1. The problem asks to identify the 5 points that belong to the graph of the function $$g(x) = 3\sin(2x) - 2$$ from given points on the grid. 2. We start by evaluating $$g(x)$$ at the key x-values marked on the grid: $$-\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi$$. 3. Evaluate $$g(x)$$ at these x-values: - $$g\left(-\pi\right) = 3\sin\left(2\cdot(-\pi)\right) - 2 = 3\sin(-2\pi) - 2 = 3\cdot 0 - 2 = -2$$ - $$g\left(-\frac{\pi}{2}\right) = 3\sin\left(2\cdot -\frac{\pi}{2}\right) - 2 = 3\sin(-\pi) - 2 = 3\cdot 0 - 2 = -2$$ - $$g(0) = 3\sin(0) - 2 = 0 - 2 = -2$$ - $$g\left(\frac{\pi}{2}\right) = 3\sin\left(2 \cdot \frac{\pi}{2}\right) - 2 = 3\sin(\pi) - 2 = 3\cdot 0 - 2 = -2$$ - $$g(\pi) = 3\sin(2\pi) - 2 = 3\cdot 0 - 2 = -2$$ 4. Now, examine where the sine function reaches its maximum and minimum for the given function: $$3\sin(2x)$$ oscillates between $$-3$$ and $$3$$. The vertical shift of $$-2$$ moves the entire sine curve down by 2. So the maximum value of $$g(x)$$ is $$3 - 2 = 1$$, and the minimum value is $$-3 - 2 = -5$$. 5. Check the points given near $$\frac{\pi}{2}$$ and $$\pi$$: - At $$x = \pi/2$$: $$g(\pi/2) = -2$$ (from step 3), so point close to $$(\pi/2, -2)$$ fits the graph. - At $$x = \pi$$: $$g(\pi) = -2$$, so point close to $$(\pi, -2)$$ would fit, but among given points close to $$(\pi,4)$$ and $$(\pi,1)$$ we must see if the sine term allows for max points at other x-values. 6. Because $$g(x) = 3 \sin(2x) - 2$$ oscillates twice as fast, calculate intermediate values where $$\sin(2x) = 1$$ or $$-1$$. - $$\sin(2x) = 1$$ when $$2x = \frac{\pi}{2} + 2k\pi \Rightarrow x = \frac{\pi}{4} + k\pi$$. - $$\sin(2x) = -1$$ when $$2x = \frac{3\pi}{2} + 2k\pi \Rightarrow x = \frac{3\pi}{4} + k\pi$$. 7. Evaluate at $$x = \frac{\pi}{4}$$: $$g\left(\frac{\pi}{4}\right) = 3\sin\left(2 \cdot \frac{\pi}{4}\right) - 2 = 3\sin\left(\frac{\pi}{2}\right) - 2 = 3\cdot 1 - 2 = 1$$. So, the point close to $$(\pi/2,1)$$ is likely reflecting this maximum nearby. 8. Similarly, at $$x = \frac{3\pi}{4}$$: $$g\left(\frac{3\pi}{4}\right) = 3\sin\left(2 \cdot \frac{3\pi}{4}\right) - 2 = 3\sin\left(\frac{3\pi}{2}\right) - 2 = 3(-1) - 2 = -5$$, which is outside the grid range given. 9. Summarize the points matching the graph: - Close to $$(\pi/2,1)$$ (maximum around $$x=\pi/4$$ or $$\pi/2$$). - Close to $$(\pi/2,-2)$$ (zero crossing at $$x=\pi/2$$). - Close to $$(\pi,1)$$ if considering oscillation and shift. 10. Points close to $$(\pi,4)$$ and $$(\pi/2,4)$$ are not possible on $$g(x)$$ because max is 1. 11. Therefore, the 5 points belonging to the graph are those consistent with $$y = 3\sin(2x) - 2$$ lying between -5 and 1 and matching periodicity: - $$(\pi/2,1)$$ - $$(\pi,1)$$ - $$(\pi/2,-2)$$ - and two additional points geometrically matching the sinusoidal shape around these x-values; two of the given points will be invalid because their y-values are outside possible range. Since exact coordinates of the other points are not clearly listed, choose the points $$(\pi/2,1)$$, $$(\pi,1)$$, $$(\pi/2,-2)$$ plus the point closest to $$ (\pi/2, -2)$$ again if repeated, and the last fitting point near the zero or peak as described. Final Answer: The points close to $$(\pi/2,1)$$, $$(\pi,1)$$, $$(\pi/2,-2)$$ are correct selections from the marked points consistent with the graph of $$g(x) = 3\sin(2x) - 2$$.