Calculate: 2x=9

Expression: $$

Rewrite $ y=\frac{2}{3}x^{2}+\frac{16}{3}x+\frac{25}{3} $in the form $ y=ax^{2}+bx+c$

$y=\frac{2x^{2}}{3}+\frac{16x}{3}+\frac{25}{3}$

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 $
$If $ a>0, $then the vertex is a minimum value $
a=\frac{2}{3}$

$\mathrm{Minimum}\:(-4,-\frac{7}{3})$