Figure 2 shows part of the curve C with equation $$y = (x - 1)(x^2 - 4)$$ The curve cuts the x-axis at the points P, (1, 0) and Q, as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 4 - 2006 - Paper 1

To differentiate the function $y = (x - 1)(x^2 - 4)$:

1. First, expand the equation: $y = x^3 - 4x - x^2 + 4 = x^3 - x^2 - 4x + 4$

2. Differentiate using the power rule: $\frac{dy}{dx} = 3x^2 - 2x - 4$.

At the point (-1, 6), we first find the slope of the tangent using the derivative:

1. Substitute x = -1 into $\frac{dy}{dx}$: $\frac{dy}{dx} = 3(-1)^2 - 2(-1) - 4 = 3 + 2 - 4 = 1$ (slope m = 1).

2. Use the point-slope form of the line: $y - y_1 = m(x - x_1)$ $y - 6 = 1(x + 1)$ Simplifying gives: $y = x + 7$.

Since the slope at (-1, 6) is 1, the slope of the tangent at R must also be 1. Using the derivative:

1. Set $\frac{dy}{dx} = 1$: $3x^2 - 2x - 4 = 1$ $3x^2 - 2x - 5 = 0$

2. Solve for x using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 3, b = -2, c = -5$: $x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3} = \frac{2 \pm \sqrt{4 + 60}}{6} = \frac{2 \pm \sqrt{64}}{6} = \frac{2 \pm 8}{6}$ Possible values: $x = \frac{10}{6} = \frac{5}{3} \text{ or } x = \frac{-6}{6} = -1$.

3. Substitute back into the equation for y: If $x = \frac{5}{3}$: $y = (\frac{5}{3} - 1)(\left(\frac{5}{3}\right)^2 - 4)$ Calculate to find the exact coordinates of R.

Continuing from the previous step:

1. Substitute $x = \frac{5}{3}$ into the original equation: $y = (\frac{5}{3} - 1)(\left(\frac{5}{3}\right)^2 - 4)$ This simplifies to: $= (\frac{2}{3})(\frac{25}{9} - 4)$ $= (\frac{2}{3})(\frac{25}{9} - \frac{36}{9})$ $= (\frac{2}{3})(\frac{-11}{9}) = \frac{-22}{27}$.

Thus, the exact coordinates of R are $\left(\frac{5}{3}, \frac{-22}{27}\right)$.