5181c87b-7889-42a3-a59a-567375ce4079.html

<h4>Activity (15 minutes)</h4>
<p>This activity prompts students to again compare a linear function with an exponential one, but this time without a
  context, and the exponential function grows much more slowly over a long period of time. Even if students predict that
  the exponential function will grow more quickly because it is exponential, they still need to decide which one reaches
  2,000 first. To do that, students may try:</p>
<ul>
  <li> continuing the table of values with increasingly larger values.</li>
  <li> using graphing technology.</li>
  <li> finding when \(f(x)=2,000\) and substituting this value \((x=1,000)\) into \(g(x)\).</li>
</ul>
<p>Making graphing and spreadsheet technology available gives students an opportunity to choose appropriate tools
  strategically.</p>
<h4>Launch</h4>
<p>Present the two equations that define two functions \(f\)&nbsp;and \(g\): \(f(x)=2x\) and \(g(x)=(1.01)^x\). Give
  students a moment to share what they notice and wonder about the functions. Ask questions such as:</p>
<ul>
  <li>&ldquo;What is happening in the first function? What about in the second function?&rdquo;</li>
  <li>&ldquo;Which function do you think will reach a value of 2,000 first?&rdquo;</li>
</ul>
<br>
<div class="os-raise-extrasupport">
  <div class="os-raise-extrasupport-header">
    <p class="os-raise-extrasupport-title"> Support for English Language Learners</p>
  </div>
  <div class="os-raise-extrasupport-body">
    <p>
Ask students to prepare a visual display that shows their mathematical thinking in response to the question,
    &ldquo;Which function do you think will reach a value of 2,000 first?&rdquo; Invite previously identified students
    to share their representations. In groups of 2, ask students to compare their strategies and discuss how the
    strategies are the same or different. Consider using the prompt: &ldquo;Which strategy do you believe is the most
    efficient in determining a solution?&rdquo; Listen for and amplify observations of advantages and disadvantages to
    different approaches, such as using a particular approach for a certain time interval. This will help students
    connect different approaches and determine the effectiveness of each one through partner and whole-class discussion.</p>
  </div>
</div>
<br>
<div class="os-raise-extrasupport">
  <div class="os-raise-extrasupport-header">
    <p class="os-raise-extrasupport-title">Support for Students with Disabilities</p>
  </div>
  <div class="os-raise-extrasupport-body">
    <p>
      Provide access to tools and assistive technologies, such as a graphing calculator or graphing software. Some
      students may benefit from a checklist or list of steps to be able to use the calculator or software </p>
    </div>
  </div>
<br>
<h4>Student Activity</h4>
<p>Consider the functions \(f(x)=2x\) and \(g(x)=(1.01)^x\).<br></p>
<ol class="os-raise-noindent">
  <li> Complete the table of values for the functions \(f\) and \(g\). </li>
</ol>
<table class="os-raise-midsizetable">
  <thead>
    <tr>
      <th scope="col">
        \(x\)
      </th>
      <th scope="col">
        \(f(x)\)
      </th>
      <th scope="col">
        \(g(x)\)
      </th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <td>
        1
      </td>
      <td>
      </td>
      <td>
      </td>
    </tr>
    <tr>
      <td>
        10
      </td>
      <td>
      </td>
      <td>
      </td>
    </tr>
    <tr>
      <td>
        50
      </td>
      <td>
      </td>
      <td>
      </td>
    </tr>
    <tr>
      <td>
        100
      </td>
      <td>
      </td>
      <td>
      </td>
    </tr>
    <tr>
      <td>
        500
      </td>
      <td>
      </td>
      <td>
      </td>
    </tr>
  </tbody>
</table>
<br>
<p class="os-raise-text-bold">Answer: </p>
<table class="os-raise-midsizetable">
  <thead>
    <tr>
      <th scope="col">
        \(x\)
      </th>
      <th scope="col">
        \(f(x)\)
      </th>
      <th scope="col">
        \(g(x)\)
      </th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <td>
        1
      </td>
      <td>
        2
      </td>
      <td>
        1.01
      </td>
    </tr>
    <tr>
      <td>
        10
      </td>
      <td>
        20
      </td>
      <td>
        ~1.10
      </td>
    </tr>
    <tr>
      <td>
        50
      </td>
      <td>
        100
      </td>
      <td>
        ~1.645
      </td>
    </tr>
    <tr>
      <td>
        100
      </td>
      <td>
        200
      </td>
      <td>
        ~2.7
      </td>
    </tr>
    <tr>
      <td>
        500
      </td>
      <td>
        1,000
      </td>
      <td>
        ~144.8
      </td>
    </tr>
  </tbody>
</table>
<br>
<ol class="os-raise-noindent" start="2">
  <li> Based on the table of values, which function do you think grows faster? Be prepared to show your reasoning. <br>
    <br>
    <strong>Answer:</strong> For example:&nbsp;\(f\)<em>&nbsp;</em>is growing faster for the values
    of&nbsp;\(x\)&nbsp;in the table, but as&nbsp;\(x\)&nbsp;increases, \(g\) starts to grow faster. So
    as&nbsp;\(x\)&nbsp;increases, \(g\) may eventually grow faster than \(f\).
