15. (12th "Hope Cup" Senior High School Competition Problem) Given that the function $y=f(x)$ is defined on $\mathbf{R}$ with the smallest positive period $T=5$, it is a linear function on $[0,1]$. The graph is symmetric about the origin on $[-1,1]$, and it is a quadratic function on $[1,4]$, achieving its minimum value of -5 at $x=2$. Try to find the analytical expression for $y=f(x)$.
15. Let $x \in [0,1]$, then $f(x) = kx + b$. According to the problem, $f(0) = 0$, so $b = 0$.

For $x \in [-1,0)$, $f(x) = -f(-x) = -k(-x) = kx$.
Therefore, for $x \in [-1,1]$, $f(x) = kx$.
Let $x \in [1,4]$, then $f(x) = a(x-2)^2 - 5$, so $f(1) = k = a - 5$.
Also, $f(4) = 4a - 5$, and $f(4) = f(-1) = -k$,
Thus,
$$
4a - 5 = -k.
$$

From (1) and (2), we get $a = 2$, $k = -3$.
$$
\begin{array}{l}
f(x) = \left\{\begin{array}{ll}
2(x-2)^2 - 5, & (1 \leq x \leq 4), \\
-3x, & (-1 \leq x < 1).
\end{array} \quad\right. \text{ By periodicity, we get } \\
f(x) = \left\{\begin{array}{ll}
-3(x - 5l), & 5l - 1 \leq x < 5l + 1, \\
2(x - 5l - 2)^2 - 5, & 5l + 1 \leq x \leq 5l + 4,
\end{array} (l \in \mathbf{Z}).\right.
\end{array}
$$