Lesson 14
Absolute Value Functions (Part 2)
14.1: Temperature in Toronto (10 minutes)
Student Facing
Toronto is a city at the border of the United States and Canada, just north of Buffalo, New York. Here are twelve guesses of the average temperature of Toronto, in degrees Celsius, in February 2017.
- 5
- 2
- -5
- 3
- 0
- -1
- 1.5
- 4
- -2.5
- 6
- 4
- -0.5
-
The actual average temperature of Toronto in February 2017 is 0 degree Celsius.
Use this information to sketch a scatter plot representing the guesses, \(x\), and the corresponding absolute guessing errors, \(y\).
- What rule can you write to find the output given the input?
Student Response
For access, consult one of our IM Certified Partners.
Anticipated Misconceptions
Some students may struggle to plot the data without explicitly computing the absolute guessing errors first or creating a table of values. Encourage them to take those intermediate steps if they are helpful.
Activity Synthesis
Select a student to display the completed scatter plot. Ask students:
- "How is the scatter plot for this data like the scatter plot for the absolute guessing errors from an earlier lesson?" (The points still form a V shape. The two parts of the V now meets at \((0,0)\). There are still no negative \(y\)-values.)
- "How are they different?" (The graphs are shifted horizontally toward the vertical axis. The two parts of the V now meets at \((0,0)\).)
To help students see that the "actual average temperature" is like the "actual number of items in a jar" they saw earlier, highlight that:
- Earlier, we saw that when the actual number of items is \(a\), then the absolute guessing error is "the distance of guess from \(a\)" which can be expressed as "the absolute value of (guess - \(a\))" or \(|\text{guess} - a|\).
- Here, likewise, when the actual average temperature is \(a\) and the guess is \(x\), the absolute guessing error, \(y\), is "the distance of \(x\) from \(a\)," which can be written as \(y = |x-a|\).
- When the actual temperature in Toronto is 0 degree Celsius, \(y\) is "the distance of \(x\) from 0," which can be written as \(y=|x-0|\) or simply \(y=|x|\).
14.2: The Distance Function (15 minutes)
Activity
This activity formally introduces students to the absolute value function as the function that takes an input value and gives its distance from the origin as the output. Students see two different ways to represent this function algebraically: using the absolute value notation and using the cases notation:
\(A(x)=|x|\)
\(A(x)=\begin{cases} x,& x \geq 0 \\ \text{-}x, & x < 0 \end{cases}\)
The equation in cases notation focuses on what has to happen algebraically to an input value to give its distance from the origin. Different rules apply to different intervals of the domain.
As students work, look for those who plot points for the graph of function \(A\) and those who sketch two lines. Ask students who sketch different types of graphs to share during discussion.
Launch
Arrange students in groups of 2. Give students a few minutes of quiet work time, than time to share their response with their partner. Follow with a whole-class discussion.
Design Principle(s): Optimize output (for explanation); Cultivate conversation
Student Facing
The function \(A\) gives the distance of \(x\) from 0 on the number line.
- Complete the table and sketch a graph of function \(A\).
\(x\) \(A(x)\) 8 5.6 \(\pi\) \(\frac12\) 1 0 \(\text-\frac12\) -1 -5.6 8 -
Andre and Elena are trying to write a rule for this function.
- Andre writes: \(A(x)=\begin{cases} x,& x \geq 0 \\ \text-x, & x < 0 \\ \end{cases}\)
- Elena writes: \(A(x) = |x|\)
Explain why both equations correctly represent the function \(A\).
Student Response
For access, consult one of our IM Certified Partners.
Anticipated Misconceptions
Students who interpret the “\(\text-x\)” in Andre's rule to mean "negative \(x\)" (rather than "the opposite of \(x\)") may be unsure how to use that information. Ask students to evaluate the function for specific values of \(x\) and to write down each step. For instance, when \(x\) is -2, \(A(\text{-}2)\) is \(\text{-}(\text{-}2)\), which is 2.
Activity Synthesis
Select previously identified students to display their graph of function \(A\). Ask students who plotted only the ordered pairs in the table whether the pairs that are not in the table, if plotted, would also fall on the same two lines. Emphasize that \(A\) can be shown with two lines.
Next, ask students to share their response to the last question. Students may find Elena's rule easier to explain because of their work with absolute value in recent activities and struggle to explain Andre's rule.
We can reason about Andre's equation a couple of ways:
-
By thinking about rules: The equation is that of a piecewise function because different rules are applied to different parts of the domain to give the "distance from 0" as the output:
- When the input \(x\) is positive or 0, its distance from 0 is that same number, \(x\).
