If $x^2-(a-3)x+a=0$ has atleast one positive root then 

$x^2-(a-3)x+a=0$
has at least one positive root,
$\therefore \text{ Roots are real }\Rightarrow (a-3{)}^2-4a\ge 0$
$\Rightarrow a\le 1\text{ or }a\ge 9$

Let both roots be negative or zero. Then
$f\left(0\right)=a\ge 0\text{ and }\alpha +\beta =a-3\le 0\text{ i.e },a\le 3$
At least one root is positive for
$\begin{array}{c}(-\mathrm{\infty },1]\cup [9,\mathrm{\infty })-[0,3]\\ \text{ i.e., when }a\in (-\mathrm{\infty },0)\cup [9,\mathrm{\infty })\end{array}$