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<h4>Activity (20 minutes)</h4>
<p>This activity introduces students to the average rate of change by building on what students know about rate of
change and slope.</p>
<p>Students see that finding the change in the output for every unit of change in the input can be a useful way to
generalize what happens between two function values, regardless of the behaviors of individual data points between
them. They recognize that this number tells us how, on average, one quantity is changing relative to the other, and
that it can be useful for comparing the trends in different intervals of a function.</p>
<p>In the activity, students find average rates of change by reasoning and using their knowledge of linear
relationships. Monitor for the following strategies (the examples shown are for finding the average rate of change
between 6 p.m. and 10 p.m.):</p>
<ul>
<li>Find the hourly changes in temperature and calculate the average. The temperature fell by 2°F the first hour,
then 4°F, 0°F, and 3°F in the next three hours. The average is \(\frac{2+4+0+3}4=\frac94\), or
2.25°F per hour.</li>
</ul>
<ul>
<li>Find the overall change in temperature and divide by the overall change in hours. The temperature fell 9°F in
4 hours, which means an average drop of 2.25°F per hour.</li>
<li>Find the slope of the line connecting the two end points of the interval. The slope between \((6,17)\) and
\((10,8)\) is \(\frac{8-17}{10-6}=\frac{-9}4=-2.25\), which means an average drop of 2.25°F per hour.</li>
</ul>
<p>In the activity synthesis, students generalize their reasoning and see that finding the average rate of change is
indeed equivalent to finding the slope of the line connecting the two points.</p>
<h4>Launch</h4>
<p>Ask students to keep their materials closed. Display a <a href="https://k12.openstax.org/contents/raise/resources/059a8addf80870b33a2430ddd9af8d78607384db" target="_blank">scatter plot</a> with the two temperature data points plotted, as
shown.</p>
<p>Tell students that it shows the temperature at 4 p.m. and the temperature at 10 p.m.</p>
<p><img alt="Two points. hours since noon and temperature in degrees Fahrenheit."
src="https://k12.openstax.org/contents/raise/resources/059a8addf80870b33a2430ddd9af8d78607384db"></p>
<p>Ask students:</p>
<ul>
<li>“How did the temperature change between 4 p.m. and 10 p.m.?” (It fell.)</li>
<li>“How much did it fall?” (17°F)</li>
<li>“Can we tell how fast, or at what rate, the temperature was falling? If so, how?” (It fell 17°F in
6 hours. If it was falling at a constant rate, it would be falling \(\frac{17}{6}\) or about 2.8°F per hour. If
we connect the points with a line and find its slope, it would be about –2.8.)</li>
<li>“If not, why not?” (There is not enough information about what happened between the two data points.
It might not be falling at the same rate.)</li>
</ul>
<p>Display a <a href="https://k12.openstax.org/contents/raise/resources/30a136c871c453e55714fb6db87ad6175c151e90" target="_blank">scatter plot</a> with the hourly data points plotted.</p>
<p>Give students a moment to notice and wonder something about the data and solicit a few responses. If no students
mention the temperatures changing at different rates (a lot, a little, or not at all) each hour, ask them about it.
</p>
<p><img alt="Discrete function. hours since noon and temperature in degrees Fahrenheit."
src="https://k12.openstax.org/contents/raise/resources/30a136c871c453e55714fb6db87ad6175c151e90"></p>
<p>Discuss with students: “Do the hourly data between 4 p.m. and 10 p.m. help us better characterize how fast the
temperature was falling between 4 p.m. and 10 p.m.? Could we still say that it was falling about 2.8°F an
hour?”</p>
<p>Students are likely to comment that a line connecting \((4,25)\) and \((10,8)\) approximates the distribution of
points in that interval and conclude that –2.8°F per hour is a reasonable rate to use.</p>
<p>Explain to students that it can be helpful to have a way to quantify how a quantity is changing over a particular
interval, without worrying about the smaller changes during the interval. The rate of –2.8°F per hour serves
that purpose. It is the average rate of change between 4 p.m. and 10 p.m.</p>
<h4>Student Activity </h4>
<p>The table and graph show a more complete picture of the temperature changes on the same winter day. The function
\(T\) gives the temperature in degrees Fahrenheit, \(h\) hours since noon.</p>
<p><img alt="Discrete function. hours since noon and temperature in degrees Fahrenheit."
