Consider: $$ f(x) = -2 an \left( \frac{3}{2} x \right) $$ 6.1 Write down the period of $f$ - NSC Mathematics - Question 6 - 2018 - Paper 2

To find the general solution for $t$, we equate $2$ to $-2\tan\left(\frac{3}{2}t\right)$. Rearranging gives:
$an\left(\frac{3}{2}t\right) = -1$
The solutions for $an(x) = -1$ occur at:
$x = 135^{\circ} + k \cdot 180^{\circ} \quad \text{or} \quad x = 315^{\circ} + k \cdot 180^{\circ}, \, k \in \mathbb{Z}$
Substituting $x = \frac{3}{2}t$:

$\frac{3}{2}t = 315^{\circ} + k \cdot 180^{\circ} \Rightarrow t = 210^{\circ} + k \cdot 240^{\circ} \quad (2)$
Thus, the general solutions are:
$t = 90^{\circ} + k \cdot 120^{\circ} \quad \text{and} \quad t = 210^{\circ} + k \cdot 240^{\circ}, \, k \in \mathbb{Z}$.

To sketch the graph of $f(x)$, we identify the key features:

• Asymptotes: Occur where $\tan\left( \frac{3}{2} x \right)$ is undefined. This happens at:
  $\frac{3}{2}x = 90^{\circ} + k \cdot 180^{\circ}\Rightarrow x = 60^{\circ} + k \cdot 120^{\circ}, \ k \in \mathbb{Z}$
  Calculating for $k = -1$ gives $x = -60^{\circ}$, and for $k = 0$, $x = 60^{\circ}$ and $k = 1$, $x = 180^{\circ}$.
• X-Intercept: Set $f(x) = 0$: the x-intercept is at $x = 0^{\circ}$ where $f(0) = 0$.
• Y-Intercept: Evaluate $f(0)$ which is $0$.
• Shape: The function will be negative in the intervals between asymptotes. The graph should depict all asymptotes, intercepts, and necessary endpoints along the given interval.

To determine where $f(x) \geq 2$, observe the graph to find the region above the line $y = 2$. This will typically occur between the asymptotes and will be visually accessible by checking the intervals. The key x-values can be identified where the graph crosses or intersects the line $y = 2$. One must check the segments between the asymptotes within the specified range to locate these intersections.

The transformation from $f$ to $g$ can be broken down into two main effects:

1. Horizontal Shift: The term $+60^{\circ}$ indicates a shift of the graph to the left by $40^{\circ}$ due to the inner function modification of $an$.
2. Vertical Stretch/Reflection: The negative coefficient $-2$ reflects the graph across the x-axis and vertically stretches it by a factor of $2$. Thus, the graph $g$ is derived from $f$ by a leftward translation and reflection.