A curve C has equation y = (2x² + x - 1) / (x-1). Find the equations of the asymptotes of C. Show that there is no point on C for which 1 < y < 9.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2014). The mark or mark type is shown in brackets.

1. [B1] Vertical asymptote is x = 1
2. [M1A1] Using polynomial division, y = (2x² + x - 1)/(x-1) = 2x + 3 + 2/(x-1). So, y = 2x + 3 is the oblique asymptote.
3. [M1A1] To show no points for 1 < y < 9: Rearrange the equation to x(y-1) = 2x² + x - 1, which leads to 2x² + (1 - y)x + (y - 1) = 0. For real roots of x, the discriminant must be non-negative: Δ = (1 - y)² - 4(2)(y - 1) ≥ 0.
4. [M1A1] (1 - y)² - 8(y - 1) ≥ 0 ⇒ y² - 2y + 1 - 8y + 8 ≥ 0 ⇒ y² - 10y + 9 ≥ 0. Factorising gives (y - 1)(y - 9) ≥ 0.
5. [M1A1] This implies y ≤ 1 or y ≥ 9. Therefore, there are no points on C for which 1 < y < 9. (AG) Thumbnail sketch, or similar, required.