1. The problem is to graph the function $y = \cos(3x + \frac{\pi}{2})$. 2. We start with the basic cosine function $y = \cos x$, which has a period of $2\pi$ and amplitude 1. 3. The function $y = \cos(3x + \frac{\pi}{2})$ involves two transformations: - A horizontal compression by a factor of $\frac{1}{3}$ due to the coefficient 3 inside the cosine. - A horizontal phase shift due to the $+ \frac{\pi}{2}$ inside the cosine. 4. The period of $y = \cos(3x)$ is $\frac{2\pi}{3}$ because period $= \frac{2\pi}{|b|}$ for $y=\cos(bx)$. 5. The phase shift is found by solving $3x + \frac{\pi}{2} = 0$ which gives $x = -\frac{\pi}{6}$. This means the graph shifts to the left by $\frac{\pi}{6}$. 6. So, the graph of $y = \cos(3x + \frac{\pi}{2})$ is the graph of $y = \cos(3x)$ shifted left by $\frac{\pi}{6}$. 7. In summary: - Start with $y = \cos x$. - Compress horizontally by factor 3 to get $y = \cos(3x)$. - Shift left by $\frac{\pi}{6}$ to get $y = \cos(3x + \frac{\pi}{2})$. 8. The amplitude remains 1, and the range is $[-1,1]$. Final answer: The graph is a horizontally compressed cosine wave with period $\frac{2\pi}{3}$, shifted left by $\frac{\pi}{6}$.