2. On the same diagram, draw the graphs of \( y=|3 x| \) and \( y=|x-3| \) for the domain \( -2 \leq x \leq 3 \)\n\nHence solve the equation \( |3 x|=|x-3| \).

To draw the graphs for the functions \( y = |3x| \) and \( y = |x - 3| \) over the domain \( -2 \leq x \leq 3 \), we can start by evaluating key points: For \( y = |3x| \): - When \( x = -2 \), \( y = |3 \cdot (-2)| = 6 \) - When \( x = 0 \), \( y = |3 \cdot 0| = 0 \) - When \( x = 1 \), \( y = |3 \cdot 1| = 3 \) - When \( x = 3 \), \( y = |3 \cdot 3| = 9 \) For \( y = |x - 3| \): - When \( x = -2 \), \( y = |-2 - 3| = 5 \) - When \( x = 0 \), \( y = |0 - 3| = 3 \) - When \( x = 3 \), \( y = |3 - 3| = 0 \) Now, plotting these points, we observe that the graph of \( y = |3x| \) is a V-shape that opens upward with its vertex at the origin. The graph of \( y = |x - 3| \) is another V-shape that opens upward with its vertex at \( x = 3 \). To solve the equation \( |3x| = |x - 3| \), we consider two cases based on the properties of absolute values: 1. **Case 1**: \( 3x = x - 3 \) Rearranging gives: \[ 2x = -3 \implies x = -\frac{3}{2} \] 2. **Case 2**: \( 3x = -(x - 3) \) Simplifying this results in: \[ 3x = -x + 3 \implies 4x = 3 \implies x = \frac{3}{4} \] Next, verify if these values lie within the given domain \( -2 \leq x \leq 3 \): - \( x = -\frac{3}{2} \) is within the range. - \( x = \frac{3}{4} \) is also within the range. Thus, the solutions to the equation \( |3x| = |x - 3| \) are \( x = -\frac{3}{2} \) and \( x = \frac{3}{4} \).