Answer
The graph of the function is shown below as,
Work Step by Step
The given expression is $\frac{5{{x}^{2}}}{{{x}^{2}}-4}\cdot \frac{{{x}^{2}}+4x+4}{10{{x}^{3}}}$
Now, simplify it as:
$\begin{align}
& \frac{5{{x}^{2}}}{{{x}^{2}}-4}\cdot \frac{{{x}^{2}}+4x+4}{10{{x}^{3}}}=\frac{5{{x}^{2}}}{\left( x+2 \right)\left( x-2 \right)}\cdot \frac{{{\left( x+2 \right)}^{2}}}{10{{x}^{3}}} \\
& =\frac{\left( x+2 \right)}{2x\left( x-2 \right)}
\end{align}$
The simplified equation is $f\left( x \right)=\frac{\left( x+2 \right)}{2x\left( x-2 \right)}$.
Now, use the following steps to plot the graph of the function as,
Step 1: Compute the symmetry.
Find $f\left( -x \right)$ by replacing $x$ with $-x$.
$\begin{align}
& f\left( x \right)=\frac{\left( x+2 \right)}{2x\left( x-2 \right)} \\
& f\left( -x \right)=\frac{\left( -x+2 \right)}{2\left( -x \right)\left( -x-2 \right)} \\
& =\frac{\left( -x+2 \right)}{2\left( x \right)\left( x+2 \right)}
\end{align}$
Thus, the graph is not symmetrical about the $y\text{-}$ axis or origin.
Step 2: Compute the $y\text{-}$ intercept by finding $f\left( 0 \right)$.
Then,
$\begin{align}
& f\left( x \right)=\frac{\left( x+2 \right)}{2x\left( x-2 \right)} \\
& f\left( 0 \right)=\frac{\left( 0+2 \right)}{2\left( 0 \right)\left( 0-2 \right)} \\
& =\infty
\end{align}$
Thus, the graph of the function has no $y\text{-}$ intercept.
Step 3: Compute the $x\text{-}$ intercept.
Evaluate the numerator term equal to zero.
$\begin{align}
& x+2=0 \\
& x=-2
\end{align}$
The graph passes through the point $\left( -2,0 \right)$.
Step 4: Determine the vertical asymptote.
Solve the denominator term equal to zero.
$\begin{align}
& 2x\left( x-2 \right)=0 \\
& x=0,2
\end{align}$
Thus, the function has two vertical asymptotes and their equations are $x=0$ and $x=2$.
Step 5: Find the horizontal asymptote.
The degree of the numerator and denominator term is different. There is no horizontal asymptote. The equation is $y=0$.
