Write a rule for \( g \) that represents a translation \( \frac{3}{2} \) units down and 1 unit right, followed by a reflection in the line \( y=-\frac{3}{2} \) of the graph of \( f(x)=\frac{1}{2} \) sin \( 6 x \). \( g(x)=\square \)
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To find the function \( g(x) \), we'll first incorporate the translation and then the reflection. 1. The function given is \( f(x) = \frac{1}{2} \sin(6x) \). Translating it 1 unit to the right modifies it to \( f(x-1) = \frac{1}{2} \sin(6(x-1)) = \frac{1}{2} \sin(6x - 6) \). 2. Translating it down by \( \frac{3}{2} \) results in: \[ f(x-1) - \frac{3}{2} = \frac{1}{2} \sin(6x - 6) - \frac{3}{2}. \] 3. Now, to reflect in the line \( y = -\frac{3}{2} \), we subtract from \(-\frac{3}{2}\) and then negate the entire function: The reflection transformation can be described as: \[ y' = -f(x) + b \quad \text{where} \quad b \text{ is the vertical offset.} \] For a reflection through \( y = -\frac{3}{2} \), we calculate as follows: \[ g(x) = - \left( \frac{1}{2} \sin(6x - 6) - \frac{3}{2} \right) - \frac{3}{2}. \] 4. Simplifying: \[ g(x) = -\frac{1}{2} \sin(6x - 6) + \frac{3}{2} - \frac{3}{2} = -\frac{1}{2} \sin(6x - 6). \] So, the final output is: \[ g(x) = -\frac{1}{2} \sin(6x - 6). \]