Question

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

Explanation:

Step1: Identify coordinates of vertices

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

• \( S \): From the graph, \( S(-3, -3) \) (wait, looking at the grid: S is at x=-3, y=-3? Wait, no, let's check again. Wait, the grid: S is at x=-3, y=-3? Wait, the original points: V is at (0, 3)? Wait, no, the y-axis: V is at (0, 3)? Wait, the graph shows V at (0, 3)? Wait, the y-axis has 4 at the top, so V is (0, 3)? Wait, no, the green dot for V is at (0, 3)? Wait, the user's graph: V is at (0, 3), U is at (3, 3)? Wait, no, the x-axis: U is at (3, 3)? Wait, no, the grid lines: x=0, y=3 for V; x=3, y=3 for U? Wait, no, the x-axis: 2 is marked, so U is at (3, 3)? Wait, no, let's re-express:

Looking at the graph:

• \( S \): x = -3, y = -3? Wait, no, the S is at x=-3, y=-3? Wait, the T is at x=3, y=-3? Wait, no, T is at (3, -3)? Wait, V is at (0, 3), U is at (3, 3), S is at (-3, -3), T is at (3, -3)? Wait, no, the original trapezoid: V(0, 3), U(3, 3), T(3, -3), S(-3, -3)? Wait, no, the S is at (-3, -3), T at (3, -3), V at (0, 3), U at (3, 3). Wait, let's confirm:

From the graph:

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

Yes, that makes sense. So vertices are \( S(-3, -3) \), \( T(3, -3) \), \( U(3, 3) \), \( V(0, 3) \).

Step2: Apply dilation with scale factor 3 centered at origin

Dilation rule: For a point \( (x, y) \), after dilation with scale factor \( k \) centered at origin, the new point is \( (k \cdot x, k \cdot y) \).

So:

• For \( S(-3, -3) \): New \( S' = (3 \cdot (-3), 3 \cdot (-3)) = (-9, -9) \)
• For \( T(3, -3) \): New \( T' = (3 \cdot 3, 3 \cdot (-3)) = (9, -9) \)
• For \( U(3, 3) \): New \( U' = (3 \cdot 3, 3 \cdot 3) = (9, 9) \)
• For \( V(0, 3) \): New \( V' = (3 \cdot 0, 3 \cdot 3) = (0, 9) \)

Step3: Plot the new points

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

Answer:

The dilated trapezoid \( S'T'U'V' \) has vertices at \( S'(-9, -9) \), \( T'(9, -9) \), \( U'(9, 9) \), \( V'(0, 9) \). To graph it, plot these points and connect \( S' \) to \( T' \) to \( U' \) to \( V' \) to \( S' \).