1. The problem asks to sketch the graph of the function $f(x) = \sec x$. 2. Recall that $\sec x = \frac{1}{\cos x}$. 3. Important rules: - $\sec x$ is undefined where $\cos x = 0$. - Vertical asymptotes occur at $x = \frac{\pi}{2} + k\pi$, where $k$ is any integer. - The graph has the same period as $\cos x$, which is $2\pi$. - $\sec x$ is positive where $\cos x$ is positive and negative where $\cos x$ is negative. 4. To sketch: - Identify vertical asymptotes at $x = \frac{\pi}{2} + k\pi$. - Between asymptotes, $\sec x$ forms U-shaped curves opening upwards or downwards depending on the sign of $\cos x$. - At $x = 2k\pi$, $\sec x = 1$ (minimum points). - At $x = (2k+1)\pi$, $\sec x = -1$ (maximum points). 5. Summary: - The graph has vertical asymptotes at $x = \frac{\pi}{2} + k\pi$. - The graph oscillates between $+\infty$ and $-\infty$ with period $2\pi$. Final answer: The graph of $f(x) = \sec x$ has vertical asymptotes at $x = \frac{\pi}{2} + k\pi$ and oscillates with period $2\pi$ following the reciprocal of $\cos x$.