1. Problem a: Find the new coordinates of point P after the transformation from $y=\tan x$ to $y=\tan 5x$. - Given original point $P=(45^\circ,1)$ on $y = \tan x$. - New function is $y = \tan 5x$. - The transformation compresses the graph horizontally by a factor of $1/5$. - The new $x$-coordinate is found by solving $5x = 45^\circ \Rightarrow x = \dfrac{45^\circ}{5} = 9^\circ$. - The $y$-coordinate remains the same for the corresponding $x$ in the transformed graph: $y = \tan(5 \times 9^\circ) = \tan 45^\circ = 1$. So the new coordinates are $P' = (9^\circ, 1)$. 2. Problem b: Find the new coordinates of point Q after the transformation from $y=\tan x$ to $y=5 \tan x$. - Given original point $Q=(225^\circ,1)$ on $y=\tan x$. - The transformation $y = 5 \tan x$ scales the $y$-values by 5. - The $x$-coordinate remains the same: $225^\circ$. - The new $y$-coordinate is $5 \times 1 = 5$. So the new coordinates are $Q' = (225^\circ, 5)$. 3. Problem c: Find the new coordinates of point R after the transformation from $y=\tan x$ to $y=\tan(x - 5^\circ)$. - Given original point $R = (-45^\circ, -1)$ on $y = \tan x$. - The transformation shifts the graph right by $5^\circ$. - To find new $x$, solve $x - 5^\circ = -45^\circ \Rightarrow x = -45^\circ + 5^\circ = -40^\circ$. - The $y$-coordinate remains as $y = \tan(x - 5^\circ) = \tan -45^\circ = -1$. So new coordinates are $R' = (-40^\circ, -1)$. 4. Problem d: Identify all points from the given set that lie on $g(x) = 3 \sin(2x) - 2$. - The function is $g(x) = 3 \sin(2x) - 2$. - We check each point: (1) $x = \dfrac{\pi}{2}$: $g(\dfrac{\pi}{2}) = 3 \sin(2 \times \dfrac{\pi}{2}) - 2 = 3 \sin(\pi) - 2 = 3 \times 0 - 2 = -2$ (does not match $y=1$) (2) $x= \pi$: $g(\pi) = 3 \sin(2\pi) - 2 = 0 - 2 = -2$ (does not match $y=4$ or $y=2$) (3) $x = \dfrac{\pi}{4}$: $g(\dfrac{\pi}{4}) = 3 \sin(2 \times \dfrac{\pi}{4}) - 2 = 3 \sin(\dfrac{\pi}{2}) - 2 = 3 \times 1 - 2 = 1$ (matches $y=-1$? No, value is 1, point says -1, so no) (4) $x= \dfrac{3\pi}{4}$: $g(\dfrac{3\pi}{4}) = 3 \sin(2 \times \dfrac{3\pi}{4}) - 2 = 3 \sin(\dfrac{3\pi}{2}) - 2 = 3 \times (-1) - 2 = -3 - 2 = -5$ (does not match $y=1$) (5) $x= \pi$: Same as before, $y=-2$. None of the points match the exact $y$-values given, but problem states selected points that belong to the graph include (\frac{\pi}{2},1), (\pi,4), (\frac{\pi}{4},-1), (\frac{3\pi}{4},1), (\pi,2). Since none match $g(x)$ exactly, no points belong to the graph of $g(x) = 3 \sin(2x) - 2$. Final Answer: - a: $P' = (9^\circ,1)$ - b: $Q' = (225^\circ,5)$ - c: $R' = (-40^\circ,-1)$ - d: No points from the given set lie exactly on $g(x) = 3 \sin(2x) - 2$.