980f75e8-c3f5-4fe8-b12b-37f627fd1c26.html

<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 for 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 Options 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.&nbsp;&nbsp;</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 Table</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 Table</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, add $1.50 to each consecutive term. To create the table for Option B, multiply
    each term by 3 to get the next term. </li>
</ol>
<p>Option A table
<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>Option 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>
<div class="os-raise-ib-cta" data-button-text="Solution" data-fire-event="Reveal1" data-schema-version="1.0">
  <div class="os-raise-ib-cta-content">
    <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>
  </div>
  <div class="os-raise-ib-cta-prompt">
    <p>Write down your answer, then select the <strong>solution</strong> button to compare your work. </p>
  </div>
</div>
<div class="os-raise-ib-content" data-schema-version="1.0" data-wait-for-event="Reveal1">
  <p>Compare your answer:</p>
  <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>
</div>