1. The problem asks for the new coordinates of point P(2, -3) when transformed by the functions $y = f(2x)$ and $y = f(\frac{1}{4}x)$. 2. For transformations of the form $y = f(kx)$, the x-coordinate of any point on the curve changes according to $x' = \frac{x}{k}$. The y-coordinate remains the same because the function value is evaluated at the new x. 3. (a) For $y = f(2x)$, $k=2$, so the new x-coordinate is $x' = \frac{2}{2} = 1$. The y-coordinate remains $-3$. Thus, the transformed point is $(1, -3)$. 4. (b) For $y = f(\frac{1}{4}x)$, $k=\frac{1}{4}$, so the new x-coordinate is $x' = \frac{2}{\frac{1}{4}} = 2 \times 4 = 8$. The y-coordinate remains $-3$. Thus, the transformed point is $(8, -3)$.