src="https://k12.openstax.org/contents/raise/resources/30a136c871c453e55714fb6db87ad6175c151e90"></p>
<br>
<table class="os-raise-horizontaltable">
<thead></thead>
<tbody>
<tr>
<th scope="row">
\(h\)
</th>
<td>
0
</td>
<td>
1
</td>
<td>
2
</td>
<td>
3
</td>
<td>
4
</td>
<td>
5
</td>
<td>
6
</td>
<td>
7
</td>
<td>
8
</td>
<td>
9
</td>
<td>
10
</td>
<td>
11
</td>
<td>
12
</td>
</tr>
<tr>
<th scope="row">
\(T(h)\)
</th>
<td>
18
</td>
<td>
19
</td>
<td>
20
</td>
<td>
20
</td>
<td>
25
</td>
<td>
23
</td>
<td>
17
</td>
<td>
15
</td>
<td>
11
</td>
<td>
11
</td>
<td>
8
</td>
<td>
6
</td>
<td>
7
</td>
</tr>
</tbody>
</table>
<br>
<p>In questions 1–3, find the average rate of change for the following intervals. Be prepared to show your reasoning.
</p>
<ol class="os-raise-noindent">
<li>between noon and 1 p.m.</li>
</ol>
<p><strong>Answer:</strong> 1°F per hour. The temperature rose by 1°F in 1 hour.</p>
<ol class="os-raise-noindent" start="2">
<li>between noon and 4 p.m. </li>
</ol>
<p><strong>Answer:</strong> 1.75°F per hour. The temperature increased by 7°F in 4 hours. The average increase
is \(\frac74\) or 1.75. </p>
<ol class="os-raise-noindent" start="3">
<li>between noon and midnight</li>
</ol>
<p><strong>Answer:</strong> –0.92°F an hour. I found the slope of the line connecting the two points, and it
is \(-\frac{11}{12}\).</p>
<p>In questions 4–6, use the slope formula to find the slope of the line that connects the following points.
</p>
<ol class="os-raise-noindent" start="4">
<li>\((0, 18)\) and \((1, 19)\)</li>
</ol>
<p><strong>Answer:</strong> \(\frac{19-18}{1-0} = 1\)</p>
<ol class="os-raise-noindent" start="5">
<li>\((0, 18)\) and \((4, 25)\)</li>
</ol>
<p><strong>Answer:</strong> \(\frac{25-18}{4-0} = \frac74\) </p>
<ol class="os-raise-noindent" start="6">
<li>\((0, 18)\) and \((12, 7)\)</li>
</ol>
<p><strong>Answer:</strong> \(\frac{7-18}{12-0} = \frac{-11}{12}\)</p>
<ol class="os-raise-noindent" start="7">
<li>Compare your answers to questions 1–3 with your answers to questions 4–6. Explain what this tells you about the
average rate of change and the slope of a line. </li>
</ol>
<p><strong>Answer:</strong> The average rate of change between two points is the same as the slope of the line
connecting the 2 points. </p>
<p>For questions 8 and 9, think back to Mai and Tyler’s disagreement about temperature to solve these problems.
