Answer
a) The slant asymptote of the given rational function is $y=x$.
b) Graph the function as:
Work Step by Step
(a)
Consider the rational function $f\left( x \right)=\frac{{{x}^{2}}-4}{x}$. Here, the degree of the numerator term is 2 and the degree of denominator term is 1. $x=0$ is not a factor of the function ${{x}^{2}}-4$ , since the graph of the function has a slant asymptote.
Now, divide the function ${{x}^{2}}-4$ with $x$.
So,
$\frac{{{x}^{2}}-4}{x}=x-\frac{4}{x}$
Therefore, the equation has a slant asymptote at $y=x$.
(b)
Consider the rational function $f\left( x \right)=\frac{{{x}^{2}}-1}{x}$.
Here, the degree of the numerator term is 2 and the degree of denominator term is 1. And $x=0$ is not a factor of the function ${{x}^{2}}-4$ , since the graph of the function has a slant asymptote.
Step 1:
Substitute $x=-x$
$\begin{align}
& f\left( -x \right)=\frac{{{\left( -x \right)}^{2}}-4}{\left( -x \right)} \\
& =-\frac{{{x}^{2}}-4}{x} \\
& =-f\left( x \right)
\end{align}$
The graph is symmetrical about the $y$ axis.
Step 2: To calculate the x intercepts equate $f\left( x \right)=0$.
$\begin{align}
& \frac{{{x}^{2}}-4}{x}=0 \\
& x=\pm 2
\end{align}$ ,
Step 3: To calculate the y intercepts evaluate $f\left( 0 \right)$.
$f\left( 0 \right)=\infty $
Step 4: Since the degree of the numerator is greater than the denominator, there is no horizontal asymptote.
Step 5: For the vertical asymptote, equate the denominator to 0.
$x=0$
