Graph of $y = P(t) = |t+4|$.

The graph is positioned by its corner, which occurs when the inside of the absolute value formula equals $0$. $t+4=0$, when $t=-4$.

$P$ has a global minimum of $\answer {0}$, which occurs at $\answer {-4}$. $P$ has no maximums.

$P$ decreases on $(-\infty , -4]$ and increases on $[-4, \infty )$

$$P(t) = |t+4| = \begin {cases} -(t+4) & \text {if $t<\answer {-4}$,}\\ t+4 & \text {if $t\ge \answer {-4}$}. \end {cases}$$

Let $B(r) = |r-3| + 2$.

$\blacktriangleright$ Solve $B(r) = 5$.

We can see from the graph that there are two solutions to this equation. Our first goal in solving this equation is to get the absolute term by itself.

$|r-3| + 2 = 5$

$|r-3| = 3$

There are only two numbers whose absolute value is $3$. They are $\answer {-3}$ or $\answer {3}$.

We have two possibilities, either the inside of the absolute value, $r-3=-3$ or $r-3=3$.

If $r-3=-3$, then $r=0$. If $r-3=3$, then $r=6$.

We can also turn to the definition of the absolute value function.

The function $|r-3| + 2$ can be written as

$$|r-3| + 2 = \begin {cases} -(r-3) + 2 & \text {if $t<\answer {3}$,}\\ r-3 + 2 & \text {if $t\ge \answer {3}$}. \end {cases}$$

We are looking for where $-(r-3) + 2 = 5$ and $t < 3$

or

where $(r-3) + 2 = 5$ and $t \ge 3$.

These produce the solutions $r=0$ and $r=6$