Question

graph the image of trapezoid efgh after the following glide reflection: translation 16 units left reflection across the x - axis

Explanation:

Step1: Identify Coordinates of EFGH

First, find the coordinates of each vertex of trapezoid \( EFGH \). From the graph:

• \( E(2, -4) \)
• \( F(4, -8) \)
• \( G(8, -8) \)
• \( H(12, -4) \)

Step2: Apply Translation 16 Units Left

A translation 16 units left subtracts 16 from the \( x \)-coordinate of each point.

• For \( E(2, -4) \): \( x = 2 - 16 = -14 \), \( y = -4 \) → \( E'(-14, -4) \)
• For \( F(4, -8) \): \( x = 4 - 16 = -12 \), \( y = -8 \) → \( F'(-12, -8) \)
• For \( G(8, -8) \): \( x = 8 - 16 = -8 \), \( y = -8 \) → \( G'(-8, -8) \)
• For \( H(12, -4) \): \( x = 12 - 16 = -4 \), \( y = -4 \) → \( H'(-4, -4) \)

Step3: Apply Reflection Across the x - axis

A reflection across the \( x \)-axis changes the sign of the \( y \)-coordinate (\( (x, y) \to (x, -y) \)).

• For \( E'(-14, -4) \): \( y = -(-4) = 4 \) → \( E''(-14, 4) \)
• For \( F'(-12, -8) \): \( y = -(-8) = 8 \) → \( F''(-12, 8) \)
• For \( G'(-8, -8) \): \( y = -(-8) = 8 \) → \( G''(-8, 8) \)
• For \( H'(-4, -4) \): \( y = -(-4) = 4 \) → \( H''(-4, 4) \)

Step4: Graph the New Points

Plot the points \( E''(-14, 4) \), \( F''(-12, 8) \), \( G''(-8, 8) \), and \( H''(-4, 4) \) on the coordinate plane and connect them to form the image of the trapezoid after the glide reflection.

Answer:

The image of trapezoid \( EFGH \) after the glide reflection has vertices at \( E''(-14, 4) \), \( F''(-12, 8) \), \( G''(-8, 8) \), and \( H''(-4, 4) \). (To graph, plot these points and connect them in order.)