A curve has the equation $y = 3(4 - x)^{3} + 5$.

1. Find the derivative of the curve equation

$\frac{dy}{dx} = 3 \cdot 3(4 - x)^2 \cdot (-1) = -9(4 - x)^2$

2. Set the derivative to zero to find the $x$-coordinate of the stationary point

For stationary points, $\frac{dy}{dx} = 0$
$-9(4 - x)^2 = 0$
$4 - x = 0$
$x = 4$

3. Find the corresponding $y$-coordinate

When $x = 4$,
$y = 3(4 - 4)^3 + 5 = 5$
The coordinates of the stationary point are $(4, 5)$.

1. Recall the first derivative from part (i)

From part (i), $\frac{dy}{dx} = -9(4 - x)^2$.
The stationary point is at $x = 4$.

2. Apply the first derivative test around x = 4

Let's test values of $x$ slightly less than and slightly greater than 4.
When $x = 3.9$, $\frac{dy}{dx} = -9(4 - 3.9)^2 = -9(0.1)^2 = -0.09 < 0$.
When $x = 4.1$, $\frac{dy}{dx} = -9(4 - 4.1)^2 = -9(-0.1)^2 = -0.09 < 0$.

3. Conclude the nature of the stationary point

Since the gradient is negative on both sides of the stationary point, the sign of the gradient does not change.
Therefore, the point $(4, 5)$ is a stationary point of inflexion.

1. Find the $y$-intercept

When $x = 0$,
$y = 3(4 - 0)^3 + 5$
$y = 3(64) + 5 = 197$
The $y$-intercept is $(0, 197)$.

2. Find the $x$-intercept

When $y = 0$,
$3(4 - x)^3 + 5 = 0$
$(4 - x)^3 = -\frac{5}{3}$
$4 - x = \sqrt[3]{-\frac{5}{3}}$
$x = 4 + \sqrt[3]{\frac{5}{3}} \approx 5.19$
The $x$-intercept is $(4 + \sqrt[3]{\frac{5}{3}}, 0)$.

3. Sketch the graph

Draw a cubic curve with a negative leading coefficient (decreasing overall). The curve passes through the $y$-axis at $197$, flattens out at the stationary point of inflexion $(4, 5)$, and crosses the $x$-axis at approximately $5.19$.