Part b reveals one value ($-2$) of $k$ for which the function $f(x + k)$ would result in the arrow clearing the wall. Setting up and solving an inequality will reveal all such values of $k$.

First, identify $f(x + k)$ as defined in this particular case:

$$
\begin{align}
f(x + k) &= 6 - (x + k)^2 \\
&=6-x^2 -2kx-k^2
\end{align}
$$
Substituting $2$ for $x$ yields an expression that represents all possible heights of the second arrow when it reaches the castle wall:

$$
\begin{align}
f(2+k) &= 6 - 2^2 -2k(2)-k^2 \\
&= 6 - 4 - 4k - k^2 \\
&= 2 - 4k - k^2
\end{align}
$$
In order for the second arrow to clear the wall, this expression must be greater than $5$. The values of $k$ for which this is true are found by solving the inequality:

$$
\begin{align}
2 - 4k - k^2 \gt& 5 \\
-3 - 4k - k^2 \gt& 0 \\
k^2 + 4k + 3 \lt& 0 \\
(k + 1)(k + 3) \lt& 0
\end{align}
$$
One of the factors ($k + 1$ or $k + 3$) must be negative, and the other must be positive. The sign of the first factor varies according to three cases: $k + 1$ is negative when $k \lt -1$, zero when $k = -1$, and positive when $k \gt -1$. The sign of the second factor also varies according to three cases: $k + 3$ is negative when $k \lt -3$, zero when $k = -3$, and positive when $k \gt -3$.

The lists above show that a value for $k$ that is less than $-1$ and greater than $-3$ will make $k + 1$ negative and $k + 3$ positive. Additionally, it shows that a value for $k$ that is greater than $-1$ and less than $-3$ will make $k + 1$ positive and $k + 3$ negative. However, since $k$ cannot simultaneously be greater than $-1$ and less than $-3$, only the former scenario is possible: $k \lt -1$ and $k \gt -3$, or $-3 \lt k \lt -1$.

Another approach to solving the inequality $(k + 1)(k + 3) \lt 0$ is to recognize that $k + 1$ is necessarily less than $k + 3$, which requires the following to be true if the two factors have opposite signs:

$$
\begin{align}
k+1 \lt &0 \lt k+3 \\
1 \lt -&k \lt 3 \\
-1 \gt &k \gt -3 \\
-3 \lt &k \lt -1
\end{align}
$$
Therefore the second arrow, when following a path defined by $f(x + k)$, will clear the castle wall if $k$ is greater than $-3$ and less than $-1$.