Question

Which sequence of transformations carries ABCD onto HGFE

Answer 1

If you look closely at the options, the first 3 options will not produce an image that will be in Quadrant II with the vertices coinciding HGFE.
But if you look at the 4th option, translating the points 8 units to the left will produce an image in Quadrant III. Reflecting the image across the x-axis will produce triangle HGFE.

Answer 2

Answer:

The correct option is D.

Step-by-step explanation:

It is given that the after transformation figure ABCD onto HGFE. It means HGFE is image of ABCD after transformation.

From the figure it is noticed that A(2,-1) and H(-6,1).

Rotation of 90 degree clockwise about the origin followed by the reflection across the x-axis is expressed as

$(x,y)\rightarrow (y,-x)\rightarrow (y,x)$

Therefore the image of A(2,-1) is (-1,2), which are not the coordinates of H. Option A is incorrect.

Translation 3 unit up followed by the reflection across the y-axis is expressed as

$(x,y)\rightarrow (x,y+3)\rightarrow (-x,y+3)$

Therefore the image of A(2,-1) is (-2,2), which are not the coordinates of H. Option B is incorrect.

Reflection across the x-axis followed by rotation of 90 degree counterclockwise about the origin is expressed as

$(x,y)\rightarrow (x,-y)\rightarrow (y,x)$

Therefore the image of A(2,-1) is (-1,2), which are not the coordinates of H. Option C is incorrect.

Translation 8 unit left followed by reflection across the x-axis is expressed as

$(x,y)\rightarrow (x-8,y)\rightarrow (x-8,-y)$

Therefore the image of A(2,-1) is (-6,1), which are the coordinates of H. Option D is correct.