Answer
$f(x) $ increasing on $(-\infty,7/2) $ and decreasing on $(7/2,\infty) $
$x= 7/2$ is a local maximum
Work Step by Step
Given $$y=-x^{2}+7 x-17 $$
Since
$$f'(x) = -2x+7$$
Then $f(x)$ has critical points when
\begin{align*}
f'(x)&=0\\
-2x+7&=0
\end{align*}
Then $x= 7/2$ is a critical point. To find the intervals where $f$ is increasing and decreasing, choose $x=0$, $x= 4$
\begin{align*}
f'(0)&= 7>0\\
f'(4)&= -1<0
\end{align*}
Hence, $f(x) $ is increasing on $(-\infty,7/2) $ and decreasing on $(7/2,\infty) $. Since $f(x)$ changes from increasing to decreasing, then $x= 7/2$ is a local maximum.
