Graph the image of △IJK after the following sequence of transformations:
Translation 17 units left and 5 units up
Reflection across the line y=1

Please find attached the graph of the image of ΔIJK following a translation transformation of (-17, 5), and a reflection across the line y = 5What is a transformation in geometry?A transformation is an operation that changes the location, size and shape of geometric figures.The coordinates of the vertex of ΔIJK are; I(14, 8), J(8, 5), K(14, 2)The translation transformation is 17 units left and 5 units up = T₍₋₁₇, ₅₎The location of the vertices of the image of ΔIJK following the translation transformation are; [tex]I(14,\, 8)\ \underset{\longrightarrow}{T_{(-17,\, 5)}}\ I'(-3,\, 13)[/tex][tex]J(8, 5)\ \underset{\longrightarrow}{T_{(-17,\, 5)}}\ j'(-9,\, 10)[/tex][tex]K(14, 2)\ \underset{\longrightarrow}{T_{(-17,\, 5)}}\ K'(-3,\, 7)[/tex]The coordinates of a point following a reflection across the line y = 1 are as follows;The coordinates of the vertices of the image I'J'K' relative to the line y = 1 are;I'(-3, 13), Relative point I'(-3, 12)J'(-9, 10) has a relative point at (-9, 9)K'(-3, 7) has a relative point (-3, 6)The coordinates of the vertices of the reflected image are therefore;(x, y)  [tex]\underset{\longrightarrow}{R_{y = 1}}[/tex]  (x, -(y - 1))Which gives;I'(-3, 12)  [tex]\underset{\longrightarrow}{R_{y = 1}}[/tex] I''(-3, -12 + 1) = I''(-3, -11)J'(-9, 9) [tex]\underset{\longrightarrow}{R_{y = 1}}[/tex] I''(-9, -9 + 1) = I''(-9, -8) k'(-3, 6)  [tex]\underset{\longrightarrow}{R_{y = 1}}[/tex] k''(-3, -6 + 1) = k''(-3, -5)Please find attached the image of the ΔIJK following the translation and reflection transformationLearn more about transformation of figures in geometric here:https://brainly.com/question/2523916#SPJ1