section-1-3.xml

<?xml version="1.0" encoding="UTF-8" ?>
<!-- <mathbook><book> -->

<section xml:id="graphs-of-functions"   xmlns:xi="http://www.w3.org/2001/XInclude">
<title>Graphs of Functions</title>
<subsection><title>Reading Function Values from a Graph</title>   
<p>The Dow-Jones Industrial Average (DJIA) gives the average of the stock prices of 500 major companies. The graph in <xref ref="fig-DJIA" text="type-global"/>  shows the DJIA as a function of time during the stock market correction of October 1987. The DJIA is thus <m>f(t)</m>, recorded at noon on day <m>t</m> of October.</p>
<figure xml:id="fig-DJIA"><caption></caption><image source="images/fig-DJIA"   width="60%"><description>Dow Jones Industrial Averge</description></image></figure>
<p>The values of the input variable, time, are displayed on the horizontal axis, and the values of the output variable, DJIA, are displayed on the vertical axis. There is no formula that gives the DJIA for a particular day; but it is still a function, defined by its graph. The value of <m>f(t)</m> is specified by the vertical coordinate of the point with the given <m>t</m>-coordinate.
</p>

<example xml:id="example-DJIA">
    <ol label="a">
        <li>The coordinates of point <m>P</m> in <xref ref="fig-DJIA" text="type-global"/> are <m>(15, 2412)</m>. What do the coordinates tell you about the function <m>f</m>?</li>
        <li>If the DJIA was 1726 at noon on October 20, what can you say about the graph of <m>f</m>?</li>
    </ol>
<solution>
    <ol label="a">
        <li>The coordinates of point <m>P</m> tell us that <m>f(15) = 2412</m>, so the DJIA was 2412 at noon on October 15.</li>
        <li>We can say that <m>f(20) = 1726</m>, so the point <m>(20, 1726)</m> lies on the graph of <m>f</m>. This point is labeled <m>Q</m> in <xref ref="fig-DJIA" text="type-global"/>.</li>
    </ol>
</solution>
</example>

<p>Thus, the coordinates of each point on the graph of the function represent a pair of corresponding
values of the two variables. </p>

<assemblage><title>Graph of a Function</title>
<p>The point <m>(a, b)</m> lies on the graph of the function <m>f</m> if and only if <m>f(a)=b</m>.
</p></assemblage>

<exercise><statement>
    <p>
       The water level in Lake Huron alters unpredictably over time. The graph in <xref ref="fig-Lake-Huron" text="type-global"/> gives the average water level, <m>L(t)</m>, in meters in the year <m>t</m> over a 20-year period. (Source: The Canadian Hydrographic Service) 
       <figure xml:id="fig-Lake-Huron"><caption></caption><image source="images/fig-Lake-Huron"  width="80%"><description>graph of Lake Huron water level</description></image> </figure>
       <ol label="a">
           <li>
               <p>The coordinates of point <m>H</m> in <xref ref="fig-Lake-Huron" text="type-global"/> are <m>(1997, 176.98)</m>. What do the coordinates tell you about the function <m>L</m>?</p>
           </li>
           <li>
               <p>The average water level in <m>2004</m> was <m>176.11</m> meters. Write this fact in function notation. What can you say about the graph of <m>L</m>?</p>
           </li>
       </ol>
    </p>
</statement>
<answer><p>
    <ol label="a">
        <li><p><m>L(1997) = 176.98</m>; the average water level was <m>176.98</m> meters in <m>1997</m>.</p></li>
        <li><p><m>L(2004) = 176.11</m>. The point <m>(2004, 176.11)</m> lies on the graph of <m>L</m>.</p></li>
    </ol>
</p></answer>
</exercise>

<p>
    Another way of describing how a graph depicts a function is as follows:
</p>

<assemblage><title>Functions and Coordinates</title>
<p>
    Each point on the graph of the function <m>f</m> has coordinates <m>(x, f (x))</m> for some value of <m>x</m>.
</p>
</assemblage>

