elisegayezm

2022-09-07

Asymptotes of a rational function

We have a function

$f(x)={\displaystyle \frac{2{x}^{4}+4{x}^{3}+3{x}^{2}+4x-4}{{x}^{3}-{x}^{2}-6x}}$

How would I systematically go about finding the asymptotes of this function? I know how to find the asymptotes of for example log functions or functions with a square root in it, but I don't really know how to find them for this function.

We have a function

$f(x)={\displaystyle \frac{2{x}^{4}+4{x}^{3}+3{x}^{2}+4x-4}{{x}^{3}-{x}^{2}-6x}}$

How would I systematically go about finding the asymptotes of this function? I know how to find the asymptotes of for example log functions or functions with a square root in it, but I don't really know how to find them for this function.

Jase Powell

Beginner2022-09-08Added 11 answers

The problem can be simplified by dividing the numerator and denominator by (x+2):

$\frac{2{x}^{3}+3x-2}{{x}^{2}-3x}}\phantom{\rule{0ex}{0ex}$

Vertical asymptotes occur where y tends to infinity, which happens at the roots of the denominator:

${x}^{2}-3x=x(x-3)=0\phantom{\rule{0ex}{0ex}}$

There are two roots so the vertical asymptotes are x=0 and x=3.

Horizontal asymptotes occur where x tends to infinity, which becomes clear by dividing the numerator and the denominator by the highest power of x:

$\underset{x\to \mathrm{\infty}}{lim}{\displaystyle \frac{2{x}^{3}+3x-2}{{x}^{2}-3x}}=\underset{x\to \mathrm{\infty}}{lim}{\displaystyle \frac{2+3/{x}^{2}-2/{x}^{3}}{1/x-3/{x}^{2}}}=\mathrm{\infty}\phantom{\rule{0ex}{0ex}}$

There is no limit so there are no horizontal asymptotes.

Oblique asymptotes occur where the graph approaches the line (mx+b), which becomes clear by dividing the numerator and denominator:

$\frac{2{x}^{3}+3x-2}{{x}^{2}-3x}}=2x+6\text{remains}21x-2\phantom{\rule{0ex}{0ex}$

There is a whole part so the oblique asymptote is y=2x+6.

$\frac{2{x}^{3}+3x-2}{{x}^{2}-3x}}\phantom{\rule{0ex}{0ex}$

Vertical asymptotes occur where y tends to infinity, which happens at the roots of the denominator:

${x}^{2}-3x=x(x-3)=0\phantom{\rule{0ex}{0ex}}$

There are two roots so the vertical asymptotes are x=0 and x=3.

Horizontal asymptotes occur where x tends to infinity, which becomes clear by dividing the numerator and the denominator by the highest power of x:

$\underset{x\to \mathrm{\infty}}{lim}{\displaystyle \frac{2{x}^{3}+3x-2}{{x}^{2}-3x}}=\underset{x\to \mathrm{\infty}}{lim}{\displaystyle \frac{2+3/{x}^{2}-2/{x}^{3}}{1/x-3/{x}^{2}}}=\mathrm{\infty}\phantom{\rule{0ex}{0ex}}$

There is no limit so there are no horizontal asymptotes.

Oblique asymptotes occur where the graph approaches the line (mx+b), which becomes clear by dividing the numerator and denominator:

$\frac{2{x}^{3}+3x-2}{{x}^{2}-3x}}=2x+6\text{remains}21x-2\phantom{\rule{0ex}{0ex}$

There is a whole part so the oblique asymptote is y=2x+6.

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