Question

write the coordinates of the vertices after a translation 3 units up.

Explanation:

Step1: Identify original coordinates

First, find the original coordinates of vertices \( C \), \( D \), and \( E \) from the graph.

• \( C \): Let's assume from the grid, \( C = (5, -9) \) (wait, maybe better to re - check. Wait, the y - axis: looking at the graph, the lower part, \( C \) is at \( x = 5 \), \( y=-9 \)? Wait, no, maybe I misread. Wait, the grid: the positive y is up, negative down. Let's see, point \( D \) is at \( (5,1) \)? Wait, no, the blue dot for \( D \) is at \( x = 5 \), \( y = 1 \)? Wait, no, the vertical line from \( x = 5 \), and \( D \) is at \( y = 1 \), \( C \) is at \( x = 5 \), \( y=-9 \), and \( E \) is at \( x = 9 \), \( y=-9 \). Wait, maybe the original coordinates:

Let's correctly identify:

• \( D \): \( (5, 1) \) (since it's on \( x = 5 \), \( y = 1 \))
• \( C \): \( (5, -9) \) (same \( x = 5 \), \( y=-9 \))
• \( E \): \( (9, -9) \) ( \( x = 9 \), \( y=-9 \))

Step2: Apply translation (3 units up)

Translation 3 units up means we add 3 to the y - coordinate of each vertex. The translation rule for a point \( (x,y) \) is \( (x,y + 3) \).

• For \( D=(5,1) \): New \( y = 1+3 = 4 \), so \( D'=(5,4) \)
• For \( C=(5,-9) \): New \( y=-9 + 3=-6 \), so \( C'=(5,-6) \)
• For \( E=(9,-9) \): New \( y=-9 + 3=-6 \), so \( E'=(9,-6) \)

Wait, maybe I made a mistake in original coordinates. Let's re - examine the graph. The y - axis: the top is 10, bottom is - 10. The point \( D \) is at \( x = 5 \), \( y = 1 \) (since it's 1 unit above the x - axis). Point \( C \) is at \( x = 5 \), \( y=-9 \) (9 units below x - axis), and \( E \) is at \( x = 9 \), \( y=-9 \). So when we translate 3 units up, we add 3 to y - coordinate.

Answer:

If the original vertices are \( C(5, - 9) \), \( D(5,1) \), \( E(9, - 9) \), then after translation 3 units up:
\( C'(5,-6) \), \( D'(5,4) \), \( E'(9,-6) \)