Suppose that $f(x)=a(x-b)(x-c)$ is a quadratic function where $a, b$ and $c$ are distinct positive integers less than 10. For each choice of $a, b$ and $c$, the function $f(x)$ has a minimum value. What is the minimum of these possible minimum values?
Since $a>0$, each parabola of the given form opens upwards and so its minimum occurs at its vertex.

The $x$-coordinate of the vertex will be the average of the $x$-coordinates of the roots, which in this case are $x=b$ and $x=c$. Thus, the $x$ coordinate of the vertex is $x=\frac{b+c}{2}$.

Therefore, the minimum value of $f(x)$ is the value of $f(x)$ when $x=\frac{b+c}{2}$, which is

$$
f\left(\frac{b+c}{2}\right)=a\left(\frac{b+c}{2}-b\right)\left(\frac{b+c}{2}-c\right)=a\left(\frac{-b+c}{2}\right)\left(\frac{b-c}{2}\right)=-\frac{a(b-c)^{2}}{4}
$$

Since $a$ and $(b-c)^{2}$ are non-negative, then $-\frac{a(b-c)^{2}}{4}$ is minimized when $a(b-c)^{2}$ is maximized. Since $a, b$ and $c$ are distinct positive integers less than 10 , then $a \leq 9$ and $-8 \leq b-c \leq 8$.

We suppose, without loss of generality, that $b>c$ and so $0<b-c \leq 8$.

If $b-c=8$, then we must have $b=9$ and $c=1$. Since $a, b$ and $c$ are distinct, then we cannot have both $a=9$ and $b-c=8$.

The maximum of $a(b-c)^{2}$ could occur when $a=9$ and $b-c=7$ (the maximum possible value if $a=9$ ) or when $a=8$ and $b-c=8$. Any other combination of values of $a$ and $b-c$ cannot give the maximum because we could increase one value or the other.

When $a=9$ and $b-c=7$ (which means $b=8$ and $c=1$ ), the value of $a(b-c)^{2}$ is 441 .

When $a=8$ and $b-c=8$ (which means $b=9$ and $c=1$ ), the value of $a(b-c)^{2}$ is 512 .

Therefore, the minimum of the minimum values of $f(x)$ is $-\frac{512}{4}$ which equals -128 .