reflect triangle ABC over the line Y equals to translate the image right three and up to. Then what are the coordinates of the verticals in the image

ANSWER
[tex]\begin{gathered} A^{\prime}^{\prime}(-2,4) \\ B^{\prime}^{\prime}(-3,7) \\ C^{\prime}^{\prime}(0,6) \end{gathered}[/tex]EXPLANATION
First, let us find the coordinates of the vertices of the triangle:
[tex]\begin{gathered} A(-5,2) \\ B(-6,-1) \\ C(-3,0) \end{gathered}[/tex]Now, we have to reflect the points over the line y = 2.
To do this, find the distance between the y-coordinate of each vertex and y = 2 and add it to 2. That becomes the new y-coordinate of the point while its x-coordinate remains the same.
Therefore, the coordinates become:
[tex]\begin{gathered} A(-5,2)\rightarrow A^{\prime}(-5,(2-2)+2)\Rightarrow A^{\prime}(-5,2) \\ B(-6,-1)\rightarrow B^{\prime}(-6,(2-(-1)+2)\Rightarrow B^{\prime}(-6,5) \\ C(-3,0)\rightarrow C^{\prime}(-3,(2-0)+2)\Rightarrow C^{\prime}(-3,4) \end{gathered}[/tex]Now, we have to translate the points 3 units right and 2 units up. To do that, add 3 units to the x-coordinates and add 2 units to the y-coordinates of A'B'C':
[tex]\begin{gathered} A^{\prime}(-5,2)\rightarrow A^{\prime}^{\prime}(-5+3,2+2)\rightarrow A^{\prime}^{\prime}(-2,4) \\ B^{\prime}(-6,5)\rightarrow B^{\prime}^{\prime}(-6+3,5+2)\rightarrow B^{\prime}^{\prime}(-3,7) \\ C^{\prime}(-3,4)\rightarrow C^{\prime}^{\prime}(-3+3,4+2)\rightarrow C^{\prime}^{\prime}(0,6) \end{gathered}[/tex]That is the answer.