33f9d858-d7ed-458d-9e19-d293a9ce66b2.html

<h4>Activity (20 minutes)</h4>
<p>In this activity, students graph functions in vertex form and deepen their understanding of the connections between a
  quadratic expression in this form and the corresponding graph.</p>
<p>Students have previously observed that the graph of a function expressed in the form of \(y=a(x&minus;h)^2+k\) opens
  up when \(a\) is positive and opens down when \(a\) is negative. This means that they could sketch a rough graph that
  goes through a particular vertex and determine the direction of the opening. The work here reinforces and builds on
  that earlier observation. Students identify two additional points that are on the graph and use them to help sketch a
  curve. They also reason about whether the vertex represents the minimum or maximum of the function by analyzing the
  relative values of the coordinate pairs on either side of the vertex.</p>
<p>There are multiple opportunities for students to look for and make use of structure here.</p>
<ul class="os-raise-noindent">
  <li> Given the coordinates of the vertex and of one other point, they may reason that the graph opens up or down by
    comparing the \(y\)-values of the two points. For example, if the vertex is \((4,10)\) and one point is \((0,18)\),
    they may conclude that the vertex must be the minimum of the graph simply because the other known point has a
    greater \(y\)-value. </li>
  <li> After identifying the coordinates of one point on the graph, students may use the symmetry of the graph, the axis
    of symmetry, of a quadratic function to quickly find the coordinates of a second point. </li>
  <li> When making sense of Priya's reasoning in the last question, they might notice that in the expression defining
    \(p\), the \(-(x&minus;4)^2\) portion has a value of 0 at the vertex (when \(x\) is 4), because \(-(4&minus;4)^2\)
    is \(0^2\) or 0. For other values of \(x\), \((x&minus;4)^2\) will always be positive because squaring a number
    always gives a positive result. But because of the negative coefficient, \(-(x&minus;4)^2\) will always be negative
    or less than 0. This means that for all values of \(x\) other than 4, the value of \(p\) would be less than
    \(p(4)\). </li>
</ul>
<p>As the focus here is on graphing a function by identifying points on the graph and by using structure, technology is
  not an appropriate tool.</p>
<h4>Launch</h4>
<p>Display the two equations defining \(p\) and \(q\) for all to see. Give students a moment to think about how the
  equations are alike and how they are different. Invite students to share their comments.</p>
<p>If not mentioned by students, ask: "Where are the vertices of the graphs located?" (All of the equations have an
  \((x&minus;4)^2\) in them and a 10 constant term. This means that the vertex for each graph is at the same location,
  at \((4,10)\).)</p>
<p>Remind students that, earlier in the unit, we learned that the vertex of a graph represents the maximum or the
  minimum value of a function. Ask students:</p>
<ul class="os-raise-noindent">
  <li> "How can we tell if the vertex of the graph of \(p\) represents the maximum or the minimum?" (If the graph opens
    upward, the vertex is the minimum. If it opens downward, it is the maximum.) </li>
  <li> "How can we tell if the graph of \(p\) opens upward or downward?" (In an earlier lesson, we saw that when a
    quadratic function is expressed in the form of \(y=a(x&minus;h)^2+k\) and the coefficient \(a\) is negative, the
    graph opens downward. This means the vertex represents the maximum of the function. Or we could plot some coordinate
    pairs to help us visualize the graph.) </li>
</ul>
<p>If no students mentioned finding additional points on each graph to determine whether the graph opens up or down, ask
  them about it.</p>
<p>Because this activity is designed to be completed without technology, ask students to put away any devices.</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>
  </div>
  <div class="os-raise-extrasupport-body">
    <p class="os-raise-extrasupport-name">MLR 1 Discussion Supports: Writing, Conversing</p>
    <p>Use this routine to help students improve their writing by providing them with multiple opportunities to clarify their explanations through conversation. Invite students to meet with one to two partners to share and get feedback on their explanation of how Priya might have reasoned about whether the vertex is the minimum or maximum. Students should first check to see if they agree with each other about the values that go in the table. Give two to three minutes to revise their initial draft based on feedback from their peers.</p>
    <p class="os-raise-text-italicize">Design Principle(s): Support sense-making; Optimize output (for explanation)</p>
    <p class="os-raise-extrasupport-title">Provide support for students</p>
      <p>
        <a href="https://k12.openstax.org/contents/raise/resources/a5ae5bd09b27a5f53239a539c6009c19c92f7db7" target="_blank">Distribute graphic organizers</a>
           to the students to assist them with participating in this routine.
