1. **State the problem:** We need to find a function of the form $f(x) = A \sec(Bx)$ that matches the given graph. 2. **Identify vertical asymptotes:** The graph has vertical asymptotes at $x = \frac{\pi}{6}$ and $x = \frac{\pi}{2}$. For $f(x) = A \sec(Bx)$, vertical asymptotes occur where $\cos(Bx) = 0$. 3. **Find $B$ using asymptotes:** The cosine function has zeros at $Bx = \frac{\pi}{2} + k\pi$, for integers $k$. Given two consecutive asymptotes at $x = \frac{\pi}{6}$ and $x = \frac{\pi}{2}$, the distance between them is: $$\frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}$$ The period between zeros of cosine is $\frac{\pi}{B}$, so: $$\frac{\pi}{B} = \frac{\pi}{3} \implies B = 3$$ 4. **Find amplitude $A$ using maximum value:** The graph peaks near $y = 6$ and dips near $y = -6$, so the amplitude $A = 6$. 5. **Write the function:** $$f(x) = 6 \sec(3x)$$ 6. **Summary:** - Vertical asymptotes at $x = \frac{\pi}{6}, \frac{\pi}{2}$ imply $B=3$. - Maximum and minimum values imply $A=6$. **Final answer:** $$f(x) = 6 \sec(3x)$$