Solve the equation −(x+2)2 + 9 = 0 and find the coordinates of the x-intercept C and turning point A for the graph of y = 3x2 − 6x.

$$-(x+2)^2 = -9$$ $$(x+2)^2 = 9$$

$$x+2 = \pm\sqrt{9}$$ $$x+2 = \pm3$$

For $x+2 = 3$:

$$x = 3 - 2$$ $$x = 1$$

For $x+2 = -3$:

$$x = -3 - 2$$ $$x = -5$$

The solutions are $x=1$ and $x=-5$.

$$\boxed{x=1, x=-5}$$

$$3x = 0 \implies x = 0$$ $$x - 2 = 0 \implies x = 2$$

The graph shows that C is the x-intercept other than the origin. Therefore, the x-coordinate of C is 2.
The coordinates of C are $(2, 0)$.

$$\boxed{C=(2,0)}$$

b)(ii) Find the coordinates of A.
Point A is the turning point (vertex) of the parabola $y = 3x^2 - 6x$.
For a quadratic equation $y = ax^2 + bx + c$, the x-coordinate of the vertex is given by $x = \frac{-b}{2a}$.
In this equation, $a=3$, $b=-6$, and $c=0$.

$$x = \frac{-(-6)}{2(3)}$$ $$x = \frac{6}{6}$$ $$x = 1$$

$$y = 3(1)^2 - 6(1)$$ $$y = 3(1) - 6$$ $$y = 3 - 6$$ $$y = -3$$

The coordinates of A are $(1, -3)$.

$$\boxed{A=(1,-3)}$$