The curve $C$ is defined by the parametric equations $x = t^2$ and $y = 2t + t^3$, where $-1 \le t \le 1$.

1. Find the coordinates of the endpoints and intercepts by evaluating the parametric equations at key values of $t$.

Evaluate $x = t^2$ and $y = 2t + t^3$ at the boundaries and at $t=0$:
When $t = -1$: $x = (-1)^2 = 1$, $y = 2(-1) + (-1)^3 = -3$. Endpoint is $(1, -3)$.
When $t = 0$: $x = 0^2 = 0$, $y = 2(0) + 0^3 = 0$. Origin is $(0, 0)$.
When $t = 1$: $x = 1^2 = 1$, $y = 2(1) + 1^3 = 3$. Endpoint is $(1, 3)$.

2. Determine the gradient of the curve to identify its behaviour at the origin.

$\frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 2 + 3t^2$
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2 + 3t^2}{2t}$
As $t \to 0$, the denominator approaches $0$ while the numerator approaches $2$. Therefore, $\frac{dy}{dx} \to \pm\infty$, which indicates that the curve has a vertical tangent at the origin $(0,0)$.

3. Sketch the graph using the identified key features.

Observe that replacing $t$ with $-t$ gives $x(-t) = x(t)$ and $y(-t) = -y(t)$, showing the curve is symmetric about the $x$-axis.
Draw a continuous, smooth curve starting at $(1, -3)$ in the fourth quadrant, curving up to pass through the origin $(0,0)$ where it is vertical (tangent to the $y$-axis), and continuing symmetrically into the first quadrant to end exactly at $(1, 3)$.

1. Differentiate the parametric equations with respect to $t$.

$x = t^2 \implies \frac{dx}{dt} = 2t$

$y = 2t + t^3 \implies \frac{dy}{dt} = 2 + 3t^2$

2. Find the gradient of the tangent, $\frac{dy}{dx}$, at $t = \frac{1}{2}$.

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2 + 3t^2}{2t}$

When $t = \frac{1}{2}$,

$\frac{dy}{dx} = \frac{2 + 3\left(\frac{1}{2}\right)^2}{2\left(\frac{1}{2}\right)} = \frac{2 + \frac{3}{4}}{1} = \frac{11}{4}$

3. Calculate the coordinates of the point at $t = \frac{1}{2}$.

$x = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$

$y = 2\left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)^3 = 1 + \frac{1}{8} = \frac{9}{8}$

4. Determine the gradient of the normal and formulate the equation of the normal line.

Gradient of normal $m = -\frac{1}{\frac{dy}{dx}} = -\frac{4}{11}$

Using the point-slope form $y - y_1 = m(x - x_1)$:

$y - \frac{9}{8} = -\frac{4}{11}\left(x - \frac{1}{4}\right)$

$y = -\frac{4}{11}x + \frac{1}{11} + \frac{9}{8}$

$y = -\frac{4}{11}x + \frac{8}{88} + \frac{99}{88}$

$y = -\frac{4}{11}x + \frac{107}{88}$