Alternative Format — Lesson 5: Modelling Periodic Behaviour

Let's Start Thinking

Round and Round We Go

Ferris wheels are popular attractions that offer their riders a bird's eye view of their host cities.

The Wiener Riesenrad is one of the oldest Ferris wheels in the world that is still in operation at the time of this recording. Originally built in 1897 in Vienna, Austria, it stands \(64.75\) metres tall with a diameter of \(60.69\) metres and makes one revolution in about \(5\) minutes.

Wiener Riesenrad, Vienna, Austria

The Tianjin Eye has the distinction of being the only Ferris wheel built over a bridge in the city of Tianjin, China. Passengers board the wheel from beneath the bridge to reach a height of \(120\) metres at a speed of \(30\) minutes per revolution.

Tianjin Eye, Tianjin, China

In this lesson, we will learn about how to model periodic behavior, like a person's height above the ground while riding a Ferris wheel. We will make connections between the features of the physical phenomena and the properties of the corresponding sinusoidal functions.

Citations

1. [information: Wiener Riesenrad] Official Travel Guide, Austria. (n.d.). Vienna's giant ferris wheel. Retrieved from https://www.austria.info/us/austria/stunning-and-surprising-art-architecture/vienna-s-giant-ferris-wheel
2. [facts: Wiener Riesenrad] Wiener Riesenrad. (n.d.). Technical description. Retrieved from https://www.wienerriesenrad.com/en/giant-ferris-wheel/technical-data
3. [facts: Tianjin Eye] Travel China Guide. (n.d.). Tianjin Eye. Retrieved from https://www.travelchinaguide.com/attraction/tianjin/eye.htm

Try This

During a ride on a Ferris wheel, Angel reached a maximum height of \(60\) m and a minimum height of \(2\) m. The graph of her distance above the ground reached over time is shown.

Use the graph to write a cosine function to estimate Angel's distance above the ground, in metres, in terms of time, in minutes.

Sinusoidal Equations From a Graph

Explore This 1

Description

How are the values of \(a\), \(b\), \(h\), and \(k\) related to the graph of a sinusoidal function?

Example 1

Can you determine an equation for the function to match? That is, can you create a sinusoidal function of the form \(y=a \cos \left(b(x-h)^{\circ}\right)+k\) that matches the dashed red line?

Starting Graph: \(a=1\), \(b=1\), \(h= 0\), \(k=0\)

First Transformation: \(a=4 \), \(b=1\), \(h= 0\), \(k=0\)

Second Transformation: \(a=4 \), \(b=1\), \(h= -90\), \(k=0\)

Third Transformation: \(a=4 \), \(b=1.5\), \(h= -90\), \(k=0\)

Example 2

Can you determine an equation for the function to match? That is, can you create a sinusoidal function of the form \(y=a \sin \left(b(x-h)^{\circ}\right)+k\) that matches the dashed red line?

Starting Graph: \(a=1\), \(b=1\), \(h= 0\), \(k=0\)

First Transformation: \(a=1 \), \(b=1\), \(h= 0\), \(k=-3\)

Second Transformation: \(a=2 \)​​​​​​, \(b=1\), \(h= 0\), \(k=-3\)

Third Transformation: \(a=2 \)​​​​​​​​​​​​​​, \(b=1\), \(h= 30\), \(k=-3\)

Fourth Transformation: \(a=2 \)​​​​​​​​​​​​​​, \(b=0.5\), \(h= 30\)​​​​​​​, \(k=-3\)

Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0.

Interactive Version

Transformations of Sinusoidal Functions

Explore This 1 Summary

Here's an example of a function to match you may have seen in this activity with the sine function as a function to change. 

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

You may have used the amplitude of the function to match to decide if a vertical stretch is needed to determine the value of \(a\). Recall the amplitude \(= |a|\).

You might have used the period of the function to match to decide if a horizontal stretch is needed to determine the value of \(b\). Recall the period \(=\dfrac{360}{|b|}\).

You might have used the axis of the function to match to decide if a vertical translation was needed to determine the value of \(k\). Recall the axis is determined by the line \(y=k\).

Lastly, you might have used a key point, such as an axis point or a maximum or a minimum on the function to match, to decide if a phase shift was needed to determine the value of \(h\).

