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This lesson will introduce the transformations of the graphs of the sine, cosine, and tangent function. These trigonometric functions will be used to model the movement of a variety of attractions in an amusement park. Their transformations will be determined by comparing the parameters of the function rules with their parent functions.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.
Topics Related to Transformations of Functions

Translation of a Function
Horizontal Stretch and Shrink of a Function
Vertical Stretch and Shrink of a Function
Reflection of a Function

Topics Related to Trigonometric Functions

Explore

Stretching and Shrinking a Graph

The graphs of sine, cosine, and tangent functions are shown in the following applet. Investigate how their graphs are transformed as the values

a

and

b

change.

Think about the following questions!

How is each graph transformed as the value of $a$ changes?
How is each graph transformed as the value of $b$ changes?

Discussion

Stretching and Shrinking Trigonometric Functions

Like other functions, the parent functions of sine, cosine, and tangent can be transformed. The graphs of $y = a sin (b x),$ $y = a cos (b x),$ and $y = a tan (b x)$ represent stretch or shrink transformations of their parent functions.

Vertical Shrink or Stretch

The graph of a parent trigonometric function can be vertically stretched or shrunk by multiplying the function rule by a constant, positive number $a .$ If $a > 1,$ the function will be stretched. Conversely, the function will be shrunk if $0 < a < 1 .$

Vertical Stretch or Shrink
Parent Function	$a > 1$	$0 < a < 1$
$y = sin x$	$y = a sin x$ Stretch parent function vertically by a factor of $a$	$y = a sin x$ Shrink parent function vertically by a factor of $a$
$y = cos x$	$y = a cos x$ Stretch parent function vertically by a factor of $a$	$y = a cos x$ Shrink parent function vertically by a factor of $a$
$y = tan x$	$y = a tan x$ Stretch parent function vertically by a factor of $a$	$y = a tan x$ Shrink parent function vertically by a factor of $a$

Examine how the sine and cosine graphs change when the factor

a

changes.

Vertical stretch or shrink by changing the value of $a$ for sine and cosine functions

As shown, the factor

a

affects the amplitudes of the graphs of sine and cosine functions. Therefore, the range of these functions also changes based on this value. Still, the

x -

intercepts — the zeros of the functions — do not change. Since the tangent function does not have an amplitude, its graph will be examined separately.

vertical shrink and stretch of tangent function graph

As shown, when the tangent function is multiplied by a number greater than

1,

the function gets steeper. When the function is multiplied by a number between

0

and

1,

it gets flatter. However, the

x -

intercepts of the tangent function do not change.

Horizontal Shrink or Stretch

The graph of a trigonometric function can be horizontally stretched or shrunk by multiplying the input of the function by a positive number $b .$ If $b > 1,$ the graph shrinks horizontally by a factor of $b .$ Conversely, if $0 < b < 1,$ the graph stretch horizontally by a factor of $b .$

Horizontal Stretch or Shrink
Parent Function	$0 < b < 1$	$b > 1$
$y = sin x$	$y = sin (b x)$ Stretch parent function horizontally by a factor of $b$	$y = sin (b x)$ Shrink parent function horizontally by a factor of $b$
$y = cos x$	$y = cos (b x)$ Stretch parent function horizontally by a factor of $b$	$y = cos (b x)$ Shrink parent function horizontally by a factor of $b$
$y = tan x$	$y = tan (b x)$ Stretch parent function horizontally by a factor of $b$	$y = tan (b x)$ Shrink parent function horizontally by a factor of $b$

Examine the following applet to see how the graphs are affected as the value of

b

changes.

horizontal stretch and shrink of the trigonometric functions

As shown, the value of

b

affects the period of the graph. Recall that the number of cycles in a

2 π

interval represents the period of the graph and is found by the formula

p = \frac{2 π}{∣ b ∣} .

Therefore, the number of cycles in a

2 π

interval increases as the value of

b

increases, and it decreases as the value of

b

decreases.

Example

Ferris Wheel

Dominika and her best friend Paulina go to an amusement park to have fun after school. They first decide to ride on the Ferris wheel.

The following sine function models the height

h

in meters of their seat above the ground after the time

t

in seconds.

h = 20 sin (0.16 t)

a Identify the amplitude and period of this function. Round the answer to one decimal place if necessary.

b Identify the transformations done on the parent sine function and then graph the given function.

