Finding Gradients (Edexcel IGCSE Further Maths)

Spec point — 9A

Finding Gradients

How do I find the gradient of a curve at a point?

• 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 $\mathrm{f} ( x ) = x^{2} + 3 x - 4$

  • then $\mathrm{f} ' ( x ) = 2 x + 3$
  • the gradient of $y = \mathrm{f} ( x )$ when $x=1$ is $\mathrm{f}^{'} ( 1 ) = 2 ( 1 ) + 3 = 5$
  • the gradient of $y = \mathrm{f} ( x )$ when $x=-2$ is $\mathrm{f}^{'} ( - 2 ) = 2 ( - 2 ) + 3 = - 1$

• 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

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

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

This is a 'powers of $x$' derivative

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

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

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

Substitute $x=2$ into $\mathrm{f} ( x )$* *

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

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

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

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

Final answer: gradient$=-7$