There are many different sequences of decisions that can lead to a correct match.

Let's discuss two possible sequences.

Matching the Graph — Sequence 1

In this first sequence, we will focus on the order of operations for the transformations.

1. Examine reflections and stretches to select values for \(a\) and \(b\) since these involve multiplication. Correct!​​
2. Examine translations to select values for \(h\) and \(k\), which appear as addition or subtraction in the function equation.Correct!

Recall we are trying to create an equation of the form \(y=a \sin(b(x-h)^\circ)+k\)​​​.

We start with \(y=\sin(x^{\circ})\).

The amplitude of the function to match is \(2\). So \(a\) could be \(+2\) or \(-2\). Let's continue with \(y=-2\sin(x^{\circ})\).

\(y=-2\sin(x^{\circ})\)

Now let's look at the period. Using the two minimum points to define one cycle, the cycle starts when \(x=-150\) and ends when \(x=570\).

So the period is \(150 +570=720\). 

And there is a horizontal stretch by a factor of \(2\). This means that the value of \(b\) could be \(+0.5\) or \(-0.5\).

Let's continue with \(y=-2\sin(0.5x^{\circ})\).

\(y=-2\sin(0.5x^{\circ})\)

We are finished with the reflections and stretches, so we will now work on the translations.

The axis of the function to match is \(y=-3\). So \(k\) must be \(-3\).

So we will continue with our function to change to be \(y=-2\sin(0.5x^{\circ})-3\).

\(y=-2\sin(0.5x^{\circ})-3\)

Lastly, let's look at a key point on the function to change to determine the phase shift. A maximum point on the function to change is \((540,-1)\). If this is shifted to the left to the maximum point on the function to match at \((210,-1)\), this would be a shift of \(330\) to the left. This means that \(h =-330\). That is, \(y=-2\sin(0.5(x-(-330))^{\circ})-3\).

We have now finished with the translations and we have matched the functions.

Therefore, \(y=-2\sin(0.5(x+330)^{\circ})-3\). This is an equation for the function to match.

It is important to note that earlier we made some choices about whether the values of \(a\) or \(b\) were positive or negative. And we were given the sine function instead of a cosine function as our base graph. All of these have some effect on what phase shift will be needed to determine the function to match.

Matching the Graph — Sequence 2

Let's look at another decision sequence that also leads to a correct match. In this match, we will exploit the fact that the origin is not affected by reflections and stretches to first decide on the translations for our function to match. After we know where the image of the origin will be, then we will select the remaining parameters.

1. Examine translations to select values for \(h\) and \(k\).
2. Examine reflections and stretches to select values for \(a\) and \(b\).​

Recall we are trying to create an equation of the form \(y=a \sin(b(x-h)^\circ)+k\)​​​.

\(y=\sin(x^{\circ})\)

The function to match has a horizontal axis at \(y =-3\).

Thus, \(k\) must be \(-3\). 

\(y=\sin(x^{\circ})-3\)

To select the phase shift, notice that the function to change has an axis point at the \(y\)-intercept at \((0, -3)\), and the function to match has an axis point at \((30, -3)\). Shifting to the right by \(30\) aligns these points.

So we can use \(h=30\).

\(y=\sin((x-30)^{\circ})-3\)

We are finished with the translations and will now continue with the remaining parameters.

The amplitude of the function to match is \(2\), and an \(a\) value of \(+2\) keeps both functions increasing through the common point at \((30, -3)\).

So we will use \(a=2\).

\(y=2\sin((x-30)^{\circ})-3\)

Finally, we see that the function to change goes through a whole cycle in the time that the function to match goes through half a cycle.

So we need a horizontal stretch by a factor of \(2\), or the value of \(b\) could be \(0.5\). 

\(y=2\sin(\class{add13=bg3}{0.5}(x-30)^{\circ})-3\)

We have now finished with all the transformations and we have matched the functions using a different sequence of decisions than we used previously.

The equation of \(y=2\sin(\class{add13=bg3}{0.5}(x-30)^{\circ})-3\) is another equation for the function to match. The choices we make and the order we make them can affect the equation of the model for a sinusoidal function.

For sinusoidal functions, there is more than one equation that will produce the same graph. In fact, there are infinitely many of them.

Properties of Sinusoidal Functions

The properties of a sinusoidal function can be determined from its equation.

