Answer
f is decreasing in $(-\infty, \frac{-9}{2})$ and f is increasing in $(\frac{-9}{2},+\infty)$
Work Step by Step
$f(x)=x^{2}+9x+8$
$f'(x)=2x+9$
$f'(x)=0 \rightarrow 2x+9=0 \rightarrow x=\frac{-9}{2}$
a>0 so we have: f is decreasing in $(-\infty, \frac{-9}{2})$ and f is increasing in $(\frac{-9}{2},+\infty)$
