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We are taught at school that every time we have a function like that:

$y = b + \frac{a}{f(x)}$

Then, one of the asymptotes will be y = b

For example: $y = (x + 1) + \frac{4}{x - 2}$

Then, the asymptotes will be:

$y = (x + 1)$

$x = 2$

I wonder why y = x +1 is an asymptote since there is no indetermination around that

Does somebody know how to prove that for me?

Thanks in advance

Matheus Minguini
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  • For the slant asymptote, isn't it obvious that the vertical separation between the graphs approaches zero as $|x| \rightarrow \infty$? The difference between $(x+1) + \frac{4}{x-2}$ and $(x+1)$ is $\frac{4}{x-2},$ and this difference clearly approaches zero as $x \rightarrow -\infty$ and as $x \rightarrow +\infty.$ For example, for $x=10^9,$ calculate $y$-coordinate of $y = (x+1) + \frac{4}{x-2}$ and the $y$-coordinate of $y=(x+1).$ Using the distance formula, how far apart are these two points? What does this suggest about how far apart the graphs are in the vicinity of $x=10^9$? – Dave L. Renfro Aug 24 '20 at 16:31
  • @DaveL.Renfro, yeah, x = 2 is an indetermination for sure, otherwise, we would be dividing by zero. I just still don't get why is it B is always also an asymptote though – Matheus Minguini Aug 24 '20 at 19:36
  • Look at the graphs from $x=-5$ to $x=5$ and from $x=-10$ to $x=10$ and from $x=-20$ to $x=20$. Maybe this will help, as well as this. – Dave L. Renfro Aug 24 '20 at 20:06

2 Answers2

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There is no question that when $f(x)=0$ there is a vertical asymptote at that $x$.

But also when $f(x) \rightarrow \infty$ the equation has an asymptote at $y=b$

Another way to look at it, is to investigate the function with the substitution $z = \frac{1}{x-2}$ since $x=2$ is a vertical asymptote. Then look at the behavior near $z \rightarrow 0$ which is the same as looking at $x \rightarrow \pm \infty$.

We have $y =3 + 4z + \frac{1}{z}$ with $z$ near zero, which makes the terms proportional to $z$ being zero. Then back substitute $x$ to get the asymptote line.

$ y = 3 + \frac{1}{z} = x + 1$

JAlex
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  • thanks for putting time and effort commenting, I really appreciate! Still, I don't quite understand the example of yours! I still don't understand intuitively why x + 1 is a 'line' untouchable by the whole function – Matheus Minguini Aug 27 '20 at 13:18
  • @MatheusMinguini did you plot y(x) as well as y=x+1 to see actually what is going on? – JAlex Aug 27 '20 at 14:06
  • @MatheusMinguini with these functions you can devise a coordinate transformation for $x$ and $y$ to bring the equation on the form $y' = \frac{1}{x'}$ which has asymptotes at $x'=0$ and $y'=0$. Now bring those asymptotes back into the $x$,$y$ world. – JAlex Aug 27 '20 at 14:16
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We are taught at school that every time we have a function like that: $$y = b + \frac{a}{f(x)}$$ Then, one of the asymptotes will be $y = b$

That is true only if $a,b$ are constants and $f(x)$ is a polinomial (or more in general, a function that tends to infinity with $x$).

You cannot expect that to be true in general (for example, it could happen that $f(x)= a$).

leonbloy
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