      </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>
    <p class="os-raise-extrasupport-name">Representation: Internalize Comprehension</p>
   </div>
  <div class="os-raise-extrasupport-body">
    <p>Activate or supply background knowledge by creating a display that highlights the following information: a graph of the form \(y=a{(x-h)}^2+k\) opens up when \(a\) is positive and opens down when \(a\) is negative; examining \(y\)-values of two points can help to determine if the graph opens up or down; if coordinates of one point are found on the graph, the symmetry of the graph may be used to find another point. Engage students in finding the vertex, deciding whether the graph opens up or down, and finding two more points. The display can be completed during the synthesis.</p>
    <p class="os-raise-text-italicize">Supports accessibility for: Memory; Conceptual processing</p>
  </div>
</div>

<br>
<h4>Student Activity</h4>
<p>Examine these two equations that define quadratic functions.</p>
<p>\(p(x)= -(x&minus;4)^2+10\)</p>
<p>\(q(x)=\frac {1}{2}(x&minus;4)^2+10\)</p>
<ol class="os-raise-noindent">
  <li>The graph of \(p\) passes through \((0,-6)\) and \((4,10)\), as shown on the coordinate plane.</li>
</ol>
<p><img height="404" src="https://k12.openstax.org/contents/raise/resources/70b01f6147ee71f3375b8b75b04e230921ad040c"
    width="428"></p>
<ol type="a">
  <li> Find the coordinates of another point on the graph of \(p\). Explain or show your reasoning. </li>
</ol>
<p><strong>Answer:</strong> \((8,-6)\). If we find \(p(8)\) and substitute 8 for \(x\) in the expression, we have:
  \(p(8)=-(8&minus;4)^2+10\), which is equal to \(-16+10\) or -6.</p>
<ol start="2" type="a">
  <li> Use the graphing tool or technology outside the course. Plot \(p(x)= -(x&minus;4)^2+10\) and the points \((0,
    -6)\), \((4, 10)\), and \((8, -6)\) to label the graph. </li>
</ol>
<p>Select the <strong>solution</strong> button to compare your work.</p>
<p><strong>Answer:</strong></p>
<p><img height="305" src="https://k12.openstax.org/contents/raise/resources/f1115944b5a4693784d28b587b77e28a2b9bf5ad"
    width="300"></p>
<p>The graph of a quadratic function is symmetrical across the vertical line containing the vertex. If a point on the
  graph that is 4 units to the left of the line containing the vertex has a \(y\)-value of -6, there is a point to the
  right of the line containing the vertex that also has a \(y\)-value of -6.</p>
<p>A line containing the vertex is called the parabola&rsquo;s axis of symmetry.</p>
<ol class="os-raise-noindent" start="2">
  <li>Find the \(x\)- and \(y\)-coordinates of the vertex and sketch the graph.</li>
</ol>
<ol type="a">
  <li> What is the \(x\)-coordinate of the vertex of \(q(x)=\frac {1}{2}(x&minus;4)^2+10\)? </li>
</ol>
<p><strong>Answer:</strong> 4</p>
<ol start="2" type="a">
  <li> What is the the \(y\)-coordinate of the vertex of \(q(x)=\frac {1}{2}(x&minus;4)^2+10\)? </li>
</ol>
<p><strong>Answer:</strong>10</p>
<ol start="3" type="a">
  <li> Use the graphing tool or technology outside the course. On the same coordinate plane containing the points and
    equation from question 1, plot the coordinates of the vertex of the function \(q\). </li>
</ol>
<p>Select the <strong>solution</strong> button to compare your work.</p>
<p><strong>Answer:</strong></p>
<p><img height="305" src="https://k12.openstax.org/contents/raise/resources/3313f94d34630ae5e05743629e176ab8ea73d7d8"
    width="300"></p>
<ol start="4" type="a">