- When the input \(x\) is negative, its distance from 0 is the opposite of the number, \(\text-x\).
-
By using the graph: The two halves of the graph are lines with different slopes.
- If we cover up the left side of the vertical axis and see only positive values of \(x\), we see a line with a slope of 1 and that represents \(y=x\).
- If we cover up the right side and see only negative values of \(x\), we see a line with a slope of -1 and that represents \(y=\text{-}x\)
Explain to students that:
- Function \(A\) is the absolute value function. It gives the distance of an input value from a certain value, the origin, in this case.
- The graph of function \(A\) is a V shape with the two lines of points converging at \((0,0)\), which is the minimum of the graph. We call this point the vertex of the graph. It is the point where the graph changes direction.
- The absolute value function is also a piecewise function because different rules are applied to different parts of the domain to get the output.
Supports accessibility for: Conceptual processing; Language
14.3: Moving Graphs Around (10 minutes)
Activity
In this activity, students expand their awareness of absolute value functions by analyzing the graphs and equations of several absolute value functions. The graphs have the same V shape but not the same vertex. The expressions that define the functions have a constant term added to or subtracted from the input, \(x\), or from the absolute value of the input, \(|x|\). Students observe how each constant term affects the graph and consider possible explanations.
To make sense of the placement of each graph on the coordinate plane, students may:
- Find some input-output pairs for each function and verify that the graph contains those pairs of values.
- For \(j\) and \(k\): Recall that adding a constant term to (or subtracting one from) a linear equation shifts the graph vertically. They then think of \(|x|+2\) and \(|x|-2\) the same way: adding 2 moves the graph of \(|x|\) up 2 units and subtracting 2 moves it down.
-
For \(g\) and \(h\):
- Interpret the expressions \(|x-2|\) and \(|x+2|\) in terms of finding the absolute error of a guess, \(x\), from the target numbers 2 and -2, respectively. They may recall that the vertex of the absolute error graph is at the target number on the \(x\)-axis.
- Reason that the absolute value of any expression cannot be negative. This means the vertex of each graph, which is also its minimum, will have a value of 0. They then look for the \(x\)-value at which \(|x-2|\) is 0 and at which \(|x+2|=0\).
- Think of each function as a piecewise function. For \(g(x)=|x-2|\), the two pieces are two linear functions, defined by \(x-2\) and its opposite, \(\text-(x-2)\). Their graphs meet at \(x=2\). For \(h(x)=|x+2|\), the two linear functions are defined by \(x+2\) and \(\text-(x+2)\). Their graphs meet at \(x=\text-2\).
Monitor for students who use these or other strategies and invite them to share during discussion. All strategies except the first involve looking for and making use of structure (MP7).
Dynamic graphing technology can be very useful for observing how the parameters of a function are related to the features of its graph and can help students generalize their observations. Consider giving students individual access to dynamic graphing technology, and then giving them time to explain the behaviors of the graphs. If students don't have individual access, projecting the applet in the digital launch would be helpful.
Launch
Dynamic graphing technology is great for showing how graphs move when their equations change. One of the simplest ways to show the changes is by using sliders that control the values of the parameters. This process is usually very easy in online graphing calculators such as Desmos. If using Desmos, demonstrate the following process to students.
- When a letter other than \(x\), \(y\), or \(e\) is typed into an input field, Desmos will offer a slider. For example, typing the equation \( f(x)=|x+a|\) will prompt a choice to create a slider for \(f\). Because this is not what we want here, ignore the prompt and keep typing. (Once an equal sign is entered, \(f\) is recognized as the name of the function.)
- When Desmos offers a slider for \(a\), hit the return key or click on the blue button that says \(a\) to create the slider then finish inputting the expression.
- Repeat the process for \(g(x)=|x|+b\). (The absolute value bars are accessible on the virtual keyboard.)
- To edit the upper and lower limits and the step by which the slider moves, click on the upper or lower limit of that slider. A small editing window will open where changes can be made.
If you will be using graphing technology other than Desmos for this activity, you may need to prepare alternate instructions.
Supports accessibility for: Organization; Conceptual processing; Attention
Student Facing
Graph the functions \( f(x)=|x+a|\) and \(g(x)=|x|+b\). Experiment with different values of \(a\) and \(b\) and observe what happens to the graphs.
- How does changing \(a\) change the graph?
- How does changing \(b\) change the graph?
Student Response
For access, consult one of our IM Certified Partners.
Student Facing
Are you ready for more?
- Look for the minimum of each graph. Each point represents the least value of the function.
- For function \(f\), what value of \(x\) gives the least output value?