</p>
<ol class="os-raise-noindent" start="8">
<li>Use the <span class="os-raise-ib-tooltip" data-store="storename" data-schema-version="1.0">average rate of
change</span> to show which time period—4 p.m. to 6 p.m. or 6 p.m. to 10 p.m.—experienced a faster temperature
drop. </li>
</ol>
<p><strong>Answer:</strong> The temperature dropped faster between 4 p.m. and 6 p.m. For example: In the first interval,
the temperature fell 8°F in 2 hours, which is an average drop of 4°F an hour. In the second, it fell 9°F in 4 hours,
which is an average drop of 2.25°F per hour.</p>
<ol class="os-raise-noindent" start="9">
<li>Use the slope formula to show which line has the greater slope—the line connecting the points that represent the
temperature at 4 p.m. and 6 p.m. or the line connecting the points that represent the temperature at 6 p.m. and 10
p.m. </li>
</ol>
<p><strong>Answer:</strong> The slope of the line representing the temperature between 4 p.m. and 6 p.m. has a greater
slope. The slope of the line connecting the points representing the temperature between 4 p.m. and 6 p.m. is
\(\frac{17-25}{6-4} = -4\). The slope of the line connecting the points representing the temperature between 6 p.m.
and 10 p.m. is \(\frac{8-17}{10-6} = \frac{9}{4}=-2.25\).</p>
<ol class="os-raise-noindent" start="10">
<li>Compare your answers to question 8 and question 9.</li>
</ol>
<p><strong>Answer:</strong> Finding the average rate of change and using the slope formula give the same results. The
average rate of change and the slope of the line between hours 4 and 6 is greater than the average rate of change and
the slope of the line between hours 6 and 10. </p>
<h4>Student Facing Extension</h4>
<h5>Are you ready for more?</h5>
<ol class="os-raise-noindent">
<li>Over what interval did the temperature decrease the most rapidly?</li>
<p> <strong>Answer: </strong> between 5 p.m. and 6 p.m.</p>
<li>Over what interval did the temperature increase the most rapidly?</li>
<p> <strong>Answer: </strong> between 3 p.m. and 4 p.m.</p>
</ol>
<h4>Anticipated Misconceptions</h4>
<p>It is possible that students will not understand that a graph is also read from left to right. This could be a
problem when determining intervals of increase and decrease. Often students forget that the \(y\)-values are on top of
the formula for average rate of change, and the \(x\)-values are on the bottom. One way to check themselves if they
are mixing up the numerator and denominator is for them to check the reasonableness of the answer. Sometimes this
helps them find their mistakes on their own.</p>
<h4>Activity Synthesis</h4>
<p>Select students to share how they determined the average rates of change for the specified intervals. Arrange the
presentations in the order listed in the Activity Narrative.</p>
<p>Then, point out that all strategies involve finding how much the temperature (the output) changed per unit of change
in time (the input), and that this is done by division.</p>
<p>Display a graph with two points labeled \((a,f(a))\) and \((b,f(b))\). Ask students how we might find the average
rate of change between the two points.</p>
<p>Make sure students can generalize their work and see that:</p>
<p>average rate of \(\mathrm{change} = \frac{f(b)-f(a)}{b-a}\)</p>
<p><img alt="Discrete function graph."
src="https://k12.openstax.org/contents/raise/resources/4a685e6943e7ca22e4d6b3cfba2d0b5f34495bbb"></p>
<br>
<div class="os-raise-extrasupport">
<div class="os-raise-extrasupport-header">
<p class="os-raise-extrasupport-title">Support for English Language Learners</p>
<p class="os-raise-extrasupport-name">MLR 7 Compare and Connect: Representing, Conversing</p>
</div>
<div class="os-raise-extrasupport-body">
<p>Prior to the whole-class discussion, invite students to create a visual display that shows how they determined the
average rates of change for the specified intervals. Students should consider how to represent their strategy so
that other students will be able to understand their solution method. Students may wish to add notes or details to
their displays to help communicate their reasoning and thinking. Begin the whole-class discussion by selecting and
arranging 2–4 displays for all to see. Give students 1–2 minutes of quiet time to think individually to
interpret the displays before inviting the authors to share their strategies as described in the activity narrative.
This will help students make connections between different ways to find the average rate of change.</p>
<p class="os-raise-text-italicize">Design Principle(s): Cultivate conversation; Maximize meta-awareness</p>
<p class="os-raise-extrasupport-title">Learn more about this routine</p>
<p>
<a href="https://www.youtube.com/watch?v=PF8fRA107OA;&rel=0" target="_blank">View the instructional video</a>
and
<a href="https://k12.openstax.org/contents/raise/resources/94a1159e7b81493c647515711f325771076d99b8" target="_blank">follow along with the materials</a>
to assist you with learning this routine.
