The curve C₁ has equation $$y = x^3(x + 2)$$ (a) Find \(\frac{dy}{dx}\) - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 1

To find where C₁ meets the x-axis, we set (y = 0):

$x^3(x + 2) = 0$.

This gives:

1. (x = 0)
2. (x + 2 = 0 \Rightarrow x = -2)

Therefore, C₁ meets the x-axis at the points ((0, 0)) and ((-2, 0)). The sketch should indicate that the curve intersects the x-axis at these points, with appropriate curvature.

We have already calculated (\frac{dy}{dx} = 4x^3 + 6x^2). Now, let’s evaluate the gradient at the x-intercepts.

1. At (x = 0): $\frac{dy}{dx} = 4(0)^3 + 6(0)^2 = 0$

2. At (x = -2): $\frac{dy}{dx} = 4(-2)^3 + 6(-2)^2 = 4(-8) + 6(4) = -32 + 24 = -8$

Thus, the gradients of C₁ at the points ((0, 0)) and ((-2, 0)) are 0 and -8 respectively.

To find the points where C₂ meets the axes, we will set (y = 0) and evaluate:

1. For the x-intercepts: $0 = (x - k^2)(x - k + 2)$ This gives intercepts at (x = k^2) and (x = k - 2).

2. For the y-intercept, set (x = 0): $y = (0 - k^2)(0 - k + 2) = k^2(k - 2)$ Therefore, the coordinates of where C₂ meets the axes should be noted as ((k^2, 0)) and ((k - 2, 0)) and ((0, k^2(k - 2))). The sketch should reflect a horizontal translation.