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I am trying to find all of the asymptotes of a rational function, and the answer choices I am given are confusing. Because the degree of the function in the numerator was greater than the degree of the function in the denominator, I divided the numerator by the denominator using synthetic division to get the slant asymptote.

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

After dividing by x=2 (or (x-2)) $y=2x+8$

Answer Choices

A. $y=x+6, x=2$

B. $x=-x=-3$

C. $y=1, x=2$

D. $y=x+6, x=-2$

Image of the question

Xander Henderson
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RyuJV
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  • These answers are not asymptotes, most of them are only individual points? – Jalog_the_Martian Apr 26 '22 at 19:01
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    @Jalog_the_Martian There was a typesetting problem. (Fixed, but there is still a missing value in B.) – Théophile Apr 26 '22 at 19:02
  • Even now C is still just a point, I think the original author needs to carefully rewrite the question as even with your clean typesetting the options are messy. However I shall attempt an answer ignoring the options – Jalog_the_Martian Apr 26 '22 at 19:04
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    @Jalog_the_Martian While it is not clear, I believe that we are meant to interpret each equation as a line in the plane. Thus (C) corresponds to the horizontal line $y=1$ and the vertical line $x=2$ (meaning that the function has a vertical 'tote at $x=2$, and a horizontal 'tote at $y=1$). – Xander Henderson Apr 26 '22 at 19:13
  • Ah yeah, that does make sense. Well spotted. – Jalog_the_Martian Apr 26 '22 at 19:15
  • The original text of the question had $x = -x = 3$ for (B), which would be vertical asymptotes at $x=\pm 3$, I think. I've made that edit. – Xander Henderson Apr 26 '22 at 19:15
  • I now added the image of the question to show the equation. Sorry for the misformatting – RyuJV Apr 26 '22 at 21:27
  • I think option c should be two lines, $x=-1$ and $x=3$. The equation $x=-x$ is really just the line $x=0$, and the set $x=-x=3$ is empty. – 1Rock Apr 26 '22 at 21:38
  • @1Rock I think that there are commas or spacing missing in that problem. It is really poorly presented. – Xander Henderson Apr 26 '22 at 21:47

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I am struggling to follow your options but let us simply discuss the asymptotes.

You have made a mistake in your synthetic division. For $ x \to ±\infty$ You should get this: $$ \textrm{lim}_{x \to ±\infty} f(x)= \textrm{lim}_{x \to ±\infty} \frac{(x+6)(x-2)+15}{x-2} =\textrm{lim}_{x \to ±\infty} \frac{(x+6)(x-2)}{x-2} + \textrm{lim}_{x \to ±\infty} \frac{15}{x-2} \\ = x+6 $$

So the slant asymptote is $y=x+6$.

Now we also have a horizontal asymptote when the denominator is $0$. This simply given by $x-2=0 \implies x=2$.

Jalog_the_Martian
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