Answer

a Amplitude:

20

Period:

39.3

b Transformations: A vertical stretch by a factor of

20

and a horizontal stretch by a factor of

0.16

of the graph of the parent sine function.

Graph:

$graph of $h=20\sin(0.16t)$$

Hint

a In a function in the form

y = a sin (b x),

∣ a ∣

represents the amplitude and the period is given by

\frac{2 π}{∣ b ∣} .

b Identify the value of

a

to see if the function was vertically transformed. Then look at the value of

b

to see if a horizontal transformation was applied.

Solution

a Notice that the given function is in the form of

y = a sin (b x) .

y = a sin (b x) ⇓ h = 20 sin (0.16 t)

In this form,

∣ a ∣

represents the amplitude and

\frac{2 π}{∣ b ∣}

represents the period of the function. With this in mind, start by identifying the amplitude.

Amplitude : ∣ a ∣ \Rightarrow ∣ 20 ∣ = 20

The amplitude of the function is

20

meters. In the context of this problem, the amplitude represents the radius of the Ferris wheel. Now, find the period of the graph by substituting

b = 0.16

into the period formula.

p = \frac{2 π}{∣ b ∣}

Substitute

$b = 0.16$

p = \frac{2 π}{∣ 0 . 1 6 ∣}

AbsPos

$∣ 0.16 ∣ = 0.16$

p = \frac{2 π}{0 . 1 6}

UseCalc

Use a calculator

p = 39.269908 \dots

RoundDec

Round to $1$ decimal place(s)

p \approx 39.3

The period of the graph is about

39.3

seconds. Notice that the approximate value

40

can also be seen in the time bar of the graph because every seat returns to its initial place after about

40

seconds.

b To graph the function, the types of the transformations applied to the parent sine function should be identified.

y = sin x \Rightarrow h = 20 sin (0.16 t)

Note that the variables need to be changed from

(x, y)

(t, h) .

The axis names of the coordinate plane also need to be changed. Consider the table to review what the values of

a

and

b

represent.

Transformations of $h = sin t$ to $h = a sin (b t)$
Vertical Stretch or Shrink	$If a > 1, h = sin t stretches vertically by a h = a sin t$
Vertical Stretch or Shrink	$If 0 < a < 1, h = sin t shrinks vertically by a h = a sin t$
Horizontal Stretch or Shrink	$If b > 1, h = sin t shrinks horizontally by b h = sin (b t)$
Horizontal Stretch or Shrink	$If 0 < b < 1, h = sin t stretches horizontally by b h = sin (b t)$

In the given function rule, the value of

a

is greater than

1,

which represents a vertical stretch by a factor of

20 .

The value of

b

is less than

1

and greater than

0,

which represents a horizontal stretch by a factor of

0.16 .

This information can be used to transform the graph of the parent sine function to

h = 20 sin (0.16 t) .

$the graphs of the functions $h=\sin t,$ $h=\sin (0.16t),$ and $h=20\sin (0.16t)$$

Example

Townscape

After going on the Ferris wheel, Dominika and Paulina go up to the big red triangular tower to see the townscape from the viewing deck.

A red tower with two girl figures on its balcony in a green field

Paulina wonders how to find the height of the tower. Dominika says that the height

y,

in meters, of a tower like this one can be modeled by the following tangent function, where

θ

is the angle indicated.

y = 10 tan (1.5 θ)

a Identify the amplitude, period, and asymptotes of the tangent function that represents the height of the towers. Round the answer to one decimal place if necessary.

b Apply the appropriate transformations to the parent tangent function to graph the given function.

c Find the height of the tower if

θ = 5 0^{\circ} .

Round the answer to two decimal places.

Answer

a Amplitude: No amplitude

Period: $\frac{2 π}{3}$

Asymptotes: Odd multiples of

\frac{π}{3}

b Graph:

$graph of the function Y=10 \tan(1.5 \theta)$

37.32

meters

Hint

a Recall the formulas for the period and asymptotes of a tangent function written in the form

y = a tan (b x) .

b What do the values of

a

and

b

represent?

c Substitute the given angle into the given tangent function.

