Given the coordinates of the triangle END
[tex]\begin{gathered} E\rightarrow(-5,-1)_ \\ N\operatorname{\rightarrow}(-3,-2) \\ \begin{equation*} D\rightarrow(-3,2) \end{equation*} \end{gathered}[/tex]
Firstly the triangle was translated 2units horizontally to the right then shifted downward by 3units.
[tex]\begin{gathered} E\rightarrow(-5+2,-1-3)=(-3,-4) \\ N\rightarrow(-3+2,-2-3)=(-1,-5) \\ D\rightarrow(-3+2,2-3)=(-1,-1) \end{gathered}[/tex]
The transformed image was then rotated 90degrees counterclockwise about the origin
When rotating a point 90 degrees counterclockwise about the origin our point A(x,y) becomes A'(-y,x).
Hence,
[tex]\begin{gathered} E^{\prime}\rightarrow(-(-4),-3)=(4,-3) \\ N\rightarrow(-(-5),-1)=(5,-1) \\ D\rightarrow(-(-1),-1)=(1,-1) \end{gathered}[/tex]
Therefore, the coordinates of the transformed image is
[tex]\begin{gathered} E^{\prime}\rightarrow(4,-3) \\ N\rightarrow(5,-1) \\ D\rightarrow(1,-1) \end{gathered}[/tex]
The graph of the image will be shown below
