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Draw two perpendicular lines, one horizontal (x-axis) and one vertical (y-axis), intersecting at the origin (0,0).

Using a ruler, mark every 2 cm along both axes. Each 2 cm interval represents 4 units.
Label the x-axis from -16 to 16 and the y-axis from -16 to 16. For example, at 2 cm from the origin, mark 4; at 4 cm, mark 8; and so on. Do the same for negative values.

2) Draw the original triangle $\triangle ABC$

The given vertices are:

• $A = (2,2)$
• $B = (8,-2)$
• $C = (6,6)$

Draw straight lines to connect A to B, B to C, and C to A to form $\triangle ABC$.

3) Draw the image $\triangle A'B'C'$ of $\triangle ABC$ after a $180^\circ$ rotation about the origin

A $180^\circ$ rotation about the origin transforms a point $(x,y)$ to $(-x,-y)$.

• $A(2,2) \to A'(-2,-2)$
• $B(8,-2) \to B'(-8,-(-2)) = B'(-8,2)$
• $C(6,6) \to C'(-6,-6)$ The coordinates of $\triangle A'B'C'$ are:
• $A' = \boxed{(-2,-2)}$
• $B' = \boxed{(-8,2)}$
• $C' = \boxed{(-6,-6)}$

Plot the points $A'$, $B'$, and $C'$ on your graph. Connect them with straight lines to form $\triangle A'B'C'$.

4) Draw the image $\triangle A^2B^2C^2$ of $\triangle ABC$ under the translation vector $(-8,5)$

A translation by vector $(h,k)$ transforms a point $(x,y)$ to $(x+h, y+k)$. Here, $(h,k) = (-8,5)$.

• $A(2,2) \to A^2(2+(-8), 2+5) = A^2(-6,7)$
• $B(8,-2) \to B^2(8+(-8), -2+5) = B^2(0,3)$
• $C(6,6) \to C^2(6+(-8), 6+5) = C^2(-2,11)$ The coordinates of $\triangle A^2B^2C^2$ are:
• $A^2 = \boxed{(-6,7)}$
• $B^2 = \boxed{(0,3)}$
• $C^2 = \boxed{(-2,11)}$

Plot the points $A^2$, $B^2$, and $C^2$ on your graph. Connect them with straight lines to form $\triangle A^2B^2C^2$.

5) Draw the image $\triangle A^3B^3C^3$ of $\triangle A'B'C'$ under the reflection in the line $y=0$

A reflection in the line $y=0$ (the x-axis) transforms a point $(x,y)$ to $(x,-y)$. We are reflecting $\triangle A'B'C'$.

• $A'(-2,-2) \to A^3(-2, -(-2)) = A^3(-2,2)$
• $B'(-8,2) \to B^3(-8, -2)$
• $C'(-6,-6) \to C^3(-6, -(-6)) = C^3(-6,6)$ The coordinates of $\triangle A^3B^3C^3$ are:
• $A^3 = \boxed{(-2,2)}$
• $B^3 = \boxed{(-8,-2)}$
• $C^3 = \boxed{(-6,6)}$

Plot the points $A^3$, $B^3$, and $C^3$ on your graph. Connect them with straight lines to form $\triangle A^3B^3C^3$.