曲线 $C$ 的参数方程为
$x = t + 2e^t, \quad y = 2t + e^{-t}, \quad t > -2.$

1. Find the coordinates of the start point of the curve.

Evaluate $x$ and $y$ at the boundary $t = -2$:
$x = -2 + 2e^{-2}$
$y = -4 + e^{-(-2)} = -4 + e^2$
The curve starts at the point $(-2 + 2e^{-2}, -4 + e^2)$. Since $t > -2$, this point should be represented with a hollow (open) circle.

2. Find the exact coordinates of the minimum turning point.

Differentiate $y$ with respect to $t$ to locate the stationary point:
$\frac{dy}{dt} = 2 - e^{-t}$
Set $\frac{dy}{dt} = 0$:
$2 - e^{-t} = 0 \implies e^{-t} = 2 \implies t = -\ln 2$
Substitute $t = -\ln 2$ into $x$ and $y$:
$x = -\ln 2 + 2e^{-\ln 2} = -\ln 2 + 2\left(\frac{1}{2}\right) = 1 - \ln 2$
$y = 2(-\ln 2) + e^{\ln 2} = -2\ln 2 + 2 = 2 - 2\ln 2$
The minimum point is exactly $(1 - \ln 2, 2 - 2\ln 2)$.

3. Determine the coordinates of the $y$-intercept using a graphing calculator.

At the $y$-intercept, $x = 0$:
$t + 2e^t = 0$
Using a graphing calculator, we find the root to be $t \approx -0.85260$.
Substitute this value into $y$:
$y = 2(-0.85260) + e^{0.85260} \approx 0.641$
The $y$-intercept is $(0, 0.641)$.

4. Sketch the curve with all key features labelled.

Plot the curve on the $x$-$y$ plane:

• Mark an open circle at the start point $(-2 + 2e^{-2}, -4 + e^2)$.
• Draw the curve passing downwards through the $y$-axis at $(0, 0.641)$.
• Ensure the curve reaches a minimum at $(1 - \ln 2, 2 - 2\ln 2)$ and then increases smoothly as $x$ increases.

1. Find the coordinates of the point $P$ where $t = 0$.

Substitute $t = 0$ into the parametric equations:
$x = 0 + 2e^0 = 2$
$y = 2(0) + e^{-0} = 1$
The coordinates of $P$ are $(2, 1)$.

2. Find the derivatives of $x$ and $y$ with respect to $t$.

$\frac{dx}{dt} = 1 + 2e^t$
$\frac{dy}{dt} = 2 - e^{-t}$

3. Calculate the gradient of the tangent at $t=0$.

At $t = 0$:
$\frac{dx}{dt} = 1 + 2(1) = 3$
$\frac{dy}{dt} = 2 - 1 = 1$
Using the chain rule, the gradient of the tangent is:
$m_t = \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1}{3}$

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

The gradient of the normal is:
$m_n = -\frac{1}{m_t} = -3$
Using the point-gradient formula for the normal at $P(2, 1)$:
$y - 1 = -3(x - 2)$
$y - 1 = -3x + 6$
$y = -3x + 7$