## Content

### The tangent line to a graph

Given a graph \(y= f(x)\), we have seen how to calculate the gradient of a tangent line to this graph. We can go further and find the equation of a tangent line.

Consider the tangent line to the graph \(y=f(x)\) at \(x=a\). This line has gradient \(f'(a)\) and passes through the point \((a, f(a))\). Once we know a point on the line and its gradient, we can write down its equation:

\[ y - f(a) = f'(a)\,(x-a). \](See the module Coordinate geometry .)

#### Example

Find the equation of the tangent line to the graph \(y=\dfrac{1}{2} x^2\) at \(x=3\).

#### Solution

Letting \(f(x) = \dfrac{1}{2} x^2\), we have \(f'(x) = x\), so \(f(3) = \dfrac{9}{2}\) and \(f'(3) = 3\). Thus the tangent line has gradient 3 and passes through \((3, \dfrac{9}{2})\), and is given by

\[ y - \dfrac{9}{2} = 3(x-3) \]or, equivalently,

\[ y = 3x - \dfrac{9}{2}. \]