Differentiate with respect to x

$\frac{d y}{d x} = - 3 x^{2} - 12 x - 9$

[1]

Any turning points are when the gradient ( $\frac{d y}{d x}$ ) is equal to zero

$- 3 x^{2} - 12 x - 9 = 0$

[1]

Use any quadratic method to solve for x. There is a common factor of -3 so start by dividing all terms by -3
(notice that it is a cubic graph, so there should be two turning points and this matches the fact that $\frac{d y}{d x} = 0$ is quadratic

$x^{2} + 4 x + 3 = 0 (x + 3) (x + 1) = 0 x = - 3, x = - 1$

[1]

We meed the full coordinates so substitute each x value into the original equation to find y

$x = - 3, y = 1 - {(- 3)}^{3} - 6 {(- 3)}^{2} - 9 (- 3) = 1$

therefore there is a point at $(- 3, 1)$

$x = - 1, y = 1 - {(- 1)}^{3} - 6 {(- 1)}^{2} - 9 (- 1) = 5$

therefore there is a point at $(- 1, 5)$

x = -3 is further left than x = -1. We can see from the diagram that the minimum point B is the point with the smallest (furthest left) x-value, where x = -3
Likewise, C is the maximum point where x = -1

B is the minimum point at $(- 3, 1)$ [1]
C is the maximum point at $(- 1, 5)$ [1]

Differentiation (Edexcel IGCSE Maths)

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