have the following values at $x=0$ :

$y=asin(b*0)=asin0=a*0=0$

$y=acos(b*0)=acos0=a*1=a!=0$

This tells you that the graph of $y=asinbx$ passes through $(0,0)$ regardless of the values of $a$ and $b$ , and the graph of $y=acosbx$ never passes through $(0,0)$ regardless of the values of $a$ and $b$ . So, for example, if you are given a graph passing through the origin and are asked to determine which function it represents, you know right away that it is not in the form $y=acosbx$ .

You need to be careful about the sign of $a$ . You might determine that a function has an amplitude of $4$ , for example. Though it is possible that $a=4$ , it is also possible that $a=-4$ . Here is an example of each of these two possibilities.

A graph shows the function Y equals 4 sine X. The curve has a hill and valley from 0 to 2 pi. The maximum is 4 and the minimum is negative 4. The curve completes 1 cycle on an interval from 0 and 2 pi.

You will need to compare the graph to that of $y=sinx$ or $y=cosx$ to see if, in addition to any stretching or shrinking, there has been a reflectionA mirror-image of a graph. If the reflection is over the $x$ -axis, then the part of the original graph that was below the $x$ -axis will be above the $x$ -axis, and vice versa. over the $x$ -axis. The graph above on the right can be thought of as the result of stretching and reflecting the graph of $y=sinx$ across the $x$ -axis. If there has been a reflection, then the value of $a$ will be 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 form $y=asinbx$ or $y=acosbx$ whose graph is shown below.
	First, observe that the graph passes through the origin, so you are looking for a function of the form $y=asinbx$ .
	Next, observe that the maximum value of the function is $2$ and the minimum is $-2$ , so the amplitude is $2$ . The graph has the same “orientation” as $y=sinx$ . (It starts with a hill to the right of the $y$ -axis.) This implies that $a$ is positive, and in particular, $a=2$ .
	Finally, observe that the graph shows two cycles and that one entire cycle is contained in the interval $[0, (2pi)/3]$ . Therefore, the period is $(2pi)/3$ . Since $(2pi)/3=(2pi)/\|b\|$ , $b=+-3$ . According to our process, once you have determined if $a$ is positive or negative, you can always choose a positive value of $b$ . So $b=3$ .
	Combine these three pieces of information.
Answer	$y=2sin3x$

Example
Problem	Determine a function of the form $y=asinbx$ or $y=acosbx$ whose graph is shown below.
	First, observe that the graph does not pass through the origin, but rather crests, reaching a maximum when $x = 0$ , so you are looking for a function of the form $y=acosbx$ .
	Next, observe that the maximum value of the function is $1/2$ and the minimum is $-1/2$ , so the amplitude is $1/2$ . The graph has the same “orientation” as $y=cosx$ . (It has a hill with the $y$ -axis running through the middle.) This implies that $a$ is positive, and in particular, $a=1/2$ .
	Finally, observe that the graph shows two cycles and that one entire cycle is contained in the interval $[0,pi]$ . Therefore, the period is $pi$ . Because $pi=(2pi)/\|b\|$ , $b=+-2$ . According to our process, once you have determined if $a$ is positive or negative, you can always choose a positive value of $b$ . So $b=2$ .
	Combine these three pieces of information.
Answer	$y=1/2cos2x$

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

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

A graph shows a curve that completes 1 cycle on an interval from negative 2 pi to 2 pi. It starts on the x-axis at (negative 2 pi, 0), forms a hill with a maximum value at (negative pi, 1), passes through the origin, decreases to a valley with a minimum value at (pi, negative 1), and returns to the y-axis at (2 pi, 0).

A) $y=-sinx$

B) $y=sin(1/2 x)$

C) $y=-sin2x$

D) $y=-sin(1/2 x)$

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Remember that when writing a function you can use the notation $f(x)$ in place of the variable $y$ .

Which of the following graphs represents $f(x)=3cos(2/3x)$ ?

A graph shows a curve that has a maximum of 3, a minimum of negative 3, and it crosses the y-axis at the origin. It completes 1 cycle on an interval from negative 3 pi over 2 to 3 pi over 2.

A graph shows a curve that has a maximum of 3, a minimum of negative 3, and it crosses the y-axis at (0, 3). It completes 1 cycle on an interval from negative 3 pi over 2 to 3 pi over 2.

A graph shows a curve that has a maximum of 3, a minimum of negative 3, and it crosses the y-axis at (0, 3). It completes 2 cycles on an interval from negative 4 pi over 3 to 4 pi over 3.

A graph shows a curve that has a maximum of two-thirds, a minimum of negative two-thirds, and crosses the y-axis at (0, two-thirds). It completes 4 cycles on an interval from negative 4 pi over 3 to 4 pi over 3.

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Summary

The functions $y=asinbx$ and $y=acosbx$ are periodic functions: their graphs have a repeating pattern of hills and valleys that continues in both directions forever. The height of the hill or the depth of the valley is called the amplitude, and is equal to $|a|$ . Any one full pattern in the graph is called a cycle, and the length of an interval over which a cycle occurs is called the period. The period is equal to the value $(2pi)/|b|$ .

You can use this information to graph any of these functions by starting with the basic graph of $y=sinx$ or $y=cosx$ 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 it based on the signs of $a$ and $b$ .

You can also start with a graph, determine the values of $a$ and $b$ , and then determine a function that it represents.