The minimum value of $(x-2)(x-9)$ is

1. Let $f(x)=(x-2)(x-9)$. Expanding the polynomial gives:
$$f(x)=x^2-11x+18$$
This is a quadratic expression of the form $a{x}^2+bx+c$ with $a=1$, $b=-11$, and $c=18$. Since $a>0$, the parabola opens upwards and has a minimum value at its vertex where $x=-\frac{b}{2a}$:
$$x=-\frac{-11}{2(1)}=\frac{11}{2}$$
Substitute $x=\frac{11}{2}$ back into the factored expression to compute the minimum value:
$$\text{Minimum Value}=\left(\frac{11}{2}-2\right)\left(\frac{11}{2}-9\right)$$
$$\text{Minimum Value}=\left(\frac{7}{2}\right)\left(\frac{-7}{2}\right)=-\frac{49}{4}$$