20. Given $f(x)=a x^{2}+b x$, satisfying $1 \leqslant f(-1) \leqslant 2$ and $2 \leqslant f(1) \leqslant 4$, find the range of $f(-2)$.
20. From $f(1)=a+b, f(-1)=a-b$, we can solve for:
$$
a=\frac{1}{2}(f(1)+f(-1)), b=\frac{1}{2}(f(1)-f(-1)).
$$

Substituting the above two equations into $f(x)=a x^{2}+b x$, and simplifying, we get
$$
f(x)=f(1)\left(\frac{x^{2}+x}{2}\right)+f(-1)\left(\frac{x^{2}-x}{2}\right) \text {, }
$$

Thus, $f(-2)=f(1)+3 f(-1)$.
Given that $2 \leqslant f(1) \leqslant 4,1 \leqslant f(-1) \leqslant 2$, it follows that $5 \leqslant f(-2) \leqslant 10$.