Answer:
Reflection across x = -2; clockwise rotation of 90° about the origin
Step-by-step explanation:
Assume that the mapping is like that in Fig. 1.
1. Reflection about x = -2
It looks like the first transformation is a reflection about the line x = -2
When you reflect a shape about a line, each point in the image is the same distance from the line of reflection as the corresponding point in the pre-image.
The rule for reflection about a line x = h is (x,y) ⟶ (2h - x, y}
We can use the rule to calculate the coordinates of the first image.
[tex]\begin{array}{cc}\textbf{A} & \textbf{A"} \\(-5,5) & (1,5) \\(-2,5) & (-2,5) \\(-1,1) & (-3,1) \\(-4, 1) & (0,1) \\\end{array}[/tex]
2. Rotation 90° clockwise about the origin
It looks like the second transformation is a 90° clockwise rotation about the origin.
The rule is (x,y) ⟶ (y,-x).
In other words, interchange x and y and make x negative.
[tex]\begin{array}{cc}\textbf{A"} & \textbf{A'} \\(1,5) & (5,-1) \\(-2,5) & (5,2) \\(-3,1) & (1,3) \\(0, 1) & (1,0) \\\end{array}[/tex]
These are the coordinates of P'Q'R'S'.
Figure 2 shows the reflection of A across the line x = -2 to give A" and its rotation to give A'.
