(a) (i) Ans: 2
From the graph, when \( x = 2 \), the corresponding \( y \)-value is \( 2 \).
(ii) Ans: \( x = -2, 1, 2 \)
The graph intersects the horizontal line \( y = 5 \) at \( x = -2 \), \( x = 1 \), and \( x = 2 \).
(iii) Ans: 4
The graph touches \( y = 4 \) at exactly two points, so \( k = 4 \).
(iv) Ans: asymptote
The line \( x = 0 \) (the y-axis) is a vertical asymptote because the graph approaches infinity as \( x \) approaches 0.
(b) (i)
Plot the line \( y = x – 2 \) by connecting points like \((-3, -5)\) and \((3, 1)\).
(ii) Ans: \( x = -1 \)
The intersection of \( f(x) \) and \( y = x – 2 \) occurs at \( x = -1 \).
(c) Ans: \( c = 2 \)
Using the point \( (1, -1) \) on the graph, substitute into \( f(x) = x^2 – \frac{c}{x} \): \(-1 = 1 – c \), solving gives \( c = 2 \).
(d) Ans: \( p = -1 \), \( q = -2 \)
Substitute \( f(x) = x^2 – \frac{2}{x} \) into \( f(x) = x – 2 \), rearrange to get \( x^3 – x^2 – 2x = 2 \), so \( p = -1 \) and \( q = -2 \).