  <li> Find two other points that are on the graph of \(q\). Try to identify points that are symmetrical over the axis
    of symmetry. </li>
</ol>
<p><strong>Answer: </strong></p>
<ul>
  <li> One possible answer could be \((2,12)\) and \((6,12)\). Substituting 2 for \(x\), we have: \(q(2)=\frac
    {1}{2}(2&minus;4)^2+10\), which is 12, so \((2,12)\) is one point on the graph. Substituting 6 for \(x\), we have:
    \(q(6)=\frac {1}{2} 1(6&minus;4)^2+10\), which equals 12, so \((6,12)\) is another point. </li>
</ul>
<ul class="os-raise-noindent">
  <li> Another possible answer could be \((0,18)\) and \((8,18)\). Substituting 0 for \(x\), we get \(q(0)=18\), so
    \((0,18)\) is one point on the graph. This point is 4 units to the left and 8 units up from the vertex. Because the
    graph has a vertical line of symmetry, or axis of symmetry, through the vertex, it also goes through a point that is
    4 units to the right and 8 units up, which is \((8,18)\). </li>
</ul>
<ol start="5" type="a">
  <li> Use the graphing tool or technology outside the course. Sketch and label the graph of \(q\). Explain or show your
    reasoning. </li>
</ol>
<p>Select the <strong>Solution</strong> button to compare your work.</p>
<p><strong>Answer:</strong></p>
<p><img height="282" src="https://k12.openstax.org/contents/raise/resources/e5fbb242593d9cc5cbf800151cadcf53bc7e57ff"
    width="299"></p>
<ol class="os-raise-noindent" start="3">
  <li>Recall that \(p(x)= -(x&minus;4)^2+10\). Pilar says, "Once I know the vertex is \((4,10)\), I can find out,
    without graphing, whether the vertex is the maximum or the minimum of function \(p\). I would just compare the
    coordinates of the vertex with the coordinates of a point on either side of it."</li>
</ol>
<p>Complete the table by answering questions a and b.</p>
<table class="os-raise-horizontaltable">
  <thead></thead>
  <tbody>
    <tr>
      <th scope="row">
        \(x\)
      </th>
      <td>
        \(3\)
      </td>
      <td>
        \(4\)
      </td>
      <td>
        \(5\)
      </td>
    </tr>
    <tr>
      <th scope="row">
        \(p(x)\)
      </th>
      <td>&nbsp;</td>
      <td>
        \(10\)
      </td>
      <td>&nbsp;</td>
    </tr>
  </tbody>
</table>
<br>
<ol type="a">
  <li> Evaluate \(p(3)\). </li>
</ol>
<p><strong>Answer:</strong> \(p(3)=-(3&minus;4)^2+10=-(-1)^2+10=-1+10=9\).</p>
<ol start="2" type="a">
  <li> Evaluate \(p(5)\). </li>
</ol>
<p><strong>Answer:</strong> \(p(5)=-(5&minus;4)^2+10=-(1)^2+10=-1+10=9\).</p>
<ol start="3" type="a">
  <li> Explain how Pilar might have reasoned whether the vertex is the minimum or maximum. </li>
</ol>
<p><strong>Answer:</strong> The values of \(p(3)\) and \(p(5)\) are both less than 10. The vertex of a quadratic graph
  represents the minimum or the maximum of the function. Because the two points on either side of the vertex of \(p\)
  have a lesser \(y\)-value, the vertex must be the maximum.</p>
<h4>Student Facing Extension</h4>
<h5>Are you ready for more?</h5>
<ol class="os-raise-noindent">
  <li>Choose the equation for a quadratic function whose graph has the vertex at \((2,3)\) and contains the point
    \((0,-5)\).</li>
</ol>
<ul class="os-raise-noindent">
  <li> \(y=-2(x&minus;2)^2+3\) </li>
  <li> \(y=-3(x&minus;2)62+2\) </li>
  <li> \(y=-5(x&minus;2)^2\) </li>
  <li> \(y=2(x&minus;2)^2+3\) </li>
</ul>
<p><strong>Answer:</strong> \(y=-2(x&minus;2)^2+3\).</p>
<ol class="os-raise-noindent" start="2">
