The equation of a curve is $y = 2x^3 + 6x^2 - 18x + 5$.

1. Find the first derivative of the curve equation.

$\frac{dy}{dx} = \frac{d}{dx}(2x^3 + 6x^2 - 18x + 5) = 6x^2 + 12x - 18$

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

For stationary points, $\frac{dy}{dx} = 0$:
$6x^2 + 12x - 18 = 0$
$x^2 + 2x - 3 = 0$
$(x + 3)(x - 1) = 0$
$x = -3$ or $x = 1$

3. Find the corresponding $y$-coordinates.

When $x = -3$, $y = 2(-3)^3 + 6(-3)^2 - 18(-3) + 5 = -54 + 54 + 54 + 5 = 59$
When $x = 1$, $y = 2(1)^3 + 6(1)^2 - 18(1) + 5 = 2 + 6 - 18 + 5 = -5$
The coordinates of the stationary points are $(-3, 59)$ and $(1, -5)$.

4. Find the second derivative to determine the nature of the stationary points.

$\frac{d^2y}{dx^2} = \frac{d}{dx}(6x^2 + 12x - 18) = 12x + 12$

5. Apply the second derivative test to each stationary point.

When $x = -3$, $\frac{d^2y}{dx^2} = 12(-3) + 12 = -24 < 0$.
Therefore, $(-3, 59)$ is a maximum point.

When $x = 1$, $\frac{d^2y}{dx^2} = 12(1) + 12 = 24 > 0$.
Therefore, $(1, -5)$ is a minimum point.

1. Find the expression for the gradient of the curve.

Let the gradient be $m$.
$m = \frac{dy}{dx} = 6x^2 + 12x - 18$

2. Differentiate the gradient expression and equate to zero to find the value of $x$ for the minimum gradient.

$\frac{dm}{dx} = \frac{d^2y}{dx^2} = 12x + 12$
For a minimum gradient, $\frac{dm}{dx} = 0$:
$12x + 12 = 0$
$x = -1$

3. Calculate the minimum gradient.

Substitute $x = -1$ into the gradient expression:
$m = 6(-1)^2 + 12(-1) - 18$
$m = 6 - 12 - 18 = -24$
(Note: $\frac{d^2m}{dx^2} = 12 > 0$, confirming it is a minimum.)