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PR1 parentheses fix (#380)
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AbbyANoble authored Oct 22, 2024
1 parent a7ac207 commit 57d1b95
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8 changes: 4 additions & 4 deletions source/precalculus/source/04-PR/01.ptx
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<ol marker="A." cols="1">
<li><p> The minimum value appears to occur near <m> (0, 8) </m>. </p></li>
<li><p> The minimum value appears to occur near <m> (-\dfrac {1}{5}, 10) </m>. </p></li>
<li><p> The minimum value appears to occur near <m> \left(-\dfrac {1}{5}, 10\right) </m>. </p></li>
<li><p> The minimum value appears to occur near <m> (2, -4) </m>. </p></li>
<li><p> There is no minimum value of this function.</p></li>
</ol>
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<ol marker="A." cols="1">
<li><p> The maximum value appears to occur near <m> (-2, 44) </m>. </p></li>
<li><p> The maximum value appears to occur near <m> (-\dfrac {1}{5}, 10) </m>. </p></li>
<li><p> The maximum value appears to occur near <m> \left(-\dfrac {1}{5}, 10\right) </m>. </p></li>
<li><p> The maximum value appears to occur near <m> (2, -4) </m>. </p></li>
<li><p> There is no maximum value of this function.</p></li>
</ol>
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<observation xml:id="standard-form-aos">
<statement>
<p> Just as with the vertex form of a quadratic, we can use the standard form of a quadratic to find the <term>axis of symmetry</term> <idx><h>quadratic function</h><h>axis of symmetry</h></idx> and the <term>vertex</term> <idx><h>quadratic function</h><h>vertex from standard form</h></idx>by using the values of <m>a, b </m>, and <m> c </m>. Given the standard form of a quadratic, the axis of symmetry is the vertical line <m> x=\dfrac {-b}{2a}</m> and the vertex is at the point <m> (\dfrac{-b}{2a},f(\dfrac{-b}{2a}))</m>.
<p> Just as with the vertex form of a quadratic, we can use the standard form of a quadratic to find the <term>axis of symmetry</term> <idx><h>quadratic function</h><h>axis of symmetry</h></idx> and the <term>vertex</term> <idx><h>quadratic function</h><h>vertex from standard form</h></idx>by using the values of <m>a, b </m>, and <m> c </m>. Given the standard form of a quadratic, the axis of symmetry is the vertical line <m> x=\dfrac {-b}{2a}</m> and the vertex is at the point <m> \left(\dfrac{-b}{2a},f\left(\dfrac{-b}{2a}\right)\right)</m>.
</p>
</statement>
</observation>
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<title>Videos</title>
<p>It would be great to include videos down here, like in the Calculus book!</p>
</subsection> -->
</section>
</section>

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