  <li>Use the graphing tool or technology outside the course. Sketch a graph of your function.</li>
</ol>
<p><strong>Answer:</strong></p>
<p><img height="190" src="https://k12.openstax.org/contents/raise/resources/262459fc82a95393d77c4500ef427aa048739f2b"
    width="268"></p>
<h4>Anticipated Misconceptions</h4>
<p>Students may be unsure about what input value to choose to find additional points on each graph. Without telling
  students a specific value to use, encourage them to choose an \(x\)-value that is simple to evaluate and would help
  them sketch the graph.</p>
<h4>Activity Synthesis</h4>
<p>Invite students to share their graphs and how they went about finding the coordinates of one other point on the graph
  of \(p\) and two other points on the graph of \(q\). Highlight explanations that make use of the symmetry of the graph
  to identify an additional point on the graph once one point (aside from the vertex) is known.</p>
<p>Then, select students to explain their analysis of Pilar's reasoning. Make sure students see that the vertex of the
  graph of \(p\) cannot be the minimum value of the function (and thus cannot have an upward-opening graph) because
  there are other values of \(p\) that are less than 10. If the vertex was the minimum, no other values of \(p\) would
  be less than at \(p(4)\).</p>
<p>Likewise, once we see that \((0,-6)\) is on the graph of \(p\) and its \(y\)-value is less than that of the vertex,
  we can reason that the vertex \((4,10)\) represents the maximum of the function and the graph would open downward.</p>
<p>Consider showing a table such as these to clarify the input-output relationship in each function.</p>
<p>Function \(p\)</p>
<table class="os-raise-horizontaltable">
  <thead></thead>
  <tbody>
    <tr>
      <th scope="row">
        \(x\)
      </th>
      <td>
        \(0\)
      </td>
      <td>
        \(1\)
      </td>
      <td>
        \(2\)
      </td>
      <td>
        \(3\)
      </td>
      <td>
        \(4\)
      </td>
      <td>
        \(5\)
      </td>
      <td>
        \(6\)
      </td>
    </tr>
    <tr>
      <th scope="row">
        \(-(x&minus;4)^2+10\)
      </th>
      <td>
        \(-6\)
      </td>
      <td>
        \(1\)
      </td>
      <td>
        \(6\)
      </td>
      <td>
        \(9\)
      </td>
      <td>
        \(10\)
      </td>
      <td>
        \(9\)
      </td>
      <td>
        \(6\)
      </td>
    </tr>
  </tbody>
</table>
<br>
<p>Function \(q\)
<table class="os-raise-horizontaltable">
  <thead></thead>
  <tbody>
    <tr>
      <th scope="row">
        \(x\)
      </th>
      <td>
        \(0\)
      </td>
      <td>
        \(1\)
      </td>
      <td>
        \(2\)
      </td>
      <td>
        \(3\)
      </td>
      <td>
        \(4\)
      </td>
      <td>
        \(5\)
      </td>
      <td>
        \(6\)
      </td>
    </tr>
    <tr>
      <th scope="row">
        \(\frac {1}{2}(x&minus;4)^2+10\)
      </th>
      <td>
        \(18\)
      </td>
      <td>
        \(14 \frac {1}{2}\)
      </td>
      <td>
        \(12\)
      </td>
      <td>
        \(10 \frac {1}{2}\)
      </td>
      <td>
        \(10\)
      </td>
      <td>
        \(10 \frac {1}{2}\)
      </td>
      <td>
        \(12\)
      </td>
    </tr>
  </tbody>
</table>
<br>
<p>If time permits, discuss with students how we can also determine whether the vertex is the maximum or minimum by
  studying the structure of the squared term in the vertex form. Let's take function \(q\) as an example. Ask students:
</p>
