Answer
$ y=\ \frac{1}{36}$ is the minimum, when $ x=\frac{1}{18}$
Work Step by Step
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}$.
