1. **State the problem:** We need to determine the equation of a trigonometric function based on its graph. 2. **Analyze the graph features:** The graph shows two repeating vertical-curving branches with vertical asymptotes at $x = -4\pi$, $x = -2\pi$, $x = 0$, and $x = 2\pi$. This pattern suggests a function with vertical asymptotes at multiples of $\pi$ or $2\pi$. 3. **Identify the function type:** Vertical asymptotes at regular intervals and the shape of the graph resembling cubic-like arcs suggest the function is a tangent or cotangent function, which have vertical asymptotes where their denominators are zero. 4. **Recall the formulas:** - Tangent function: $$y = a \tan(bx + c) + d$$ with vertical asymptotes where $bx + c = \frac{\pi}{2} + k\pi$, $k \in \mathbb{Z}$. - Cotangent function: $$y = a \cot(bx + c) + d$$ with vertical asymptotes where $bx + c = k\pi$, $k \in \mathbb{Z}$. 5. **Match asymptotes:** The asymptotes are at $x = -4\pi, -2\pi, 0, 2\pi$, which are multiples of $2\pi$. This matches the cotangent function's asymptotes at $k\pi$ if $b=\frac{1}{2}$, because then vertical asymptotes occur at $x$ where $\frac{1}{2}x = k\pi \Rightarrow x = 2k\pi$. 6. **Write the function:** The function is likely $$y = a \cot\left(\frac{x}{2}\right) + d$$. 7. **Determine $a$ and $d$:** The graph is symmetric about the origin and has no vertical shift, so $d=0$. The amplitude $a$ affects the steepness; assuming standard cotangent shape, $a=1$. 8. **Final equation:** $$ y = \cot\left(\frac{x}{2}\right) $$ This matches the given asymptotes and the shape of the graph.