  </li>
</ol>
<ol class="os-raise-noindent" start="3">
  <li> Which function do you think will reach a value of 2,000 first? Show your reasoning. If you get stuck, consider
    increasing \(x\) by 100 a few times and record the function values in the table. <br>
    <br>
    <strong>Answer:</strong> When \(x\) is 600, \(f(600)=1,200\) and \(g(600) \approx 392\), so&nbsp;\(f\) takes a
    larger value. When&nbsp;\(x\)&nbsp;is 700, \(f(700)=1,400\) and \(g(700) \approx 1,059\), so \(f\) still takes a
    larger value. But when&nbsp;\(x\)&nbsp;is 800, \(f(800)=1,600\) and \(g(800) \approx 2,865\), so \(g\) takes a
    larger value.
  </li>
</ol>
<h4>Student Facing Extension</h4>
<h5>Are you ready for more?</h5>
<p>Consider the functions \(g(x)=x^5\) and \(f(x)=5^x\). While it is true that \(f(7)&gt;g(7)\), for example, it is hard
  to check this using mental math. Find a value of \(x\) for which properties of exponents allow you to conclude that
  \(f(x)&gt;g(x)\) without a calculator.<br><br>
  <strong>Answer:&nbsp;</strong>Your answer may vary, but here are some samples: One example is \(x=25\). We have
  \(f(25)=5^{25}\), whereas \(g(25)=25^5=(5^2)^5=5^{10}\), which is certainly smaller than \(5^{25}\).
</p>
<h4>Activity Synthesis</h4>
<p>Select students who used contrasting approaches to share. The exponential function \(g\)&nbsp;<span>grows
    sufficiently slowly that it looks like it will not catch up with or ever overtake the linear function \(f\). This is
    true when using a table of values as well as when using a graph. Invite students who used contrasting methods to
    share, in the order shown in the Activity Narrative.</span></p>
<p>Emphasize that the table needs to be continued for a long time to identify when the values of \(g\) become greater
  than those of \(f\). Similarly, the window of the graph needs to be chosen carefully. A very efficient method is to
  identify that \(f(1,000)=2,000\) and then evaluate \(g(1,000)\). Since \(g(1,000)\) is much greater than 2,000, \(g\)
  reaches 2,000 first.</p>
<p>If not already illustrated by students who used graphing, graph the functions in Desmos and set the domain values to
  \( 0 \leq x \leq 7 \) and the range values to \(-100 \leq y \leq 20000\). </p>
<p>Though for quite a while it doesn&rsquo;t seem like the values of \(g\)&nbsp;would ever catch up with those
  of&nbsp;\(f\)<em>,</em> if the growth is allowed to take place long enough, exponential eventually surpasses linear.
</p>
<h3> 5.14.3: Self Check&nbsp;</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>Terence looked down the second column of the table below and noticed that
  \(\frac{3}{1}=\frac{9}{3}=\frac{27}{9}=\frac{81}{27}\). Because of his observation, he claimed that the input-output
  pairs in this table could be modeled with an exponential function. Explain why Terence is correct or incorrect.&nbsp;
</p>
<table class="os-raise-skinnytable">
  <thead>
    <tr>
      <th scope="col">
        \(x\)
      </th>
      <th scope="col">
        \(T(x)\)
      </th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <td>
        0
      </td>
      <td>
        1
      </td>
    </tr>
    <tr>
      <td>
        1
      </td>
      <td>
        3
      </td>
    </tr>
    <tr>
      <td>
        4
      </td>
      <td>
        9
      </td>
    </tr>
    <tr>
      <td>
        13
      </td>
      <td>
        27
      </td>
    </tr>
    <tr>
      <td>
        40
      </td>
      <td>
        81
      </td>
    </tr>
  </tbody>
</table>
<br>
<br>
<table class="os-raise-textheavytable">
  <thead>
    <tr>
      <th scope="col">
        Answers
      </th>
      <th scope="col">
        Feedback
      </th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <td>
        Terence is correct: Terms in a table that are connected by a factor are exponential.
      </td>
      <td>
        Incorrect. Let&rsquo;s try again a different way: Terms in an exponential table are connected by a factor, but
        the terms are not consecutive. If you look at this table, the<em> \(x\)</em>-values are not increasing by 1. The
        answer is Terence is incorrect. This table represents a linear relationship since the ratio of change of each
        row is consistent.
