Project 2

Link to the Interactive Graph: https://www.desmos.com/calculator/i4l2puerry

Introduction

In this investigation, we explored function composition using equations, tables of values and graphs in the Desmos application. The domain of a function is defined as “all inputs for which the expression defining a function has a value, or for which the function makes sense in a given context” (National Governor’s Association, 2010). We discovered it is possible to use the domain of an existing function to anticipate the domain when combined with another to create a new function.

The entire investigation is divided into 3 tasks: Combining Functions, Inverses and Moving Stuff Around. Each task presents a different perspective of building new functions from existing functions. The standards addressed in this investigation include:

F-BF1b – Combine standard function types using arithmetic operations.
F-BF3 – Identify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology.
F-BF4a – Solve an equation of the form f(x) = c for a simple function f that has an inverse, and write an expression for the inverse.
F-TF5 – Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline.

Combining functions

We started this task by creating a graph of two functions, $f(x)=\frac{1}{x-2}$ and $g(x)=\sqrt{x}$ , with the intent of exploring compositions. Desmos recognizes multiple inputs when creating a given expression. For example, we discovered the obelus could be used to represent division and a horizontal fraction bar, while typing the letters “sqrt” from a keyboard will produce this symbol:

Next, we generated a table of values for functions $f(x)$ and $g(x)$ by selecting the wheel icon (“Edit [a] List”), then choosing the table icon (“Convert to a Table”) (Figure 1).

When observing the domain of or sets of possible x-values, of $f(x)$ , we noticed the table displayed values for x except where $x=2$ . At this value, the table shows $f(x)$ is undefined. To determine if there are any domain values between 1 and 2 or between 2 and 3 for which f(x) may be undefined, we added additional values to the table. As we came closer to $x=2$ , we noticed that the values only decreased in magnitude, but never reached a point where they were undefined (Figure 2). We also entered additional negative and positive values – beyond − $2\le x\le 3$ – to see if there were any other points at which the domain was undefined.

While investigating the table values, we could also see how relates to the graph of f(x), the curve approaches, but never touches, $x=2$ . Following this line of thinking we determine the domain of $f(x)$ would be all real numbers, except $x=2$ (Figure 3).

When observing the domain of $g(x)$ , using the Desmos generated table of values, we were able to determine the function is undefined at $x=-2$ and -1, but for values greater than 0, a corresponding value was given. In relating the table of values to the graph of $g(x)$ , it can be seen that the curve starts at the origin. The domain, therefore, would be all real numbers greater than or equal to 0 (Figure 4).

We then graphed the compositions of $f(x)$ and $g(x)$ , and created a table of values for each function to evaluate and compare their domains. Prior to beginning this part of the investigation, it is important that students understand symbolic notation for a composition. The small circle, ∘, is a composite function operator which lets us know that $f \circ g(x)$ can also be represented as $f(g(x))$ . Both representations mean “the function of f composed with g”, and we will enter the second form into Desmos. The same process was used to rewrite $g \circ f(x)$ , or “the function of g composed with f” as $g(f(x))$ .

With $f(g(x))$ , the table of values was similar that of $g(x)$ in that the function is undefined at $x=-2$ and -1, but for values greater than 0, it appeared a corresponding value was listed. Looking at the graph of this function, for the values $0\le x\le 4$ , this was true, however, the curve approached, but never touched $x=4$ . In this way, the graph of $f(g(x))$ resembled $f(x)$ . We determined the domain of $f(g(x))$ would be all real numbers greater than or equal to 0 except 4 (Figure 5).

With $g(f(x))$ , we analyzed the Desmos generated table of values to determine whether the function is undefined at $-2\le x\le 2$ . We then added values of x greater than 2 began to populate the table with values of $g(f(x))$ which were not undefined. Much like $g(x)$ , we determined there are no negative values in the domain of $g(f(x))$ . When viewing the graph, we were able to see the curved approached, but never touched $x=2$ . In this way, $g(f(x))$ was similar to $f(x)$ . The domain of $g(f(x))$ is all real numbers greater than 2 (Figure 6).

Through this task, we could generalize the domain of $f(g(x))$ followed a structure similar to $f(x)$ (all real numbers except where a particular value/values are undefined), and the values specific to the function changed because of $g(x)$ . In the same manner, the domain of $g(f(x))$ followed a structure similar to $g(x)$ (all positive real numbers), and the values specific to the function changed because of $f(x)$ .

