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 the point $A(1 ; 2)$ on the graph of $f$, we first substitute the point into the equation:
$-2 \tan\left(\frac{3}{2} \cdot 1\right) = 2.$
This simplifies to:
$\tan\left(\frac{3}{2}\right) = -1.$
We equate this to the standard tangent solutions:
$t = 135^{\circ} + k \cdot 180^{\circ} , k \in \mathbb{Z} \; \text{or} \; t = 315^{\circ} + k \cdot 180^{\circ}, k \in \mathbb{Z}$.

The graph of the function $f(x)$ from $x = -120^{\circ}$ to $x = 180^{\circ}$ should include:

• Asymptotes at $x = 60^{\circ}$ and $x = 180^{\circ}$, where the function is undefined.
• The $y$-intercept at $(0, 0)$ as $f(0) = -2 \tan(0) = 0$.
• The general shape of the graph shows a negative slope between the asymptotes with the intercepts and endpoints clearly marked.

From the graph, we find that $f(x) \geq 2$ occurs when:

• $x$ is in the interval where the graph is above the line $y = 2$.
  This occurs specifically in the intervals $[-60^{\circ}, -30^{\circ}]$ and $[60^{\circ}, 90^{\circ}]$.

The transformation from the graph of $f(x) = -2 \tan\left(\frac{3}{2} x\right)$ to $g(x) = -2 \tan\left(\frac{3}{2} x + 60^{\circ}\right)$ involves:

• A horizontal translation of $40^{\circ}$ to the left, which shifts the entire graph horizontally.
• This transformation skews the graph and affects the placement of the asymptotes and intercepts.