      </td>
    </tr>
    <tr>
      <td>
        Terence is correct: This table can be modeled by \(y=1(3)^x\).
      </td>
      <td>
        Incorrect. Let&rsquo;s try again a different way: This table would be represented by this equation if the terms
        were consecutive. The \(x\)<i>-</i>values are not increasing by 1 each time. The answer is Terence is incorrect.
        This table represents a linear relationship since the ratio of change of each row is consistent.
      </td>
    </tr>
    <tr>
      <td>
        Terence is incorrect: This table represents points that do not have an exponential or linear relationship
        because the \(x\) and \(y\) change by different amounts.
      </td>
      <td>
        Incorrect. Let&rsquo;s try again a different way: The table represents a linear relationship, not an exponential
        relationship. The answer is Terence is incorrect. This table represents a linear relationship since the ratio of
        change of each row is consistent.
      </td>
    </tr>
    <tr>
      <td>
        Terence is incorrect: This table represents a linear relationship since the ratio of change of each row is
        consistent.
      </td>
      <td>
        That&rsquo;s correct! Check yourself: Since the \(x\)-values are not increasing by 1 each time, you need to
        check the ratio of the rate of change. Since \(\frac{2}{1}=\frac{6}{3}=\frac{18}{9}=\frac{54}{27}\), this is a
        linear relationship.
      </td>
    </tr>
  </tbody>
</table>
<br>
<h3>5.14.3: Additional Resources</h3>
<p class="os-raise-text-bold"><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></p>
<h4>Determining and Comparing Linear and Exponential Functions</h4>
<p>Mr. Smith has an apple orchard. He hires his daughter, Lucy, to pick apples and offers her two payment options:</p>
<p>Option A: $1.50 per bushel of apples picked</p>
<p>Option B: 1 cent for coming to work, 3 cents picking one bushel, 9 cents for picking two bushels, 27 cents for
  picking three bushels, and so on, with the amount of money tripling for each additional bushel picked</p>
<ol class="os-raise-noindent">
  <li> Which option is linear, and which option is exponential? How do you know? </li>
  <li> Create a table to model each scenario. </li>
  <li> If she picks 6 bushels, which option is better? </li>
  <li> If she picks 12 bushels, which option is better? </li>
  <li> How many bushels does she need to pick for Option B to be better than Option A? </li>
</ol>
<p>Let&rsquo;s first think about linear and exponential functions in general. Let&rsquo;s also consider Option A and B
  when \(x\) represents the bushels of apples picked and \(f(x)\) represents the total amount she earns based on \(x\)
  bushels of apples.</p>
<ol class="os-raise-noindent">
  <li> Since Option A is $1.50 per bushel picked, it is linear. For Option B, the amount of money is tripling for each
    bushel picked, so it is exponential. </li>
</ol>

<table class="os-raise-doubleheadertable">
  <thead>
    <tr>
      <th scope="col"></th>
      <th scope="col">Linear Model</th>
      <th scope="col">Exponential Model</th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <th scope="row">General Form</th>
      <td>
        \(f(x)=ax+b\)
      </td>
      <td>
        \(f(x)=a(b)^x\)
      </td>
    </tr>
    <tr>
      <th scope="row">Meaning of Parameters \(a\) and \(b\)</th>
      <td>
        \(a\) is the slope of the line or the constant rate of change; \(b\) is the \(y\)-intercept or the
          \(f(x)\) value at \(x=0\).
      </td>
      <td>
        \(a\) is the \(y\)-intercept or the \(f(x)\) value when \(x=0\); \(b\) is the base or the constant quotient
          of change.
      </td>
    </tr>
    <tr>
      <th scope="row">Example</th>
      <td>
        \(f(x)=1.50x\)
      </td>
      <td>
        \(f(x)=.01(3)^x\)
      </td>
    </tr>
    <tr>
      <th scope="row"> Rule for Finding \(f(x+1)\) from \(f(x)\)</th>
      <td>
        Starting at \((0,0)\), to find \(f(x+1)\), add 1.5 to \(f(x)\).
      </td>
      <td>
        Starting at \((0,0.1)\), to find \(f(x+1)\), multiply \(f(x)\) by 3.