- For function \(g\), what value of \(x\) gives the least output value?
- Another function is defined by \(m(x)=|x+11.5|\).
- What value of \(x\) produces the least output of function \(m\)? Be prepared to explain how you know.
- Describe the graph of \(m\).
Student Response
For access, consult one of our IM Certified Partners.
Launch
Give students a few minutes of quiet think time. Provide access to graphing technology, if requested.
Ask students to not only observe how the addition or subtraction of 2 affects each graph, but also be prepared to offer an explanation as to why it makes sense that the graph is where it is. Consider demonstrating what a possible explanation could look or sound like, using function \(f\) as an example. Or, consider giving prompts such as:
- It makes sense that the graph of \(\underline {\hspace{0.5in}}\) is located \(\underline {\hspace{0.5in}}\) because . . . .
- I know that the vertex of graph of \(\underline {\hspace{0.5in}}\) belongs at \(\underline {\hspace{0.5in}}\) because . . . .
Follow with a whole-class discussion.
Supports accessibility for: Organization; Conceptual processing; Attention
Student Facing
Here are equations and graphs that represent five absolute value functions.
Notice that the number 2 appears in the equations for functions \(g,h,j\), and \(k\). Describe how the addition or subtraction of 2 affects the graph of each function.
Then, think about a possible explanation for the position of the graph. How can you show that it really belongs where it is on the coordinate plane?
Student Response
For access, consult one of our IM Certified Partners.
Student Facing
Are you ready for more?
-
Mark the minimum of each graph in the activity. Each point you marked represents the least output value of the function.
In each function, what value of \(x\) gives that minimum output value?
-
- Another function is defined by \(m(x)=|x+11.5|\). What value of \(x\) produces the least output of function \(m\)? Be prepared to explain how you know.
- Describe or sketch the graph of \(m\).
Student Response
For access, consult one of our IM Certified Partners.
Activity Synthesis
Select students to share their explanations, spotlighting the strategies shown in the Activity Narrative.
If time is limited, focus the discussion on this question: "In the case of \(g(x)=|x-2|\) and \(h(x)=|x+2|\), why does subtracting 2 from \(x\) move the graph to the right and adding 2 to \(x\) move it to the left?"
If no students made a connection to their work on finding absolute guessing errors, ask questions such as:
- "In a guessing game, if the target number is 2 and a guess is 10, how do we find the absolute error?" (Find the difference of 10 and 2, or subtract 2 from 10, and find the absolute value, or \(|10-2|\).) "What if the guess is \(x\)?" (Subtract 2 from \(x\) and find the absolute value, or \(|x-2|\).)
- "If we plot the guesses and absolute errors, where would the vertex of the V-shape be?" (At \((2,0)\)).
- "If the target number is -2 and a guess is \(10\), how do we find the absolute error?" (Find the difference between 10 and -2, or subtract -2 from 10, and find the absolute value, or \(|10– (\text-2)|\), which is \(|10+2|\).) "What if the guess is \(x\)?" (Subtract -2 from \(x\) and find the absolute value, or \(|x-(\text-2)|\) or \(|x+2|\).)
- "It the target number is -2 and we plot the guesses and absolute errors, where would the vertex of this graph be?" (At \((\text-2,0))\).)
Another way to reason about the graphs of \(g\) and \(h\) is by using the structure of the equation. We know from earlier work that the vertex of a graph of absolute errors is always on the horizontal axis, so its output value is 0. Ask students:
- "For what value of \(x\) is \(|x|=0\)?" (0)
- "For what value of \(x\) is \(|x-2|=0\)? How do you know?"(2, because \(|2-2|=|0|=0\))
- "For what value of \(x\) is \(|x+2|=0\)? How do you know?"(-2, because \(|\text-2+2|=|0|=0\))
If time permits and if not already mentioned by students, remind students that each function and can be seen as a piecewise function with two parts, each part being a linear function.
Graphing the two linear functions gives two lines that intersect on the horizontal axis. For \(g\), the two lines meet at \((2,0)\). For \(h\), they meet at \((\text-2,0)\).
Design Principle(s): Cultivate conversation; Maximize meta-awareness
14.4: More Moving Graphs Around (15 minutes)
Optional activity
This optional activity gives students a chance to test the observations or apply the generalizations they made in the previous activity. They match equations and graphs of other absolute value functions, as well as sketch the graph of an equation without a match.
Students also encounter new cases in which a constant term is added or subtracted both before and after absolute value is applied to the input, resulting in a graph that shifts both vertically and horizontally relative to the graph of \(f(x)=|x|\).