Solution

a The amplitude, period, and asymptotes of the given tangent function will be identified one at a time. Start by identifying whether the tangent function is given in the form

y = a tan (b θ) .

y = a tan (b θ) ⇓ y = 10 tan (1.5 θ)

Remember that a tangent function has no amplitude because it has no maximum and minimum values. Now recall the formulas for the period and asymptotes of a tangent function.

	$y = a tan (b θ)$
Period	$\frac{π}{∣ b ∣}$
Asymptotes	$\frac{π}{2 ∣ b ∣}$

Next, substitute

b = 1.5

into the corresponding formulas to identify the period and asymptotes.

Period : Asymptote : \frac{π}{∣ 1 . 5 ∣} \Rightarrow \frac{2 π}{3} \frac{π}{2 ∣ 1 . 5 ∣} \Rightarrow \frac{π}{3}

The period of the function is

\frac{2 π}{3} .

Recall that the period is the distance between any two consecutive vertical asymptotes. Since there is no translation, the branch that passes through the origin has one asymptotes at

\frac{π}{3}

and the other at

\frac{2 π}{3}

units in the negative direction.

Second Asymptote : \frac{π}{3} - \frac{2 π}{3} = - \frac{π}{3}

The second asymptote of this branch is located at

- \frac{π}{3} .

The next asymptotes can be found by moving

\frac{2 π}{3}

units to the left or right.

Positive Asymptotes : Negative Asymptotes : + \frac{2 π}{3} \Rightarrow π, \frac{5 π}{3}, \frac{7 π}{3} \dots - \frac{2 π}{3} \Rightarrow - π, - \frac{5 π}{3}, - \frac{7 π}{3} \dots

In other words, the asymptotes can be located at the odd multiples of

\frac{π}{3} .

b To graph the given tangent function, recall what the values of

a

and

b

represent for

y = a tan (b θ) .

Transformations of $y = tan θ$ to $y = a tan (b θ)$
Vertical Stretch or Shrink	$If a > 1, y = tan θ stretches vertically by a y = a tan θ$
Vertical Stretch or Shrink	$If 0 < a < 1, y = tan θ shrinks vertically by a y = a tan θ$
Horizontal Stretch or Shrink	$If b > 1, y = tan θ shrinks horizontally by b y = tan (b θ)$
Horizontal Stretch or Shrink	$If 0 < b < 1, h = sin θ stretches horizontally by b h = sin (b θ)$

According to the table, since

a = 10

is greater than

1,

it represents a vertical stretch. Additionally, because

b = 1.5

is also greater than

1,

it represents a horizontal shrink. With this in mind, the graph of the function

y = 10 tan (1.5 θ)

will be plotted as a transformation of the parent tangent function by considering its period and asymptotes.

The graph of the function

y = 10 tan (1.5 θ)

is a vertical stretch by a factor

10

and a horizontal stretch by a factor of

1.5

of the graph of parent tangent function.

b This time the height of the tower will be found for the given angle

θ .

To do so,

θ = 5 0^{\circ}

will be substituted into the given tangent function, which will then be solved for

y .

y = 10 tan (1.5 θ)

Substitute

$θ = 5 0^{\circ}$

y = 10 tan (1.5 (5 0^{\circ}))

Multiply

y = 10 tan (7 5^{\circ})

UseCalc

Use a calculator

y = 37.320508 \dots

RoundDec

Round to $2$ decimal place(s)

y = 37.32

The red tower is about

37.32

meters tall if

θ = 5 0^{\circ} .

Explore

Translating a Trigonometric Function Graph

Observe how the graphs of sine, cosine, or tangent functions are translated horizontally and vertically by changing the values of

h

and

k .

translations of sine, cosine, and tangent functions

Consider the following questions!

How does each graph change when the value of $h$ changes?
How does each graph change when the value of $k$ changes?

Discussion

Translations of Trigonometric Functions

Trigonometric functions can be translated vertically or horizontally like the other functions. Next, these translations will be examined one at a time.

Vertical Translation

Let $f (x)$ be a parent trigonometric function. Then, $f (x) + k$ will translate the parent function vertically. If $k > 0,$ then the graph moves $k$ units up. However, if $k < 0,$ then the graph moves $k$ units down.

