## Question 8:

### Part (i)
$x = \frac{2}{3}$ and $y = \frac{1}{9}$ | B1, B1 **[2]** | $-1$ if any others given. Accept min 2 s.f. accuracy

### Part (ii)
Large positive $x$, $y \to \frac{1}{9}^+$ (e.g. consider $x = 100$) | M1 | Approaches horizontal asymptote, not inconsistent with their (i)

Large negative $x$, $y \to \frac{1}{9}^-$ (e.g. consider $x = -100$) | A1 | Correct approaches

| E1 **[3]** | Reasonable attempt to justify approaches

### Part (iii)
Curve with $x = \frac{2}{3}$ shown with correct approaches | B1(ft) | —

$y = \frac{1}{9}$ shown with correct approaches (from below on left, above on right) | B1(ft) | 1 for each branch, consistent with horizontal asymptote in (i) or (ii)

$(2, 0),\ (-2, 0)$ and $(0, -1)$ shown | B1(ft), B1, B1 **[5]** | Both $x$ intercepts and $y$ intercept. Give these marks if coordinates shown in workings, even if not shown on graph

### Part (iv)
$-1 = \frac{x^2-4}{(3x-2)^2} \Rightarrow -9x^2+12x-4 = x^2-4$ | — | —

$\Rightarrow 10x^2 - 12x = 0$ | — | —

$\Rightarrow 2x(5x-6) = 0$ | — | —

$\Rightarrow x = 0$ or $x = \frac{6}{5}$ | M1 | Reasonable attempt at solving inequality

From sketch: $y \geq -1$ for $x \leq 0$ and $x \geq \frac{6}{5}$ | A1 | Both values – give for seeing $0$ and $\frac{6}{5}$, even if inequalities are wrong

| B1 | For $x \leq 0$

| F1 **[4]** | Lose only one mark if any strict inequalities given

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