<ul class="os-raise-noindent">
  <li> "In the expression defining \(q\), what is the value of \((x&minus;4)^2\) when \(x\) is 4?" (0) </li>
  <li> "What is the value of \((x&minus;4)^2\) when \(x\) is less than 4, say, when it is 1? Is it greater or less than
    0?" (At \(x=1\), \((x&minus;4)^2\) is 9, which is greater than 0.) </li>
  <li> "What about when \(x\) is greater than 4, say, when it is 5?" (At \(x=5\), \((x&minus;4)^2\) is 1, which is also
    greater than 0.) </li>
  <li> "If the squared term \((x&minus;4)^2\) is 0 when \(x\) is 4, and greater than 0 when \(x\) is any other number,
    does \(x=4\) give the minimum or the maximum point on the graph?" (The minimum) </li>
  <li> "Does multiplying \((x&minus;4)^2\) by 12 change the fact that the minimum is at \(x=4\)? Why or why not?" (No.
    Multiplying \((x&minus;4)^2\) by 12 halves all the values but they will still be positive or greater than 0. Zero
    times 12 is still 0.) </li>
  <li> "What about adding 10 to \((x&minus;4)^2\)?" (No. Adding 10 increases the value of \((x&minus;4)^2\) by 10,
    regardless of the input.) </li>
</ul>
<h3>7.16.2: 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 equation of a quadratic function is \(y=(x-1)^2+5\) and the point \((3,9)\) also lies on the parabola. What is
  another point that lies on the parabola?</p>
<table class="os-raise-textheavytable">
  <thead>
    <tr>
      <th scope="col">
        Answers
      </th>
      <th scope="col">
        Feedback
      </th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <td>
        \((-5, 9)\)
      </td>
      <td>
        Incorrect. Let&rsquo;s try again a different way: The vertex is at \((1,5)\). Use this to find a point symmetric
        to \((3,9)\) on the parabola. The \(x\)-coordinate will be 2 units to the left of 1. The answer is \((-1,9)\).
      </td>
    </tr>
    <tr>
      <td>
        \((3, 1)\)
      </td>
      <td>
        Incorrect. Let&rsquo;s try again a different way: The vertex is at \((1,5)\). Use this to find a point symmetric
        to \((3,9)\) on the parabola. The \(x\)-coordinate will be 2 units to the left of 1. The answer is \((-1,9)\).
      </td>
    </tr>
    <tr>
      <td>
        \((-1,9)\)
      </td>
      <td>
        That&rsquo;s correct! Check yourself: Because parabolas are symmetric and \((3, 9)\) is two units to the right
        of \(x=1\), there must be another point on the parabola two units to the left of \(x=1\) with the same
        \(y\)-coordinate as \((3, 9)\). That point is \((-1, 9)\).
      </td>
    </tr>
    <tr>
      <td>
        \((-1, 5)\)
      </td>
      <td>
        Incorrect. Let&rsquo;s try again a different way: The \(x\) -coordinate is correct. To find the \(y\)-value,
        substitute -1 into the equation. The point will be symmetric to \((3,9)\) on the parabola. The answer is
        \((-1,9)\).
      </td>
    </tr>
  </tbody>
</table>
<br>
<h3>7.16.2: 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>Using Key Points to Graph Quadratics</h4>
<p>We saw that vertex form is especially helpful for finding the vertex of a graph of a quadratic function. For example,
  we can tell that the function \(p\) given by \(p(x)=(x-3)^2+1\) has a vertex at \((3,1)\).</p>
<p><img height="147" src="https://k12.openstax.org/contents/raise/resources/06ea9e2b84d0ec1d445d1a7db4cf7f8177977f4a"
    width="247"></p>
<p>We also noticed that, when the squared expression \((x-3)^2\) has a positive coefficient, the graph opens upward.
