Let $L_i(t) = -2t + 1$.
Let $L_o(t) = \frac {1}{2}t + 2$.

Let $K(x)$ be a piecewise defined function defined by

$$K(x) = \begin {cases} 2|x+6| - 4 & \, \text { on } \, [-8,-3) \\ -6 & \, \text { on } \, [-3,1) \\ -\frac {3}{2}(x-4)(x-6) & \, \text { on } \, (4,8] \end {cases}$$

This time, we will compose three functions: $T = L_o \circ K \circ L_i$.

What will be domain (horizontal) effects and range (vertical) effects? The values $L_i$ will become the inputs to $K$. Therefore, $L_i$ affects the domain of $K$ and moves the graph horizontally. $L_o$ will take the range values from $K$ and move the graph vertically.

$\blacktriangleright$ Domain - horizontal : $L_i(t) = -2t + 1$

• The leading coefficient for $L_i$ is negative. The graph is going to be reflected horizontally.
  The parabola will be on the left and the “Vee” on the right.
• The leading coefficent is $-2$, which speeds up the input into $K$, which compresses the graph horizontally.
• Finally, $L_i$ is adding $1$, which is going into $K$, which shifts the graph to the left.

$\blacktriangleright$ Range - vertical : $L_o = \frac {1}{2}t + 2$

• The leading coefficient for $L_o$ is $\frac {1}{2}$. The graph is going to be compressed vertically.
  
• Finally, the graph will be shifted up $2$.

Graph of $y = T(m) = (L_o \circ K \circ L_i)(m)$.

Note: All of the endpoints remained the same type.

• The short arm of the “Vee” is a solid dot on both graphs.
• The long arm of the “Vee” is a hollow dot on both graphs.
• The end of the horizontal line segment nearest the “Vee” is a solid dot on both graphs.
• The end of the horizontal line segment nearest the parabola is a hollow dot on both graphs.
• The inside endpoint on the parabola is hollow on both graphs.
• The outside endpoint on the parabola is solid on both graphs.

All horizontal measurements are half what they were, so that when $L_i$ multiplies them by $2$, they get back to their original size for input into $K$.

All height measurements are half what the were to begin with.

We can trace the solid left enpoint of the line segment: $(-3, K(-3)) = (-3, -6)$ on the graph of $K$ through the transformations:

$$\begin{align*} \text{Start :} & \text{ } & (-3, K(-3)) \\ \text{Start :} & \text{ } & (-3, -6) \\ \text{Horz Shift Left $1$ :} & \text{ } & (-4,-6) \\ \text{Horz Reflection :} & \text{ } & (4, -6) \\ \text{Horz Compress Factor $\tfrac{1}{2}$ :} & \text{ } & (2, -6) \\ \text{Vert Compress Factor $\tfrac{1}{2}$:} & \text{ } & (2, -3) \\ \text{Vert Shift Up $2$ :} & \text{ } & (2, -1) \\ \text{Composition :} & \text{ } & (2, -1) \end{align*}$$

The solid endpoint of the horizontal line segment will be $(2, -1)$.