Find $\frac{{{\text{d}}y}}{{{\text{d}}x}}$.

Determine the equation of the normal to the curve at the point $x = 3$ in the form $ax + by + c = 0$ where $a,{\text{ }}b,{\text{ }}c \in \mathbb{Z}$.

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = {(1 - x)^{ - 2}}\;\;\;\left( { = \frac{1}{{{{(1 - x)}^2}}}} \right)$     (M1)A1

[2 marks]

gradient of Tangent $= \frac{1}{4}$     (A1)

gradient of Normal $=  - 4$     (M1)

$y + \frac{1}{2} =  - 4(x - 3)$ or attempt to find $c$ in $y = mx + c$     M1

$8x + 2y - 23 = 0$     A1

[4 marks]

Total [6 marks]