Chapter 7.4, Problem 110E

To determine

The graph of the given function and hence find its asymptotes.

Answer to Problem 110E

The given function has two asymptotes. One vertical asymptote $x=-1$ and one oblique asymptote $y=x-1$ and it has no horizontal asymptote.

Explanation of Solution

Given: The given function is $f(x)=\frac{x^2-36}{x+1}$ .

Concept Used: Consider any rational function $f(x)=\frac{g(x)}{h(x)}$ .

Vertical asymptotes of the function are given by $h(x)=0$

Horizontal asymptotes of the function are given by $y=\underset{x\to \infty }{{\displaystyle \lim }}f(x)$ .

Oblique asymptotes of the function are given by the linear part in the quotient after long division. Hence, if $f(x)=\frac{g(x)}{h(x)}=q(x)+\frac{r(x)}{h(x)}$ then the oblique asymptote is given by $y=q(x)$ . Note that oblique asymptotes of rational functions exists only when $\left|g(x)\right|>\left|h(x)\right|$ where $\left|F(x)\right|$ denotes the degree of $F(x)$ .

Graph: The graph of the given function is-

  

Interpretation: The given function is $f(x)=\frac{x^2-36}{x+1}$ .

The points through which the given curve passes through are-

x	$f(x)$
0	-36
1	$-\raisebox{1ex}{$35$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
3	$\raisebox{1ex}{$-27$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$
6	0
-2	32
-3	$\raisebox{1ex}{$27$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
-6	0

Vertical asymptote of the function is given by-

  $x+1=0$

  $\Rightarrow x=-1$

Horizontal asymptote of the function is given by-

  $y=\underset{x\to \infty }{{\displaystyle \lim }}f(x)$

  $\Rightarrow y=\underset{x\to \infty }{{\displaystyle \lim }}\frac{x^2-36}{x+1}$

  $\Rightarrow y=\underset{x\to \infty }{{\displaystyle \lim }}\frac{2x}{1}$ (Using L’Hospital’s rule)

  $\Rightarrow y=\infty$

Since the limit doesn’t exist, so the given curve has no horizontal asymptote.

By long division the given function simplifies to-

  $\frac{x^2-36}{x+1}$

  $=x-1-\frac{35}{x+1}$

Thus, the given curve has an oblique asymptote $y=x-1$ .

Thus the given function has one vertical asymptote $x=-1$ and one oblique asymptote $y=x-1$ .