The parametric equations of a curve $C_1$ are $x=3t^3+2t^2$ and $y=-t^3+2t^2+1$ for $-1 \le t \le 1$. Another curve $C_2$ has equation $x^2+(y-1)^2=k$ where $k\in\mathbb{R}^+$.

1. Find the exact coordinates of the endpoints of $C_1$.

Substitute $t = -1$ and $t = 1$ into the parametric equations:

When $t = -1$:
$x = 3(-1)^3 + 2(-1)^2 = -3 + 2 = -1$
$y = -(-1)^3 + 2(-1)^2 + 1 = 1 + 2 + 1 = 4$
Endpoint 1 is $(-1, 4)$.

When $t = 1$:
$x = 3(1)^3 + 2(1)^2 = 3 + 2 = 5$
$y = -(1)^3 + 2(1)^2 + 1 = -1 + 2 + 1 = 2$
Endpoint 2 is $(5, 2)$.

2. Find the exact coordinates of the axial intercepts.

For $y$-intercepts, set $x = 0$:
$3t^3 + 2t^2 = 0$
$t^2(3t + 2) = 0$
$t = 0$ or $t = -\frac{2}{3}$ (both fall within the domain $[-1, 1]$)

When $t = 0$, $y = -(0)^3 + 2(0)^2 + 1 = 1$.
When $t = -\frac{2}{3}$, $y = -(-\frac{2}{3})^3 + 2(-\frac{2}{3})^2 + 1 = \frac{8}{27} + \frac{8}{9} + 1 = \frac{59}{27}$.
The $y$-intercepts are $(0, 1)$ and $(0, \frac{59}{27})$.

For $x$-intercepts, set $y = 0$:
$-t^3 + 2t^2 + 1 = 0 \implies t^3 - 2t^2 - 1 = 0$
Using a graphing calculator, the only real root is $t \approx 2.21$, which is outside the domain $[-1, 1]$.
Thus, there are no $x$-intercepts.

3. Determine the shape of the curve and sketch it.

Using a graphing calculator in parametric mode for $-1 \le t \le 1$:

• The curve begins at $(-1, 4)$.
• It moves downwards, crossing the $y$-axis at $(0, \frac{59}{27})$.
• It slightly enters the first quadrant, then curves back to form a sharp cusp at the $y$-intercept $(0, 1)$ (where $t=0$).
• Finally, it sweeps upwards and to the right to terminate at $(5, 2)$.

(Sketch the continuous curve accordingly, clearly marking the exact coordinates found in Steps 1 and 2.)

1. Identify the geometrical shape, its centre, and its radius from the Cartesian equation.

The equation $x^2 + (y-1)^2 = k$ is of the form $(x-a)^2 + (y-b)^2 = r^2$, which represents a circle.

Centre $(a, b) = (0, 1)$
Radius $r = \sqrt{k}$ (since $k \in \mathbb{R}^+$)

1. Substitute the parametric equations of $C_1$ into the Cartesian equation of $C_2$.

$(3t^3 + 2t^2)^2 + (-t^3 + 2t^2 + 1 - 1)^2 = 16$

2. Expand and simplify the equation into a single polynomial in $t$.

$9t^6 + 12t^5 + 4t^4 + (-t^3 + 2t^2)^2 = 16$
$9t^6 + 12t^5 + 4t^4 + t^6 - 4t^5 + 4t^4 = 16$
$10t^6 + 8t^5 + 8t^4 - 16 = 0$
$5t^6 + 4t^5 + 4t^4 - 8 = 0$

3. Solve for $t$ using a graphing calculator, accounting for the restricted domain.

Using a graphing calculator to solve $5t^6 + 4t^5 + 4t^4 - 8 = 0$:
$t \approx 0.90825 \quad \text{or} \quad t \approx -1.25$
Since the domain for $C_1$ is $-1 \le t \le 1$, we reject $t \approx -1.25$.
$t = 0.90825$

4. Substitute $t$ back into the parametric equations to find the coordinates of the intersection point.

$x = 3(0.90825)^3 + 2(0.90825)^2 \approx 3.90$
$y = -(0.90825)^3 + 2(0.90825)^2 + 1 \approx 1.90$
The coordinates of the intersection point are $(3.90, 1.90)$ (to 3 s.f.).

1. Relate intersections to the geometrical properties of $C_1$ and $C_2$.

From part (b), $C_2$ describes a family of circles centred at $(0, 1)$ with radius $\sqrt{k}$.
The number of intersections corresponds to the number of points on $C_1$ that are at a squared distance of $k$ from $(0, 1)$.

2. Determine the squared distances from the centre of $C_2$ to key points on $C_1$.

From part (a), $C_1$ originates from $(0,1)$ at $t=0$, and its endpoints are $(-1, 4)$ at $t=-1$ and $(5, 2)$ at $t=1$.
Evaluating the squared distance from $(0,1)$ to the endpoints:
At $t=-1$, distance$^2 = (-1-0)^2 + (4-1)^2 = 10$.
At $t=1$, distance$^2 = (5-0)^2 + (2-1)^2 = 26$.
As $t$ moves away from $0$ towards $-1$ or $1$, the distance strictly increases monotonically along the curve's two "branches".

3. Deduce the possible number of intersections as $k$ varies.

Since $k \in \mathbb{R}^+$, $C_2$ has radius $> 0$.

• For $0 < k \le 10$, the circle intersects both the left branch and the right branch of $C_1$, giving $2$ points of intersection.
• For $10 < k \le 26$, the circle only intersects the longer right branch (from $t=0$ to $t=1$), giving $1$ point.
• For $k > 26$, the circle is entirely outside the span of $C_1$, giving $0$ points.
  Hence, the possible numbers of points of intersection are 0, 1, and 2.