8. (a) Factorise completely 9x - 4x³ (b) Sketch the curve C with equation y = 9x - 4x³ Show on your sketch the coordinates at which the curve meets the x-axis - Edexcel - A-Level Maths Pure - Question 10 - 2015 - Paper 1

To sketch the curve, we will find the x-intercepts by setting y = 0:

$0 = 9x - 4x^3$

This is the same as factoring we did in part (a). The x-intercepts occur when:

$x(3 - 2x)(3 + 2x) = 0$

Setting each factor equal to zero, we solve:

1. $x = 0$
2. $$$$

ightarrow x = rac{3}{2}$3.$3 + 2x = 0 ightarrow x = -rac{3}{2}$$

Thus, the curve meets the x-axis at the coordinates (0, 0), (1.5, 0), and (-1.5, 0). While sketching, the curve will exhibit a cubic behavior, starting from the left below the x-axis, crossing at these intercepts, and continuing upwards.

Be sure to label these coordinates accordingly on your graph.

To find the length of AB, we first determine the coordinates of points A and B where:

• A has an x-coordinate of -2,
• B has an x-coordinate of 1.

Now we find the corresponding y-coordinates:

1. For A: $y(-2) = 9(-2) - 4(-2)^3 = -18 + 32 = 14$ Thus, point A is at (-2, 14).

2. For B: $y(1) = 9(1) - 4(1)^3 = 9 - 4 = 5$ Thus, point B is at (1, 5).

Now, we apply the distance formula to find the length of AB:

$AB = ext{distance} = riangle x^2 + riangle y^2$

where:

• $riangle x = x_B - x_A = 1 - (-2) = 3$
• $riangle y = y_B - y_A = 5 - 14 = -9$

Thus, we have:

$AB = ext{sqrt}[(3)^2 + (-9)^2] = ext{sqrt}[9 + 81] = ext{sqrt}[90] = ext{sqrt}[9 imes 10] = 3 ext{sqrt}[10]$

This shows that length AB can be written as:

$AB = k ext{sqrt}[10]$

where k = 3.