2019 H2 Maths Paper 2, Question 2

Factorising the denominator,

so $y = \frac{2 - x}{(3x + 8)(x - 1)}$ is undefined at $x = - \frac{8}{3}$ and at $x = 1$, giving the vertical asymptotes

Since $\deg\left( \text{numerator} \right) = 1 < 2 = \deg\left( \text{denominator} \right)$, $y \rightarrow 0$ as $x \rightarrow \pm \infty$, so

is a horizontal asymptote.

Axis crossings: when $x = 0$, $y = \frac{2}{- 8} = - \frac{1}{4}$, giving $\left( 0,\ - \frac{1}{4} \right)$; when $y = 0$, $2 - x = 0$, giving $x = 2$, so $(2,\ 0)$.

To know which way each branch turns, we differentiate by the quotient rule:

Setting the numerator to $0$ and solving (by GC), the turning points are at $x = 2 \pm \frac{\sqrt{42}}{3}$, i.e. at

both to 3 s.f. This gives three branches: one to the left of $x = - \frac{8}{3}$ rising from near $y = 0$ to $\text{+}\infty$; one between the asymptotes, from $- \infty$ up to the maximum $( - 0.160, - 0.248)$ and back down to $- \infty$; and one to the right of $x = 1$, falling from $\text{+}\infty$ through $(2,0)$ to the minimum $(4.16, - 0.0334)$ and rising back towards $y = 0$.

Explanation: reading off the sketch in (i), the curve lies above the $x$-axis exactly on the branch $x < - \frac{8}{3}$, and on the right-hand branch between the asymptote $x = 1$ and the intercept $x = 2$; everywhere else the curve lies on or below the $x$-axis.

so $\frac{x - 2}{3x^{2} + 5x - 8} > 0$ exactly where $\frac{2 - x}{3x^{2} + 5x - 8} < 0$, i.e. on the complement of the set found in (ii) (excluding the asymptotes and the intercept $x = 2$, where the expression is $0$ or undefined).