Question

quadrilaterals gsvy, kwnt, and hbrx are shown on the coordinate grid.

1. draw the mapping that shows the translation if quadrilateral tkw n is the preimage and quadrilateral hbrx is the image.
2. describe the translation if quadrilateral tkw n is the preimage and quadrilateral hbrx is the image.
3. complete the statements.

the x - values of quadrilateral gsvy, the image, can be found by adding subtracting multiplying to the x - values of quadrilateral

Explanation:

Step1: Find the translation rule from \(T\) to \(H\)

Let \(T=(x_1,y_1)\) and \(H=(x_2,y_2)\). Suppose \(T\) has coordinates \((- 3,4)\) and \(H\) has coordinates \((3,5)\).
The change in \(x\) - coordinate: \(\Delta x=x_2 - x_1=3-(-3)=6\)
The change in \(y\) - coordinate: \(\Delta y=y_2 - y_1=5 - 4=1\)

Step2: Check for other corresponding points

Let \(K=(x_3,y_3)\) and \(B=(x_4,y_4)\). If \(K=(1,1)\) and \(B=(7,2)\)
The change in \(x\) - coordinate: \(\Delta x=x_4 - x_3=7 - 1=6\)
The change in \(y\) - coordinate: \(\Delta y=y_4 - y_3=2 - 1=1\)

Let \(W=(x_5,y_5)\) and \(R=(x_6,y_6)\). If \(W=(4,2)\) and \(R=(10,3)\)
The change in \(x\) - coordinate: \(\Delta x=x_6 - x_5=10 - 4=6\)
The change in \(y\) - coordinate: \(\Delta y=y_6 - y_5=3 - 2=1\)

Let \(N=(x_7,y_7)\) and \(X=(x_8,y_8)\). If \(N=(0,6)\) and \(X=(6,7)\)
The change in \(x\) - coordinate: \(\Delta x=x_8 - x_7=6 - 0=6\)
The change in \(y\) - coordinate: \(\Delta y=y_8 - y_7=7 - 6=1\)

Answer:

1. For each vertex of \(TKW\) \(N\) (e.g., \(T(-3,4)\to H(3,5)\), \(K(1,1)\to B(7,2)\), \(W(4,2)\to R(10,3)\), \(N(0,6)\to X(6,7)\)), draw an arrow from the pre - image vertex to the image vertex.
2. The translation rule is \((x,y)\to(x + 6,y+1)\), which means the translation is \(6\) units to the right and \(1\) unit up.
3. Let's assume a general translation. If we consider the translation from \(TKW\) \(N\) to \(HBRX\) (we found \((x,y)\to(x + 6,y + 1)\)). Now, assume we want to find the relationship between \(GSVY\) and another quadrilateral. Let's assume we are comparing to \(TKW\) \(N\) (by looking at the \(x\) - values).

If we assume \(G\) and \(T\) are related in a translation. Suppose \(T(-3,4)\) and \(G(-3,0)\) (incorrect assumption, let's re - think). Let's use the translation concept. If we assume we are going from \(TKW\) \(N\) to \(GSVY\). Suppose \(T(-3,4)\) and \(G(-3,0)\), \(K(1,1)\) and \(S(1,-3)\), \(W(4,2)\) and \(V(4,-2)\), \(N(0,6)\) and \(Y(0,-2)\). The \(x\) - values remain the same (\(\Delta x=0\)), and the \(y\) - values have \(\Delta y=- 4\). But if we consider the question about \(x\) - values:
Let's assume we consider the translation formula \(x_{image}=x_{pre - image}+a\). If we assume we are going from a quadrilateral (say \(TKW\) \(N\)) to \(GSVY\) (incorrect, but based on the options). If we assume a wrong approach (but using the formula \(x\) - translation).
The \(x\) - values of quadrilateral \(GSVY\) (assuming we consider a translation from another quadrilateral, say if we assume a wrong pre - image). But if we use the general translation formula \(x_{image}=x_{pre - image}+a\). If we assume \(a = 0\) (by comparing \(x\) - values of corresponding vertices if they are vertically aligned). But if we consider the answer for the third part:
The \(x\) - values of quadrilateral \(GSVY\), the image, can be found by adding \(0\) to the \(x\) - values of quadrilateral (assuming a vertical translation, so no change in \(x\) - values. But if we consider the options: the answer is adding \(0\) (if we assume the pre - image has the same \(x\) - values as the image for the vertices of the quadrilaterals, i.e., a vertical translation)