Question

write the coordinates of the vertices after a translation 4 units right and 1 unit up.

Explanation:

Step1: Identify original coordinates

From the graph, vertex \( D \) has coordinates \( (-6, -3) \), and the other vertex (let's call it \( E \)) has coordinates \( (-4, -7) \). Wait, actually, looking at the graph, the two vertices: let's re - check. The first vertex (D) is at \( x=-6 \), \( y = - 3 \) (since it's 6 units left on x - axis and 3 units down on y - axis). The second vertex (let's say E) is at \( x=-4 \), \( y=-7 \) (4 units left on x - axis and 7 units down on y - axis).

Step2: Apply translation for x - coordinate

For a translation 4 units right, we add 4 to the x - coordinate of each vertex.
For vertex \( D(-6,-3) \): New \( x=-6 + 4=-2 \)
For vertex \( E(-4,-7) \): New \( x=-4 + 4 = 0 \)

Step3: Apply translation for y - coordinate

For a translation 1 unit up, we add 1 to the y - coordinate of each vertex.
For vertex \( D(-6,-3) \): New \( y=-3 + 1=-2 \)
For vertex \( E(-4,-7) \): New \( y=-7 + 1=-6 \)

Wait, maybe I misidentified the vertices. Let's re - examine the graph. The first vertex (the one labeled D) is at \( (-6, - 3) \) (x=-6, y=-3) and the other vertex (let's call it F) is at \( (-4, - 7) \) (x=-4, y=-7). After translation 4 units right (add 4 to x) and 1 unit up (add 1 to y):

For \( D(-6,-3) \):
New \( x=-6 + 4=-2 \)
New \( y=-3+1 = - 2 \)
So new coordinates of D: \( (-2,-2) \)

For the other vertex (let's say F) at \( (-4,-7) \):
New \( x=-4 + 4=0 \)
New \( y=-7 + 1=-6 \)
So new coordinates of F: \( (0,-6) \)

Wait, maybe there is a third vertex? Wait, the graph shows a triangle? Wait, maybe I missed a vertex. Wait, the first vertex is at \( (-6, - 3) \), then a vertical line down? Wait, no, the graph has two visible vertices? Wait, maybe the original vertices are \( D(-6,-3) \) and \( E(-4,-7) \), and maybe a third vertex? Wait, perhaps the original figure is a triangle with vertices, let's assume the three vertices: Wait, maybe the first vertex is \( (-6, - 3) \), the second is \( (-6, - 7) \) (vertical line) and the third is \( (-4, - 7) \). Let's correct that.

Let's re - identify the vertices correctly. Let's say the three vertices of the triangle are:
Vertex 1: \( A(-6,-3) \)
Vertex 2: \( B(-6,-7) \) (since it's directly below A, same x, y differs by 4)
Vertex 3: \( C(-4,-7) \) (to the right of B, same y)

Now apply translation: 4 units right (add 4 to x) and 1 unit up (add 1 to y)

For vertex \( A(-6,-3) \):
New \( x=-6 + 4=-2 \)
New \( y=-3 + 1=-2 \)
So \( A'(-2,-2) \)

For vertex \( B(-6,-7) \):
New \( x=-6 + 4=-2 \)
New \( y=-7 + 1=-6 \)
So \( B'(-2,-6) \)

For vertex \( C(-4,-7) \):
New \( x=-4 + 4=0 \)
New \( y=-7 + 1=-6 \)
So \( C'(0,-6) \)

Answer:

If the original vertices are \( D(-6,-3) \), \( E(-6,-7) \), \( F(-4,-7) \), the new coordinates after translation are \( (-2,-2) \), \( (-2,-6) \), \( (0,-6) \) respectively.

(If we consider only the two visible vertices: For the vertex at \( (-6,-3) \), new coordinates are \( (-2,-2) \); for the vertex at \( (-4,-7) \), new coordinates are \( (0,-6) \))