$y=\frac{2x^{2}}{3}+\frac{16x}{3}+\frac{25}{3}$
The parabola params are:$a=\frac{2}{3},b=\frac{16}{3},c=\frac{25}{3}$
$x_{v}=-\frac{b}{2a}$$x_{v}=-\frac{(\frac{16}{3})}{2(\frac{2}{3})}$
Simplify $-\frac{\frac{16}{3}}{2(\frac{2}{3})}:{\quad}-4$$x_{v}=-4$
Plug in $ x_{v}=-4 $to find the $ y_{v} $value$y_{v}=-\frac{7}{3}$
Therefore the parabola vertex is$(-4,-\frac{7}{3})$
If $ a<0, $then the vertex is a maximum value $$\mathrm{Minimum}\:(-4,-\frac{7}{3})$