= Solution
The graph of $F$ is the symmetric tent with vertices
$$
(-1,1-\mu),\qquad(0,1),\qquad(1,1-\mu).
$$
If $0<\mu<1$, then $F\geq0$ and
$$
F^2(x)=1-\mu+\mu^2|x|,
$$
so $F^2$ is an upward-pointing V with its minimum at $(0,1-\mu)$ and equal endpoint values $1-\mu+\mu^2$.
If $1<\mu<2$, the zeros of $F$ are $\pm1/\mu$ and
$$
F^2(x)=
\begin{cases}
1-\mu+\mu^2|x|,&|x|\leq1/\mu,\\
1+\mu-\mu^2|x|,&1/\mu\leq|x|\leq1.
\end{cases}
$$
Thus its five vertices are
$$
\boxed{(-1,1+\mu-\mu^2),\quad(-1/\mu,1),\quad(0,1-\mu),\quad(1/\mu,1),\quad(1,1+\mu-\mu^2).}
$$
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