Answer
$f(x) $ is decreasing on $(-\infty,-0.6) $ and increasing on $(-0.6,\infty) $ .
$x= -0.6$ is a local minimum
Work Step by Step
Given $$y=5 x^{2}+6 x-4$$
Since
$$f'(x) = 10x+6$$
Then $f(x)$ has critical points when
\begin{align*}
f'(x)&=0\\
10x+6&=0
\end{align*}
Then $x= -0.6$ is a critical point. To find the interval where $f$ is increasing and decreasing, choose $x=0$, $x= -1$
\begin{align*}
f'(0)&= 6>0\\
f'(-1)&= -4<0
\end{align*}
Hence, $f(x) $ is decreasing on $(-\infty,-0.6) $ and increasing on $(-0.6,\infty) $. Since $f(x)$ changes from decreasing to increasing, then $x= -0.6$ is a local minimum.