Conversely, knowing the properties of a sinusoidal function, we can write its equation.

Reflections in the \(y\)-Axis and Sinusoidal Functions

In a previous lesson, we observed the following:

• When the sine function is reflected in the \(y\)-axis, the effect is the same as reflection in the \(x\)-axis.

• When the cosine function is reflected in the \(y\)-axis, the effect is the same as having no reflection.

As a result of these facts, many applications of sinusoidal functions strictly use positive values for the parameter \(b\).

Modelling Sinusoidal Functions

To write an equation to model the graph of a sinusoidal function, we need to identify the values of the parameters \(a\), \(b\), \(h\), and \(k\).

Example 1

Solution — Part A

To determine the equation, we need to find the values for the parameters \(a\), \(b\), \(h\), and \(k\) for the general sine function, \(y=a\sin\left(b\left(x-h\right)^\circ\right)+k\).

Recall that the parameters are related to the properties of the graph.

So looking at the graph, we can see that the maximum value of the function is \(4\) and the minimum value is \(-1\).

The amplitude of the function is a maximum minus the minimum all over \(2\):

\(\dfrac{\text{max}-\text{min}}{2}=\dfrac{4-\left(-1\right)}{2}=2.5\)

So we can let \(a=2.5\).

We know that we can calculate the period using the formula \(360^\circ\) divided by the absolute value of \(b\). But from the graph, we can see that the period is \(120^\circ\).

So equating these two expressions and assuming that \(b\) is positive, we can solve for \(b\):

\(b=\dfrac{360^\circ}{120^\circ}=3\)

We find that \(b = 3\).

The axis of the graph is \(y = 1.5\), so \(k=1.5\).

The values of \(a\), \(b\), and \(k\) are straightforward to find, but the phase shift requires more careful consideration. 

We are working with a sine function, and we have chosen \(a\) and \(b\) to be positive, so there are no reflections.

The \(y\)-intercept of the base sine function is an increasing axis point, so we need to find its image on the given graph that is also an increasing axis point.

If we use the point \((30, 1.5)\), our phase shift would be \(30\) to the right, and \(h\) would be equal to \(30\). 

We summarize our findings as follows:

• Amplitude:

  \(\dfrac{\text{max}-\text{min}}{2}=\dfrac{4-\left(-1\right)}{2}=2.5\); \(a=2.5\).

• Period: \(\dfrac{360^\circ}{\lvert b\rvert}=120^\circ\); if \(b\gt0\), then \(b=\dfrac{360^\circ}{120^\circ}=3\).
• Axis: \(y=1.5\); \(k=1.5\).
• Phase Shift: For a sine graph, when \(a,b\gt0\), the image of the \(y\)-intercept is an increasing axis point. Using \(\left(30,1.5\right)\), \(h=30\).

Therefore, a sine function that represents the graph could be \(y=2.5\sin\left(3\left(x-30\right)^\circ\right)+1.5\).

Solution — Part B

Recall part b): Determine the equation of a cosine function that represents the graph.

To determine a cosine function, the amplitude, period, and axis are the same as in part a), so \(a=2.5\), \(b=3\), and \(k=1.5\).

We only need to find the phase shift. So again, we will focus on a key point. 

The \(y\)-intercept of the base cosine function is a maximum point. Since \(a\) and \(b\) are positive, we need to find a maximum point on the given graph to be its image. Using the point \(\left(-60,4\right)\), our phase shift would be \(60^\circ\) to the left, and \(h=-60\).

• That is, for a cosine graph, when \(a,b\gt0\), the image of the \(y\)-intercept is a maximum point. Using \(\left(-60,4\right)\), \(h=-60\).

So a cosine function that represents the graph is \(y=2.5\cos\left(3\left(x+60\right)^\circ\right)+1.5\).

Notice that we could have used the point \((60, 4)\) as the image of the maximum point instead, and gotten a slightly different, but equally correct equation.

Again, we find that a cosine function that represents the graph is \(y=2.5\cos\left(3\left(x+60\right)^\circ\right)+1.5\).

Try This Revisited

Solution

To determine an equation of Angel's distance above the ground, \(d\) in metres, we will identify the properties of the function, where \(t\) represents the time in minutes after the ride begins.

