How do you graph $f\left( x \right)=\dfrac{{{x}^{3}}+1}{{{x}^{2}}-4}?$

Hint: As the above given fraction mentioned in question is in form of a fraction so it a type of rational fractions,
While solving it make sure you are not equating the denominator of the fractions with zero, as the denominator of rational fraction will not be equal to zero.
Solve, The denominator cannot be equal to zero (0), solve expressions given in numerator and denominator separately, with the help of algebraic equations, and determine the value of the expression.

Complete step by step solution: We know that, $f\left( x \right)=\dfrac{{{x}^{3}}+1}{{{x}^{2}}-4}$
Here, denominator $\ne 0$
Then we will be able to draw the graph when
${{x}^{2}}-4\ne 0$
$\therefore {{x}^{2}}-{{2}^{2}}\ne 0$
Using the formula, ${{a}^{2}}-{{b}^{2}}=\left( a-b \right).\left( a+b \right)$
$\Rightarrow \therefore \left( x-2 \right)\left( x+2 \right)\ne 0$
$\Rightarrow x=2$ and $x=-2$
We name the two straight lines $x=2$ and $x=-2$ vertical means points asymptotes of $f\left( x \right).$
i.e. that the cause of $f\left( x \right)$ zero crosses this lines.
Numerator $={{x}^{3}}+1$
$f\left( x \right)=0$
Numerator $=0$
${{x}^{3}}+1=0$
${{x}^{3}}=0-1$
${{x}^{3}}=-1$
$x=\sqrt[3]{-1}$
$x=-1$
Then:- $\left( -1,0 \right)$ will be the point of cross.
Hence in this way we can graph
$f\left( x \right)={{x}^{3}}+1$
${{x}^{2}}-4$

Additional Information:
Rational functions are those functions which are in the form of p/q but value of q will never be equal to zero.
$\left( 1 \right)$ Rational function are functions, which are created by dividing the function. Formally, they are represented as $f\left( x \right),g\left( x \right)$
Where $f\left( x \right)$ and $g\left( x \right)$ are both functions.
For example: $\dfrac{2{{x}^{2}}+3x-5}{5x-7}$ is rational.
At function where, $f\left( x \right)=2{{x}^{2}}+3x-5$ and $g\left( x \right)=5x-7$
$\left( 2 \right)$ Asymptotes: A rational number cannot touch the Asymptotes, if you make $x=0$ in the function. In a calculator, you may get a divide by $0$ error. This happens when rational numbers touch vertical asymptotes.

Note:
$\left( 1 \right)$ Remember, the denominator cannot equal to $0.$ because if you take denominators equal to $0$ in the calculator you may get an error.
$\left( 2 \right)$ Solve the numerator and denominator of $f\left( x \right)$ separately because it gets easy to solve and find the point of $f\left( x \right)$