<example xml:id="example-function-graph">
    <p><xref ref="fig-function" text="type-global"/> shows the graph of a function <m>g</m>.
    <ol label="a">
        <li>Find <m>g(-2)</m> and <m>g(5)</m>.</li>
        <li>For what value(s) of <m>t</m> is <m>g(t) = -2</m>?</li>
        <li>What is the largest, or maximum, value of <m>g(t)</m>? For what value of <m>t</m> does the function take on its maximum value?</li>
        <li>On what intervals is <m>g</m> increasing?</li>
    </ol></p>
    <figure xml:id="fig-function"><caption></caption><image source="images/fig-function"   width="40%"><description>graph of a function</description></image></figure>

<solution>
    <ol label="a">
        <li>To find <m>g(-2)</m>, we look for the point with <m>t</m>-coordinate <m>-2</m>. The point <m>(-2, 0)</m> lies on the graph of <m>g</m>, so <m>g(-2) = 0</m>. Similarly, the point <m>(5, 1)</m> lies on the graph, so <m>g(5) = 1</m>.</li>
        <li>We look for points on the graph with <m>y</m>-coordinate <m>-2</m>. Because the points <m>(-5, -2)</m>, <m>(-3, -2)</m>, and <m>(3, -2)</m> lie on the graph, we know that <m>g(-5) = -2</m>, <m>g(-3) = -2</m>, and <m>g(3) = -2</m>. Thus, the <m>t</m>-values we want are <m>-5</m>, <m>-3</m>, and <m>3</m>.</li>
        <li>The highest point on the graph is <m>(1, 4)</m>, so the largest <m>y</m>-value is <m>4</m>. Thus, the maximum value of <m>g(t)</m> is <m>4</m>, and it occurs when <m>t = 1</m>.</li>
        <li>A graph is increasing if the <m>y</m>-values get larger as we read from left to right. The graph of <m>g</m> is increasing for <m>t</m>-values between <m>-4</m> and <m>1</m>, and between <m>3</m> and <m>5</m>. Thus, <m>g</m> is increasing on the intervals <m>(-4, 1)</m> and <m>(3, 5)</m>.</li>
    </ol>
</solution>
</example>

<exercise xml:id="exercise-function-graph"><statement>
    Refer to the graph of the function <m>g</m> shown in <xref ref="fig-function" text="type-global"/> in <xref ref="example-function-graph" text="type-global"/>.
    <ol label="a">
        <li>Find <m>g(0)</m>.</li>
        <li>For what value(s) of <m>t</m> is <m>g(t) = 0</m>?</li>
        <li>What is the smallest, or minimum, value of <m>g(t)</m>? For what value of <m>t</m> does the function take on its minimum value?</li>
        <li>On what intervals is <m>g</m> decreasing?</li>
    </ol>  </statement>  
    <answer><p>
        <ol label="a" cols="2">
            <li><p><m>3</m></p></li>
            <li><p><m>-2, 2, 4</m></p></li>
            <li><p><m>-3</m>; <m>t = -4</m></p></li>
            <li><p><m>(-5, -4)</m> and <m>(1, 3)</m></p></li>
        </ol>
    </p></answer>
</exercise>
<technology>
    <title><!--<image source="images/icon-GC.jpg"  width="8%"><description>Graphing Calculator</description></image>-->Finding Coordinates with a Graphing Calculator</title>
    <p>We can use the <c>TRACE</c> feature of the calculator to find the coordinates of points on a graph. For example, graph the equation <m>y = -2.6x - 5.4</m> in the window</p>
        <p><sidebyside>
            <tabular>
                <row>
                    <cell>Xmin<m>=-5</m></cell>
                    <cell></cell>
                    <cell>Xmax<m>=4.4</m></cell>
                </row>
                <row>
                    <cell>Ymin<m>=-20</m></cell>
                    <cell></cell>
                    <cell>Ymax<m>=15</m></cell>
                </row>
            </tabular>
        </sidebyside></p>
    <figure xml:id="fig-GC-trace"><caption></caption><image source="images/fig-GC-trace"   width="40%"><description>GC trace</description></image></figure>
    <p>Press <c>TRACE</c>, and a “bug” begins flashing on the display. The coordinates of the bug appear at the bottom of the display, as shown in <xref ref="fig-GC-trace" text="type-global"/>. Use the left and right arrows to move the bug along the graph. You can check that the coordinates of the point <m>(2, -10.6)</m> do satisfy the equation <m>y = -2.6x - 5.4</m>.</p>
    <p>The points identified by the Trace bug depend on the window settings and on the type of calculator. If we want to find the <m>y</m>-coordinate for a particular <m>x</m>-value, we enter the <m>x</m>-coordinate of the desired point and press <c>ENTER</c>.</p>
</technology>

