A line of symmetry of a figure is about which the reflection of the figure is the figure itself. The diagram below shows the only line of symmetry of the figure.

The correct response is one.

The reflection of the point at  about the -axis is the point at ; therefore, the image of  is . Since these two points have the same -coordinate, the distance between them is the absolute value of the difference between their -coordinates:

Refer to the figure below, which shows the reflection of the given hexagon about  ; we will call the image of , call the image of , and so forth.

Now, refer to the figure below, which shows the reflection of the image about  ; we will call  the image of , call  the image of , and so forth.

Note that the vertices coincide with those of the original hexagon, and that the images of the points are in the same clockwise order as the original points. Since coincides with , coincides with , and so forth, a clockwise rotation of five-sixth of a complete turn - that is, ,

is required to make each point its own image under the three transformations.

Refer to the figure below, which shows the reflection of about ; we will call this image .

Note that  coincides with . Now, refer to the figure below, which shows the reflection of  about ; we will call this image - the final image - 

Note that  coincides with , making this the correct response.

A dilation is the stretching or shrinking of a figure.

A rotation is the turning of a a figure about a point. 

A reflection is the flipping of a figure across a line.

A translation is is the sliding of a figure in a direction. 

With a translation, the image is not only congruent to its original size and shape, but its orientation remains the same. A translation fits this figure best because the shape seems to move upward and rightward without changing size, shape, or orientation.

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 four units and down six units, we set and ; we can replace with , or , and with , or . The equation of the image can be written as

Simplify by distributing:

Collect like terms:

Add 26 to both sides:

,

the correct choice.

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

right two units and up five units, we set  and ; we can therefore replace  with and  with . The equation of the image can be written as 

This can be rewritten by applying the binomial square pattern as follows:

Collect like terms; the equation becomes

Subtract 100 from both sides:

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

right four units and down two units, we set  and ; we can therefore replace  with , and  with , or . The equation of the image can be written as

The expression at right can be simplified. First, use the distributive property on the middle expression:

Now, simplify the first expression by using the binomial square pattern:

Collect like terms on the right:

Subtract 2 from both sides:

,

the equation of the image.

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 .