Given the coordinates of the parent function
[tex]\begin{gathered} A\rightarrow(-2,-1) \\ B\rightarrow(1,3) \\ C\rightarrow(2,-4) \end{gathered}[/tex]Then, the triangle is translated 2 units right and 3 units up. This will then be
[tex]\begin{gathered} A^{\prime}\rightarrow(-2+2,-1+3)=(0,2) \\ B^{\prime}\rightarrow(1+2,3+3)=(3,6) \\ C^{\prime}\rightarrow(2+2,-4+3)=(4,-1) \end{gathered}[/tex]Hence, the coordinates of the vertices of triangle A'B'C' are
[tex]A^{\prime}(0,2),B^{\prime}(3,6),C^{\prime}(4,-1)[/tex]