</subsection>

<subsection><title>Constructing the Graph of a Function</title>
<p>Although some functions are defined by their graphs, we can also construct graphs for
functions described by tables or equations. We make these graphs the same way we graph
equations in two variables: by plotting points whose coordinates satisfy the equation.</p>
    
<example xml:id="example-graph-square-root">
<p>Graph the function <m>f(x) = \sqrt{x + 4}</m>.</p>
<solution>
    <p>Choose several convenient values for <m>x</m> and evaluate the function to find the corresponding <m>f(x)</m>-values. For this function we cannot choose <m>x</m>-values less than <m>-4</m>, because the square root of a negative number is not a real number. 
    <me>f(\alert{-4}) =\sqrt{\alert{-4} + 4}=\sqrt{0}= 0</me>
    <me>f(\alert{-3}) =\sqrt{\alert{-3} + 4}=\sqrt{1}= 1</me>
    <me>f(\alert{0}) =\sqrt{\alert{0} + 4}=\sqrt{4}=2</me>
    <me>f(\alert{2}) =\sqrt{\alert{2} + 4}=\sqrt{6}\approx 2.45</me>
    <me>f(\alert{5}) =\sqrt{\alert{5} + 4}=\sqrt{9}=3</me>
    The results are shown in the table.</p>
    <sidebyside valigns="middle middle">
    <tabular top="major" halign="center" right="minor" left="minor" bottom="minor"> <row bottom="minor">
        <cell><m>x</m></cell>
        <cell><m>f(x)</m></cell>
    </row>
    <row>
        <cell><m>-4</m></cell>
        <cell><m>0</m></cell>
    </row>
    <row>
        <cell><m>-3</m></cell>
        <cell><m>1</m></cell>
    </row>
    <row>
        <cell><m>0</m></cell>
        <cell><m>2</m></cell>
    </row>
    <row>
        <cell><m>2</m></cell>
        <cell><m>\sqrt{6}</m></cell>
    </row>
    <row>
        <cell><m>5</m></cell>
        <cell><m>3</m></cell>
        </row>
  </tabular>
 
    <figure xml:id="fig-sq-root"><caption></caption><image source="images/fig-sq-root"><description>square root graph</description></image></figure>
    </sidebyside>
</solution>
</example>

<technology>
    <title><!--<image source="images/icon-GC.jpg"  width="8%"><description>Graphing Calculator</description></image>-->Using a Calculator to Graph a Function</title>
    <p>We can also use a graphing calculator to obtain a table and graph for the function in <xref ref="example-graph-square-root" text="type-global"/>. We graph a function just as we graphed an equation. For this function, we enter <m>Y_1 = \sqrt{~^~}(X+4)</m>
    and press <c>ZOOM</c> <m>6</m> for the standard window.  The calculator’s graph is shown in <xref ref="fig-GC-sq-root" text="type-global"/>.</p>
    <figure xml:id="fig-GC-sq-root"><caption></caption><image source="images/fig-GC-sq-root"   width="40%"><description>GC square root graph</description></image></figure>
</technology>