</p>
<p class="os-raise-extrasupport-title">Provide support for students</p>
<p>
<a href="https://k12.openstax.org/contents/raise/resources/0b8a1a4ac3425e84a1d5452b3a5dffa38deb6b13" target="_blank">Distribute graphic organizers</a>
to the students to assist them with participating in this routine.
</p>
</div>
</div>
<br>
<h3>4.8.2: Self Check </h3>
<p class="os-raise-text-bold"><em>After the activity, students will answer the following question to check their
understanding of the concepts explored in the activity.</em></p>
<p class="os-raise-text-bold">QUESTION:</p>
<p>The temperature \(t\), in degrees Fahrenheit, of a cup of coffee that was poured and then forgotten for \(m\) minutes
is represented in the table below</p>
<table class="os-raise-horizontaltable">
<thead></thead>
<tbody>
<tr>
<th scope="row">Time in minutes</th>
<td>
0
</td>
<td>
15
</td>
<td>
25
</td>
<td>
40
</td>
</tr>
<tr>
<th scope="row">Temperature</th>
<td>
185°F
</td>
<td>
155°F
</td>
<td>
140°F
</td>
<td>
120°F
</td>
</tr>
</tbody>
</table>
<br>
<p>What was the average rate of change from when the cup was poured to 40 minutes later?</p>
<table class="os-raise-textheavytable">
<thead>
<tr>
<th scope="col">Answers</th>
<th scope="col">Feedback</th>
</tr>
</thead>
<tbody>
<tr>
<td>
–2 degrees per minute
</td>
<td>
Incorrect. Let’s try again a different way: This is the rate of change of the first two values in the
table. You need to use the first and last value from the table to find the answer. The answer is –1.625
degrees per minute.
</td>
</tr>
<tr>
<td>
–0.615 degrees per minute
</td>
<td>
Incorrect. Let’s try again a different way: Remember, the \(y\)-values go on top of the formula and the
\(x\)-values go on the bottom. The answer is –1.625 degrees per minute.
</td>
</tr>
<tr>
<td>
–1.333 degrees per minute
</td>
<td>
Incorrect. Let’s try again a different way: This is the rate of change between the last two values in the
table. You need to use the first and last value from the table to find the answer. The answer is –1.625
degrees per minute.</p>
</td>
</tr>
<tr>
<td>
–1.625 degrees per minute
</td>
<td>
That’s correct! Check yourself: The average rate of change formula is \(\frac{f(b)-f(a)}{b-a}\). For this
situation, it would be \(\frac{120-185}{40-0}\), which is \(\frac{-65}{40}\). The answer is –1.625 degrees
per minute.
</td>
</tr>
</tbody>
</table>
<br>
<h3> 4.8.2: Additional Resources</h3>
<p><strong><em>The following content is available to students who would like more support based on their experience with
the self check. Students will not automatically have access to this content, so you may wish to share it with
those who could benefit from it. </em></strong></p>
<h4> Finding Rates of Change </h4>
<p>Jay was racing bikes with a friend on the way to the park. A table and a graph of the times are shown below.</p>
<table class="os-raise-midsizetable">
<caption>Jay</caption>
<thead>
<tr>
<th scope="col">Time (minutes)</th>
<th scope="col">Distance (miles)</th>
</tr>
</thead>
<tbody>
<tr>
<td>
0
</td>
<td>
0
</td>
</tr>
<tr>
<td>
5
</td>
<td>
0.84
</td>
</tr>
<tr>
<td>
10
</td>
<td>
1.86
</td>
</tr>
<tr>
<td>
15
</td>
<td>
3.00
</td>
</tr>
<tr>
<td>
20
</td>
<td>
4.27
</td>
</tr>
<tr>
<td>
25
</td>
<td>
5.67
</td>
</tr>
</tbody>
</table>
<br>
<p><img
alt="A SCATTERPLOT THAT SHOWS THE RELATIONSHIP BETWEEN TIME IN MINUTES (X-AXIS) AND DISTANCE IN MILES (Y-AXIS). PLOTTED POINTS INCLUDE (0, 0), (5, 0.84), (10, 1.86), (15, 3), (20, 4.27), AND (25, 5.67)."