Domain

From the graph, we can see the ride begins when \(t=0\) and it appears to end when \(t=20\), so the domain is \(\left\{t\in\mathbb{R}\mid 0 \leq t\leq 20\right\}\).

Amplitude

Angel's maximum distance above the ground was about \(60\) m and her minimum distance was about \(2\) m. The amplitude of the graph is \(\dfrac{60-2}{2}=29\), so \(\lvert a \rvert=29\).

Period

Angel boarded the Ferris wheel at the minimum distance above the ground when \(t=0\) and the graph shows that Angel's distance was at the minimum \(3\) more times in \(20\) minutes. The period is \(\dfrac{20}{3}\) so \(\lvert b \rvert=360\div \dfrac{20}{3}=54\).

Axis

From the maximum and minimum distances, Angel's average distance above the ground is \(\dfrac{60+2}{2}=31\) m. So \(k=31\).

Phase Shift

For a cosine function, the \(y\)-intercept is a maximum. Its image on Angel's graph could be determined with one of the following options.

Check Your Understanding 1

Question — Version 1

The graph can be represented by an equation in the form \(y=a \cos \left(b(x-h)^{\circ}\right)+k\). Find the values of \(a\), \(b\), \(h\), and \(k\).

Answer — Version 1

• \(a = -2\)
• \(b = 0.8\)
• \(h = 30\)
• \(k = -1\)

Feedback — Version 1

The base function is \(y=\cos \left(x^{\circ}\right)\).

The axis is \(y=-1\), so \(k=-1\).

A minimum point is \((30,-3)\), so \(a\) will be negative.

The amplitude is \(2\), so \(a=-2\).

The period is \(450^{\circ}\), so \(b=\dfrac{360^{\circ}}{450^{\circ}}=0.8\).

One equation that would represent the function is \(y=-2 \cos \left(0.8(x-30)^{\circ}\right)-1\).

Question — Version 2

The graph can be represented by an equation in the form \(y=a \sin \left(b(x-h)^{\circ}\right)+k\). Find the values of \(a\), \(b\), \(h\), and \(k\).

Answer — Version 2

• \(a = 3\)
• \(b = 2\)
• \(h = 60\)
• \(k = -2\)

Feedback — Version 2

The base function is \(y=\sin \left(x^{\circ}\right)\).

Since \((60,-2)\) is an axis point and the function is increasing, \(h=60\) and \(k=-2\) and \(a\) will be positive.

Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0.

The amplitude is \(3\), so \(a=3\).

The period is \(180^{\circ}\), so \(b=\dfrac{360^{\circ}}{180^{\circ}}=2\).

One equation that would represent the function is \(y=3 \sin \left(2(x-60)^{\circ}\right)-2\).

Interactive Version

Transformations of Sinusoidal Functions

Sinusoidal Equations From Properties

Periodic behaviour can be described in words as well as with equations and graphs.

Example 2

Solution — Part A

To write an equation using a cosine function, we need to identify the values of \(a\), \(b\), \(h\), and \(k\) from the given properties.

The amplitude is \(5\) so \(a\) could be \(5\) or \(-5\).

We can assume that \(b\) is positive since a reflection in the \(y\)-axis has no effect on the cosine function.

We are told the period is \(90^\circ\). The period could also be calculated by the formula \(\dfrac{360^\circ}{\lvert b\rvert}\). Equating these, we can solve for \(b\), \(b\gt0\). 

The function \(y=5\cos\left(4x^\circ\right)\) has the correct amplitude and period, but it has a maximum at the point \(\left(0,5\right)\) instead of at \(\left(0,2\right)\) as needed.

A vertical translation \(3\) units down would move the maximum into the correct position. No phase shift is needed so \(h=0\) and \(k=-3\).

Therefore, the cosine function \(y=5\cos\left(4x^\circ\right)-3\) would represent the function described.

Solution — Part B

Recall part b): Write an equation for the function using a sine function.

To write an equation using a sine function, we will use the same values for \(a\), \(b\), and \(k\) found in part a). However, we will need to find the value for \(h\).

The sine function \(y=5\sin\left(4x^\circ\right)\) has a maximum at the point \(\left(22.5,5\right)\), one quarter of the period to the right of the \(y\)-axis. To move this point to the maximum at \(\left(0,2\right)\) as needed, we need a phase shift of \(22.5^\circ\) to the left and a translation \(3\) units down so \(h=-22.5\) and \(k=-3\).

