The graph shows part of the curve $y = -3(2x-4)^{-1/3}$. The point $P(-2,\ 1.5)$ lies on the curve and the normal to the curve at $P$ meets the x-axis at $Q$.

1. Differentiate the equation of the curve to find $\frac{dy}{dx}$.

$y = -3(2x-4)^{-\frac{1}{3}}$

$\frac{dy}{dx} = -3 \left(-\frac{1}{3}\right)(2x-4)^{-\frac{4}{3}} \cdot (2)$

$\frac{dy}{dx} = 2(2x-4)^{-\frac{4}{3}}$

2. Find the gradient of the tangent at $P(-2, 1.5)$.

Substitute $x = -2$ into $\frac{dy}{dx}$:
$\frac{dy}{dx} = 2(2(-2)-4)^{-\frac{4}{3}}$
$\frac{dy}{dx} = 2(-8)^{-\frac{4}{3}}$
$\frac{dy}{dx} = 2 \left(\frac{1}{(-8)^{\frac{4}{3}}}\right) = 2 \left(\frac{1}{16}\right) = \frac{1}{8}$

3. Find the gradient of the normal at $P$.

Gradient of normal $= -\frac{1}{\text{Gradient of tangent}}$
Gradient of normal $= -\frac{1}{\frac{1}{8}} = -8$

4. Find the equation of the normal and the coordinates of $Q$.

Equation of normal at $P(-2, 1.5)$:
$y - 1.5 = -8(x - (-2))$
$y - 1.5 = -8(x + 2)$

Since $Q$ lies on the x-axis, substitute $y = 0$:
$0 - 1.5 = -8(x + 2)$
$-1.5 = -8x - 16$
$8x = -14.5$
$x = -\frac{14.5}{8} = -1.8125$

Coordinates of $Q$ are $(-1.8125, 0)$.

1. Find the gradient of the curve at $P(-2, 1.5)$ to determine the equation of the normal.

$\frac{dy}{dx} = -3 \left(-\frac{1}{3}\right)(2x-4)^{-4/3}(2) = 2(2x-4)^{-4/3}$
At $x = -2$, $\frac{dy}{dx} = 2(-4-4)^{-4/3} = 2(-8)^{-4/3} = 2 \left(\frac{1}{(-2)^4}\right) = \frac{2}{16} = \frac{1}{8}$

2. Find the coordinates of $Q$ (from part a).

Gradient of normal $= -\frac{1}{1/8} = -8$
Equation of normal at $P(-2, 1.5)$:
$y - 1.5 = -8(x - (-2))$
$y = -8x - 16 + 1.5 = -8x - 14.5$
At $x$-axis, $y = 0$:
$0 = -8x - 14.5 \implies x = -\frac{14.5}{8} = -1.8125$
So $Q$ is $(-1.8125, 0)$.

3. Calculate the area of the triangular region bounded by the normal, the line $x=-2$, and the $x$-axis.

Base of triangle $= -1.8125 - (-2) = 0.1875$
Height of triangle $= 1.5$
Area of triangle $= \frac{1}{2} \times 0.1875 \times 1.5 = 0.140625$

4. Calculate the area under the curve from $x = -2$ to $x = 0$.

$\int_{-2}^{0} -3(2x-4)^{-1/3} dx = \left[ \frac{-3(2x-4)^{2/3}}{\frac{2}{3} \times 2} \right]_{-2}^{0}$
$= \left[ -\frac{9}{4}(2x-4)^{2/3} \right]_{-2}^{0}$
$= -\frac{9}{4}(-4)^{2/3} - \left( -\frac{9}{4}(-8)^{2/3} \right)$
$= -\frac{9}{4}(16)^{1/3} + \frac{9}{4}(4)$
$= 9 - \frac{9}{4}(2^{4/3}) \approx 3.33036$

5. Find the total shaded area.

Total Area $= 0.140625 + 3.33036 = 3.470985 \approx 3.471$ (to 3 d.p.)