Finding Gradients (Edexcel IGCSE Further Pure Maths)

Revision Note

Written by: Roger B

Reviewed by: Dan Finlay

The gradient of a curve at a point is the gradient of the tangent to the curve at that point
- Find the gradient by substituting the value of $x$ at that point into the derivative of the curve
For example, if $f (x) = x^{2} + 3 x - 4$
- then $f' (x) = 2 x + 3$
- the gradient of $y = f (x)$ when $x = 1$ is $f^{'} (1) = 2 (1) + 3 = 5$
- the gradient of $y = f (x)$ when $x = - 2$ is $f^{'} (- 2) = 2 (- 2) + 3 = - 1$
Your exam calculator cannot find a derivative function in terms of $x$
- But it may be able to find the numerical value of a derivative at a point
- You can use this to check your work

A function $f$ is defined by $f (x) = x^{3} - 6 x^{2} + 5 x - 12$ .

(a) Find $f^{'} (x)$ .

This is a 'powers of $x$ ' derivative

$f^{'} (x) = 3 x^{3 - 1} - 6 (2 x^{2 - 1}) + 5$

$f^{'} (x) = 3 x^{2} - 12 x + 5$

(b) Show that the curve $y = f (x)$ goes through the point $(2, - 18)$ , and find the gradient of the tangent to the curve at that point.

Substitute $x = 2$ into $f (x)$

$\begin{array}{rcl} f (2) & = & {(2)}^{3} - 6 {(2)}^{2} + 5 (2) - 12 \\ = & 8 - 24 + 10 - 12 \\ = & - 18 \end{array}$

$f (2) = - 18$ so the curve goes through $(2, - 18)$

Now substitute $x = 2$ into $f^{'} (x)$ to find the gradient

$\begin{array}{rcl} f' (2) & = & 3 {(2)}^{2} - 12 (2) + 5 \\ = & 12 - 24 + 5 \\ = & - 7 \end{array}$

gradient $= - 7$

Last updated: 20 October 2024

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