Find the coordinates of the stationary points on the graph of $y = 2x^3 - 9x^2 + 13$. Hence, determine the nature of the stationary point(s).

1. Find the first derivative of the function

$\frac{dy}{dx} = \frac{d}{dx}(2x^3 - 9x^2 + 13) = 6x^2 - 18x$

2. Set the derivative to zero to find the $x$-coordinates of the stationary points

For stationary points, $\frac{dy}{dx} = 0$:
$6x^2 - 18x = 0$
$6x(x - 3) = 0$
$x = 0 \text{ or } x = 3$

3. Find the corresponding $y$-coordinates

When $x = 0$:
$y = 2(0)^3 - 9(0)^2 + 13 = 13$

When $x = 3$:
$y = 2(3)^3 - 9(3)^2 + 13$
$y = 2(27) - 9(9) + 13$
$y = 54 - 81 + 13 = -14$

4. State the coordinates of the stationary points

The coordinates of the stationary points are $(0, 13)$ and $(3, -14)$.

1. Find the second derivative of the function

From part (a), $\frac{dy}{dx} = 6x^2 - 18x$.
$\frac{d^2y}{dx^2} = 12x - 18$

2. Determine the nature of the stationary point at $x = 0$

At $x = 0$, $\frac{d^2y}{dx^2} = 12(0) - 18 = -18$.
Since $\frac{d^2y}{dx^2} < 0$, the point $(0, 13)$ is a maximum point.

3. Determine the nature of the stationary point at $x = 3$

At $x = 3$, $\frac{d^2y}{dx^2} = 12(3) - 18 = 18$.
Since $\frac{d^2y}{dx^2} > 0$, the point $(3, -14)$ is a minimum point.