Question

graph the image of trapezoid stuv after a dilation with a scale factor of 3, centered at the origin.

Explanation:

Step1: Identify Coordinates

First, find the coordinates of the vertices of trapezoid \( STUV \):

• \( S(-3, -3) \) (wait, looking at the graph: S is at (-3, -3)? Wait, no, let's check the grid. The grid has x from -10 to 10, y from -10 to 10. Let's re - identify:
• \( V(0, 3) \)? Wait, no, the green dot for V: looking at the y - axis, it's at (0, 3)? Wait, the original graph: V is at (0, 3)? Wait, the user's graph: V is at (0, 3)? Wait, no, the original points:
• \( S(-3, -3) \)? Wait, no, let's look again. The points:
• \( S \): x = - 3, y=-3? Wait, the grid lines: each square is 1 unit. So:
• \( V(0, 3) \)? Wait, no, the green dot for V: on the y - axis, y = 3? Wait, the user's graph: V is at (0, 3), U is at (3, 3)? Wait, no, U is at (3, 3)? Wait, no, the x - coordinate for U: between 2 and 4? Wait, no, let's re - check:
• \( V(0, 3) \), \( U(3, 3) \), \( T(3, - 3) \), \( S(-3, - 3) \)? Wait, no, the original graph: S is at (-3, -3), T is at (3, -3), V is at (0, 3), U is at (3, 3). Wait, maybe I misread. Let's do it properly:
• From the graph:
• \( S(-3, -3) \) (x=-3, y = - 3)
• \( T(3, -3) \) (x = 3, y=-3)
• \( U(3, 3) \) (x = 3, y = 3)
• \( V(0, 3) \) (x = 0, y = 3)

Step2: Apply Dilation Rule

The rule for dilation centered at the origin with scale factor \( k \) is \( (x,y)\to(kx,ky) \). Here, \( k = 3 \).

• For \( S(-3,-3) \):
• New coordinates \( S'=3\times(-3),3\times(-3)=(-9,-9) \)
• For \( T(3,-3) \):
• New coordinates \( T'=3\times3,3\times(-3)=(9,-9) \)
• For \( U(3,3) \):
• New coordinates \( U'=3\times3,3\times3=(9,9) \)
• For \( V(0,3) \):
• New coordinates \( V'=3\times0,3\times3=(0,9) \)

Step3: Plot the New Points

Plot the points \( S'(-9,-9) \), \( T'(9,-9) \), \( U'(9,9) \), \( V'(0,9) \) and connect them to form the dilated trapezoid.

Wait, maybe I made a mistake in the original coordinates. Let's re - identify the original coordinates correctly:

Looking at the graph:

• \( S \): x=-3, y = - 3? Wait, no, the grid: each square is 1 unit. Let's check the original points:

• \( S \): x=-3, y=-3? Wait, the y - coordinate for S: below the x - axis, at y=-3? And x=-3.

• \( T \): x = 3, y=-3.

• \( U \): x = 3, y = 3.

• \( V \): x = 0, y = 3.

Yes, that makes sense. So applying dilation with scale factor 3 centered at the origin:

• For a point \( (x,y) \), the new point \( (x',y')=(3x,3y) \)

So:

• \( S(-3,-3)\to S'(-9,-9) \)

• \( T(3,-3)\to T'(9,-9) \)

• \( U(3,3)\to U'(9,9) \)

• \( V(0,3)\to V'(0,9) \)

Then, to graph the dilated trapezoid, plot these four new points and connect them in the same order as the original trapezoid (S - T - U - V - S).

Answer:

To graph the image of trapezoid \( STUV \) after a dilation with a scale factor of 3 centered at the origin:

1. Find the original coordinates of the vertices:

• \( S(-3,-3) \), \( T(3,-3) \), \( U(3,3) \), \( V(0,3) \)

1. Apply the dilation rule \( (x,y)\to(3x,3y) \):

• \( S'(-9,-9) \)
• \( T'(9,-9) \)
• \( U'(9,9) \)
• \( V'(0,9) \)

1. Plot the points \( S' \), \( T' \), \( U' \), \( V' \) and connect them to form the dilated trapezoid.