HiSET - Translations | Practice Hub

Translation

The graph on the left shows an object in the Cartesian plane. A transformation is performed on it, resulting in the graph on the right.

Which of the following transformations best fits the graphs?

Translation

Dilation

Rotation about the origin

Reflection in the x-axis

Reflection in the y-axis

Explanation

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.

Consider regular Hexagon .

On this hexagon, perform the translation $F\rightarrow B$ . Then perform a $120^{\circ }$ clockwise rotation on the image with center at .

Let be the image of under these transformations, be the image of , and so forth. Under these images, which point on the original hexagon does fall?

Explanation

The translation $F\rightarrow B$ 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 $120 ^{\circ }$ - 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.

Consider regular Hexagon .

On this hexagon, perform the translation $F\rightarrow C$ . Then perform a $180^{\circ }$ rotation on the image with center at .

Let be the image of under these transformations, be the image of , and so forth. Under these images, which point on the original hexagon does fall?

Explanation

The translation $F\rightarrow C$ 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 $180 ^{\circ }$ - 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.

Consider regular Hexagon .

On this hexagon, perform the translation $A\rightarrow B$ . Then perform a $120^{\circ }$ rotation on the image with center at . Let be the image of under these transformations, and so forth.

Which of the following correctly shows Hexagon relative to Hexagon ?

Hexagons

Explanation

The translation $A\rightarrow B$ 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 $120 ^{\circ }$ - 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

Hexagons

Consider regular Hexagon .

On this hexagon, perform the translation $A\rightarrow B$ . Then reflect the hexagon about $\overleftrightarrow{BD}$ . Let be the image of under these transformations, and so forth.

Which point on Hexagon is the image of under these transformations?

Explanation

If the image is reflected about $\overleftrightarrow{BD}$ , 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 .

Translate the graph of the equation

left four units and down six units. Give the equation of the image.

Explanation

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.

Translate the graph of the equation

$x^{2}+ y^{2}= 100$

right two units and up five units. Give the equation of the image.

$x^{2}- 4x + y^{2}-10y-71=0$

$x^{2}- 2x + y^{2}-5y-71=0$

$x^{2}+ 4x + y^{2}+10y-71=0$

$x^{2}+ 2x + y^{2}+5y-71=0$

None of the other choices gives the correct response.

Explanation

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

$x^{2}+ y^{2}= 100$

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

$(x-2)^{2}+( y-5)^{2}= 100$

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

$x^{2}- 2 (x) (2)+ 2^{2}+ y^{2}- 2 (5)(x)+ 5^{2} = 100$

$x^{2}- 4x+4+ y^{2}-10y+2 5 = 100$

Collect like terms; the equation becomes

$x^{2}- 4x + y^{2}-10y+29 = 100$

Subtract 100 from both sides:

$x^{2}- 4x + y^{2}-10y+29- 100 = 100 - 100$

$x^{2}- 4x + y^{2}-10y-71=0$

Translate the graph of the equation

$y = x^{2}+ 5x - 7$

right four units and down two units. Give the equation of the image.

$y= x^{2} -3x-13$

$y= x^{2} -3x-9$

$y = x^{2}+ 13x + 27$

$y = x^{2}+ 13x + 29$

$y = x^{2}+ 13x -9$

Explanation

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

$y = x^{2}+ 5x - 7$

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

$y+2 = (x-4)^{2}+ 5(x-4) - 7$

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

$y+2 = (x-4)^{2}+ 5x-5 (4) - 7$

$y+2 = (x-4)^{2}+ 5x-20 - 7$

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

$y+2 = x^{2} -2 (x)(4) +4^{2}+ 5x-20 - 7$

$y+2 = x^{2} -8x+16+ 5x-20 - 7$

Collect like terms on the right:

$y+2 = x^{2} -8x+ 5x+16 -20 - 7$

$y+2 = x^{2} -3x-11$

Subtract 2 from both sides:

$y+2 -2= x^{2} -3x-11-2$

$y= x^{2} -3x-13$ ,

the equation of the image.

Translate the graph of the equation

left three units and down five units. Give the equation of the image.

None of the other choices gives the correct response.

Explanation

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:

$y+5 = | 3x+3 \cdot 3 - 8 |$

Collect like terms:

Subtract 5 from both sides:

the correct equation of the image.

On the coordinate plane, let , , and be located at the origin, , and . Construct the median of $\bigtriangleup OAB$ from and let the foot of the median be . On the triangle, perform the translation $A \rightarrow M$ . Where is the image of ?

Explanation

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.