In making the matches (and before using technology to check their graphs), students are likely to reason in different ways and to rely on structure to varying degrees (MP7). For instance, students may:
- Evaluate each function at different input values and then match the input-output pairs to the coordinates on the graphs.
- Start with the graph of \(|x|\) and then shift sideways for addition or subtraction inside the absolute value symbols, and then shift the graph vertically for addition or subtraction after absolute value is applied.
- Think about the \(x\)-value that would produce the smallest possible output and relate that ordered pair to the vertex of the graph.
Monitor for students using different strategies and invite them to share during discussion.
Students will have opportunities to explore transformations of functions and their graphs in a later unit and more formally in a future course.
Launch
Provide access to devices that can run Desmos or other graphing technology.
Student Facing
Here are five equations and four graphs.
- Equation 1: \(y=|x-3|\)
- Equation 2: \(y=|x-9|+3\)
- Equation 3: \(y=|x|-6\)
- Equation 4: \(y=|x+3|\)
- Equation 5: \(y=|x+3|-6\)
- Match each equation with a graph that represents it. One equation has no match.
- For the equation without a match, sketch a graph on the blank coordinate plane.
- Use graphing technology to check your matches and your graph. Revise your matches and graphs as needed.
Student Response
For access, consult one of our IM Certified Partners.
Activity Synthesis
Select previously identified students to share their strategies for making a match. If students use the strategies listed in the Activity Narrative, order their presentation as shown.
If time permits, ask students to use a strategy that they find effective to describe the graphs of \(f(x) = |x-3|+5\) and \(y = |x+4|+7\).
Design Principle(s): Support sense-making
Lesson Synthesis
Lesson Synthesis
Display the graph of \(A(x)=|x|\) for all to see, along with the two equations for the function, \(A(x)=|x|\) and \(\displaystyle A(x)=\begin{cases} x,& x\geq 0 \\ \text{-}x,& x < 0 \end{cases} \)
Ask questions such as:
- "None of the points on the graph of this absolute value function lie below the \(x\)-axis. Why is that?" (The absolute value of a number is never negative.)
- "Suppose we know that \(A(x)\) is 4. How do we know what value or values of \(x\) would give an output of 4?" (We look at numbers that are 4 units from 0. There are two numbers that meet this requirement: 4 and -4.)
- "How do we use the equation \(A(x)=|x|\) to find the function value at \(x=\text{-}5\)?" (\(A(\text-5) = |\text-5|\), which is 5.)
- "How do we use the equation that uses the cases notation to find the function value at \(x=\text-5\)?" (Because -5 is less than 0, we use the rule for \(x<0\) and find \(A(\text{-}5)\), which gives \(\text{-}(\text{-}5)\) or 5.)
14.5: Cool-down - Elevations of Places (5 minutes)
Cool-Down
For access, consult one of our IM Certified Partners.
Student Lesson Summary
Student Facing
In a guessing game, each guess can be seen as an input of a function and each absolute guessing error as an output. Because absolute guessing error tells us how far a guess is from a target number, the output is distance.
Suppose the target number is 0.
- We can find the distance of a guess, \(x\), from 0 by calculating \(x-0\). Because distance cannot be negative, what we want to find is \(|x-0|\), or simply \(|x|\).
- If function \(f\) gives the distance of \(x\) from 0, we can define it with the equation:
\(f(x)=|x|\)
Function \(f\) is the absolute value function. It gives the distance of \(x\) from 0 by finding the absolute value of \(x\).
The graph of function \(f\) is a V shape with the two lines converging at \((0,0)\).
We call this point the vertex of the graph. It is the point where a graph changes direction, from going down to going up, or the other way around.
We can also think of a function like \(f\) as a piecewise function because different rules apply when \(x\) is less than 0 and when it is greater than 0.
Suppose we want to find the distance between \(x\) and 4.
- We can find the difference between \(x\) and 4 by calculating \(x-4\). Distance cannot be negative, so what we want is the absolute value of that difference: \(|x-4|\).
- If function \(p\) gives the distance of \(x\) from 4, we can define it with the equation:
\(p(x)= |x - 4|\)
Now suppose we want to find the distance between \(x\) and -4.
- We can find the difference of \(x\) and -4 by calculating \(x-(\text-4)\), which is equal to \(x+4\). Distance cannot be negative, so let's find the absolute value: \(|x+4|\).
- If function \(q\) gives the distance of \(x\) from -4, we can define it with the equation:
\(q(x)= |x + 4|\)
Notice that the graphs of \(p\) and \(q\) are like that of \(f\), but they have shifted horizontally.