If the graph of an equation is translated to the right units, and upward  units, the equation of the image can be found by replacing  with  and  with  in the equation of the original graph. 

Since we are moving the graph of the equation

left three units and down five units, we set and ; we can therefore replace  with , or , and  with , or . The equation of the image can be written as

We can simplify the expression on the right by distributing:

Collect like terms:

Subtract 5 from both sides:

,

the correct equation of the image.

The midpoint of a segment with endpoints at  and is located at

Substitute the coordinates of  and  in this formula to find that midpoint of is located at

, or .

Substitute the coordinates of and to find that midpoint of is located at

, or .

To perform the translation , or, equivalently,

,

on a point, it is necessary to add

to its -coordinate, and

to its -coordinate.

Therefore, the -coordinate of the image of  under this translation is

;

its -coordinate is

The image of is located at .

The translation  on a figure is the translation that shifts a figure so that the image of , which we will call , coincides with . All other points shift the same distance in the same direction. Below shows the image of the given hexagon under this translation, with the image of  marked as :

If the image is reflected about , the new image is the original hexagon. Calling  the image of  under this reflection, we get the following:

, the image of  under these two transformations, coincides with .

The translation  on a figure is the translation that shifts a figure so that the image of , which we will call , coincides with . All other points shift the same distance in the same direction. Below shows the image of the given hexagon under this translation:

If this new hexagon is rotated clockwise  - one third of a turn - about , and call  the image of , and so forth, the result is as follows:

Removing the intermediate markings, we see that the correct response is 

The translation  on a figure is the translation that shifts a figure so that the image of  coincides with . All other points shift the same distance in the same direction. Below shows the image of the given hexagon under this translation, with  the image of :

If this new hexagon is rotated  - one half of a turn - about  - the image is the original hexagon, but the vertices can be relabeled. Letting  be the image of  under this rotation, and so forth:

 coincides with  in the original hexagon, making  the correct response.

By definition, a median of a triangle has as its endpoints one vertex and the midpoint of the opposite side. Therefore, the endpoints of the median from are itself, which is at , and , which itself is the midpoint of the side with origin  and , which is , as its endpoints.

The midpoint of a segment with endpoints at  and is located at

,

so, substituting the coordinates of and in the formula, we see that is

, or .

See the figure below:

To perform the translation , or, equivalently,

,

on a point, it is necessary to add

and

to the - and - coordinates, respectively. Therefore, the image of  is located at 

,

or

.

The translation  on a figure is the translation that shifts a figure so that the image of  coincides with . All other points shift the same distance in the same direction. Below shows the image of the given hexagon under this translation, with the image of  marked as :

If this new hexagon is rotated clockwise  - one third of a turn - about  - the image is the original hexagon, but the vertices can be relabeled. Letting  be the image of  under this rotation, and so forth:

 coincides with  in the original hexagon, making  the correct response.

Rotating a point 

geometrically in the plane about the origin  is equivalent to negating the coordinates of the point algebraically to obtain

.

Thus, since our initial point was

we negate both coordinates to get

as the rotation about the origin by .

The diagram below superimposes the two figures:

The transformation moves the black diagonal to the position of the red diagonal, and, consequently, points  and  to points  and , respectively. This constitutes two-tenths of a complete turn clockwise, or a clockwise rotation of 

A counterclockwise rotation of  is  ofa complete rotation. Observe the following diagram:

In the right figure, the question mark has been turned one-eighth of a complete turn counterclockwise. This is the correct orientation.