Part (a): Set the given expression for y equal to -2 and solve for x. Part (b): For the function y = x(x − 2), find (i) the coordinates where the curve cuts the x-axis and (ii) the minimum value of th

$$1 - 2x = 0 \implies 2x = 1 \implies x = \frac{1}{2}$$ $$1 + x = 0 \implies x = -1$$

The values of $x$ are $\boxed{x = \frac{1}{2} \text{ or } x = -1}$.

Set the function equal to zero.

$$x(x - 2) = 0$$

The coordinates are $\boxed{(0, 0) \text{ and } (2, 0)}$.

b)(ii) What is the maximum value of the function?
The function is $y = x(x - 2) = x^2 - 2x$. This is a parabola that opens upwards (since the coefficient of $x^2$ is positive). Therefore, it has a minimum value, not a maximum value (the function increases indefinitely towards positive infinity). Assuming the question meant to ask for the minimum value:

For a quadratic $y = ax^2 + bx + c$, the x-coordinate of the vertex is $x = -\frac{b}{2a}$.
Here, $a=1$ and $b=-2$.

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

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

The minimum value of the function is $\boxed{-1}$.