1. The problem is to analyze and graph the function $$y = 5 \sin\left(\frac{5}{2}(x+2)\right) - 2$$. 2. The general form of a sine function is $$y = A \sin(B(x - C)) + D$$ where: - $A$ is the amplitude (height from the center line to peak), - $B$ affects the period (length of one cycle), - $C$ is the horizontal shift (phase shift), - $D$ is the vertical shift. 3. For our function: - Amplitude $A = 5$ (the sine wave oscillates 5 units above and below the midline), - $B = \frac{5}{2}$, - Horizontal shift $C = -2$ (since inside the sine is $x + 2$), - Vertical shift $D = -2$. 4. The period $T$ of a sine function is given by $$T = \frac{2\pi}{|B|}$$. Calculate: $$T = \frac{2\pi}{\frac{5}{2}} = 2\pi \times \frac{2}{5} = \frac{4\pi}{5}$$. 5. The graph oscillates between $-2 - 5 = -7$ and $-2 + 5 = 3$ vertically. 6. The phase shift moves the graph 2 units to the left. 7. The function can be graphed using these parameters to show the sine wave with amplitude 5, period $\frac{4\pi}{5}$, shifted left by 2, and shifted down by 2.