1. **State the problem:** We need to sketch the graph of the function $$f(x) = 2 \cos\left(x - \frac{\pi}{2}\right)$$ and understand its properties. 2. **Recall the cosine function properties:** The general form of a cosine function is $$f(x) = A \cos(B(x - C)) + D$$ where: - $A$ is the amplitude (height of peaks), - $B$ affects the period (length of one cycle), - $C$ is the horizontal shift (phase shift), - $D$ is the vertical shift. 3. **Identify parameters for our function:** - Amplitude $A = 2$ (so the graph oscillates between $-2$ and $2$), - $B = 1$ (so the period is $\frac{2\pi}{1} = 2\pi$), - Phase shift $C = \frac{\pi}{2}$ (shifted right by $\frac{\pi}{2}$), - Vertical shift $D = 0$ (centered on $y=0$). 4. **Period and key points:** The period is $2\pi$, so the function repeats every $2\pi$ units. Key points for cosine normally at $0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$ shift to the right by $\frac{\pi}{2}$: - Start at $x = \frac{\pi}{2}$ with $f\left(\frac{\pi}{2}\right) = 2 \cos(0) = 2$ (maximum), - At $x = \pi$, $f(\pi) = 2 \cos\left(\pi - \frac{\pi}{2}\right) = 2 \cos\left(\frac{\pi}{2}\right) = 0$, - At $x = \frac{3\pi}{2}$, $f\left(\frac{3\pi}{2}\right) = 2 \cos(\pi) = -2$ (minimum), - At $x = 2\pi$, $f(2\pi) = 2 \cos\left(\frac{3\pi}{2}\right) = 0$, - At $x = \frac{5\pi}{2}$, $f\left(\frac{5\pi}{2}\right) = 2 \cos(2\pi) = 2$ (maximum again). 5. **Sketch details:** - The graph oscillates between $-2$ and $2$. - The wave is shifted right by $\frac{\pi}{2}$. - The x-axis labels range from $-2\pi$ to $2\pi$. - The y-axis ranges from $-5$ to $5$ for clarity. **Final answer:** The graph is a cosine wave with amplitude 2, period $2\pi$, shifted right by $\frac{\pi}{2}$, centered on $y=0$.