section-2-3.xml

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

<section xml:id="transformations"   xmlns:xi="http://www.w3.org/2001/XInclude">
<title>Transformations of Graphs</title><introduction>
<p>
    Models for real situations are often variations of the basic functions introduced in <xref ref="basic-functions" text="type-global" />. In this section, we explore how certain changes in the formula for a function affect its graph. In particular, we will compare the graph of <m>y = f (x)</m> with the graphs of <me>y = f (x) + k, y = f (x + h), \text{ and  }  y = a f (x)</me> for different values of the constants <m>k</m>, <m>h</m>, and <m>a</m>. Such variations are called <term>transformations</term><idx>transformations</idx> of the graph.
</p>
</introduction>
<subsection><title>Vertical Translations</title> 
<xref ref="fig-parabolas" text="type-global"/> shows the graphs of <m>f (x) = x^2 + 4</m>, <m>g(x) = x^2 - 4</m>, and the basic parabola, <m>y = x^2</m>. By comparing tables of values, we can see exactly how the graphs of <m>f</m> and <m>g</m> are related to the basic parabola.
<sidebyside widths="35% 50%"> 
<figure xml:id="fig-parabolas"><caption></caption><image source="images/fig-parabolas"><description>parabolas</description></image></figure>
    
<p><paragraphs><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>~0~</m></cell>
            <cell><m>~~1~~</m></cell>
            <cell><m>~2~</m></cell>
        </row>
        <row>
            <cell><m>y=x^2</m></cell>
            <cell><m>4</m></cell>
            <cell><m>1</m></cell>
            <cell><m>0</m></cell>
            <cell><m>1</m></cell>
            <cell><m>4</m></cell>
        </row>
         <row>
            <cell><m>f(x)=x^2+4</m></cell>
            <cell><m>8</m></cell>
            <cell><m>5</m></cell>
            <cell><m>4</m></cell>
            <cell><m>5</m></cell>
            <cell><m>8</m></cell>
        </row>
    </tabular></p>

<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>0</m></cell>
            <cell><m>1</m></cell>
            <cell><m>2</m></cell>
        </row>
        <row>
            <cell><m>y=x^2</m></cell>
            <cell><m>4</m></cell>
            <cell><m>1</m></cell>
            <cell><m>0</m></cell>
            <cell><m>1</m></cell>
            <cell><m>4</m></cell>
        </row>
         <row>
            <cell><m>g(x)=x^2-4</m></cell>
            <cell><m>0</m></cell>
            <cell><m>-3</m></cell>
            <cell><m>-4</m></cell>
            <cell><m>-3</m></cell>
            <cell><m>0</m></cell>
        </row>
    </tabular></p></paragraphs></p>
    </sidebyside>
<p>
    Each <m>y</m>-value in the table for <m>f (x)</m> is four units greater than the corresponding <m>y</m>-value for the basic parabola. Consequently, each point on the graph of <m>f (x)</m> is four units higher than the corresponding point on the basic parabola, as shown by the arrows. Similarly, each point on the graph of <m>g(x)</m> is four units lower than the corresponding point on the basic parabola.
</p>
<p>
    The graphs of <m>y = f (x)</m> and <m>y = g(x)</m> are said to be <term>translations</term><idx>translations</idx> of the graph of <m>y = x^2</m>. They are shifted to a different location in the plane but retain the same size and shape as the original graph. In general, we have the following principles.
</p>

<assemblage><title>Vertical Translations</title><p></p>
<p>
    Compared with the graph of <m>y = f (x)</m>,
    <ol>
        <li>The graph of <m>y=f(x)+k</m> (<m>k\gt 0</m>) is shifted <em>upward </em> <m>k</m> units.</li>
        <li>The graph of <m>y=f(x)-k</m> (<m>k\gt 0</m>) is shifted <em>downward </em> <m>k</m> units.</li>
    </ol>
</p>
</assemblage>

