## Calculus (3rd Edition)

$y=0$
Given $$y=x^{2}-6 x+9$$ So, we have \begin{aligned} y&=x^{2}-6 x+9\\ &=(x-3)(x-3)\\ &=(x-3)^2 \\ \end{aligned} The smallest value for a square power is zero. So, we see that $y=0$ is the minimum, when $x=3$.