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      <text>10.2.15 (Example of finding normal to a curve)</text>
    </name>
    <questiontext format="moodle_auto_format">
      <text><![CDATA[<h3>Example</h3>

<p>Find the equation of the normal to the curve \(y = \frac{8}{\sqrt{x}}\) at the point where \(x = 4\).</p>

<h3>Solution</h3>

<p>When \(x = 4\), \(y = \frac{8}{\sqrt{4}} = \frac{8}{2} = 4\). Thus the point where the curve intersects the normal is \((4,4)\).</p>
<p>Now, \(f(x) = 8x^{1/2}\), thus \(f'(x) = -4x^{-3/2}\) and so \(f'(4) = -4 \cdot 4^{-3/2} = -\frac{1}{2}\).</p>
<p>Therefore the gradient of the normal at the point \((4,4)\) is given by 
\[
-\frac{1}{f'(4)} = -\frac{1}{-1/2} = 2.
\]</p>
<p>So, using the formula above, the equation of the normal is
\[
y = 2(x - 4) + 4 \implies y = 2x -4.
\]</p>]]></text>
    </questiontext>
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