Inverses

Using the same graph, we then added the function $x=f(y)$ and compared it to the graph of $f(x)$ . Desmos did not automatically provide the option to generate a table of values, so using the graph as our sole comparison we noticed two things:

On the graph of $f(x)$ , the curve approaches $x=2$ and $y=0$ (the x-axis) but never touches either line.
On the graph of $x=f(y)$ , the curve approaches $x=0$ (the y-axis) and $y=2$ but never touches either line.

Since the x- and y- values are reversed in these situations, we could visually determine $x=f(y)$ is the inverse of $f(x)$ .

Let’s redefine the inverse

Trying to create a new table using x and $f(y)$ as headers was not allowed by the Desmos application, but we were able to create a table by redefining the function as $h(x)$ . Using what we learned about reversing the x− and y− values, we will rewrite $f(x)$ as its inverse:

First, we know $f(x)=y$ , so when writing the inverse of $f(x)$ , or $x=\frac {1}{f(y)-2}$
We solve for y, but as this is a new function, we will represent y as $h(x)=\frac {1}{x}+2$

The equation, table and graph of $h(x)$ are shown in Figure 7.

It is worth noting that the data entered into the list bar can be moved up or down as needed. Since we were analyzing $f(x)$ and $h(x)$ , $f(x)$ was moved closer to $h(x)$ . By adding a third column to the table of $f(x)$ , we were able to compare the two functions in another way. While $f(x)$ is undefined at $x=2$ , the domain values less than 2 have negative corresponding range values which approach 0, but never reach 0. The domain values greater than 2 have positive corresponding range values which also approach 0, but never reach 0. The function $h(x)$ is undefined at $x=0$ , and the domain values less than or greater than 0, the corresponding range values approach, but never reach 2. This also demonstrates an inverse relationship (Figure 8).

The graphs of $f \circ h(x)$ and $h\circ f(x)$ appear as the line $y=x$ . The domain of $f \circ h(x)$ and $h\circ f(x)$ would be all real numbers. Through these graphs, we can see $f(x)$ is symmetric to $h(x)$ as both functions are reflected across the line $y=x$ (Figure 9).

Figure 8 (left) Figure 9 (right)

For the function $g(x)$ , we defined the inverse as $j(x)$ . The equation, table and graph of $j(x)$ are shown in Figure 10. Looking at the table of values, we can easily see the inverse relationship through the points (4, 2) from $g(x)$ and (2, 4) from $j(x)$ .

Because the domain of $g(x)$ is all real numbers greater than 0, we also defined the domain of $j(x)$ using these constraints. The graphs of $g \circ j(x)$ and $j \circ g(x)$ appear as the line $y=x$ . The domain of $g\circ j(x)$ and $j \circ g(x)$ would be all real numbers greater than 0. Through these graphs, we can see $g(x)$ is symmetric to $j(x)$ as both functions are reflected across the line $y=x$ (Figure 11).

Moving stuff around

Next, we were tasked to graph three functions: $f(x)=sin(x)$ , $t(x)=x+k$ , and $s(x)=ax$ . First, we were asked to graph four different compositions of $f(x)=sin(x)$ . By graphing $t \circ f(x)$ , $s \circ f(x)$ , $f \circ t(x)$ , and $f \circ s(x)$ , we were then able to explore how the different compositions compare to the original function $f(x)$ . By looking at these differences between the composition and the original function we can isolate how the graph of $f(x)$ will change when the variables in $t(x)$ and $s(x)$ are manipulated individually using the sliders.

The first composition we were asked to graph was $t \circ f(x)$ (Figure 12). After graphing $t \circ f(x)$ the first thing we noticed was that the graph of $t \circ f(x)$ appeared to be the same as the graph of $f(x)$ but translated vertically. To check this we then looked to our table of values to help justify our guess. Looking at the table for $x=1$ , we noticed that for $k=1$ from the function $t(x)$ , the difference in value between $t \circ f(x)$ and $f(x)$ was extremely close to 1 (1.00000002).

When we started changing the value of k using the slider we then observed that as we changed the value of $k$ that the values of $t \circ f(x)$ and f(x) would always have a difference of $k$ (Figure 13). From this we concluded that, depending on the value of $k$ from the function $t(x)$ , the graph of the composition would be vertically translated by $k$ compared to $f(x)$ .

