Question

graph the image of parallelogram qrst after a translation 3 units right and 5 units down.

Explanation:

Step1: Identify original coordinates

First, find the coordinates of each vertex of parallelogram \( QRST \):

• \( Q(-8, -1) \)
• \( R(0, -1) \)
• \( S(1, 6) \) (Wait, correction: From the graph, \( S \) is at \( (1, 6) \)? Wait, no, looking at the grid: \( S \) is at \( (1, 6) \)? Wait, the y - axis: \( S \) is on \( y = 6 \), x - coordinate: the grid lines, \( R \) is at \( (0, -1) \), \( S \) is at \( (1, 6) \)? Wait, no, let's re - check:

Looking at the graph:

• \( Q \): x = - 8, y=-1 (since it's on the line \( y=-1 \), x=-8)
• \( R \): x = 0, y=-1 (on \( y = - 1 \), x = 0)
• \( T \): x=-7, y = 6 (since it's horizontal from \( S \), \( S \) is at \( (1, 6) \)? Wait, no, the horizontal line from \( T \) to \( S \): \( T \) is at \( x=-7 \), \( y = 6 \), \( S \) is at \( x = 1 \), \( y = 6 \)? Wait, maybe I misread. Let's do it properly:

From the graph:

• \( Q \): (-8, -1) (x=-8, y=-1)
• \( R \): (0, -1) (x = 0, y=-1)
• \( S \): (1, 6)? No, wait, the vertical line from \( R \) to \( S \): \( R \) is (0, -1), \( S \) is (0, 6)? Wait, no, the purple lines: \( T \) is connected to \( Q \), \( S \) is connected to \( R \). So \( T \) has coordinates (-7, 6) (x=-7, y = 6), \( S \) has coordinates (1, 6)? No, maybe the correct coordinates are:

Looking at the grid:

• \( Q \): (-8, -1)
• \( R \): (0, -1)
• \( S \): (1, 6)? No, let's check the x - axis and y - axis. The x - axis has grid lines at integers, y - axis too. So \( T \) is at (-7, 6), \( S \) is at (1, 6), \( R \) is at (0, -1), \( Q \) is at (-8, -1). Wait, maybe a better way: the translation rule is \( (x,y)\to(x + 3,y-5) \) (3 units right means add 3 to x, 5 units down means subtract 5 from y).

Step2: Apply translation to each vertex

• For \( Q(-8,-1) \):

New x - coordinate: \( - 8+3=-5 \)
New y - coordinate: \( - 1-5=-6 \)
So \( Q'(-5,-6) \)

• For \( R(0,-1) \):

New x - coordinate: \( 0 + 3=3 \)
New y - coordinate: \( - 1-5=-6 \)
So \( R'(3,-6) \)

• For \( S(1,6) \)? Wait, no, let's re - identify \( S \). Wait, the original \( S \): looking at the graph, \( S \) is at (1, 6)? No, the vertical line from \( R(0,-1) \) to \( S \): \( R \) is (0, -1), \( S \) is (0, 6)? Wait, that makes more sense. So \( S(0,6) \)

Then for \( S(0,6) \):
New x - coordinate: \( 0+3 = 3 \)
New y - coordinate: \( 6-5 = 1 \)
So \( S'(3,1) \)

• For \( T \): \( T \) is at (-7,6) (since it's horizontal from \( S(0,6) \), so x=-7, y = 6)

New x - coordinate: \( - 7+3=-4 \)
New y - coordinate: \( 6-5 = 1 \)
So \( T'(-4,1) \)

Step3: Plot the new points

Plot \( Q'(-5,-6) \), \( R'(3,-6) \), \( S'(3,1) \), \( T'(-4,1) \) and connect them to form the translated parallelogram.

Answer:

The translated parallelogram \( Q'R'S'T' \) has vertices \( Q'(-5,-6) \), \( R'(3,-6) \), \( S'(3,1) \), \( T'(-4,1) \). (To graph, plot these points and connect them in order.)