Question

1. draw the mapping that shows the translation if quadrilateral tkw n is the pre - image and quadrilateral hbrx is the image.
2. describe the translation if quadrilateral tkw n is the pre - image and quadrilateral hbrx is the image.

the translation is 6 units to the right and 1 unit up.

1. complete the statements.

the x - values of quadrilateral gsvy, the image, can be found by adding subtracting multiplying to the x - values of quadrilateral xrbh, the pre - image. the y - values of quadrilateral gsvy, the image, can be found by adding subtracting units to the y - values of quadrilateral xrbh, the pre - image.

1. describe the translation if quadrilateral xrbh is the pre - image and quadrilateral gsvy is the image.

Explanation:

Step1: Analyze the translation from \( TKW \) to \( HBRX \)

Let's take a point from the pre - image \( T \) (say \( T(-5,3)\)) and its corresponding point in the image \( H \) (say \( H(4,4)\)).
The change in \( x \) - value: \( 4-(-5)=9 \). The change in \( y \) - value: \( 4 - 3=1 \).

Step2: Analyze the translation from \( XRBH \) to \( GSVY \)

Let's take a point from the pre - image \( X \) (say \( X(11,5)\)) and its corresponding point in the image \( G \) (say \( G(- 4,-2)\)).
The change in \( x \) - value: \( -4-11=-15 \). The change in \( y \) - value: \( -2 - 5=-7 \).

Answer:

1. For each vertex \( (x,y) \) of \( TKW \), the mapping is \( (x,y)\to(x + 9,y + 1) \) to get the vertices of \( HBRX \).
2. The translation is \( 9 \) units to the right and \( 1 \) unit up.
3. The \( x \) - values of quadrilateral \( GSVY \), the image, can be found by subtracting \( 15 \) to the \( x \) - values of quadrilateral \( XRBH \), the pre - image. The \( y \) - values of quadrilateral \( GSVY \), the image, can be found by subtracting \( 7 \) units to the \( y \) - values of quadrilateral \( XRBH \), the pre - image.
4. The translation is \( 15 \) units to the left and \( 7 \) units down.