(i) The vertical asymptote occurs where the denominator of \( f(x) = 1 – \frac{1}{x-2} \) is zero:

\[ x – 2 = 0 \implies x = 2 \]

(ii) The horizontal asymptote is found by evaluating the limits as \( x \to \pm \infty \):

\[ \lim_{x \to \infty} f(x) = 1 – \frac{1}{x-2} \approx 1 – 0 = 1 \]

\[ \lim_{x \to -\infty} f(x) = 1 – \frac{1}{x-2} \approx 1 – 0 = 1 \]

Thus, the horizontal asymptote is \( y = 1 \).

Answer:
(i) \( x = 2 \)
(ii) \( y = 1 \)

(i) The \( y \)-intercept occurs at \( x = 0 \):

\[ f(0) = 1 – \frac{1}{0 – 2} = 1 – \left(-\frac{1}{2}\right) = 1 + \frac{1}{2} = \frac{3}{2} \]

Coordinates: \( \left(0, \frac{3}{2}\right) \).

(ii) The \( x \)-intercept occurs when \( f(x) = 0 \):

\[ 1 – \frac{1}{x – 2} = 0 \implies \frac{1}{x – 2} = 1 \implies x – 2 = 1 \implies x = 3 \]

Coordinates: \( (3, 0) \).

Answer:
(i) \( \left(0, \frac{3}{2}\right) \)
(ii) \( (3, 0) \)

Sketch the graph of \( y = f(x) \), incorporating:
– Vertical asymptote at \( x = 2 \).
– Horizontal asymptote at \( y = 1 \).
– Intercepts at \( \left(0, \frac{3}{2}\right) \) and \( (3, 0) \).
– Behavior: As \( x \to 2^- \), \( f(x) \to -\infty \); as \( x \to 2^+ \), \( f(x) \to \infty \). As \( x \to \pm \infty \), \( f(x) \to 1 \).