Answer
$$
f(x)=2x+\frac{10}{x}
$$
The domain is
$$(-\infty, 0) \cup(0, \infty)$$
The graph is symmetric about the origin.
The critical numbers are: $x =\sqrt{5} $ and $x=-\sqrt{5}$ .
Critical point:
$$
(\sqrt{5} , 4\sqrt{5} ) , \quad (-\sqrt{5}, - 4\sqrt{5})
$$
The function $f$ Increasing on
$$
(-\infty, -\sqrt{5}) \quad \text{and} \quad(\sqrt{5},\infty)
$$
Decreasing on $$(-\sqrt{5} , \sqrt{5}).$$
Relative minimum at $\sqrt{5},$ and relative maximum at $ -\sqrt{5}$
There are no inflection points.
$f(x)$ is concave downward on $(-\infty, 0)$ .
$f(x)$ is concave upward on $(0, \infty )$ .
Since $f(x)$ is never zero, so there are no x-intercepts.
Since $f(x)$ does not exist for $x=0 $ so there is no y-intercept.
Vertical asymptotic at $x=0$
$$
\lim\limits_{x \to 2}f(x)=\infty.
$$
Work Step by Step
$$
f(x)=2x+\frac{10}{x}=2x+10x^{-1}
$$
since $f(x)$ does not exist when $x=0,$ the domain is
$$(-\infty, 0) \cup(0, \infty)$$
$$
\begin{aligned}
f(-x) &=2(-x)+\frac{10}{(-x)}\\
&=-\left(2 x+10 x^{-1}\right)\\
&=-f(x)
\end{aligned}
$$
The graph is symmetric about the origin.
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x)&= 2-10 x^{-2} \\
&= 2-\frac{10}{x^{2}} \\
&=\frac{2\left(x^{2}-5\right)}{x^{2}}
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $f^{\prime} =0 $. Now, solve the equation $f^{\prime} =0 $ to get
$$
\begin{aligned}
f^{\prime}(x) & =\frac{2\left(x^{2}-5\right)}{x^{2}}=0 \\
\Rightarrow \quad\quad\quad\quad\\
x=\pm \sqrt{5}
\end{aligned}
$$
So,
the critical numbers are: $x =\sqrt{5} $ and $x=-\sqrt{5}$ .
To find the critical points substituting in $f$ as follows:
$$
f(\sqrt{5})=2(\sqrt{5})+\frac{10}{(\sqrt{5})}= 4\sqrt{5}
$$
$$
f(-\sqrt{5})=2(-\sqrt{5})+\frac{10}{(-\sqrt{5})}=- 4\sqrt{5}
$$
So, Critical point:
$$
(\sqrt{5} , 4\sqrt{5} ) , \quad (-\sqrt{5}, - 4\sqrt{5})
$$
Now, we can use the first derivative test. Check the sign of $f^{\prime}(x)$. in the intervals
$$
(-\infty, -\sqrt{5}), \quad ( -\sqrt{5}, 0), \quad ( 0 , \sqrt{5}) \quad \text {and }\quad ( \sqrt{5}, \infty ) .
$$
(1)
Test a number in the interval $(-\infty, -\sqrt{5}) $ say $-3$:
$$
\begin{aligned}
f^{\prime}(-3) &=\frac{2\left((-3)^{2}-5\right)}{(-3)^{2}} \\
&=\frac{8}{9} \approx 0.8 \\
& \gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(-\infty, -\sqrt{5})$.
(2)
Test a number in the interval $( -\sqrt{5}, 0) $ say $-1$:
$$
\begin{aligned}
f^{\prime}(-1) &=\frac{2\left((-1)^{2}-5\right)}{(-1)^{2}} \\
&=-8 \\
& \lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $( -\sqrt{5}, 0) $.
(3)
Test a number in the interval $( 0, \sqrt{5}) $ say $1$:
$$
\begin{aligned}
f^{\prime}(1) &=\frac{2\left((1)^{2}-5\right)}{(1)^{2}} \\
&=-8 \\
& \lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $( 0, \sqrt{5}) $.
(4)
Test a number in the interval $( \sqrt{5}, \infty) $ say $ 3$:
$$
\begin{aligned}
f^{\prime}(3) &=\frac{2\left((3)^{2}-5\right)}{(3)^{2}} \\
&=\frac{8}{9} \approx 0.8 \\
& \gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $( \sqrt{5}, \infty)$.
So, the function $f$ Increasing on
$$
(-\infty, -\sqrt{5}) \quad \text{and} \quad(\sqrt{5},\infty)
$$
Decreasing on $$(-\sqrt{5} , \sqrt{5}).$$
and we find that:
relative minimum at $\sqrt{5},$ and relative maximum at $ -\sqrt{5}$
Set $f^{\prime\prime}(x)$ equal to $0$ and solve for $x$.
$$
\begin{aligned}
f^{\prime \prime}(x) &=20 x^{-3}=\frac{20}{x^{3}} \\
\Rightarrow\quad\quad\quad\quad\quad\\
f^{\prime \prime}(x) \text { is never equal to zero.}
\end{aligned}
$$
so
There are no inflection points.
(Recall that $f(x)$ does not exist at $x=0$)
Now, we can use the second derivative test. Check the sign of $f^{\prime\prime}(x)$. in the intervals
$$
(-\infty, 0), \quad \quad ( 0 ,\infty ).
$$
$$
\begin{array}{l}
f^{\prime \prime}(-1)=20 (-1)^{-3} =-20 \lt 0 \\
f^{\prime \prime}(1)=20 (1)^{-3} =20 \gt 0 \\
\end{array}
$$
Test a number in the interval $(-\infty, 0)$ we find $f^{\prime\prime}(x)$ is negative in these interval, so $f(x)$ is concave downward on $(-\infty, 0)$ .
Test a number in the interval $(0, \infty )$ we find $f^{\prime\prime}(x)$ is positive in these interval, so $f(x)$ is concave upward on $(0, \infty )$ .
Since $f(x)$ is never zero, so there are no x-intercepts.
y-intercept:
Since $f(x)$ does not exist for $x=0 $ so there is no y-intercept.
Vertical asymptotic at $x=0$
$$
\lim\limits_{x \to 2}f(x)=\infty
$$
