Determine the equation of the graph and the coordinates of its turning point.

The graph meets the x-axis at $x = -4$ and $x = 2$. These are the roots of the quadratic equation.
The general factored form of a quadratic equation is $y = a(x - x_1)(x - x_2)$, where $x_1$ and $x_2$ are the roots.
Substituting the given roots:

$$y = a(x - (-4))(x - 2)$$ $$y = a(x + 4)(x - 2)$$

From the graph, the parabola crosses the y-axis (where $x=0$) at $y = -8$. This is the point $(0, -8)$.
Substitute this point into the equation from Step 1:

$$-8 = a(0 + 4)(0 - 2)$$ $$-8 = a(4)(-2)$$ $$-8 = -8a$$ $$a = 1$$

Substitute $a=1$ back into the factored form and expand:

$$y = 1(x + 4)(x - 2)$$ $$y = x^2 - 2x + 4x - 8$$ $$y = x^2 + 2x - 8$$

The equation of the graph is $\boxed{y = x^2 + 2x - 8}$.

b) (ii) Coordinates of the turning point

For a quadratic equation in the form $y = ax^2 + bx + c$, the x-coordinate of the turning point (vertex) is given by the formula $x = \frac{-b}{2a}$.
From our equation $y = x^2 + 2x - 8$, we have $a=1$ and $b=2$.

$$x = \frac{-2}{2(1)}$$ $$x = -1$$

Substitute the x-coordinate ($x = -1$) back into the equation of the graph:

$$y = (-1)^2 + 2(-1) - 8$$ $$y = 1 - 2 - 8$$ $$y = -1 - 8$$ $$y = -9$$

The coordinates of the turning point are $\boxed{(-1, -9)}$.