Answer
a) The slant asymptote of the given function is $y=x$.
b)
The graph of the function is:
Work Step by Step
(a)
Consider the rational function $f\left( x \right)=\frac{{{x}^{2}}-x+1}{x-1}$.
Here, the degree of the numerator term is 2 and degree of the denominator term is 1. And $x=0$ is not a factor of the function ${{x}^{2}}+x-6$ , since the graph of the function has a slant asymptote.
Now, divide the function ${{x}^{2}}+x-6$ with $x-1$.
$\frac{{{x}^{2}}-x+1}{x-1}=x+\frac{1}{x-1}$
Therefore, the equation has a slant asymptote at $y=x$.
(b)
Consider the rational function $f\left( x \right)=\frac{{{x}^{2}}-x+1}{x-1}$.
Here, the degree of the numerator term is 2 and degree of the denominator term is 1. And $x=0$ is not a factor of the function ${{x}^{2}}-x+1$ , since, the graph of the function has a slant asymptote.
Substitute $x=-x$
$\begin{align}
& f\left( -x \right)=\frac{{{\left( -x \right)}^{2}}-\left( -x \right)+1}{\left( -x \right)-1} \\
& =\frac{{{x}^{2}}+x+1}{-x-1}
\end{align}$
The graph is not symmetrical about the $y$ axis.
Compute the $y$ intercept.
Put $x=0$
$\begin{align}
& f\left( x \right)=\frac{{{x}^{2}}-x+1}{x-1} \\
& f\left( 0 \right)=\frac{{{\left( 0 \right)}^{2}}-\left( 0 \right)+1}{\left( 0 \right)-1} \\
& =-1
\end{align}$
The $y$ intercept is $y=-1$.
Compute the $x$ intercept.
Put $p\left( x \right)=0$ then,
$\begin{align}
& {{x}^{2}}-x+1=0 \\
& x=\frac{1+\sqrt{1-4}}{2} \\
& =\frac{1+\sqrt{3}i}{2},\frac{1-\sqrt{3}i}{2}
\end{align}$
There is no $x$ intercept.
Now, compute the vertical asymptotes.
Set $q\left( x \right)=0$
$\begin{align}
& x-1=0 \\
& x=1
\end{align}$
Thus, the graph has a vertical asymptote as $x=1$.
Determine the horizontal asymptotes.
There is no horizontal asymptote because the degree of the numerator term is not equal to the degree of the denominator term.
