The curve $C_1$ has equation $$y = x^3(x + 2)$$ (a) Find $\frac{dy}{dx}$ - Edexcel - A-Level Maths Pure - Question 9 - 2012 - Paper 1

To find where the curve meets the x-axis, we set $y = 0$:

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

The solutions are:

1. $x = 0$
2. $x + 2 = 0 \Rightarrow x = -2$

Thus, the points are $(0, 0)$ and $(-2, 0)$. The sketch should show these points and the overall shape of the curve, touching the x-axis at $(0, 0)$ and crossing it at $(-2, 0)$.

The gradient at the points where $C_1$ meets the x-axis is found by evaluating $\frac{dy}{dx}$ at those points:

1. At $x = 0$:
   $\frac{dy}{dx} = 4(0)^3 + 6(0)^2 = 0$
2. At $x = -2$:
   $\frac{dy}{dx} = 4(-2)^3 + 6(-2)^2 = 4(-8) + 6(4) = -32 + 24 = -8$

Therefore, the gradients are $0$ at $(0, 0)$ and $-8$ at $(-2, 0)$.

To find the x-intercepts, we set $y = 0$:

$(x - k^2)(x - k + 2) = 0$

This gives us the intercepts at:

1. $x = k^2$
2. $x = k - 2$

For the y-intercept, we set $x = 0$:

$y = (0 - k^2)(0 - k + 2) = -k^2(2 - k)$

The sketch should illustrate the intercepts at $(k^2, 0)$, $(k - 2, 0)$, and $(0, -k^2(2 - k))$. The overall shape of the curve should reflect that $k > 2$, indicating a downward-opening parabola.