## Precalculus (6th Edition) Blitzer

a) The slant asymptote of the given function is $y=x$. b) The graph of the function is:
(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.