Question

graph the image of square uvwx after the following glide reflection: translation 14 units down reflection across the line x = 1

Explanation:

Step1: Apply translation

For a point $(x,y)$ in the square, after translating 14 units down, the new - point is $(x,y - 14)$. Let the coordinates of the vertices of square $UVWX$ be $U(x_U,y_U)$, $V(x_V,y_V)$, $W(x_W,y_W)$, $X(x_X,y_X)$. After translation, the new coordinates are $U'(x_U,y_U - 14)$, $V'(x_V,y_V - 14)$, $W'(x_W,y_W - 14)$, $X'(x_X,y_X - 14)$.

Step2: Apply reflection

The formula for reflecting a point $(x,y)$ across the line $x = a$ is $(2a - x,y)$. Here $a = 1$. For a point $(x,y)$ obtained after translation, after reflecting across the line $x = 1$, the new point is $(2\times1 - x,y)=(2 - x,y)$.
Let's assume the original coordinates of $U=(6,2)$, $V=(6,10)$, $W=(12,10)$, $X=(12,2)$.
After translation:
$U_1=(6,2 - 14)=(6,-12)$
$V_1=(6,10 - 14)=(6,-4)$
$W_1=(12,10 - 14)=(12,-4)$
$X_1=(12,2 - 14)=(12,-12)$
After reflection across $x = 1$:
$U_2=(2 - 6,-12)=(-4,-12)$
$V_2=(2 - 6,-4)=(-4,-4)$
$W_2=(2 - 12,-4)=(-10,-4)$
$X_2=(2 - 12,-12)=(-10,-12)$
Plot the new - vertices $(-4,-12)$, $(-4,-4)$, $(-10,-4)$, $(-10,-12)$ to get the image of the square.

Answer:

Graph the square with vertices $(-4,-12)$, $(-4,-4)$, $(-10,-4)$, $(-10,-12)$ on the given coordinate - plane.