Next, we were asked to graph $s \circ f(x)$ and compare it to $f(x)$ . The first thing we noticed was that when the $a$ from our function, $s(x)$ , is equal to one, that the graph of $s \circ f(x)$ and $f(x)$ perfectly matched together. This was reflected in our table of values for $s \circ f(x)$ and $f(x)$ by the fact that the values were exactly the same at $x=1$ , $x=2$ , and so on. Furthermore, when we changed the value of a for different values we observed another relationship. For example, at $a=2$ we noticed in our table that the values of $s \circ f(x)$ appeared to be twice that of the values found in the $f(x)$ column (Figure 14). To check this, we looked at the values for both functions at $x=1$ . At $x=1$ , the value of $s \circ f(x)$ was 1.682942 while the value of $f(x)$ was .84147098. When we divided the value of $s \circ f(x)$ by the value of $f(x)$ we found that indeed the value of $s \circ f(x)$ was extremely close to being twice (2.000000048) that of f(x) at $a=2$ . From this we concluded that depending on the value of $a$ , that the values of $s \circ f(x)$ would be multiplied $a$ times the original values of $f(x)$ . We can see this conclusion in our graph by seeing how changing the value of $a$ affects the amplitude of $s \circ f(x)$ . At $a=2$ , we can see that the amplitude of $f(x)$ to be 1 while the amplitude of $s \circ f(x)$ to be 2.

When we change the slider for $a$ , we can see the amplitude of $s \circ f(x)$ match the value of $a$ . From this we concluded that depending on the value of $a$ , that the amplitude of our composition $s \circ f(x)$ would be a times larger or smaller compared to $f(x)$ (Figure 15).

We were then asked to graph $f \circ t(x)$ and compare it to $f(x)$ . After graphing it, we looked at $f \circ t(x)$ for different values of $k$ , from our function $t(x)=x+k$ . What we noticed when we pressed the play button on our slider for $k$ was that as the value of $k$ increased, the graph of $f \circ t(x)$ appeared to be shifting horizontally to the left. Then, as the values for $k$ became negative, the graph of $f \circ t(x)$ looked to be shifting horizontally to the right (Figure 16).

This was initially confusing since when we had positive values of $k$ for $t \circ f(x)$ the graph was vertically shifted upwards by the value of $k$ . This behavior was mirrored when we had negative values of $k$ and the graph of $t \circ f(x)$ vertically shifted down by the value of $k$ .

Thus, we expected the behavior of $k$ for $f \circ t(x)$ to follow this trend by having the graph of $f \circ t(x)$ shift horizontally to the right with a positive $k$ value and shift horizontally to the left with a negative $k$ value. To help us better understand what is going on with the composition of $f \circ t(x)$ , we wrote out the composition with an equation. $f \circ t(x)=f(t(x))$ . $f(t(x))=sin(x+k)$ .

Looking at our table of values for $f \circ t(x)$ and f(x) at $k=1$ we soon noticed a pattern. The y-values of $f \circ t(x)$ at any given x were the same as the values of y for $f(x)$ but one more row down. This is the same as saying that $f(t(x))=sin(x+1)$ since our original function was $f(x)=sin(x)$ . Knowing this, the seemingly opposite behavior we observed early begins to make sense.

When we have a positive value for $k$ , we are adding that positive value to our $x$ in $f \circ t(x)$ . As a result, the graph shifts left since we are getting the same y-values as $f(x)$ , just at a smaller x-value, which explains why our graphs are the same besides the horizontal shift. For example, with $k=3$ , at $x=1$ we will be finding the value of $f(t(x))=sin(4)$ . Since there is no horizontal shift in our original function, this is the same as finding $f(x)=sin(x)$ where $x=4$ . Since $f \circ t(x)$ has the same y-value as $f(x)$ but at $x=1$ , the graph will look like it has shifted left, thereby explaining our horizontal shift to the left for positive values of $k$ (Figure 17).

This behavior has a similar reasoning for negative values of $k$ . For negative values of $k$ the situation is the opposite of what we found for positive values of $k$ . Where for positive values of $k$ we saw $f \circ t(x)$ shift left due to having the same y-values as f(x) at a larger x-value. For negative values of $k$ we can see that $f \circ t(x)$ shifts right due to have the same y-values as $f(x)$ at a smaller x-value (Figure 18).

