Question

graph the image of δfgh after a reflection over the x-axis.

Explanation:

Step1: Identify coordinates of F, G, H

From the graph, let's find the coordinates:

• \( F \): Let's assume the grid has each square as 1 unit. Looking at \( F \), it's at \( (5, -10) \)? Wait, no, looking at the y-axis, the bottom is -10, and F is at (5, -10)? Wait, no, the green triangle: F is at (5, -10)? Wait, no, the y-coordinate for F: looking at the graph, F is at (5, -10)? Wait, no, let's check again. Wait, the x-axis: F is at x=5, y=-10? Wait, no, the H is at (6, -6), F is at (5, -10), G is at (9, -10). Wait, maybe I misread. Wait, the y-axis: the positive y is up, negative down. So F: let's see, the x-coordinate: from the grid, F is at (5, -10)? Wait, no, the green points: F is at (5, -10)? Wait, H is at (6, -6), G is at (9, -10). Wait, maybe. Let's confirm:

Looking at the graph:

• \( F \): x=5, y=-10 (since it's on the line y=-10, x=5)
• \( G \): x=9, y=-10 (on y=-10, x=9)
• \( H \): x=6, y=-6 (x=6, y=-6)

Step2: Apply reflection over x-axis

The rule for reflection over the x-axis is \( (x, y)
ightarrow (x, -y) \).

So:

• For \( F(5, -10) \): Reflect over x-axis: \( (5, -(-10)) = (5, 10) \)
• For \( G(9, -10) \): Reflect over x-axis: \( (9, -(-10)) = (9, 10) \)
• For \( H(6, -6) \): Reflect over x-axis: \( (6, -(-6)) = (6, 6) \)

Step3: Graph the new points

Plot \( F'(5, 10) \), \( G'(9, 10) \), \( H'(6, 6) \) and connect them to form the reflected triangle.

Answer:

The image of \( \triangle FGH \) after reflection over the x - axis has vertices at \( F'(5, 10) \), \( G'(9, 10) \), and \( H'(6, 6) \). (To graph, plot these points and connect them.)