class="img-fluid atto_image_button_text-bottom" height="298"
src="https://k12.openstax.org/contents/raise/resources/e2be91d74e2b7550eb702312e03a47f9fe97b6f3" width="300"></p>
<p>What is the average rate of change of the first 10 minutes of the race? </p>
<p>To find the average rate of change of the first 10 minutes of the race, you would look at the first data value on the
table and the third data value on the table, since it corresponds to 10 minutes. </p>
<p>Remember the average rate of change formula is \(\frac{f(b)-f(a)}{b-a}\), where \((b, f(b))\) typically represents
your second data point and \((a, f(a))\) represents your first data point.</p>
<p>For this situation, it would be \(( 0, 0)\) or \(a=0\) and \(f(a) =0\) and \((10,1.86)\) or \(b= 10\) and
\(f(b)=1.86\).</p>
<p>Substituting the points into the average rate of change formula results in the following:</p>
<p>\(\frac{f(b)-f(a)}{b-a}\), which is \(\frac{1.86}{10}\) or 0.186 mile per minute. </p>
<p>Notice that the answer is positive, and the points on the graph are higher each time.</p>
<h4>Try It: Finding Rates of Change </h4>
<p>Jay was racing bikes with a friend on the way to the park. A table and a graph of the times are shown below.</p>
<table class="os-raise-midsizetable">
<caption>Jay</caption>
<thead>
<tr>
<th scope="col">Time (minutes)</th>
<th scope="col">Distance (miles)</th>
</tr>
</thead>
<tbody>
<tr>
<td>
0
</td>
<td>
0
</td>
</tr>
<tr>
<td>
5
</td>
<td>
0.84
</td>
</tr>
<tr>
<td>
10
</td>
<td>
1.86
</td>
</tr>
<tr>
<td>
15
</td>
<td>
3.00
</td>
</tr>
<tr>
<td>
20
</td>
<td>
4.27
</td>
</tr>
<tr>
<td>
25
</td>
<td>
5.67
</td>
</tr>
</tbody>
</table>
<br>
<p><img
alt="A SCATTERPLOT THAT SHOWS THE RELATIONSHIP BETWEEN TIME IN MINUTES (X-AXIS) AND DISTANCE IN MILES (Y-AXIS). PLOTTED POINTS INCLUDE (0, 0), (5, 0.84), (10, 1.86), (15, 3), (20, 4.27), AND (25, 5.67). "
class="img-fluid atto_image_button_text-bottom" height="300"
src="https://k12.openstax.org/contents/raise/resources/69e4be4d341573e713fabfb59f2ef379252f1a05" width="300"></p>
<p>What was Jay’s average rate of change from the beginning to 25 minutes later?</p>
<p>Write down your answer, then select the <strong>solution</strong> button to compare your work.</p>
<h5>Solution</h5>
<p>Here is how to find the average rate of change of the first 25 minutes of the race:</p>
<p>Remember, the average rate of change formula is \(\frac{f(b)-f(a)}{b-a}\), where \((b, f(b))\) typically represents
your second data point and \((a, f(a))\) represents your first data point.</p>
<p>For this situation, it would be \((0, 0)\) or \(a=0\) and \(f(a) =0\) and \((25,5.67)\) or \(b= 25\) and
\(f(b)=5.67\).</p>
<p>Substituting the points into the average rate of change formula results in the following:</p>
<p>\(\frac{f(b)-f(a)}{b-a}=\frac{5.67}{25}\), which is 0.227 mile per minute.</p>