## 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,

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

Interactive 2 ##### Exercise 17

What is the equation of the tangent line to the graph of $$y = \sqrt{9 - x^2}$$ at $$x = \dfrac{3 \sqrt{2}}{2}$$?

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