<example xml:id="example-translations">
    <p>Graph the following functions.
        <ol label="a">
            <li><m>g(x) = \abs{x} + 3</m></li>
            <li><m>h(x) = \dfrac{1}{x}- 2</m></li>
        </ol>
    </p>
<solution>
    <ol label="a">
        <li><p>The table shows that the <m>y</m>-values for <m>g(x)</m> are each three units greater than the corresponding <m>y</m>-values for the absolute value function. The graph of <m>g(x) = \abs{x} + 3</m> is a translation of the basic graph of <m>y = \abs{x}</m>, shifted upward three units, as shown in <xref ref="fig-translate-abs" text="type-global" />.</p> <p></p>
        <sidebyside widths="35% 50%" valigns="middle middle">
        <p><figure xml:id="fig-translate-abs"><caption></caption><image source="images/fig-translate-abs"><description>translate absolute value</description></image></figure></p>
        <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>0</m></cell>
            <cell><m>1</m></cell>
            <cell><m>2</m></cell>
        </row>
        <row>
            <cell><m>y=\abs{x}</m></cell>
            <cell><m>2</m></cell>
            <cell><m>1</m></cell>
            <cell><m>0</m></cell>
            <cell><m>1</m></cell>
            <cell><m>2</m></cell>
        </row>
         <row>
            <cell><m>g(x)=\abs{x}+3</m></cell>
            <cell><m>5</m></cell>
            <cell><m>4</m></cell>
            <cell><m>3</m></cell>
            <cell><m>4</m></cell>
            <cell><m>5</m></cell>
        </row>
        </tabular></p>
        </sidebyside></li>
        
        <li><p>The table shows that the <m>y</m>-values for <m>h(x)</m> are each two units smaller than the corresponding <m>y</m>-values for <m>y = \dfrac{1}{x}</m>. The graph of <m>h(x) = \dfrac{1}{x} - 2</m> is a translation of the basic graph of <m>y = \dfrac{1}{x}</m>, shifted downward two units, as shown in <xref ref="fig-translate-reciprocal" text="type-global" />.</p> <p></p>
        <sidebyside widths="38% 50%" valigns="middle middle">
        <p><figure xml:id="fig-translate-reciprocal"><caption></caption><image source="images/fig-translate-reciprocal"><description>translate reciprocal</description></image></figure></p>
        <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>\dfrac{1}{2}</m></cell>
            <cell><m>1</m></cell>
            <cell><m>2</m></cell>
        </row>
        <row>
            <cell><m>y=\dfrac{1}{x}</m></cell>
            <cell><m>\dfrac{-1}{2}</m></cell>
            <cell><m>-1</m></cell>
            <cell><m>2</m></cell>
            <cell><m>1</m></cell>
            <cell><m>\dfrac{1}{2}</m></cell>
        </row>
         <row>
            <cell><m>h(x)=\dfrac{1}{x}-2</m></cell>
            <cell><m>\dfrac{-5}{2}</m></cell>
            <cell><m>-3</m></cell>
            <cell><m>0</m></cell>
            <cell><m>-1</m></cell>
            <cell><m>\dfrac{-3}{2}</m></cell>
        </row>
        </tabular></p>
        </sidebyside>
        </li>
    </ol>
</solution>
</example>

<exercise xml:id="exercise-translate-abs"><statement>
    <ol label="a">
        <li>Graph the function <m>f (x) = \abs{x} + 1</m>.</li>
        <li>How is the graph of <m>f</m> different from the graph of <m>y = \abs{x}</m>?</li>
    </ol>
</statement>
<answer><p><ol label="a" cols="2">
    <li><p><image source="images/fig-in-ex-ans-2-3-1"  width="50%"><description>shifted absolute value</description></image> </p></li>
    <li><p>Translate <m>y =\abs{x}</m> one unit up.</p></li>
</ol></p></answer>
</exercise>

<example xml:id="example-electrical">
    <sidebyside widths= "50% 45%">
    <p>
        The function <m>E = f (h)</m> graphed in <xref ref="fig-electrical" text="type-global"/> gives the amount of electrical power, in megawatts, drawn by a community from its local power plant as a function of time during a 24-hour period in 2002. Sketch a graph of <m>y = f (h) + 300</m> and interpret its meaning.
    </p>
    <figure xml:id="fig-electrical"><caption></caption><image source="images/fig-electrical"  width="95%"><description>electrical power</description></image></figure>
</sidebyside>
<solution>
    <p>The graph of <m>y = f (h) + 300</m> is a vertical translation of the graph of <m>f</m>,</p>
    <sidebyside widths= "45% 50%"><p>
          as shown in <xref ref="fig-translate-electrical" text="type-global"/>. At each hour of the day, or for each value of <m>h</m>, the <m>y</m>-coordinate is 300 greater than on the graph of <m>f</m>. So at each hour, the community is drawing 300 megawatts more power than in 2002.
    </p>
        <figure xml:id="fig-translate-electrical"><caption></caption><image source="images/fig-translate-electrical"  width="100%"><description>electrical power</description></image></figure>
    </sidebyside>
</solution></example>

<exercise xml:id="exercise-swamp-cooler"><statement>
    An evaporative cooler, or swamp cooler, is an energy-efficient type of air conditioner used in dry climates. A typical swamp cooler can reduce the temperature inside a house by 15 degrees. <xref ref="fig-swamp-cooler" text="type-global" />a shows the graph of <m>T = f (t)</m>, the temperature inside Kate’s house <m>t</m> hours after she turns on the swamp cooler. Write a formula in terms of <m>f</m> for the function <m>g</m> shown in <xref ref="fig-swamp-cooler" text="type-global" />b and, give a possible explanation of its meaning.
    <figure xml:id="fig-swamp-cooler"><caption></caption><image source="images/fig-swamp-cooler"  width="100%"><description>swamp cooler graphs</description></image></figure>
</statement>
<answer><p><m>g(t) = f (t) + 10</m>. The outside temperature was <m>10\degree</m> hotter.</p></answer>
</exercise>

</subsection>

<subsection><title>Horizontal Translations</title>
    
<p>
    Now consider the graphs of <m>f (x) = (x + 2)^2</m> and <m>g(x) = (x - 2)^2</m> shown in <xref ref="fig-translate-parabs" text="type-global" />. Compared with the graph of the basic function <m>y = x^2</m>, the graph of <m>f (x) = (x + 2)^2</m> is shifted two units to the <em>left</em>, as shown by the arrows. You can see why this happens by studying the function values in the table. </p>

<figure xml:id="fig-translate-parabs"><caption></caption><image source="images/fig-translate-parabs"  width="40%"><description>swamp cooler graphs</description></image></figure>

    <p>Locate a particular <m>y</m>-value for <m>y = x^2</m>, say, <m>y = 4</m>. You must move two units to the left in the table to find the same <m>y</m>-value for <m>f (x)</m>, as shown by the arrow. In fact, each <m>y</m>-value for <m>f (x)</m> occurs two units to the left when compared to the same <m>y</m>-value for <m>y = x^2</m>.
    <figure xml:id="fig-table-translate-left"><caption></caption><image source="images/fig-table-translate-left"  width="70%"><description>table for left translation</description></image></figure>
</p>
<p>
    <figure xml:id="fig-table-translate-right"><caption></caption><image source="images/fig-table-translate-right"  width="70%"><description>table for right translation</description></image></figure>
    Similarly, the graph of <m>g(x) = (x - 2)^2</m> is shifted two units to the <em>right</em> compared to the graph of <m>y = x^2</m>. In the table for <m>g</m>, each <m>y</m>-value for <m>g(x)</m> occurs two units to the right of the same <m>y</m>-value for <m>y = x^2</m>. In general, we have the following principle.
</p>

<assemblage><title>Horizontal Translations</title><p></p>
<p>
    Compared with the graph of <m>y = f (x)</m>,
    <ol>
        <li>The graph of <m>y = f (x + h)~ ~</m> (<m>h \gt 0</m>) is shifted <m>h</m> units to the <em>left</em>.</li>
        <li>The graph of <m>y = f (x - h)~ ~</m> (<m>h \gt 0</m>) is shifted <m>h</m> units to the <em>right</em>.</li>
    </ol>
</p></assemblage>

<note><p>At first, the direction of a horizontal translation may seem counterintuitive.  Look again at the tables above to help you see how the shift occurs.</p></note>

<example xml:id="example-horizontal-translations">
    <p>
        Graph the following functions.
        <ol label="a">
            <li><m>g(x) =\sqrt{x + 1}</m></li>
            <li><m>h(x) = \dfrac{1}{(x - 3)^2}</m></li>
        </ol>
    </p>
    <solution>
        <ol label="a">
            <li><p>Consider the table of values for the function.</p><p>
            <sidebyside><tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
                <row bottom="minor">
                    <cell><m>x</m></cell>
                    <cell><m>-1</m></cell>
                    <cell><m>0</m></cell>
                    <cell><m>1</m></cell>
                    <cell><m>2</m></cell>
                    <cell><m>3</m></cell>
                </row>
                <row>
                    <cell><m>y=\sqrt{x}</m></cell>
                    <cell>undefined</cell>
                    <cell><m>0</m></cell>
                    <cell><m>1</m></cell>
                    <cell><m>1.414</m></cell>
                    <cell><m>1.732</m></cell>
                </row>
                <row>
                    <cell><m>y=\sqrt{x+1}</m></cell>
                    <cell><m>0</m></cell>
                    <cell><m>1</m></cell>
                    <cell><m>1.414</m></cell>
                    <cell><m>1.732</m></cell>
                    <cell><m>2</m></cell>
                </row>
            </tabular></sidebyside></p>
            <sidebyside widths="50% 35%"><p> The table shows that each <m>y</m>-value for <m>g(x)</m> occurs one unit to the left of the same <m>y</m>-value for the graph of <m>y=\sqrt{x}</m>. Consequently, each point on the graph of <m>y = g(x)</m> is shifted one unit to the left of <m>y =\sqrt{x}</m>, as shown in <xref ref="fig-shift-sqrt" text="type-global" />.</p><figure xml:id="fig-shift-sqrt"><caption></caption><image source="images/fig-shift-sqrt"><description>horizontal shift of square root</description></image></figure></sidebyside></li>

            <li><p>Consider the table of values for the function.</p><p>
            <sidebyside><tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
                <row bottom="minor">
                    <cell><m>x</m></cell>
                    <cell><m>-1</m></cell>
                    <cell><m>0</m></cell>
                    <cell><m>1</m></cell>
                    <cell><m>2</m></cell>
                    <cell><m>3</m></cell>
                    <cell><m>4</m></cell>
                </row>
                <row>
                    <cell><m>y=\dfrac{1}{x}</m></cell>
                    <cell><m>1</m></cell>
                    <cell>undefined</cell>
                    <cell><m>1</m></cell>
                    <cell><m>\dfrac{1}{4}</m></cell>
                    <cell><m>\dfrac{1}{9}</m></cell>
                    <cell><m>\dfrac{1}{16}</m></cell>
                </row>
                <row>
                    <cell><m>y=\dfrac{1}{(x-3)^2}</m></cell>
                    <cell><m>\dfrac{1}{16}</m></cell>
                    <cell><m>\dfrac{1}{9}</m></cell>
                    <cell><m>\dfrac{1}{4}</m></cell>
                    <cell><m>1</m></cell>
                    <cell>undefined</cell>
                    <cell><m>1</m></cell>
                </row>
            </tabular></sidebyside></p>
            <sidebyside widths="50% 45%"><p>The table shows that each <m>y</m>-value for <m>h(x)</m> occurs three units to the right of the same <m>y</m>-value for the graph of <m>y =\dfrac{1}{x^2}</m>. Consequently, each point on the graph of <m>y = h(x)</m> is shifted three units to the right of <m>y =\dfrac{1}{x^2}</m>, as shown in <xref ref="fig-shift-inv-sq" text="type-global" />.</p><figure xml:id="fig-shift-inv-sq"><caption></caption><image source="images/fig-shift-inv-sq"><description>horizontal shift of inverse square</description></image></figure></sidebyside>
        </li>
        </ol>
    </solution>
</example>

<exercise xml:id="exercise-shift-abs"><statement>
    <ol label="a">
        <li><p>Graph the function <m>f (x) = \abs{x + 1}</m>.</p></li>
        <li><p>How is the graph of <m>f</m> different from the graph of <m>y = \abs{x}</m>?</p></li>
    </ol>
</statement>
<answer><p><ol label="a" cols="2">
    <li><p><image source="images/fig-in-ex-ans-2-3-3"  width="50%"><description>shifted absolute value</description></image> </p></li>
    <li><p>Translate <m>y =\abs{x}</m> one unit left.</p></li>
</ol></p></answer>
</exercise>

<example xml:id="example-shift-bell-curve">
    <sidebyside widths= "50% 40%">
    <p>The function <m>N = f (p)</m> graphed in <xref ref="fig-shift-bell-curve" text="type-global"/> gives the number of people who have a given eye pressure level <m>p</m> from a sample of 100 people with healthy eyes, and the function <m>g</m> gives the number of people with pressure level <m>p</m> in a sample of 100 glaucoma patients.</p>
    <figure xml:id="fig-shift-bell-curve"><caption></caption><image source="images/fig-shift-bell-curve"><description>shift of a bell-shaped curve</description></image></figure>
    </sidebyside>
    <ol label="a">
        <li>Write a formula for <m>g</m> as a transformation of <m>f</m>.</li>
        <li>For what pressure readings could a doctor be fairly certain that a patient has glaucoma?</li>
    </ol>
    <solution>
        <ol label="a">
            <li>The graph of <m>g</m> is translated <m>10</m> units to the right of <m>f</m>, so <m>g(p) = f (p - 10)</m>.</li>
            <li>Pressure readings above <m>40</m> are a strong indication of glaucoma. Readings between <m>10</m> and <m>40</m> cannot conclusively distinguish healthy eyes from those with glaucoma.</li>
        </ol>
    </solution>
</example>

<exercise xml:id="exercise-shift-surge-curve"><statement>
    The function <m>C = f (t)</m> in <xref ref="fig-shift-surge-curve" text="type-global"/> gives the caffeine level in Delbert's bloodstream at time <m>t</m> hours after he drinks a cup of coffee, and <m>g(t)</m> gives the caffeine level in Francine's bloodstream. Write a formula for <m>g</m> in terms of <m>f</m>, and explain what it tells you about Delbert and Francine.
    <figure xml:id="fig-shift-surge-curve"><caption></caption><image source="images/fig-shift-surge-curve"  width="50%"><description>shift of caffeine surge curve</description></image></figure>
</statement>
<answer><p><m>g(t) = f (t - 3)</m>. Francine drank her coffee <m>3</m> hours after Delbert drank his.</p></answer>
</exercise>

<example xml:id="example-translate-cubic">
    <p>Graph <m>f (x) = (x + 4)^3 + 2</m>.</p>
    <solution>
        <p>
            We identify the basic graph from the structure of the formula for <m>f (x)</m>. In this case, the basic graph is <m>y = x^3</m>, so we begin by locating a few points on that graph, as shown in <xref ref="fig-translate-cubic" text="type-global"/>. </p> 
            <p>We will perform the translations separately, following the order of operations. First, we sketch a graph of <m>y = (x + 4)^3</m> by shifting each point on the basic graph four units to the left. We then move each point up two units to obtain the graph of <m>f (x) = (x + 4)^3 + 2</m>. All three graphs are shown in <xref ref="fig-translate-cubic" text="type-global"/>.
        </p>
        <figure xml:id="fig-translate-cubic"><caption></caption><image source="images/fig-translate-cubic"  width="50%"><description>translations of cubic curve</description></image></figure>
    </solution>
</example>

<exercise xml:id="exercise-translate-abs2"><statement>
    <ol label="a">
        <li>Graph the function <m>f (x) = \abs{x - 2} - 1</m>.</li>
        <li>How is the graph of <m>f</m> different from the graph of <m>y=\abs{x}</m>?</li>
    </ol>
</statement>
<answer><p><ol label="a" cols="2">
    <li><p><image source="images/fig-in-ex-ans-2-3-5"  width="50%"><description>shifted absolute value</description></image> </p></li>
    <li><p>Translate <m>y =\abs{x}</m> one unit down and two units right.</p></li>
</ol></p></answer>
</exercise>
</subsection>

<subsection><title>Scale Factors</title>
<p>
    We have seen that <em>adding</em> a constant to the expression defining a function results in a translation of its graph. What happens if we <em>multiply</em> the expression by a constant? Consider the graphs of the functions
    <me>
        f(x)= 2x^2 \text{, } ~g(x)= \frac{1}{2}x^2 \text{, and } ~h(x) = -x^2
    </me>
    shown in <xref ref="fig-2x-sq" text="type-global"/>, <xref ref="fig-half-x-sq" text="type-global"/>, and <xref ref="fig-neg-x-sq" text="type-global"/>, and compare each to the graph of <m>y = x^2</m>.
</p>
<sidebyside widths="30% 50%">
<figure xml:id="fig-2x-sq"><caption></caption><image source="images/fig-2x-sq"><description>2 x-sq and basic parabola</description></image></figure>
<tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
    <row bottom="minor">
        <cell><m>x</m></cell>
        <cell><m>y=x^2</m></cell>
        <cell><m>f(x)=2x^2</m></cell>
    </row>
    <row>
        <cell><m>-2</m></cell>
        <cell><m>4</m></cell>
        <cell><m>8</m></cell>
    </row>
    <row>
        <cell><m>-1</m></cell>
        <cell><m>1</m></cell>
        <cell><m>2</m></cell>
    </row>
    <row>
        <cell><m>0</m></cell>
        <cell><m>0</m></cell>
        <cell><m>0</m></cell>
    </row>
    <row>
        <cell><m>1</m></cell>
        <cell><m>1</m></cell>
        <cell><m>2</m></cell>
    </row>
    <row>
        <cell><m>2</m></cell>
        <cell><m>4</m></cell>
        <cell><m>8</m></cell>
    </row>
</tabular>
</sidebyside>
<p>
    Compared to the graph of <m>y = x^2</m>, the graph of <m>f (x) = 2x^2</m> is expanded, or stretched, vertically by a factor of <m>2</m>. The <m>y</m>-coordinate of each point on the graph has been doubled, as you can see in the table of values, so each point on the graph of <m>f</m> is twice as far from the <m>x</m>-axis as its counterpart on the basic graph <m>y = x^2</m>.
</p>
<sidebyside widths="30% 50%">
<figure xml:id="fig-half-x-sq"><caption></caption><image source="images/fig-half-x-sq"><description>half x-sq and basic parabola</description></image></figure>
<tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
    <row bottom="minor">
        <cell><m>x</m></cell>
        <cell><m>y=x^2</m></cell>
        <cell><m>g(x)=\frac{1}{2}x^2</m></cell>
    </row>
    <row>
        <cell><m>-2</m></cell>
        <cell><m>4</m></cell>
        <cell><m>2</m></cell>
    </row>
    <row>
        <cell><m>-1</m></cell>
        <cell><m>1</m></cell>
        <cell><m>\frac{1}{2}</m></cell>
    </row>
    <row>
        <cell><m>0</m></cell>
        <cell><m>0</m></cell>
        <cell><m>0</m></cell>
    </row>
    <row>
        <cell><m>1</m></cell>
        <cell><m>1</m></cell>
        <cell><m>\frac{1}{2}</m></cell>
    </row>
    <row>
        <cell><m>2</m></cell>
        <cell><m>4</m></cell>
        <cell><m>2</m></cell>
    </row>
</tabular>
</sidebyside>
<p>
    The graph of <m>g(x) = \frac{1}{2}x^2</m> is compressed vertically by a factor of <m>\frac{1}{2}</m>; each point is half as far from the <m>x</m>-axis as its counterpart on the graph of <m>y = x^2</m>.
</p>

<sidebyside widths="30% 50%">
<figure xml:id="fig-neg-x-sq"><caption></caption><image source="images/fig-neg-x-sq"><description>negative x-sq and basic parabola</description></image></figure>
<tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
    <row bottom="minor">
        <cell><m>x</m></cell>
        <cell><m>y=x^2</m></cell>
        <cell><m>h(x)=-x^2</m></cell>
    </row>
    <row>
        <cell><m>-2</m></cell>
        <cell><m>4</m></cell>
        <cell><m>-4</m></cell>
    </row>
    <row>
        <cell><m>-1</m></cell>
        <cell><m>1</m></cell>
        <cell><m>-1</m></cell>
    </row>
    <row>
        <cell><m>0</m></cell>
        <cell><m>0</m></cell>
        <cell><m>0</m></cell>
    </row>
    <row>
        <cell><m>1</m></cell>
        <cell><m>1</m></cell>
        <cell><m>-1</m></cell>
    </row>
    <row>
        <cell><m>2</m></cell>
        <cell><m>4</m></cell>
        <cell><m>-4</m></cell>
    </row>
</tabular>
</sidebyside>
<p>
    The graph of <m>h(x) = -x^2</m> is flipped, or reflected, about the <m>x</m>-axis; the <m>y</m>-coordinate of each point on the graph of <m>y = x^2</m> is replaced by its opposite.
</p>

<p>
    In general, we have the following principles.
</p>

<assemblage><title>Scale Factors and Reflections</title>
<p>
    Compared with the graph of <m>y = f (x)</m>, the graph of <m>y = a f (x)</m>, where <m>a \ne 0</m>, is
    <ol>
        <li>stretched vertically by a factor of <m>\abs{a}</m> if <m>\abs{a}\gt 1</m>,</li>
        <li>compressed vertically by a factor of <m>\abs{a}</m> if <m>0\lt\abs{a}\lt 1</m>, and</li>
        <li>reflected about the <m>x</m>-axis if <m>a\lt 0</m>.</li>
    </ol>
</p>
</assemblage>

<example xml:id="example-scale">
    <p>
        Graph the following functions.
        <ol label="a">
            <li><m>g(x) = 3\sqrt[3]{x}</m></li>
            <li><m>h(x) =\dfrac{-1}{2}\abs{x}</m></li>
        </ol>
    </p>
    <solution>
        <ol label="a">
            <li>The graph of <m>g(x) = 3\sqrt[3]{x}</m> is a vertical expansion of the basic graph <m>y = \sqrt[3]{x}</m> by a factor of <m>3</m>, as shown in <xref ref="fig-scale-cube-root" text="type-global"/>. Each point on the basic graph has its <m>y</m>-coordinate tripled.
            <figure xml:id="fig-scale-cube-root"><caption></caption><image source="images/fig-scale-cube-root"  width="40%"><description>scale cube root</description></image></figure>
        </li>
        <li>
            The graph of <m>h(x) = \dfrac{-1}{2}\abs{x}</m> is a vertical compression of the basic graph <m>y = \abs{x}</m> by a factor of <m>\dfrac{1}{2}</m>, combined with a reflection about the <m>x</m>-axis. You may find it helpful to graph the function in two steps, as shown in <xref ref="fig-scale-abs" text="type-global"/>.
            <figure xml:id="fig-scale-abs"><caption></caption><image source="images/fig-scale-abs"  width="40%"><description>scale absolute value</description></image></figure>
        </li>
        </ol>
    </solution>
</example>

<exercise xml:id="scale-abs"><statement>
    <ol label="a">
        <li>Graph the function <m>f (x) = 2\abs{x}</m>.</li>
        <li>How is the graph of <m>f</m> different from the graph of <m>y =\abs{x}</m>?</li>
    </ol>
</statement>
<answer><p><ol label="a" cols="2">
    <li><p><image source="images/fig-in-ex-ans-2-3-6"  width="50%"><description>shifted absolute value</description></image> </p></li>
    <li><p>Stretch <m>y =\abs{x}</m> vertically by a factor of 2.</p></li>
</ol></p></answer>
</exercise>

<example xml:id="example-alcohol-level">
    <p>
        The function <m>A = f (t)</m> graphed in <xref ref="fig-surge-alcohol" text="type-global"/> gives a person's blood alcohol level <m>t</m> hours after drinking a martini. Sketch a graph of <m>g(t) = 2 f (t)</m> and explain what it tells you.
    </p>
    <figure xml:id="fig-surge-alcohol"><caption></caption><image source="images/fig-surge-alcohol"  width="45%"><description>alcohol level after alcohol</description></image></figure>
    <solution>
        <p>
            To sketch a graph of <m>g</m>, we stretch the graph of <m>f</m> vertically by a factor of <m>2</m>, as shown in <xref ref="fig-surge-alcohol2" text="type-global"/>. At each time <m>t</m>, the person’s blood alcohol level is twice the value given by <m>f</m>. The function <m>g</m> could represent a person's blood alcohol level <m>t</m> hours after drinking two martinis.
        </p>
    <figure xml:id="fig-surge-alcohol2"><caption></caption><image source="images/fig-surge-alcohol2"  width="50%"><description>alcohol level after more alcohol</description></image></figure>
    </solution>
</example>

<exercise xml:id="daylight-hours"><statement>
    <p>If the Earth were not tilted on its axis, there would be 12 daylight hours every day all over the planet. But in fact, the length of a day in a particular location depends on the latitude and the time of year. </p> 
    <p>The graph in <xref ref="fig-daylight-hours" text="type-global"/> shows <m>H = f (t)</m>, the length of a day in Helsinki, Finland, <m>t</m> days after January 1, and <m>R = g(t)</m>, the length of a day in Rome. Each is expressed as the number of hours greater or less than 12. Write a formula for <m>f</m> in terms of <m>g</m>. What does this formula tell you?</p>
    <figure xml:id="fig-daylight-hours"><caption></caption><image source="images/fig-daylight-hours"  width="75%"><description>daylight hours at two latitudes</description></image></figure>
</statement>
<answer><p><m>f (t)\approx 2g(t)</m>. On any given day, the number of daylight hours  varies from <m>12</m> hours about twice as much in Helsinki as it does in Rome.</p></answer>
</exercise>

</subsection>
<xi:include href="./summary-2-3.xml" /> <!-- summary  -->
<xi:include href="./section-2-3-exercises.xml" /> <!-- exercises  -->

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