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

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

Let $C(z) = (L \circ K)(z) = L(K(z))$ with the implied domain.

What is $L$ going to do to this graph?

$L$ is a linear function, which means it is not going to change the shapes. There will be a “Vee”. There will be a horizontal line segment. There will be a parabola. $L$ is the outside function, which means it cannot affect horizontal features. The “Vee” will be on the left. The parabola will be on the right. The horizontal line segment will be in the middle.

$L$ is going to affect the vertical measurements.

• The leading coefficient for $L$ is negative. The graph is going to be reflected vertically.
  The “Vee” will open down. The parabola will open up.
• This reflecting will happen after the graph is move $3$ up.
• Finally, the graph will be shifted down $2$.

Graph of $y = (L \circ K)(m)$.

Note: All of the endpoints are 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.

We can also think of $(L \circ K)(m)$ algebraically:

$$(L \circ K)(m) = -(K(m)+3) - 2$$

Let $a$ be in the domain of $L \circ K$. We can think of an existing value of $K(a)$. $3$ is added to this value. The result is negated and then $2$ is subtracted.

Graphically, the coordinates of a point begin as $(a, K(a))$. Then they go through the same steps:

• $(a, K(a))$
• $(a, K(a) + 3)$
• $(a, -(K(a) + 3))$
• $(a, -(K(a) + 3) - 2)$