# 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 $a=\frac{2}{3}$

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

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