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      <text>11.1.2 Finding the gradient at a specific point</text>
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      <text><![CDATA[<h4>Finding the gradient at a specific point</h4>

Here, we shall consider a simple function to illustrate the calculation of a gradient. Look
at the graph of the function \(y(x) = x^2\) shown in the figure below. Notice that the gradient of the graph changes as we move from point to point. In some places the gradient is positive; at others it is negative. The gradient is greater at some points than at others. In fact the gradient changes from
point to point as we move along the curve.<p><br></p><p style="text-align: center;"><img src="@@PLUGINFILE@@/11-1-fig-4-x-squared.svg" alt="" width="352" height="267" role="presentation" class="img-responsive atto_image_button_text-bottom"><br></p><p style="text-align: center;"><br></p><p style="text-align: left;">Inspect the graph carefully and make the following observations:&nbsp;<br></p><p>(a) \(A\) is the point with coordinates \( (1, 1)\).&nbsp;</p><p>(b) \(B\) is the point with coordinates \( (4, 16)\).&nbsp;</p><p>(c)  We can calculate the gradient of the line \(AB\) from the formula
\[ \text{gradient} = \dfrac{ \text{difference between } y \text{ coordinates}}{ \text{difference between } x \text{ coordinates}} \]
Therefore the gradient of chord \(AB\) is equal to
\( \dfrac{16 - 1}{4 - 1} = \dfrac{15}{3} = 5\). The gradient of \(AB\) is not the same as the gradient of the graph at \(A\) but we can regard it as an approximation, or estimate of the gradient at \(A\).</p><p><br></p>

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We now carry the last task further by introducing point \(D\) at \( (2, 4)\) and point \(E\) at \( (1.5, 2.25) \) as shown in the figure below. The gradient of \(AD\) is found to be \(3\) and the gradient of \(AE\) is \(2.5\).<p><br></p>
<p style="text-align: center;"><img src="@@PLUGINFILE@@/11-1-fig-5-x-squared.svg" alt="" width="212" height="256" role="presentation" class="img-responsive atto_image_button_text-bottom"><br></p><p style="text-align: center;"><br></p><p style="text-align: left;"><br></p>

Observe that each time we carry out this procedure, and move the second point closer to \(A\), the gradient of the line drawn is getting closer and closer to the gradient of the tangent at \(A\). If we continue, the value we eventually obtain is the gradient of the tangent at \(A\) whose value is \(2\) as we will see shortly. This procedure illustrates how we define the gradient of the curve at \(A\).]]></text>
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