$27 \cdot 10$ Given the function $f(x)=a x^{2}-c$ satisfies
$$
-4 \leqslant f(1) \leqslant-1, \quad-1 \leqslant f(2) \leqslant 5 .
$$

Then, $f(3)$ should satisfy
(A) $-7 \leqslant f(3) \leqslant 26$.
(B) $-4 \leqslant f(3) \leqslant 15$.
(C) $-1 \leqslant f(3) \leqslant 20$.
(D) $-\frac{28}{3} \leqslant f(3) \leqslant \frac{35}{3}$.
(China High School Mathematics League, 1983)
[Solution 1] From the problem, we have $-4 \leqslant a-c \leqslant-1$, and $-1 \leqslant 4 a-c \leqslant 5$.
Using $a, c$ as the horizontal and vertical axes, respectively, draw the lines $c=a+1, c=a+4$, and $c=4 a-5, c=4 a+1$. The region enclosed by these four lines satisfies the conditions given in the problem.

Next, draw the family of parallel lines $c=9 a-k$ (since $f(3)=9 a-c$), and the values of $k$ are obtained at the vertices $A(0,1), C(3,7)$ of the region.

This gives $-1 \leqslant k=9 a-c \leqslant 20$, i.e.,
$$
-1 \leqslant f(3) \leqslant 20 \text {. }
$$

Therefore, the answer is $(C)$.
[Solution 2] From the problem, $f(x)=a x^{2}-c$, then
$$
\left\{\begin{array}{l}
f(1)=a-c, \\
f(2)=4 a-c,
\end{array} \text { solving gives } \left\{\begin{array}{l}
a=\frac{f(2)-f(1)}{3}, \\
b=\frac{f(2)-4 f(1)}{3} .
\end{array}\right.\right.
$$

Therefore, $\quad f(3)=9 a-c=\frac{8}{3} f(2)-\frac{5}{3} f(1)$:
Also, $-4 \leqslant f(1) \leqslant-1,-1 \leqslant f(2) \leqslant 5$, then
$$
-1 \leqslant \frac{8}{3} f(2)-\frac{5}{3} f(1) \leqslant 20 \text {, }
$$

i.e.,
$$
-1 \leqslant f(3) \leqslant 20 \text {. }
$$

Therefore, the answer is $(C)$.