Answer
See explanation.
Work Step by Step
(a) $x$ -intercept: ±2
$x$ -intercept has $f(x)=0$, and a fraction is zero if the numerator is zero.
(b) $0=x$
Vertical asymptotes are at the zeros of the denominator
$(\mathrm{c}) \lim _{x \rightarrow \pm \infty} $$f(x)=0$, so $y=0$ is a horizontal asymptote.
Horizontal asymptotes are the limits of the function
as $x \rightarrow \pm \infty$
(d) ($-\infty,-2$) and $(2,+\infty)$
We have identified critical points in (a) and (b), so evaluate on which interval the function is positive
(e) (-4,4)
The first derivative is $ x = ± 4 $; we can evaluate the first derivative as positive, so the function $f$ is increasing.
$(\mathrm{f})$ $(\sqrt{\frac{176}{5}},+\infty)$ and $(-\infty,-\sqrt{\frac{176}{5}})$
If the second derivative is positive, the function $f$ is concave up
\[
(\mathrm{g}) \pm \sqrt{\frac{176}{5}}
\]
Inflection points are the zeros of the second derivative
