Question: 3. (a) Sketch the graph of the
(i) rational function $y = -8/(x^2-4)$ and state clearly the vertical and horizontal asymptotes, the
domain and the range [4]
(ii) logarithmic function $y = \log_4(x - 7) + 9$ and state the domain and range [4]
(b) Solve the equation $\ln(6 - 13x) = 2 \ln x + \ln 5$ [2]
(c) Find the conjugate of $(i\sqrt{-9}+7i)/(1+\sqrt{-1})$ [3]
(d) Simplify $((3-\sqrt{2})^3)/(1+\sqrt{2})$, giving your answer in the form $a + b\sqrt{2}$ where $a, b \in \mathbb{Z}$ [3]
(e) (i) Expand $(1 - x/2)^{1/3}$ in ascending powers of $x$, up to and including the term in $x^3$ [3]
(ii) Use your expansion to evaluate $\sqrt[3]{0.999}$ , giving your answer to 6 decimal places [3]
(f) A ball is dropped from a height of 10 meters. Each time it hits the ground, it bounces to
80% of its previous height. Find the total distance travelled by the ball. [3]
[Total: 25 Marks]
1. **Problem Statement:**
(i) Sketch the graph of the rational function $$y = \frac{-8}{x^2 - 4}$$ and find vertical and horizontal asymptotes, domain, and range.
2. **Vertical Asymptotes:**
The denominator is zero when $$x^2 - 4 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$$.
So vertical asymptotes at $$x = -2$$ and $$x = 2$$.
3. **Horizontal Asymptote:**
As $$x \to \pm \infty$$, $$y \to 0$$ because degree of denominator is higher than numerator.
So horizontal asymptote is $$y = 0$$.
4. **Domain:**
All real numbers except where denominator is zero:
$$\text{Domain} = \{x \in \mathbb{R} : x \neq -2, x \neq 2\}$$.
5. **Range:**
Since $$y = \frac{-8}{x^2 - 4}$$, the denominator $$x^2 - 4$$ can take all real values except zero.
- When $$x^2 - 4 > 0$$ (i.e., $$|x| > 2$$), denominator positive, so $$y = -8/(positive) < 0$$.
- When $$x^2 - 4 < 0$$ (i.e., $$-2 < x < 2$$), denominator negative, so $$y = -8/(negative) > 0$$.
- As $$x \to \pm 2$$, $$y \to \pm \infty$$.
- As $$x \to \pm \infty$$, $$y \to 0^-$$.
Hence, range is all real numbers except zero:
$$\text{Range} = \{y \in \mathbb{R} : y \neq 0\}$$.
6. **Graph Sketch:**
- Vertical asymptotes at $$x = -2$$ and $$x = 2$$.
- Horizontal asymptote at $$y = 0$$.
- Function negative outside $$(-2,2)$$ and positive inside.
- Function approaches infinity near vertical asymptotes.
**Final answers:**
- Vertical asymptotes: $$x = -2, 2$$
- Horizontal asymptote: $$y = 0$$
- Domain: $$x \neq \pm 2$$
- Range: $$y \neq 0$$