Let's begin with the graph of the absolute value function $f(x) = |x|$. Since $|x|$ represents the distance of the number $x$ from $0$,
$$|x| = \begin{cases}
x && \text{if $x \geq 0$}\\
-x && \text{if $x < 0$}
\end{cases}
$$
Hence, its graph consists of the two rays $y = x, x \geq 0$ and $y = -x, x < 0$. The common endpoint of these two rays is the vertex of the graph, which is located at the origin.

Observe that $|x| = |-x|$ since the points $x$ and $-x$ are equidistant zero. Hence,
$$|4 - x| = |-(4 - x)| = |-4 + x| = |x - 4|$$
The graph of $y = |x - 4|$ is obtained from the graph of $f(x)$ by translating the graph of $f(x)$ by four units to the right. You should be able to convince yourself of this by making a table of values for $y = |x|$ and $y = |x - 4|$. Another way to see this is to write the piecewise definition @shn stated in the comments
$$|4 - x| = |x - 4| = \begin{cases}
x - 4 && \text{if $x \geq 4$}\\
4 - x && \text{if $x < 4$}
\end{cases}
$$
then graph the rays $y = x - 4, x \geq 4$ and $y = 4 -x, x < 4$.

Restricting the domain $y = |x - 4|$ to $[-3, 6]$ yields the graph of $g(x) = |x - 4|, -3 \leq x \leq 6$.

From its graph, we see that $g$ has minimum value $0$ at $x = 4$ (its vertex) and maximum value $7$ at $x = -3$. Since the function is continuous, it assumes every value between $0$ and $7$ in the interval $[-3, 4]$ and, consequently, in the interval $[-3, 6]$.