      </td>
    </tr>

  </tbody>
</table>
<br>
<div class="os-raise-d-flex os-raise-justify-content-between"><table class="os-raise-skinnytable">
<caption>Linear Model</caption>
  <thead>
    <tr>
      <th scope="col">\(x\)</th>
      <th scope="col">\(f(x)\)</th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <td> 0 </td>
      <td> 0 </td>
    </tr>
    <tr>
      <td> 1 </td>
      <td> 1.50 </td>
    </tr>
    <tr>
      <td> 2 </td>
      <td> 3.00 </td>
    </tr>
    <tr>
      <td> 3 </td>
      <td> 4.50 </td>
    </tr>
  </tbody>
</table> 

  <table class="os-raise-skinnytable">
    <caption>Exponential Model</caption>
  <thead>
    <tr>
      <th scope="col">\(x\)</th>
      <th scope="col">\(f(x)\)</th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <td> 0 </td>
      <td> 0.01 </td>
    </tr>
    <tr>
      <td> 1 </td>
      <td> 0.03 </td>
    </tr>
    <tr>
      <td> 2 </td>
      <td> 0.09 </td>
    </tr>
    <tr>
      <td> 3 </td>
      <td> 0.27 </td>
    </tr>
  </tbody>
</table></div>
<br>
<ol class="os-raise-noindent" start="2">
  <li> To create the table for Option A, added $1.50 to each consecutive term.&nbsp; To create the table for Option B,
    multiply each term by 3 to get the next term. </li>
</ol>
<p>Option A table </p>
<table class="os-raise-midsizetable">
  <thead>
    <tr>
      <th scope="col">
        Number of bushels
      </th>
      <th scope="col">
        Amount of money earned
      </th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <td>
        1
      </td>
      <td>
        1.50
      </td>
    </tr>
    <tr>
      <td>
        2
      </td>
      <td>
        3.00
      </td>
    </tr>
    <tr>
      <td>
        3
      </td>
      <td>
        4.50
      </td>
    </tr>
    <tr>
      <td>
        4
      </td>
      <td>
        6.00
      </td>
    </tr>
    <tr>
      <td>
        5
      </td>
      <td>
        7.50
      </td>
    </tr>
    <tr>
      <td>
        6
      </td>
      <td>
        9.00
      </td>
    </tr>
  </tbody>
</table>
<br>
<p> B table</p>
<table class="os-raise-midsizetable">
  <thead>
    <tr>
      <th scope="col">
        Number of bushels
      </th>
      <th scope="col">
        Amount of money earned
      </th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <td>
        1
      </td>
      <td>
        0.03
      </td>
    </tr>
    <tr>
      <td>
        2
      </td>
      <td>
        0.09
      </td>
    </tr>
    <tr>
      <td>
        3
      </td>
      <td>
        0.27
      </td>
    </tr>
    <tr>
      <td>
        4
      </td>
      <td>
        0.81
      </td>
    </tr>
    <tr>
      <td>
        5
      </td>
      <td>
        2.43
      </td>
    </tr>
    <tr>
      <td>
        6
      </td>
      <td>
        7.29
      </td>
    </tr>
  </tbody>
</table>
<br>
<ol class="os-raise-noindent" start="3">
  <li> To determine which scenario is better when she picks 6 bushels, you can look at the values in your table, or you
    can write the equation modeling each scenario and substitute in 6 for \(x\). For 6 bushels, Option A is better at
    $9.00. </li>
</ol>
<ol class="os-raise-noindent" start="4">
  <li> To determine which scenario is better when she picks 12 bushels, you can extend the table or use the graph or
    equation. Option A is $18.00 for 12 bushels, and Option B is $5,314.41 for 12 bushels, so Option B is better. </li>
</ol>
<ol class="os-raise-noindent" start="5">
  <li> To determine where Option B becomes the better option, you can extend the table, or you can graph the functions
    and find the \(x\)-value where Option B surpasses Option A. You can also use problems 3 and 4 to help you. The
    number of bushels is more than 6 but less than 12. Option B is better if you are going to pick 7 or more bushels of
    apples. </li>
</ol>
<h4>Try It: Determining and Comparing Linear and Exponential Functions</h4>
<p>Jayden has a dog-walking business. He has two plans. Plan 1 includes walking a dog once a day for a rate of $5 per
  day. Plan 2 also includes one walk a day but charges 1 cent for 1 day, 2 cents for 2 days, 4 cents for 3 days, and 8
  cents for 4 days, and it continues to double for each additional day. Mrs. Maroney needs Jayden to walk her dog every
  day for two weeks. Which plan should she choose? Show the work to justify your answer.</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 determine which of Jayden&rsquo;s plans Mrs. Maroney should choose:</p>
<p>Plan 1 is a linear plan since it&rsquo;s $5 per day. Plan 2 is doubling each day, so it is an exponential plan. Every
  other example has shown us that the exponential plan is going to be greater eventually. Since Mrs. Maroney needs
  Jayden for 2 weeks, or 14 days, we need to determine when Plan 2 becomes greater than Plan 1.</p>
<p>Plan 1 for 14 days would be:</p>
<p>\(5+5+5+5+5+5+5+5+5+5+5+5+5+5 = 5 \cdot 14 = $70\)</p>
<p>Plan 2 for 14 days would be:</p>
<p>\(.01 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 =
  .01(2)^{14} =$163.84\)</p>
<p>Plan 1 would be better for Mrs. Maroney.</p>