Chapter 2, Problem 133CR

To determine

To graph: the function $f(x)$ and check for intercept and various asymptotes.

Answer to Problem 133CR

y-intercept is $y=\frac{1}{3}$

x-intercept is $x=1.61,-0.61$

Vertical asymptote is at $x=3$

No horizontal asymptote

No holes

Slant asymptotes is $f(x)=y=x+2$

Explanation of Solution

Given information:

  $f(x)=\frac{x^2-x-1}{x-3}$

Graph: Assuming the value of x to find $f(x)$ and plot the graph

  

  

Interpretation :

To determine y-intercept put $x=0$ and solve for $y$ or $f(x)$

  $\begin{array}{c}f(x)=\frac{x^2-x-1}{x-3}\\ =\frac{0-0-1}{0^2-3}\\ =\frac{1}{3}\end{array}$

To determine x-intercept put $y=0$ and solve for $x$

  $\begin{array}{c}f(x)=\frac{x^2-x-1}{x-3}\\ 0=\frac{x^2-x-1}{x-3}\\ x^2-x-1=0\\ x=\frac{1\pm \sqrt{1^2-4\times 1\times -1}}{2\times 1}\\ =\frac{1\pm \sqrt{5}}{2}\\ =1.61,-0.61\end{array}$

To find the vertical asymptotes we have to solve the denominator by equating it equal to zero:

  $x-3=0\Rightarrow x=3$

A horizontal asymptotes is defined when the degree of the denominator is greater than or equal to degree of the numerator.

Here the degree of denominator is less than numerator therefore we can not calculate

horizontal asymptote.

Slant asymptote occur when the degree of denominator is lower than that of the numerator.

Therefore it can be calculated by solving the function in to simplest form by separating both numerator and denominator and is given as

  $f(x)=\frac{x^2-x-1}{x-3}\Rightarrow 0=x+2+\frac{5}{x^2-4}$

Therefore the slant asymptote is the straight line $f(x)=y=x+2$

Now, the degree of the denominator is less than the degree of the numerator therefore there will be no hole possible in the graph.

In the given function there is no common factor found both at numerator and denominator

Therefore there is no hole for the given function.