The extreme are found by differentiating f(x): 3x^2-12x+9 and solving for f'(x)=0. Divide f'(x)=0 through by 3:
(x-1)(x-3)=0. f(1)=5; f(3)=1. Differentiating again: f''(x)=6(x-2). f''(1)=-6 and f''(3)=6, therefore (1,5) is a maximum, and (3,1) is a minimum. The graph shows f(x)<0 when x<-0.104 (approx), crosses the vertical axis at (0,1), reaches the maximum at (1,5) dipping down to the minimum at (3,1), and rising thereafter (f(x)>0). For x>-0.104 f(x) remains positive.