Question

graph and label all pre - images and images on the coordinate planes below.
rectangles qrst with vertices q(-6, -1), r(-3, 1), s(1, -5), and t(-2, -7): (x,y)→(x + 5,y + 7)
translations
triangle cde with vertices c(2, -1), d(7, -4), and e(4, -6): (x,y)→(x - 3,y + 8)
reflections
triangle abc with vertices a(-4, 2), b(4, 7), and c(5, 1): x - axis
triangle xyz with vertices x(-5, -2), y(-3, 4), and z(-1, 1): y = x

Explanation:

Step1: Find new vertices for rectangle QRST

Apply the translation rule $(x,y)\to(x + 5,y + 7)$ to each vertex.
For $Q(-6,-1)$: $Q'=(-6 + 5,-1+7)=(-1,6)$
For $R(-3,1)$: $R'=(-3 + 5,1 + 7)=(2,8)$
For $S(1,-5)$: $S'=(1+5,-5 + 7)=(6,2)$
For $T(-2,-7)$: $T'=(-2 + 5,-7+7)=(3,0)$

Step2: Find new vertices for triangle CDE

Apply the translation rule $(x,y)\to(x - 3,y + 8)$ to each vertex.
For $C(2,-1)$: $C'=(2-3,-1 + 8)=(-1,7)$
For $D(7,-4)$: $D'=(7-3,-4 + 8)=(4,4)$
For $E(4,-6)$: $E'=(4-3,-6 + 8)=(1,2)$

Step3: Find new vertices for triangle ABC (reflection over x - axis)

The rule for reflection over the x - axis is $(x,y)\to(x,-y)$.
For $A(-4,2)$: $A'=(-4,-2)$
For $B(4,7)$: $B'=(4,-7)$
For $C(5,1)$: $C'=(5,-1)$

Step4: Find new vertices for triangle XYZ (reflection over y = x)

The rule for reflection over the line $y = x$ is $(x,y)\to(y,x)$.
For $X(-5,-2)$: $X'=(-2,-5)$
For $Y(-3,4)$: $Y'=(4,-3)$
For $Z(-1,1)$: $Z'=(1,-1)$

Answer:

Rectangle QRST: Pre - image vertices $Q(-6,-1),R(-3,1),S(1,-5),T(-2,-7)$; Image vertices $Q'(-1,6),R'(2,8),S'(6,2),T'(3,0)$
Triangle CDE: Pre - image vertices $C(2,-1),D(7,-4),E(4,-6)$; Image vertices $C'(-1,7),D'(4,4),E'(1,2)$
Triangle ABC: Pre - image vertices $A(-4,2),B(4,7),C(5,1)$; Image vertices $A'(-4,-2),B'(4,-7),C'(5,-1)$
Triangle XYZ: Pre - image vertices $X(-5,-2),Y(-3,4),Z(-1,1)$; Image vertices $X'(-2,-5),Y'(4,-3),Z'(1,-1)$