2. The point \( C(0,4) \) is translated to the right 2 units and down 4 units. a. Write a rule to describe the translation. b. What are the coordinates of the image point \( C^{\prime} \) ? \( \begin{array}{ll}\text { a. }(x, y) \rightarrow(x+2, y+4) ;(2,0) & \text { c. }(x, y) \rightarrow(x-2, y-4) ;(-2,8) \\ \text { b. }(x, y) \rightarrow(x+2, y-4) ;(2,0) & \text { d. }(x, y) \rightarrow(x-2, y+4) ;(-2,0)\end{array} \)

To translate the point \( C(0, 4) \) to the right 2 units, you add 2 to the x-coordinate. To move it down 4 units, you subtract 4 from the y-coordinate. Therefore, the rule for the translation is: \( (x, y) \rightarrow (x + 2, y - 4) \). Now, applying this rule to the original point \( C(0, 4) \), we get: \[ C' = (0 + 2, 4 - 4) = (2, 0). \] So, the coordinates of the image point \( C' \) are \( (2, 0) \). Now you have the translated point, and doesn't math feel like a fun little adventure sometimes?