Vertical Translations
Parent Function	$k > 0$	$k < 0$
$y = sin x$	$y = sin x + k,$ Translation $k$ units up	$y = sin x - k,$ Translation $k$ units down
$y = cos x$	$y = cos x + k,$ Translation $k$ units up	$y = cos x - k,$ Translation $k$ units down
$y = tan x$	$y = tan x + k,$ Translation $k$ units up	$y = tan x - k,$ Translation $k$ units down

Examine the following applet to see how the graphs of the trigonometric functions change when the value of

k

changes.

vertical translation of trigonometric functions

Notice that the periods and amplitudes of the functions do not change, but the midlines of all the functions are also translated up or down

k

units.

Horizontal Translation, or Phase Shift

A horizontal translation of a periodic function is called a phase shift. The graph of $f (x - h)$ represents a horizontal translation of $f (x)$ by $h$ units. For example, consider the parent functions of sine, cosine, and tangent functions. If $h > 0,$ the parent trigonometric function will be shifted to the right $h$ units, while if $h < 0,$ the function will be shifted to the right $h$ units.

Horizontal Translations
Parent Function	$h > 0$	$h < 0$
$y = sin x$	$y = sin (x - h)$ Translation $h$ units to the right	$y = sin (x + h)$ Translation $h$ units to the left
$y = cos x$	$y = cos (x - h)$ Translation $h$ units to the right	$y = cos (x + h)$ Translation $h$ units to the left
$y = tan x$	$y = tan (x - h)$ Translation $h$ units to the right	$y = tan (x + h)$ Translation $h$ units to the left

Examine the following trigonometric function graphs to see how they are horizontally translated as the value of

h

changes.

horizontal translations ıf the graphs sine,cosine, and tangent functions

Notice that the periods, amplitudes, and midlines of the graphs stay the same but the

x -

intercepts change according to the value of

h .

Example

Carousel

Now Dominika and Paulina are waiting to ride the carousel in the amusement park. The girls enjoy the up and down movement of wooden horses.

Paulina thinks that the movement of a rider on a horse looks like a cosine function. While talking to the operator of the ride, the girls learn that each horse has a special route. They choose horses whose routes can be represented by the following functions.

Dominika : Paulina : y = cos x y = cos (x + π) - 1

a Identify the amplitude and period of the function that represents the route of Paulina's horse.

b Select the translations which should be applied to the parent cosine function in order to get the function that represents the route of Paulina's horse.

c Which of the following is the graph of

y = cos (x + π) - 1 ?

Hint

a In a function in the form

y = a cos b (x - h) + k,

the amplitude is given by

∣ a ∣

and the period by

\frac{2 π}{b} .

b What do the values of

h

and

k

represent?

c Translate the parent function using the values of

h

and

k

to draw the graph of the given function.

Solution

a Start by recalling how to identify the amplitude and period of a cosine function in the case that it is written in the form

y = a cos b (x - h) + k .

Amplitude : ∣ a ∣ Period : \frac{2 π}{∣ b ∣}

Using this information, rewrite the given functions to identify the values of

a

and

b .

	Write in the Form $y = a cos b (x - h) + k$	Amplitude: $∣ a ∣$	Period: $\frac{2 π}{∣ b ∣}$
$y = cos x$	$y = 1 cos 1 (x - 0) + 0$	$∣ 1 ∣ = 1$	$\frac{2 π}{∣ 1 ∣} = 2 π$
$y = cos (x + π) - 1$	$y = 1 cos 1 (x - (- π)) + (- 1)$	$∣ 1 ∣ = 1$	$\frac{2 π}{∣ 1 ∣} = 2 π$

The functions have an amplitude of

1

and a period of

2 π

each.

b In a function in the form

y = a cos b (x - h) + k,

h

represents a horizontal translation and

k

a vertical translation. To identify the transformations applied to the parent cosine function, consider the rewritten form of the function representing the route of Paulina's horse.

y = a cos b (x - h) + k ⇓ y = 1 cos 1 (x - (- π)) + (- 1)

Note that

h = - π

and

k = - 1 .

Because

h < 0,

the parent function has been translated

π

units to the

left .

Next, since

k < 0,

the graph has been translated

1

unit

down .

c Now the parent cosine function will be translated

π

units to the

left

and

1

unit

down

to get the function

y = cos (x + π) - 1 .

Example

Roller Coaster

After riding on the carousel, Dominika and Paulina are looking for something more exciting when they hear the screams coming from the roller coaster. They watch the roller coaster for a while, then see the section where people are most excited and scream the loudest.

The track in this section is quite steep. The girls do some quick math and determine that the path of the roller coaster can be modeled by the following tangent function.

y = tan (x - 3 π) + 1

a Identify the period and asymptotes of the given function that represents the most exciting section of the roller coaster.

b Select the translations to apply to the parent tangent function in order to get the function that represents the most exciting section of the roller coaster.

c Which of the following is the graph of

y = tan (x - 3 π) + 1 ?

Some trigonometric function graph options

Hint

a What are the formulas for the period and asymptote of a tangent function?

b Write the given function rule in the form

y = a tan b (x - h) + k .

c Translate the parent function using the values of

h

and

k

found in the previous part.

Solution

a Begin by writing the function in the form

y = a tan b (x - h) + k .

Then match the variables with their corresponding values to identify the period and asymptotes of the given function.

y = a tan b (x - h) + k ⇓ y = 1 tan 1 (x - 3 π) + 1

Now recall the formulas for the period and asymptotes of a tangent function.

Period : Asymptote : \frac{π}{∣ b ∣} \frac{π}{∣ 2 b ∣}

Since

b = 1,

substitute this value into the formulas.

Period : \frac{π}{∣ 1 ∣} = π Asymptote : \frac{π}{∣ 2 ( 1 ) ∣} = \frac{π}{2}

The period of the given tangent function is

π

and its asymptotes are at the odd multiples of

\frac{π}{2} .

b Consider the revised form of the function written in Part A.

y = a tan b (x - h) + k ⇓ y = 1 tan 1 (x - 3 π) + 1

Here,

h = 3 π

and

k = 1

represent the horizontal and vertical translations, respectively. Consider the following table to review how these values translate the parent tangent function.

Translations of $y = tan θ$ to $y = tan (θ - h) + k$
Vertical Translation	$If k > 0, y = tan θ moves k units up y = tan (x) + k$
Vertical Translation	$If k < 0, y = tan θ moves k units down y = tan (x) - k$
Horizontal Translation	$If h > 0, y = tan θ moves h units to the right y = tan (x - h)$
Horizontal Translation	$If h < 0, y = tan θ moves h units to the left y = tan (x + h)$

From the table, it can be concluded that the function $y = tan (x - 3 π) + 1$ results from shifting the parent tangent function $3 π$ units to the right and $1$ unit up.

c Now the given function will be graphed as a translation of the parent tangent function

3 π

units to the right and

1

unit up. Recall that the tangent function has no amplitude because it has no maximum and minimum values. Use the period

π

and asymptotes at odd multiples of

\frac{π}{2}

to graph the given function.

Graph

2

matches this graph, so Graph

2

is the correct answer.

Example

Math Game

After the roller coaster, Dominika and Paulina decided to take a rest. They see a math game where each winner is awarded a teddy bear and think that it may be fun to take a look at the question.

Dominika and Paulina join the game individually and each give different answers to the question.

Which girl will win a teddy bear?

Hint

Identify the parent function and apply one transformation at a time.

Solution

To find the correct function rule to win the game, start by recalling the general form of a transformed sine function.

y = a sin b (x - h) + k

Now, recall what the variables

a,

b,

h,

and

k

represent.

Transformations of $y = sin x$
Vertical Stretch or Shrink	$y = a sin x If a > 1, y = sin x stretches vertically by a$
Vertical Stretch or Shrink	$y = a sin x If 0 < a < 1, y = sin x shrinks vertically by a$
Horizontal Stretch or Shrink	$y = sin (b x) If b > 1 y = sin x shrinks horizontally by b$
Horizontal Stretch or Shrink	$y = sin (b x) If 0 < b < 1 y = sin x stretches horizontally by b$
Vertical Translation	$y = sin (x) + k If k > 0, y = sin x moves k units up$
Vertical Translation	$y = sin (x) + k If k < 0, y = sin x moves k units down y = sin (x) - k$
Horizontal Translation	$y = sin (x - h) If h > 0, y = sin x moves h units to the right$
Horizontal Translation	$y = sin (x - h) If h < 0, y = sin x moves h units to the left$

Since the parent sine function is stretched vertically by a factor of

4,

the value of

a

4 .

A horizontal stretch by a factor of

0.5

means that the value of

b

0.5 .

A translation

3

units

up

implies that

k = 3 .

Finally, a translation

2 π

units to the

left

means that

h = - 2 π .

a = 4 b = 0.5 h = - 2 π k = 3

Now, substitute all these values into the general form of a sine function and simplify it!

y = a sin b (x - h) + k

SubstituteValues

Substitute values

y = 4 sin 0.5 (x - (- 2 π)) + 3

SubNeg

$a - (- b) = a + b$

y = 4 sin 0.5 (x + 2 π) + 3

This function rule is the same as Dominika's answer. The graphs of the parent sine function and transformed function are shown below for further exploration.

Explore

Reflections of Parent Trigonometric Functions

Examine the following graph pairs of sine, cosine, and tangent functions in the applet.

What effect does the negative sign have in these graphs?
Is it a reflection? If yes, what is the line of reflection?

Discussion

Reflecting Trigonometric Functions

Trigonometric functions can be reflected across the line called a line of reflection. This reflection can be achieved in two ways, by multiplying the function rule by

- 1

or by multiplying the inputs by

- 1 .

Consider some parent trigonometric functions.

Function y = sin x y = cos x y = tan x Reflection Across the x - axis y = - sin x y = - cos x y = - tan x

When the function rule is multiplied by

- 1,

the graph of the function is flipped over the

x -

axis because the

y -

coordinate of each point is changed to its additive inverse.

Recall that the graphs of the functions

y = a sin x,

y = a cos x,

and

y = a tan x

represent a stretch or a shrink by a factor

∣ a ∣ .

In the case that

a < 0,

the graphs represent both a stretch or a shrink by a factor of

∣ a ∣

and a reflection due to the negative sign. Now consider the second case.

Function y = sin x y = cos x y = tan x Reflection Across the y - axis y = sin - x y = cos - x y = tan - x

In this case, since every input is multiplied by

- 1,

the

x -

coordinate of each point is changed to its additive inverse and the graphs are flipped over the

y -

axis.

Note that for transformed trigonometric functions, the line of reflection becomes the midline, or

y = k,

for the following trigonometric functions.

Function y = a sin b (x - h) + k y = a cos b (x - h) + k y = a tan b (x - h) + k Reflection Line y = k y = k y = k

Example

Let's Meet Again

Dominika and Paulina have had fun during the day in the amusement park. They decide to come again together next week. To choose an appropriate day, they check the weather for the next two weeks.

a Write a sine function to model the temperature for the next two weeks.

b Explain the types of transformations applied to the parent sine function to get the function written in Part A.

Answer

a Example Function:

y = - 25 sin \frac{π}{7} + 70

b See solution.

Hint

a Plot the days and their corresponding temperatures as points on a coordinate plane. Then, use the general form of a transformed sine function,

y = a sin b (x - h) + k,

to write the function rule.

b Recall what the variables in the general sine function form represent.

Solution

a To write the sine function, start by drawing the function. To do so, plot the days and the corresponding temperatures as points in the coordinate plane.

Since the curve connecting the points looks like a sinusoid, its period, midline, and amplitude can be used to write a sine function. Note that the graph moves periodically along the

x -

axis and one full cycle is completed from

x = 0

x = 14 .

This means that the period of the function can be assumed to be

14 .

Period : 14

The maximum and minimum values of the function are

95

and

45,

respectively. The amplitude can be determined by finding half of the difference of these values.

Amplitude : \frac{9 5 - 4 5}{2} = 25

The amplitude is

25 .

Finally, the midline will be calculated as a mean of the minimum and maximum values.

Midline : \frac{9 5 + 4 5}{2} = 70

The sine function can now be written by using all of this information. Start by recalling the general transformed form of a sine function.

y = a sin b (x - h) + k

In this form,

∣ a ∣

is the amplitude. According to graph, the amplitude is

25 .

∣ a ∣ = 25 ⇓ a = 25 or a = - 25

To determine the sign of

a,

compare its graph with the graph of parent sine function.

Notice that after intersecting the midline at

x = 0,

the graph goes down to its minimum and then goes up to its maximum, unlike the parent sine function, which goes up at first and then down. Therefore, this graph represents a reflection in the midline. Therefore, the value of

a

needs to be negative.

a = - 25

Recall that the formula for the period of a sine function is

\frac{2 π}{∣ b ∣} .

The value of

b

in the formula rule can be calculated by setting the period of the graph,

p = 14 .

p = \frac{2 π}{∣ b ∣}

Substitute

$p = 14$

14 = \frac{2 π}{∣ b ∣}

Solve for

b

MultEqn

$LHS \cdot ∣ b ∣ = RHS \cdot ∣ b ∣$

14 ∣ b ∣ = 2 π

DivEqn

$LHS / 14 = RHS / 14$

∣ b ∣ = \frac{2 π}{1 4}

AbsPos

$∣ b ∣ = b$

b = \frac{2 π}{1 4}

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

b = \frac{π}{7}

Since one full cycle of the graph starts from

x = 0

and ends at

x = 14,

which is the interval of exactly one period, it can be said that there is no phase shift. Therefore,

h = 0 .

Finally, since the midline is

y = 70,

this refers that

k = 70 .

Now that all of the values for the variables are known, the sine function can be written.

y = a sin b (x - h) + k ⇓ y = - 25 sin \frac{π}{7} (x - 0) + 70 ⇕ y = - 25 sin \frac{π}{7} x + 70

The given forecast over the next two weeks can be modeled by the sine function

y = - 25 sin \frac{π}{7} x + 70 .

b To determine the types of transformations applied to the parent sine function, start by recalling what the corresponding variables in the general form of the transformed sine function represent.

Transformations of $y = sin x$
Vertical Stretch or Shrink	$y = a sin x If a > 1, y = sin x stretches vertically by a$
Vertical Stretch or Shrink	$y = a sin x If 0 < a < 1, y = sin x shrinks vertically by a$
Horizontal Stretch or Shrink	$y = sin (b x) If b > 1, y = sin x shrinks horizontally by b$
Horizontal Stretch or Shrink	$y = sin (b x) If 0 < b < 1, y = sin x stretches horizontally by b$
Vertical Translation	$y = sin (x) + k If k > 0, y = sin x moves k units up$
Vertical Translation	$y = sin (x) - k If k < 0, y = sin x moves k units down$
Horizontal Translation	$y = sin (x - h) If h > 0, y = sin x moves h units to the right$
Horizontal Translation	$y = sin (x + h) If h < 0, y = sin x moves h units to the left$
Reflection	$y = - a sin x If a < 0, y = sin x reflects in the midline y = k$

Now recall the function rule written in Part A again.

y = - 25 sin \frac{π}{7} x + 70

According to the table, it can be said that the function rule represents a vertical stretch by a factor of

25,

a horizontal stretch by a factor of

\frac{π}{7},

and a vertical translation of parent sine function

70

units up. Also, since

a = - 25

is less than

0,

there is a reflection in the midline

y = 70 .

Closure

Transformations of Cotangent, Secant, and Cosecant Functions

This lesson reviewed the transformations of sine, cosine, and tangent functions. However, the same rules can be applied to investigate the transformations of the parent functions of the remaining trigonometric functions, cotangent, secant, and cosecant. Recall the parent cotangent, secant, and cosecant function graphs.

parent cotangent, secant, and cosecant function graphs

These graphs can be stretched or shrunk vertically by a factor of

a,

stretched or shrunk horizontally by a factor of

b,

translated

h

units to the right and

k

units up. They can also be reflected if

a < 0 .

Parent Function	Transformed Form of the Function
$y = cot x$	$y = a cot b (x - h) + k$
$y = sec x$	$y = a sec b (x - h) + k$
$y = csc x$	$y = a csc b (x - h) + k$

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Catch-Up and Review

Vertical Shrink or Stretch

Horizontal Shrink or Stretch

Answer

Hint

Solution

Answer

Hint

Solution

Vertical Translation

Horizontal Translation, or Phase Shift

Hint

Solution

Hint

Solution

Hint

Solution

Answer

Hint

Solution