<exercise xml:id="exercise-cubic-graph">
    <statement><p>
        <m>f(x) = x^3 - 2</m>
        <ol label="a">
            <li><p>Complete the table of values and sketch a graph of the function.
                <tabular top="major" halign="center" right="minor" left="minor" bottom="minor"> <row bottom="minor">
                    <cell><m>x</m></cell>
                    <cell><m>-2</m></cell>
                    <cell><m>-1</m></cell>
                    <cell><m>-\frac{1}{2}</m></cell>
                    <cell><m>0</m></cell>
                    <cell><m>\frac{1}{2}</m></cell>
                    <cell><m>1</m></cell>
                    <cell><m>2</m></cell>
                    </row>
                    <row>
                    <cell><m>f(x)</m></cell>
                    <cell><m>\hphantom{000}</m></cell>
                    <cell><m>\hphantom{000}</m></cell>
                    <cell><m>\hphantom{000}</m></cell>
                    <cell><m>\hphantom{000}</m></cell>
                    <cell><m>\hphantom{000}</m></cell>
                    <cell><m>\hphantom{000}</m></cell>
                    <cell><m>\hphantom{000}</m></cell>
                     </row>
                </tabular>
            </p></li>
         <li><p>Use your calculator to make a table of values and graph the function.</p></li>
    </ol></p>
    </statement>
    <answer><p>
        <ol label="a">
            <li><p>
    <tabular top="major" halign="center" right="minor" left="minor" bottom="minor"> <row bottom="minor">
                    <cell><m>x</m></cell>
                    <cell><m>-2</m></cell>
                    <cell><m>-1</m></cell>
                    <cell><m>-\frac{1}{2}</m></cell>
                    <cell><m>0</m></cell>
                    <cell><m>\frac{1}{2}</m></cell>
                    <cell><m>1</m></cell>
                    <cell><m>2</m></cell>
                    </row>
                    <row>
                    <cell><m>f(x)</m></cell>
                    <cell><m>\alert{-10} </m></cell>
                    <cell><m>\alert{-3} </m></cell>
                    <cell><m>\alert{\frac{-17}{8}} </m></cell>
                    <cell><m>\alert{-2} </m></cell>
                    <cell><m>\alert{\frac{-15}{8}} </m></cell>
                    <cell><m>\alert{-1} </m></cell>
                    <cell><m>\alert{6} </m></cell>
                     </row>
                </tabular>
            </p></li>
            <li><p><image source="images/fig-in-ex-ans-1-3-3"  width="30%"><description>cubic</description></image> </p></li>
        </ol>
    </p></answer>
</exercise>
</subsection>

<subsection><title>The Vertical Line Test</title>
<p>In a function, two different outputs cannot be related to the same input. This restriction means that two different ordered pairs cannot have the same first coordinate. What does it mean for the graph of the function?</p>
<p>Consider the graph shown in <xref ref="fig-vertical-line-test" text="type-global"/>a. Every vertical line intersects the graph in at most one point, so there is only one point on the graph for each <m>x</m>-value. This graph represents a function. In <xref ref="fig-vertical-line-test" text="type-global"/>b, however, the line <m>x = 2</m> intersects the graph at two points, <m>(2, 1)</m> and <m>(2, 4)</m>. Two different <m>y</m>-values, <m>1</m> and <m>4</m>, are related to the same <m>x</m>-value, <m>2</m>. This graph cannot be the graph of a function.</p>

<figure xml:id="fig-vertical-line-test"><caption></caption><image source="images/fig-vertical-line-test"   width="70%"><description>vertical line test</description></image></figure>

<p>We summarize these observations as follows.</p>

<assemblage><title>The Vertical Line Test</title><line></line>
<p>A graph represents a function if and only if every vertical line intersects the graph in
at most one point.</p>
</assemblage>

<example xml:id="example-vertical-line-test">
    <p>Use the vertical line test to decide which of the graphs in <xref ref="fig-vertical-line-test2" text="type-global"/>  represent functions.</p>
    <figure xml:id="fig-vertical-line-test2"><caption></caption><image source="images/fig-vertical-line-test2"   width="90%"><description>vertical line test</description></image></figure>
<solution><p>
    <ul>
        <li><p>
    Graph (a) represents a function, because it passes the vertical line test.</p></li>
    <li><p> Graph (b) is not the graph of a function, because the vertical line at (for example) <m>x = 1</m> intersects the graph at two points.</p></li>
    <li><p> For graph (c), notice the break in the curve at <m>x = 2</m>: The solid dot at <m>(2, 1)</m> is the only point on the graph with <m>x = 2</m>; the open circle at <m>(2, 3)</m> indicates that <m>(2, 3)</m> is not a point on the graph. Thus, graph (c) is a function, with <m>f(2) = 1</m>.</p></li>
</ul>
</p></solution>
</example>

<exercise xml:id="example-vertical-line-test3">
    <statement>Use the vertical line test to determine which of the graphs in <xref ref="fig-vertical-line-test3" text="type-global"/> represent functions.
    <figure xml:id="fig-vertical-line-test3"><caption></caption><image source="images/fig-vertical-line-test3"   width="100%"><description>vertical line test</description></image></figure>
    </statement>
    <answer><p>Only (b) is a function.</p></answer>
</exercise>
</subsection>

<subsection><title>Graphical Solution of Equations and Inequalities</title>
    <p>The graph of an equation in two variables is just a picture of its solutions. When we read the coordinates of a point on the graph, we are reading a pair of <m>x</m>- and <m>y</m>-values that make the equation true. </p>

    <sidebyside widths="45% 30%" valigns="middle middle"><paragraphs>
        <p>For example, the point <m>(2, 7)</m> lies on the graph of <m>y = 2x + 3</m> shown in <xref ref="fig-function-graph" text="type-global"/> , so we know that the ordered pair <m>(2, 7)</m> is a solution of the equation <m>y = 2x + 3</m>. You can verify algebraically that <m>x = \alert{2}</m> and <m>y = \alert{7}</m> satisfy the equation: 
    <me>\text{Does }~\alert{7} = 2 (\alert{2}) + 3\text{?} ~~ \alert{Yes}</me>
    </p></paragraphs>
    <figure xml:id="fig-function-graph"><caption></caption><image source="images/fig-function-graph"><description>function graph</description></image></figure></sidebyside>

    <p>We can also say that <m>x = 2</m> is a solution of the one-variable equation <m>2x + 3 = 7</m>. In fact, we can use the graph of <m>y = 2x + 3</m> to solve the equation <m>2x + 3 = k</m> for any value of <m>k</m>. Thus, we can use graphs to find solutions to equations in one variable.</p>

<example xml:id="example-graph-to-solve">
    <p>Use the graph of <m>y = 285 - 15x</m> to solve the equation <m>150 = 285 - 15x</m>.</p>

<solution>
    <sidebyside widths="45% 50%">    
    <p>Begin by locating the point <m>P</m> on the graph for which <m>y = 150</m>, as shown in <xref ref="fig-graph-to-solve" text="type-global"/>. Now find the <m>x</m>-coordinate of point <m>P</m> by drawing an imaginary line from <m>P</m> straight down to the <m>x</m>-axis. The <m>x</m>-coordinate of <m>P</m> is <m>x = 9</m>. Thus, <m>P</m> is the point <m>(9,150)</m>, and <m>x = 9</m> when <m>y = 150</m>. The solution of the equation <m>150 = 285 - 15x</m> is <m>x = 9</m>.</p>
    <figure xml:id="fig-graph-to-solve"><caption></caption><image source="images/fig-graph-to-solve"   width="60%"><description>function graph</description></image></figure>
    </sidebyside> 
    <p>You can verify the solution algebraically by substituting <m>x = \alert{9}</m> into the equation:</p> 
    <p>Does <m>150 = 285 - 15(\alert{9})</m>? </p>
    <me>285 - 15(\alert{9}) = 285 - 135 = 150. ~~\alert{Yes}</me>
</solution>
</example>

<note>
<p>The relationship between an equation and its graph is an important one. For the
previous example, make sure you understand that the following three statements are
equivalent:
</p>
<ul>
    <li>The point <m>(9, 150)</m> lies on the graph of <m>y = 285 - 15x</m>.</li>
    <li>The ordered pair <m>(9, 150)</m> is a solution of the equation <m>y = 285 - 15x</m>.</li>
    <li><m>x = 9</m> is a solution of the equation <m>150 = 285 - 15x</m>.</li>
</ul>
</note>

<exercise xml:id="exercise-graph-to-solve"><statement>
    <ol label="a">
        <li>Use the graph of <m>y = 30 - 8x</m> shown in <xref ref="fig-graph-to-solve2" text="type-global"/> to solve the equation <me>30 - 8x = 50</me></li>
        <li>Verify your solution algebraically.</li>
    </ol>
   <figure xml:id="fig-graph-to-solve2"><caption></caption><image source="images/fig-graph-to-solve2"   width="35%"><description>function graph</description></image></figure>
    </statement>
    <answer><p><m>-2.5</m></p></answer>
</exercise>

<p>In a similar fashion, we can solve inequalities with a graph.</p>

<sidebyside widths="50% 36%" valigns="middle middle"><paragraphs><p> Consider again the graph of <m>y = 2x + 3</m>, shown in <xref ref="fig-graph-to-solve3" text="type-global"/>. We saw that <m>x = 2</m> is the solution of the equation <m>2x + 3 = 7</m>. When we use <m>x = 2</m> as the input for the function <m>f(x) = 2x + 3</m>, the output is <m>y = 7</m>. Which input values for <m>x</m> produce output values greater than <m>7</m>?</p>
<p>You can see in <xref ref="fig-graph-to-solve3" text="type-global"/> that <m>x</m>-values greater than <m>2</m> produce <m>y</m>-values greater than <m>7</m>, because points on the graph with <m>x</m>-values greater than <m>2</m> have <m>y</m>-values greater than <m>7</m>. Thus, the solutions of the inequality <m>2x + 3 \gt 7</m> are <m>x\gt 2</m>. You can verify this result by solving the inequality algebraically.</p></paragraphs>

<figure xml:id="fig-graph-to-solve3"><caption></caption><image source="images/fig-graph-to-solve3"  width="60%"><description>function graph</description></image></figure>
</sidebyside>

<example xml:id="example-graph-to-solve2">
    <p>Use the graph of <m>y = 285 - 15x</m> to solve the inequality <me>285 - 15x \gt 150</me></p>
<solution><p>
    We begin by locating the point <m>P</m> on the graph for which <m>y = 150</m> and <m>x = 9</m> (its <m>x</m>-coordinate). Now, because <m>y = 285 - 15x</m> for points on the graph, we are looking for points on the graph with <m>y</m>-coordinate greater than <m>150</m>. These points are shown in <xref ref="fig-graph-to-solve-inequality" text="type-global"/>. The <m>x</m>-coordinates of these points are the <m>x</m>-values that satisfy the inequality. From the graph, we see that the solutions are <m>x \lt 9</m>.
<figure xml:id="fig-graph-to-solve-inequality"><caption></caption><image source="images/fig-graph-to-solve-inequality"   width="40%"><description>function graph</description></image></figure>
</p></solution>
</example>

<exercise xml:id="exercise-graph-to-solve-inequality"><statement>
    <ol label="a">
        <li>Use the graph of <m>y = 30 - 8x</m> in  <xref ref="fig-graph-to-solve2" text="type-global"/> to solve the inequality <me>30 - 8x \le 50</me></li>
        <li>Solve the inequality algebraically.</li>
    </ol>
</statement> 
<answer><p><m>x\ge -2.5</m></p></answer>  
</exercise>

<p>We can also use this graphical technique to solve nonlinear equations and inequalities.</p>

<example xml:id="example-graph-to-solve-cubic">
    <p>Use a graph of <m>f(x) = -2x^3 + x^2 + 16x</m> to solve the equation 
    <me>-2x^3 + x^2 + 16x = 15</me></p>

<solution>
    <p>If we sketch in the horizontal line <m>y = 15</m>, we can see that there are three points on the graph of <m>f</m> that have <m>y</m>-coordinate <m>15</m>, as shown in <xref ref="fig-graph-to-solve-cubic" text="type-global"/>. The <m>x</m>-coordinates of these points are the solutions of the equation. 
        <figure xml:id="fig-graph-to-solve-cubic"><caption></caption><image source="images/fig-graph-to-solve-cubic"   width="40%"><description>graph of cubic</description></image></figure>
    From the graph, we see that the solutions are <m>x = -3</m>, <m>x = 1</m>, and approximately <m>x = 2.5</m>. We can verify each solution algebraically.</p> 
    <p>For example, if <m>x = \alert{-3}</m>, we have 
    \begin{align*}
        f(\alert{-3})<ampersand />-2(\alert{-3})3 + (\alert{-3})^2 + 16(\alert{-3})<backslash /><backslash />
        <ampersand />= -2(-27) + 9 - 48<backslash /><backslash />
        <ampersand />=54 + 9 - 48 = 15  
    \end{align*}
    so <m>-3</m> is a solution.  Similarly, you can check that <m>x = 1</m> and <m>x = 2.5</m> are solutions.</p>


</solution>
</example>
<exercise xml:id="exercise-graph-to-solve-quadratic"><statement>
    Use the graph of <m>y = \frac{1}{2}n^2 + 2n - 10</m> shown in <xref ref="fig-graph-to-solve-quadratic" text="type-global"/> to solve
    <me>\frac{1}{2}n^2 + 2n - 10 = 6</me>
    and verify your solutions algebraically.
<figure xml:id="fig-graph-to-solve-quadratic"><caption></caption><image source="images/fig-graph-to-solve-quadratic"   width="40%"><description>graph of cubic</description></image></figure>
</statement> 
<answer><p><m>-8, 4</m></p></answer>  
</exercise>

<technology><title><!--<image source="images/icon-GC.jpg"  width="8%"><description>Graphing Calculator</description></image>-->Using the Trace Feature</title><line></line>
    <p>You can use the Trace feature on a graphing calculator to approximate solutions to equations. Graph the function <m>f(x)</m> in <xref ref="example-graph-to-solve-cubic" text="type-global"/>  in the window</p><p>
    <sidebyside>
            <tabular>
                <row>
                    <cell>Xmin<m>=-4</m></cell>
                    <cell></cell>
                    <cell>Xmax<m>=4</m></cell>
                </row>
                <row>
                    <cell>Ymin<m>=-20</m></cell>
                    <cell></cell>
                    <cell>Ymax<m>=40</m></cell>
                </row>
            </tabular>
    </sidebyside>
    and trace along the curve to the point <m>(2.4680851, 15.512401)</m>. We are close to a solution, because the <m>y</m>-value is close to <m>15</m>. Try entering <m>x</m>-values close to <m>2.4680851</m>, for instance, <m>x = 2.4</m> and <m>x = 2.5</m>, to find a better approximation for the solution.
</p>  
<p>We can use the intersect feature on a graphing calculator to obtain more accurate estimates for the solutions of equations. </p>
</technology>

<example xml:id="example-graph-to-solve-cubic2">
    <p>Use the graph in <xref ref="example-graph-to-solve-cubic" text="type-global"/>  to solve the inequality
    <me>-2x^3 + x^2 + 16x \ge 15</me></p>
    <solution><p>
        We first locate all points on the graph that have <m>y</m>-coordinates greater than or equal to <m>15</m>. The <m>x</m>-coordinates of these points are the solutions of the inequality.</p>
        <p> <xref ref="fig-graph-to-solve-cubic2" text="type-global" /> shows the points in red, and their <m>x</m>-coordinates as intervals on the <m>x</m>-axis. The solutions are <m>x \le -3</m> and <m>1\le x \le 2.5</m>, or in interval notation, <m>(-∞, -3] ∪ [1, 2.5]</m>.
        <figure xml:id="fig-graph-to-solve-cubic2"><caption></caption><image source="images/fig-graph-to-solve-cubic2"   width="40%"><description>graph of cubic</description></image></figure>
    </p></solution>
</example>

<exercise xml:id="exercise-graph-to-solve-quadratic2">
    <statement>Use <xref ref="fig-graph-to-solve-quadratic" text="type-global"/> in <xref ref="exercise-graph-to-solve-quadratic" text="type-global" /> to solve the inequality
    <me>\frac{1}{2}n^2 + 2n - 10 \lt 6</me></statement>
    <answer><p><m>(-8, 4)</m></p></answer>
</exercise>


</subsection>

<xi:include href="./summary-1-3.xml" /> <!-- summary  -->
<xi:include href="./section-1-3-exercises.xml" /> <!-- exercises  -->

</section> 
<!-- </book>  </mathbook> -->