Part A: Determine the two different rotations that would create the image.

We are given triangle $QRS$ with vertices $Q(-2, -2), R(-6, -6), S(-5, -1)$, and its image after rotation, $Q'R'S'$, with vertices $Q'(2, -2), R'(6, -6), S'(1, -5)$.

To determine the rotations, we need to find the center of rotation and the angle of rotation. There are two possibilities for the rotations, and we will use the following steps:

Step 1: Finding the center of rotation

The center of rotation is the point that remains fixed while all other points move around it. We need to find a point equidistant from both $Q$ and $Q'$, and from $R$ and $R'$, and from $S$ and $S'$. The center of rotation lies on the perpendicular bisector of the segment joining each point and its image.

We will start by finding the midpoint of $Q$ and $Q'$, as well as the midpoint of $R$ and $R'$, and the midpoint of $S$ and $S'$.

• Midpoint of Q and Q': $\text{Midpoint of Q and Q'} = \left( \frac{-2 + 2}{2}, \frac{-2 + (-2)}{2} \right) = (0, -2)$

• Midpoint of R and R': $\text{Midpoint of R and R'} = \left( \frac{-6 + 6}{2}, \frac{-6 + (-6)}{2} \right) = (0, -6)$

• Midpoint of S and S': $\text{Midpoint of S and S'} = \left( \frac{-5 + 1}{2}, \frac{-1 + (-5)}{2} \right) = (-2, -3)$

Step 2: Analyzing the midpoints

The midpoints of $Q$ and $Q'$, and $R$ and $R'$ are clearly on a vertical line $x = 0$. Since the midpoint of $S$ and $S'$ does not align with this, we suspect that there might be two rotation centers. One is likely on the line $x = 0$, and the other could be somewhere else, as different transformations could give different rotational angles.

Part B: Explain how you know your answer to Part A is correct.

The reasoning behind the two possible rotations is based on the observation that the positions of the points change consistently in relation to each other when rotated around a specific center. The first rotation could be about the origin $(0, 0)$, and the second could involve a rotation about a different point (possibly along the line or perpendicular to $x = 0$).

Would you like me to show the exact calculation for the angles of rotation?