Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (f)/Solution 1

Vertical Asymptotes

We know that vertical asymptotes occur when the denominator is zero. Here our denominator is

$\displaystyle {}x^{2}-3$

which is zero if $\scriptstyle x=\pm {\sqrt {3}}$. Therefore we have vertical asymptotes at $\scriptstyle x=-{\sqrt {3}}$ and $\scriptstyle x={\sqrt {3}}$.

To see how the function looks near the asymptotes we could take some limits. However, from part (d) we have that the function is decreasing to the left of $\scriptstyle x=-{\sqrt {3}}$ as well as to the right. Therefore we conclude that

$\lim _{x\to -{\sqrt {3}}^{-}}f(x)=-\infty$

while

$\lim _{x\to -{\sqrt {3}}^{+}}f(x)=+\infty$

We have the exact same conclusion about the other vertical asymptote at $\scriptstyle x={\sqrt {3}}$ since it has the same decreasing properties as $\scriptstyle x={\sqrt {3}}$.

Horizontal or Slant Asymptotes

We check for horizontal or slant asymptotes. Since the degree of the numerator is larger than the degree of the denominator then we do not have any horizontal asymptotes. However, the degree of the numerator is exactly one larger than the degree of the denominator and so we do expect slant asymptotes. To figure out the slant asymptote we perform polynomial long division

${\begin{array}{cccccccccccc}&&&&&x&+&1&&&&\\&&&&&\_&\_&\_&\_&\_&\_&\_\\x^{2}&+&0x&-&3)&x^{3}&+&x^{2}&-&2x&-&3\\&&&&&x^{3}&+&0x^{2}&-&3x&&\\&&&&&\_&\_&\_&\_&\_&\_&\_\\&&&&&&&x^{2}&+&x&-&3\\&&&&&&&x^{2}&+&0x&-&3\\&&&&&\_&\_&\_&\_&\_&\_&\_\\&&&&&&&&&x&&\end{array}}$

Therefore we write $\displaystyle {}x^{3}+x^{2}-2x-3=(x^{2}-3)(x+1)+x$. So we get that

${\begin{aligned}{\frac {x^{3}+x^{2}-2x-3}{x^{2}-3}}&={\frac {(x^{2}-3)(x+1)+x}{x^{2}-3}}\\&=x+1+{\frac {x}{x^{2}-3}}\end{aligned}}$

Notice that for the last term,

$\lim _{x\to \infty }{\frac {x}{x^{2}-3}}=0.$

Therefore as x gets big we see that

${\frac {x^{3}+x^{2}-2x-3}{x^{2}-3}}$

looks like x+1. If we take x going to $\scriptstyle -\infty$ we see that the limit of the last term still vanishes and it still looks like $x+1$. Therefore, we conclude that $x+1$ is a slant asymptote to the function.

Summary

Finally then we conclude that the equations for the vertical asymptotes are

${\begin{aligned}x&=-{\sqrt {3}}\\x&={\sqrt {3}}\end{aligned}}$

while the equation of the slant asymptote is

$\displaystyle {}y=x+1.$