9. (a) Factorise completely x³ - 4x (b) Sketch the curve C with equation y = x³ - 4x, showing the coordinates of the points at which the curve meets the x-axis - Edexcel - A-Level Maths Pure - Question 10 - 2010 - Paper 2

To sketch the curve given by the equation $y = x^3 - 4x$, we first identify the points where it meets the x-axis by solving:

$x^3 - 4x = 0$

From our factorization, we have the roots at:

• x = 0
• x = 2
• x = -2

These points can be plotted on the graph, noting the shape of a cubic curve. The curve starts from the bottom left, passing through the x-axis at these points, with inflection in the regions around these x-coordinates.

We have points A (-1, y_A) and B (3, y_B) on the curve. We first calculate the y-coordinates:

For point A,
$y_A = (-1)^3 - 4(-1) = -1 + 4 = 3$
So, A is (-1, 3).

For point B,
$y_B = 3^3 - 4(3) = 27 - 12 = 15$
So, B is (3, 15).

Next, we find the slope (m) of the line connecting points A and B:

$m = \frac{y_B - y_A}{x_B - x_A} = \frac{15 - 3}{3 - (-1)} = \frac{12}{4} = 3$

Using point-slope form for the line equation: $y - y_A = m(x - x_A)$ Substituting in: $y - 3 = 3(x + 1)$ Expanding gives: $y = 3x + 6$

To find the length of segment AB, we use the distance formula:

$AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$ Substituting the coordinates of points A and B:

$AB = \sqrt{(3 - (-1))^2 + (15 - 3)^2}$ Calculating: $= \sqrt{(3 + 1)^2 + (12)^2}$
$= \sqrt{(4)^2 + (12)^2}$
$= \sqrt{16 + 144}$
$= \sqrt{160}$
$= 4\sqrt{10}$
This confirms that the length of AB can be expressed as:
$AB = k√10$ where k = 4.