Answer
minimum value: $-22$ when $x=3$.
Work Step by Step
Let's compare $f(x)=3x^2-18x+5$ to $f(x)=ax^2+bx+c$. We can see that $a=3, b=-18, c=5$. $a\gt 0$, hence the graph opens up, and so its vertex is a minimum. The minimum value is at $x=-\frac{b}{2a}=-\frac{-18}{2\cdot3}=3.$ Hence, the minimum value is $f(3)=3(3)^2-18(3)+5=-22.$