  You can use the vertex \((3, 1)\) to determine that minimum function value is 1.</p>
<p>But why does the function \(p\) take on its minimum value when \(x\) is 3?</p>
<p>Here is one way to explain it: When \(x=3\), the squared term \((x-3)^2\) equals 0, as \((3-3)^2=0^2=0\). When \(x\)
  is any other value besides 3, the squared term \((x-3)^2\) is a positive number greater than 0. (Squaring any number
  results in a positive number.) This means that the output when \(x \neq 3\) will always be greater than the output
  when \(x=3\), so the function \(p\) has a minimum value at \(x=3\).</p>
<p>This table shows some values of the function for some values of \(x\). Notice that the output is the least when
  \(x=3\) and it increases both as \(x\) increases and as it decreases.</p>
<table class="os-raise-horizontaltable">
  <thead></thead>
  <tbody>
    <tr>
      <th scope="row">
        \(x\)
      </th>
      <td>
        \(0\)
      </td>
      <td>
        \(1\)
      </td>
      <td>
        \(2\)
      </td>
      <td>
        \(3\)
      </td>
      <td>
        \(4\)
      </td>
      <td>
        \(5\)
      </td>
      <td>
        \(6\)
      </td>
    </tr>
    <tr>
      <th scope="row">
        \((x-3)^2+1\)
      </th>
      <td>
        \(10\)
      </td>
      <td>
        \(5\)
      </td>
      <td>
        \(2\)
      </td>
      <td>
        \(1\)
      </td>
      <td>
        \(2\)
      </td>
      <td>
        \(5\)
      </td>
      <td>
        \(10\)
      </td>
    </tr>
  </tbody>
</table>
<br>
<p>The squared term sometimes has a negative coefficient, for instance: \(h(x)=-2(x+4)^2\). The \(x\) value that makes
  \((x+4)^2\) equal 0 is -4, because \((-4+4)^2=0^2=0\). Any other \(x\) value makes \((x+4)^2\) greater than 0. But
  when \((x+4)^2\) is multiplied by a negative number (-2), the resulting expression, \(-2(x+4)^2\), ends up being
  negative. This means that the output when \(x-4\) will always be less than the output when \(x=-4\), so the function
  \(h\) has its maximum value when \(x=-4\).</p>
<p><img height="138" src="https://k12.openstax.org/contents/raise/resources/9eaa01d884ff6269f7f491469cb0d2a68ba299a6"
    width="215"></p>
<p>Remember that we can find the \(y\)-intercept of the graph representing any function we have seen. The
  \(y\)-coordinate of the \(y\)-intercept is the value of the function when \(x=0\). If \(g\) is defined by
  \(g(x)=(x+1)^2-5\), then the \(y\)-intercept is \((0,-4)\) because \(g(0)=(0+1)^2-5=-4\). Its vertex is at
  \((-1,-5)\). Another point on the graph with the same \(y\)-coordinate is located the same horizontal distance from
  the vertex but on the other side.</p>
<p><img height="139" src="https://k12.openstax.org/contents/raise/resources/b4c10123de9c99c891fb3ae23629cc7b14f8cc85"
    width="220"></p>
<h4>Try It: Using Key Points to Graph Quadratics</h4>
<p>A quadratic function is represented by the equation \(y=(x+3)^2-5\). If \((5, 61)\) is another point on the graph,
  find a third point on the curve.</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 another point on the curve:</p>
<p>One way to find another point on the curve is to use symmetry.</p>
<p>Since the vertex is \((-3, -5)\) and one point on the curve is \((5, 61)\), there is another point on the other side
  of the vertex. Keep in mind, the \(y\)-coordinate will remain at 61. To find the \(x\)-coordinate, we need to
  determine how far apart horizontally the vertex and the other point are located from each other. There are 8 units
  from -3 to 5, so the new point will also be 8 units from -3 but to the left. Thus, the \(x\)-coordinate will be at
  -11. So, there is another point at \((-11, 61)\).</p>