## Calculus (3rd Edition)

$y=\ \frac{1}{36}$ is the minimum, when $x=\frac{1}{18}$
Given $$y=-9x^{2}+ x$$ So, we have \begin{aligned} y&=-9x^{2}+x\\ &=-9(x^2-\frac{1}{9}x)\\ &=-9(x^2-\frac{1}{9}x+\frac{1}{18^2}-\frac{1}{18^2})\\ &=-9((x-\frac{1}{18})^2-\frac{1}{324})\\ \end{aligned} The lowest value for the squared term is zero. So, we see that $y=\frac{9}{324}=\frac{1}{36}$ is the minimum, when $x=\frac{1}{18}$.