For example, at $k=-1$ , $f(t(x))=sin(x-1)$ . Therefore at $x=1$ , $f(t(x))=sin(0)$ . For our original function, $f(x)$ , this is the same y-value as $f(0)=sin(0)$ . So, for $f \circ t(x)$ at $x=1$ , the function has the same y-value as $f(x)$ at $x=0$ . Similar to how we graphed the positive $k$ values, graphing the negative $k$ values makes $f \circ t(x)$ appear like it has been shifted horizontally to the right in comparison to f(x). In conclusion, adding positive values of $k$ shifts $f \circ t(x)$ to the left and adding negative values of $k$ to $f \circ t(x)$ shifts the composition to the right.

Next, we were asked to graph the composition $f \circ s(x)$ and compare it to $f(x)$ . What we first noticed about $f \circ s(x)$ was that when $a=1$ then the graphs of $f \circ s(x)$ and $f(x)$ match up perfectly. This is similar to what we explored about $s \circ f(x)$ . Then once we explored for different values of $a$ we discovered a pattern occurring with the graph. In our initial exploration of $f \circ s(x)$ we used the sliders to look at how the graph of $f \circ s(x)$ will change with different values of $a$ . What we noticed was that the positive values of $a$ increase the frequency of the waves in our sine function in comparison to $f(x)$ (Figure 19).

To see exactly how the frequency of $f \circ s(x)$ changed as our $a$ value increased we utilized a handy function of Desmos. By clicking on your line of interest, several grey points begin to appear on your graph. These points of interest include intersections, y-intercepts, as well as the maximum and minimum points of a graph. In order to find the frequency of the waves of our sine function, we needed to use this feature of Desmos to find the maximum point of each wave. By clicking on this grey point, Desmos will tell you it’s x and y position on the graph and keep it highlighted for comparison with other points.

Using this tool, we then checked the frequency for $a=1$ . By using the aforementioned Desmos feature and clicking on the maximum points of two consecutive waves, we found that the frequency was $2\pi$ since the difference between $\frac{5\pi}{2}$ and $\frac{\pi}{2}$ is equal to $2\pi$ (Figure 20).

Using the same method with $a=2$ we saw that the frequency becomes $\pi$ since the difference between $\frac{5\pi}{4}$ and $\frac{\pi}{4}$ is $\pi$ (Figure 21).

And finally, we start to see a pattern as we look at the frequency of $a=4$ . With $a=4$ , frequency becomes $\frac{\pi}{2}$ since the difference between $\frac{5\pi}{8}$ and $\frac{\pi}{8}$ is $\frac{\pi}{2}$ (Figure 22).

With these different values of $a$ we saw that the frequency of $f \circ s(x)$ is directly proportional to $a$ . As positive $a$ increases, the frequency of the sine function increases as well. As for negative values of $a$ , we found a similar relationship to what we explored for positive values of $a$ . By changing the slider to negative values of $a$ , we found that $f \circ s(x)$ becomes the inverse of f(x). For example, with $a=-1$ , by looking at our table of values we saw that for every $x$ , the values of $f \circ s(x)$ are inverses to those find in $f(x)$ . At $x=1$ , $f \circ s(1)=-.84147098$ and $f(1)=.84147098$ . In addition, $f \circ s(x)$ still follows the same pattern for frequency with negative values of $a$ . In conclusion, changing the value of $a$ to be either positive or negative proportionally changes the frequency of $f \circ s(x)$ in comparison to $f(x)$ . In addition, as the values of $a$ become negative, $f \circ s(x)$ becomes the inverse of $f(x)$ .

We were then asked to use our observations from the first part of the task to discuss the general sine function in the form $f(x)=asin(bx+c)+d$ . To discuss this general sine function, we first had to look back to our explorations from #1. By graphing the compositions and observing the subsequent change in table values, we were then able to isolate how changing $a$ and $k$ affected our graph. Based on what we learned from this, we can now discuss the parameters of $f(x)=a sin(bx+c)+d$ .

The variable a in the general sine function matches up to the composition $s \circ f(x)$ . Substituting in $f(x)$ into $s(x)$ we get $s\circ f(x)=a sin(x)$ . The important thing to notice here is that the $a$ is being multiplied by sin(x). From our exploration of $s\circ f(x)$ , changing the value of $a$ resulted in the amplitude of each wave of $f(x)$ changing in proportion to what $a$ was. If $a=2$ then the amplitude of $s \circ f(x)$ was found to be twice that of f(x).

