Chapter 2.6, Problem 22E

Solution

The horizontal and vertical asymptotes of $f\left(x\right)$ and describe the corresponding behaviour using limits.

The line is $y=0$ horizontal asymptotes and $x=0\text{, }x=-3$ is vertical asymptote.

Given:

The function $f\left(x\right)=\frac{x-3}{x^2+3x}$ .

Concept Used:

If a polynomial function in the form $y=\frac{f\left(x\right)}{g\left(x\right)}=\frac{\left(a_n{x}^n+\mathrm{....}\right)}{\left(b_m{x}^m+\mathrm{....}\right)}$ , then, vertical asymptotes is given by real zeros of denominator provided it should not be zero of numerator.

And,

The end behaviour asymptotes given by $\underset{x\to {\infty }^{-}}{\lim }\frac{f\left(x\right)}{g\left(x\right)}\text{ or }\underset{x\to {\infty }^{+}}{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ is horizontal asymptotes.

The condition can be concluded as,

1) If $n<m$ , the end behaviour is horizontal asymptotes that is $y=0$ ,

2) If $n=m$ the end behaviour is horizontal asymptotes is horizontal asymptotes that is $y=\frac{a_n}{b_n}$

3) If $n>m$ ,the quotient $q\left(x\right)$ of $f\left(x\right)/g\left(x\right)$ is the end behaviour asymptotes, where $q\left(x\right)$ is given by $f\left(x\right)=g\left(x\right)q\left(x\right)+r\left(x\right)$ with $r\left(x\right)$ is remainder of $f\left(x\right)/g\left(x\right)$ . Hence, no horizontal asymptotes.

Calculation:

Consider the function, $f\left(x\right)=\frac{x-3}{x^2+3x}$ .

To find vertical asymptotes, find the zeros of denominator, $x^2+3x$ :

  $x^2+3x=x\left(x+3\right)$ .

The zeros of denominator is $x=0\text{ and }x=-3$

Means, the vertical asymptotes are $x=0\text{ and }x=-3$ .

To find end behaviour asymptotes, compute, $\underset{x\to {\infty }^{-}}{\lim }f\left(x\right)$ .

  $\begin{array}{c}\underset{x\to {\infty }^{-}}{\lim }f\left(x\right)=\underset{x\to {\infty }^{-}}{\lim }\frac{x-3}{x^2+3x}\\ =\underset{x\to {\infty }^{-}}{\lim }\frac{x\left(1-\frac{3}{x}\right)}{x\left(x+3\right)}\\ =\underset{x\to {\infty }^{-}}{\lim }\frac{\left(1-\frac{3}{x}\right)}{\left(x+3\right)}\\ =0\end{array}$

Hence, end behaviour is horizontal asymptotes $y=0$ .

Conclusion:

The line is $y=0$ horizontal asymptotes and $x=0\text{, }x=-3$ is vertical asymptote.