Therefore, the sine function \(y=5\sin\left(4\left(x+22.5\right)^\circ\right)-3\) would represent the function described.

Solution — Part C

Recall part c): Use technology to graph the equations found in a) and b) to verify your answers.

We can use a graphing calculator or software, such as GeoGebra, to graph the functions we found in parts a) and b) and verify that they each have an amplitude of \(5\) units, a period of \(90^\circ\), and a maximum at \(\left(0,2\right)\).

\(y=5\cos\left(4x^\circ\right)-3\)

\(y=5\sin\left(4\left(x+22.5\right)^\circ\right)-3\)

The equations found in parts a) and b) represent sinusoidal functions having the same properties.

Let's explore how the physical features of periodic phenomena can affect sinusoidal functions.

Explore This 2

Description

Observe how changing the physical features of Windmill B affect the properties of the function measuring the height of the tip of a windmill blade.

Taller Tower

Longer Blades

Offset Blades

Faster Speed

Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0.

Interactive Version

Modelling the Motion of a Windmill Blade

Explore This 2 Summary

In the Explore This activity, you may have noticed that when the physical features of Windmill B changed compared to Windmill A,  the properties of the function measuring the height of a blade’s tip over time changed as well.

For example, if you made Windmill B’s tower taller, then the amplitude, period, and phase shift all remained the same, but the function’s axis translated vertically upward.

The table below summarizes the results of changing Windmill B’s tower height, blade length, location of a point on the path of rotation, and the speed of rotation, as compared with Windmill A.

Physical Feature of Windmill B as Compared With Windmill A
Function Property	Increase Tower Height	Increase Blade Length	Move Point on Path of Rotation	Increase Speed of Rotation
Amplitude	same	larger	same	same
Period	same	same	same	shorter time
Phase Shift	max at \(t = 0\)	max at \(t = 0\)	max at \(t \ne 0\)	max at \(t = 0\)
Axis	higher	same	same	same

Example 3

Solution — Part A

Since the equation will model the height starting at the maximum, using a cosine function will not require a phase shift or any reflections. We will use the physical features of the windmill to determine the value of the parameters of

where the independent variable \(t\) represents the time in seconds after the initial maximum and the dependent variable \(h\left(t\right)\) represents the height of the blade tip in metres.

The windmill spins \(5\) seconds per revolution so \(b=\dfrac{360}{5}=72\).

Since the tower height is \(100\) m, then \(k=100\).

The blades are \(40\) m long, so \(a=40\).

Therefore, the height of a blade tip starting at the maximum can be modelled by \(h\left(t\right)=40\cos\left(72t\right)+100,t\ge0\).

Solution — Part B

The speed of rotation for Windmill B is \(3\) seconds per revolution, so \(b=\dfrac{360}{3}=120\).

Since Windmill A and B are identical otherwise, the other parameters will be the same, \(a=40\), \(h=0\), and \(k=100\).

The height of a blade tip of Windmill B starting at the maximum can be modelled by \(h\left(t\right)=40\cos\left(120t\right)+100,t\ge0\).

Example 4

Solution

Let \(t\) represent the time in hours after high tide and \(d\left(t\right)\) represent the depth of the water in metres.

We could use a sine or a cosine function for the base function. We are modelling the depth of the water after high tide, which is a maximum, so using a cosine function would not require a phase shift, \(h=0\), or any reflections, \(a,b\gt0\).

The amplitude of the function is \(a=\dfrac{11.3-2.7}{2}=4.3\).

The period is \(12\) hours, so \(b=\dfrac{360}{12}=30\).

The average depth of the water is \(\dfrac{11.3+2.7}{2}=7\), so \(k=7\).

The depth of the water in the harbour can be estimated by the function \(d\left(t\right)=4.3\cos\left(30t\right)+7\).

Example 5

Solution — Part A

Let \(t\) represent the time in seconds after the mass reaches its lowest position and \(L\left(t\right)\) represent the length of the spring in cm.

We are modelling the length of the spring after the lowest mass position, which is a minimum, so using a cosine function with a reflection in the \(x\)-axis, \(a\lt0\), would not require a phase shift, \(h=0\).