The variable $b$ in the general sine function matches up to the composition $f \circ s(x)$ . Substituting in s(x) into f(x) we get $f \circ s(x)=sin(ax)$ . If we replace $a$ in $f \circ s(x)$ with $b$ then the general sine function that focuses on frequency matches our composition. From our exploration of $f \circ s(x)$ we learned that when $b$ was a positive value, then the frequency of $f \circ s(x)$ would change in comparison to $f(x)$ . If we had $b=2$ then the baseline frequency of $2\pi$ would increased by a factor of 2, which would result in the new frequency being $\pi$ . From this pattern we realized that the frequency of $f(x)$ was directly proportional to the value of $b$ . In addition, we observed that when the value of $b$ became negative, the graph of $f \circ s(x)$ would become the inverse of $f(x)$ .

The variable c in the general sine function matches up to the composition $f \circ t(x)$ . Substituting in $t(x)$ into $f(x)$ we get $f \circ t(x)=sin(x+k)$ . If we replace $k$ in $f \circ t(x)$ with c then $f \circ t(x)$ exactly matches the component found in the general sine function. From our exploration of $f \circ t(x)$ , we found that changing the value of $c$ to positive values made the graph of $f \circ t(x)$ shift horizontally to the left by the value of $c$ . In a similar way we also found that by adding negative values of $c$ , that the graph of $f \circ t(x)$ would be shifted horizontally to the right by the value of $c$ .

The variable $d$ in the general sine function matches up to the composition $t \circ f(x)$ . Substituting in $f(x)$ into $t(x)$ we get $t \circ f(x)=sin(x)+k$ . Replacing $k$ in $t \circ f(x)$ with $d$ then creates a match with the vertical component of the general sine function. From our exploration we found that changing the value of $d$ shifted the graph upwards for position values of $d$ . Similarly, changing the value of $d$ to be negative resulted in the graph shifting downwards by the exact value of $d$ .

Next, we were asked to discuss the “vertex form” of the quadratic function: $f(x)=a (x-h)^2+k$ . Similar to how we discussed the general sine function, we also found some correlations between the compositions we explored and the parameters of the “vertex form” of the quadratic function.

The variable a in the “vertex form” of the quadratic function matches up to the results we found for the amplitude component of $s \circ f(x)$ . From our exploration of $s \circ f(x)$ , changing the value of $a$ resulted in the amplitude of each wave of f(x) changing in proportion to what $a$ was. If $a = 2$ then the amplitude of $s\circ f(x)$ was found to be twice that of f(x). This behavior is also mirrored for $a$ in the “vertex form”. Changing the value of $a$ changes the steepness of the quadratic curve in addition to changing whether the curve opens up or down (Figure 23).

The variable h in the “vertex form” of the quadratic function matches up to the results we found for the horizontal component of $f \circ t(x)$ . From our exploration of $f \circ t(x)$ , we found that changing the value of $c$ to positive values made the graph of $f \circ t(x)$ shift horizontally to the left by the value of $c$ . In a similar way we also found that by adding negative values of $c$ , that the graph of $f \circ t(x)$ would be shifted horizontally to the right by the value of $c$ . A major difference between our investigation of the sine function and the “vertex form” of the quadratic is that the values of $h$ affect the horizontal shift of the graph much like you would expect them to. Negative values of $h$ shift the graph horizontally to the left and positive values of $h$ shift the graph horizontally to the right (Figure 24).

The variable $k$ in the “vertex form” of the quadratic function matches up to the results we found for the vertical component of $t \circ f(x)$ . From our exploration we found that changing the value of $d$ shifted the graph upwards for position values of $d$ . Similarly, changing the value of $d$ to be negative resulted in the graph shifting downwards by the exact value of $d$ . For the “vertex form” of the quadratic function this functionality of $k$ serves the same purpose as $d$ from our earlier exploration. Adding positive values of $k$ vertically shifts the graph upwards while adding negative values of $k$ vertically shifts the graph downwards (Figure 25).

While there were connections between variables $a$ , $h$ , and $k$ and the compositions found in the the compositions related to $f(x)$ , there was no connection between $f \circ s(x)$ and any other variable of the “vertex form” of the quadratic function. While in the general sine function there was a variable controlling the frequency, there is no such variable for the “vertex form” of the quadratic function. This is because there is no period for the quadratic function. For the quadratic function, there is only the one parabola rather than a curve which could be affected by a variable controlling the frequency. Looking at the “vertex form” we can see that where the variable for frequency would be, there is just an understood 1.

Finally, we then looked at examples of quadratic functions, both in vertex form and in the general form.

