quadratic

academic/misc/2024-math-calendar.html

@@ -347,7 +347,7 @@ <h1>Math Calendar 2024</h1>
         <td><a href="#jun04">04</a></td>
         <td><a href="#jun05">05</a></td>
         <td><a href="#jun06">06</a></td>
-        <td>07</td>
+        <td><a href="#jun07">07</a></td>
         <td>08</td>
         <td>09</td>
     </tr>
@@ -5070,6 +5070,19 @@ <h3 id="jun06">Jun 06</h3>
 \[\log_{10}15625+\log_{10}64=\log_{10}5^6+\log_{10}2^6=\log_{10}(5^6\cdot2^6)
 =\log_{10}10^6=6\]
 
+<h3 id="jun07">Jun 07</h3>
+
+\[\begin{align}&f(u)=u^2-2\ \text{and}\\&g(u)=(u-5)^2+3\\
+&\text{intersect at}\ (y,x)\end{align}\]
+
+<p>
+It should be more clearly stated, but it asks us to find the \(x\). To find the
+intersection point, we first find \(u\) such that \(f(u)=g(u)\).
+\[u^2-2=(u-5)^2+3\Rightarrow u^2-2=u^2-10u+25+3\Rightarrow10u=30\Rightarrow
+u=3\] Then we can evaluate \(f(3)=g(3)=7\) so the intersection point is
+\((3,7)\) so the answer would be 7.
+</p>
+
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