The spring oscillates up and down \(5\) times in \(24\) s so the period is \(24\div5=4.8\) s. Therefore,

\(b=\dfrac{360}{4.8}=75\)

The maximum length of the spring is \(6.8\) cm and the minimum length is \(4.0\) cm. The amplitude is \(\dfrac{6.8-4.0}{2}=1.4\), and, since \(a\lt0\), \(a=-1.4\). We can also calculate the average spring length, 

\(k=\dfrac{6.8+4.0}{2}=5.4\)

A model equation for the length of the spring is \(L\left(t\right)=-1.4\cos\left(75t\right)+5.4,~t\ge0\).

Solution — Part B

The graph of the spring's length over time is as follows:

Check Your Understanding 2

Question — Version 1

The depth of water in a harbour on planet Earth Prime is \(25\) m at high tide and \(11\) m at low tide. Assume that a cycle is completed every \(8\) hours.

Select all equations that could estimate the depth of water, \(y\), in metres, as a function of time, \(x\), in hours, after low tide.

1. \(y=-7 \cos (45 x)+18\)
2. \(y=7 \cos (45(x-4))+18\)
3. \(y=7 \cos (45 x)+11\)
4. \(y=7 \cos (45(x-2))+18\)

Answer — Version 1

1. \(y=-7 \cos (45 x)+18\)
2. \(y=7 \cos (45(x-4))+18\)

Feedback — Version 1

We need to identify the values of \(a\), \(b\), \(h\), and \(k\) from the given properties for the cosine function \(y=a \cos (b(x-h))+k\).

The amplitude of the function is \(\dfrac{25-11}{2}=7\).

The period is \(8\) hours, so \(b=\dfrac{360}{8}=45\).

We are using a cosine function to model the depth of the water after low tide, which is a minimum, so we could either:

• Use a vertical reflection, \(a=-7\), which would not require a phase shift, \(h=0\) ; or
• Use no reflections, \(a=7\) and \(b\gt 0\), which would require a phase shift of half a period to the right, \(h=4\).

The average depth of the water is \(\dfrac{25+11}{2}=18\), so \(k=18 \).

Therefore, the depth of the water in the harbour could be estimated by either of the following functions:

• \(y=-7 \cos (45 x)+18\)

• \(y=7 \cos (45(x-4))+18\)

Question — Version 2

Identify all equations that represent a sine function with an amplitude of \(5\) units, a period of \(720^{\circ}\), and a maximum at \((-30,2) .\)

1. \(y=5 \sin \left(\dfrac{1}{2}(x+210)^{\circ}\right)-3\)
2. \(y=5 \sin \left(2(x-210)^{\circ}\right)-3\)
3. \(y=5 \sin \left(\dfrac{1}{2}(x+105)^{\circ}\right)-3\)
4. \(y=-5 \sin \left(\dfrac{1}{2}(x+210)^{\circ}\right)-3\)

Answer — Version 2

1. \(y=5 \sin \left(\dfrac{1}{2}(x+210)^{\circ}\right)-3\)

Feedback — Version 2

We need to identify the values of \(a\), \(b\), \(h\), and \(k\) from the given properties for the sine function \(y=a \sin \left(b(x-h)^{\circ}\right)+k\).

The amplitude is \(5\) so \(a\) could be \(5\) .

The period is \(720^{\circ}\), so \(b=\dfrac{360^{\circ}}{720^{\circ}}=\dfrac{1}{2}\) .

The sine function \(y=5 \sin \left(\dfrac{1}{2} x^{\circ}\right)\) has a maximum at the point \(\left(180^{\circ}, 5\right)\), one-quarter of the period to the right of the \(y\)-axis. To move this point to \((-30,2)\) as needed, we need a phase shift of \(210^{\circ}\) to the left and a translation \(3\) units down, so \(h=-210\) and \(k=-3\).

Therefore, the sine function \(y=5 \sin \left(\dfrac{1}{2}(x+210)^{\circ}\right)-3\) would represent the function described.

Wrap-Up

Take It With You

The daily high temperature is measured in degrees Celsius. The monthly averages for Winnipeg, Manitoba, Canada are listed in the table.

Write an equation to model the daily high temperature for Winnipeg, Manitoba.