Comparing our answers with the vertex form and general form we notice that they’re equivalent to each other. For example, taking the vertex form for A) we can simplify $(x+2)^2$ to get $y = (x^2+4x+4)-3$ . Adding the constants together we end with $y = x^2+4x+1$ , which is the same as the general form Figure 25).

This also works if you have the general form and want to change it to the vertex form. Using A) as our example, we can complete the square for $y=x^2+4x+1$ to get $y = x^2+4x+2^2+1-4$ . Factoring the quantity $x^2+4x+4$ gives us $y=(x+2)^2+1-4$ . Adding together the constants leaves us with $y=(x+2)^2-3$ , which is the same as the vertex form.

And finally, we were tasked to compare how easy it is to use the standard form and the vertex form. So let’s first think about the vertex form.

Easy to use if given the vertex and other point.
Without the vertex, you have to find the general from and then convert to vertex form.
Easy to convert to the general form.

And now the general form.

System of equations can become tedious when solving for a,b, and c.
Easy to convert to the vertex form in order to find the vertex of a parabola.

In general, depending on the points that are given, using the general form or the vertex form can be easier than the other. If given the vertex and one other point of the parabola then finding the vertex form becomes extremely easy. However, if you are not given the vertex, then finding the standard form can be a little more tedious in comparison.

Teacher Discussion

Observing transformation as a result of function composition is an application of SMP 7 (Look for and make use of structure). It was in the “Combining Functions” and “Inverses” tasks that students will be able to use the key features of the function, such as the domain, to evaluate expressions, determine the appropriate input values for a new function, and identify inverse functions.

Seeing a direct correlation between changing a variable and observing a subsequent change in the composition graph provided evidence for a student’s conjectures about the compositions’ relationships to the graph of f(x). Once they observed the correlation between the changing variable and graph, students could then justify their conjecture using the values from the tables. This practice of proposing a conjecture and justifying it with reasoning and evidence is an application of SMP 3 (Construct viable arguments and critique the reasoning of others).

In the “Moving Stuff Around” task, we observed the relationships between the compositions of f(x), s(x), and t(x) and the parameters of the general sine function and vertex form of the quadratic function. To better identify these relationships, we were asked to include tables of precise values for the given functions and compositions, which is an application of SMP 6 (Precision). To complete this exploration, students could use a mathematical action technology such as Desmos as they progress through the different parts of the investigation. Using Desmos, students will be able to observe how the graphs of the different compositions changed depending on what variable from s(x) and t(x) was being manipulated. Once they observed the relationships between the compositions and the graph of f(x), they were then asked to use their observations to think about the general sine function. This portion of the activity was focused on engaging students with a high-level cognitive task by having them take their previous observations and apply it to the generalized form of the sine function. The investigation then takes the decontextualization of the students’ findings one step further by having them apply their observations to the vertex form of the quadratic function. Having students work through these activities creates a relational understanding for the structure of functions and their subsequent graphs (Skempf, 1976).

The purpose of having students redo this investigation was to have students utilize what they learned from comparing the general sine function and the quadratic function, and then apply it to comparing the vertex form of the quadratic with the standard form. Students should have identified that the different forms are equivalent to one another after some algebraic manipulation.

Lesson Plan Outline

Have students open Desmos.
Have students graph two functions. ( $f(x)= \frac{1}{x-2}$ and $g(x)=\sqrt{x}$ )
Have students examine the domains of both functions.
Have students create two compositions using these two functions.
Have students use a table of values in conjunction with the graph in order to better explain the domains of their compositions of functions.
1. Students will be able to identify values of x where the graph is undefined/not in the domain by looking at the table of values for each composition.
Have students graph four compositions of , , and .
1. Students will be able to properly set up and graph the compositions of $f(x)$ , $s(x)$ , and $t(x)$ .
Have students explore how changing the sliders affects the graph of the sine function.
Have students use a table of values to help explain why the graph changes when the sliders are changed.
1. Students will be able to utilize specific values of the table to justify their conjecture about the graph.
Have students derive a general sine function based on the relationships they observed with the compositions.
1. Students will be able to form a generalized sine function based on the structure of the compositions and the relationships they explored and justified in Desmos.

References

Alabama State Department of Education (2016). 2016 Revised Alabama Course of Study: Mathematics.

National Governor’s Association. (2010). Common Core State Standards for Mathematics. Washington, D.C. Retrieved from http://www.corestandards.org/Math/

Skemp, R. R. (1976). Relational understanding and instrumental understanding. Mathematics teaching, 77(1), 20-26.

Share this: