Below is the piecewise defined function, $T(v)$. $v$ is representing the domain values in $(-4,-1] \cup [1,7)$. $T(v)$ represents the range number paired with $v$. Therefore, $T(v)$ represents numbers in $(-9, 2]$.

$$T(v) = \begin {cases} 2v-1 & \text { if } -4 < v \leq -1 \\ -v+3 & \text { if } 1 \leq v < 7 \end {cases}$$

Graph of $y = T(v)$.

The domain of $T$ has two maximal intervals, $(-4,-1]$ and $[1,7)$. These correspond to two line segments on the graph. The endpoints give us four strategic points on the graph:

• $(-4, -9)$, which is an open point on the graph.
• $(-1, -3)$, which is a closed point on the graph.
• $(1, 2)$, which is a closed point on the graph.
• $(7, -4)$, which is an open point on the graph.

The function $T$ has no minimum value. It has a global maximum of $2$, which occurs at $1$. This is also a local maximum. There is another local maximum value of $-3$, which occurs at $-1$.

$\blacktriangleright$ $B(k) = T(k)+4$ with the induced domain.

First Question: Domain

In the definition of $B(k)$, we see $T(k)$, which means $k$ represents the domain of $T$, which is $(-4,-1] \cup [1,7)$. $k$ is also representing the domain of $B$. Therefore, the domain $B$ is also $(-4,-1] \cup [1,7)$.

The formula for $B$ is the formula for $T$ with $4$ added.

$$B(k) = \begin {cases} (2k-1)+4 = 2k+3 & \text { if } -4 < k \leq -1 \\ (-k+3)+4 = -k+7 & \text { if } 1 \leq k < 7 \end {cases}$$

Graph of $z = B(k)$.

The shape of the graph has not changed. It just slid up.

Second Question: Behavior

The function $B$ is increasing on the interval $(-4,-1]$, just like $T$.
The function $B$ is decreasing on the interval $[-1, 7)$, just like $T$.

Define $V(h)$ as

$$V(h) = \begin {cases} -2h-3 & [-6, -2] \\ -(h+3)(h-3) & (-2, 4] \\ \frac {7h}{4} - 8 & (4,6) \end {cases}$$

A graph always helps our thinking. Here is the graph of $y = V(h)$.

• The domain of $V$ is $[-6,6)$.
• $V$ has a global maximum of $9$, which occurs at $\answer {-6}$ and at $\answer {0}$.
• $V$ has a global minimum of $-7$, which occurs at $4$.
• $V$ has a local minimum of $1$, which occurs at $-2$.
• $V$ has a jump discontinuity at $-2$.
• $V$ has a jump discontinuity at $4$.
• $V$ is decreasing on $[-6, -2]$.
• $V$ is increasing on $[-2, 0]$.
• $V$ is decreasing on $\left [\answer {0}, \answer {4}\right ]$.
• $V$ is increasing on $[4, 6)$.

Let’s define a new function based on $V$.

Let $f(x) = V(x)-2$ with the induced domain.

The domain of $f$ is $\left [\answer {-6}, \answer {6}\right )$, same as $V$. The range or function values $f$ are all $\answer {2}$ less than the function values of $V$.

$$f(x) = \begin {cases} -2x-5 & [-6, -2] \\ -(x+3)(x-3)-2 & (-2, 4] \\ \frac {7x}{4} - 10 & (4,6) \end {cases}$$

• The domain of $f$ is $[-6,6)$.
• $f$ has a global maximum of $7$, one of which occurs at $-6$ and the other occurs at $\answer {0}$.
• $f$ has a global minimum of $-9$, which occurs at $4$.
• $V$ has a local minimum of $-1$, which occurs at $-2$.
• $f$ has a jump discontinuity at $-2$.
• $f$ has a jump discontinuity at $4$.
• $f$ is decreasing on $[-6, -2]$.
• $f$ is increasing on $[-2, 0]$.
• $f$ is decreasing on $\left [\answer {0}, \answer {4}\right ]$.
• $f$ is increasing on $[4, 6)$.

The places in the domain where characteristics and features occur did not changed. The maximums and minimums have all dropped by $2$. The endpoints are all still solid or hollow. Their vertical coordinates have all dropped by $2$.