图示为曲线 $y = 3e^{-\frac{1}{2}x} - 6$ 的一部分，该曲线与 $x$ 轴相交于点 $P$，与 $y$ 轴相交于点 $Q$。曲线在点 $Q$ 处的法线与 $x$ 轴交于点 $R$。求

1. Find the coordinates of $Q$ (y-intercept)

At $Q$, $x = 0$.
$y = 3e^{-\frac{1}{2}(0)} - 6 = 3(1) - 6 = -3$.
So $Q$ is $(0, -3)$.

2. Differentiate the equation of the curve to find the gradient function

$\frac{dy}{dx} = 3 \left(-\frac{1}{2}\right) e^{-\frac{1}{2}x} = -\frac{3}{2}e^{-\frac{1}{2}x}$

3. Find the gradient of the tangent and normal at $Q$

At $Q$ ($x = 0$),
Gradient of tangent, $m_t = -\frac{3}{2}e^0 = -\frac{3}{2}$.
Gradient of normal, $m_n = -\frac{1}{m_t} = \frac{2}{3}$.

4. Find the equation of the normal

Equation of normal at $(0, -3)$ with gradient $\frac{2}{3}$:
$y - (-3) = \frac{2}{3}(x - 0)$
$y = \frac{2}{3}x - 3$

1. Find the coordinates of P (x-intercept of the curve)

At $P$, $y = 0$:
$3e^{-\frac{1}{2}x} - 6 = 0$
$e^{-\frac{1}{2}x} = 2$
$-\frac{1}{2}x = \ln 2$
$x = -2\ln 2$
So $P$ is $(-2\ln 2, 0)$.

2. Find the coordinates of Q (y-intercept of the curve)

At $Q$, $x = 0$:
$y = 3e^{0} - 6 = 3 - 6 = -3$
So $Q$ is $(0, -3)$.

3. Find the gradient of the curve at Q and the equation of the normal

$\frac{dy}{dx} = 3(-\frac{1}{2})e^{-\frac{1}{2}x} = -\frac{3}{2}e^{-\frac{1}{2}x}$
At $x = 0$, $\frac{dy}{dx} = -\frac{3}{2}$.
Gradient of normal at $Q$ is $\frac{2}{3}$.
Equation of normal: $y - (-3) = \frac{2}{3}(x - 0) \implies y = \frac{2}{3}x - 3$.

4. Find the coordinates of R (x-intercept of the normal)

At $R$, $y = 0$:
$\frac{2}{3}x - 3 = 0 \implies x = \frac{9}{2}$.
So $R$ is $(\frac{9}{2}, 0)$.

5. Calculate the area bounded by the curve, x-axis and y-axis (from P to O)

Area 1 = $\int_{-2\ln 2}^{0} (0 - y) \, dx = -\int_{-2\ln 2}^{0} (3e^{-\frac{1}{2}x} - 6) \, dx$
$= -\left[ \frac{3e^{-\frac{1}{2}x}}{-1/2} - 6x \right]_{-2\ln 2}^{0}$
$= -\left[ -6e^{-\frac{1}{2}x} - 6x \right]_{-2\ln 2}^{0}$
$= -\left( (-6 - 0) - (-6e^{\ln 2} - 6(-2\ln 2)) \right)$
$= -\left( -6 - (-12 + 12\ln 2) \right)$
$= - (6 - 12\ln 2) = 12\ln 2 - 6$

6. Calculate the area of the triangle OQR and the total shaded area

Area of triangle $OQR = \frac{1}{2} \times OR \times OQ = \frac{1}{2} \times \frac{9}{2} \times 3 = \frac{27}{4}$.
Total shaded area = Area 1 + Area of triangle $OQR$
$= (12\ln 2 - 6) + \frac{27}{4}$
$= 12\ln 2 + \frac{3}{4}$