Asymptotes

Q: Hi Joel, could you run through the basics of asymptotes?

The fundamental idea of asymptotes is that it is a limit. For example, consider the function $\displaystyle y=\frac{1}{x}$ . What does it tend to as $x$ gets really, really large? The answer, of course, is $0$ , since $1$ divided by a reaaaaaaaally huge number gives us a reaaaaaaaally small number, which approaches $0$ . We can graph this function this way:

asymptotes-1-1

Since $y$ tends toward $0$ , we call $y=0$ our horizontal asymptote.

What about the $y$ ? How do we make that go to infinity? We do this by considering $1$ divided by a reaaaaaaaally small number. At the threshold, we take $1$ divided by $0$ and get an undefined number. So to get to that extreme, we let the denominator equal $0$ . In this case, $x=0$ , which just so happens to be our vertical asymptote.

BUT, what about this function, $\displaystyle y=\frac{1}{x-2}$ , which is the initial function but translated by 2 units in the positive $x$ –direction? We ask the same questions: what happens when $x$ gets really, really large and how can we make $y$ really, really large?

When $x$ gets really, really large, it follows that $x-2$ gets really, really large and $y$ equals $1$ divided by a reaaaaaaaally huge number, which gives us a reaaaaaaaally small number, which approaches $0$ . Thus the horizontal asymptote remains as $y=0$ .

How can we make $y$ go to infinity? We do this by letting the denominator equal $0$ . In this case, $x-2=0$ and $x=2$ . Hence, that is our asymptote.

asymptotes

BUT, what about this function, $\displaystyle y=\frac{1}{x-2}+3$ , which is the second function but translated by 3 units in the positive $y$ –direction? We ask the same questions: what happens when $x$ gets really, really large and how can we make $y$ really, really large?

When $x$ gets really, really large, it follows that $x-2$ gets really, really large and $\displaystyle \frac{1}{x-2}$ equals $1$ divided by a reaaaaaaaally huge number, which gives us a reaaaaaaaally small number, which approaches $0$ . Thus, overall, $\displaystyle y=\frac{1}{x-2}+3$ tends towards $\displaystyle y=3$ since the $\displaystyle \frac{1}{x-2}$ tends toward $0$ . Therefore $\displaystyle y=3$ is our horizontal asymptote.

How can we make $y$ go to infinity? We do this by letting the denominator equal $0$ . In this case, $x-2=0$ and $x=2$ . Hence, that is our asymptote.

asymptotes-1

BUT, what about this function, $\displaystyle y=x+3+\frac{1}{x-2}$ , which is the third function plus $x$ ??? We ask the same questions: what happens when $x$ gets really, really large and how can we make $y$ really, really large?

When $x$ gets really, really large, it follows that $x-2$ gets really, really large and $\displaystyle \frac{1}{x-2}$ equals $1$ divided by a reaaaaaaaally huge number, which gives us a reaaaaaaaally small number, which approaches $0$ . Thus, overall, $\displaystyle y=x+3+\frac{1}{x-2}$ tends towards $\displaystyle y=x+3$ since the $\displaystyle \frac{1}{x-2}$ tends toward $0$ . Therefore $\displaystyle y=x+3$ is our asymptote. It’s not horizontal, though. We call these slanted asymptotes oblique.

How can we make $y$ go to infinity? We do this by letting the denominator equal $0$ . In this case, $x-2=0$ and $x=2$ . Hence, that is our vertical asymptote.

asymptotes-2

BUT, what about this function, $\displaystyle y=ax+b+\frac{p}{qx+r}$ , which is the generalised third function? Bonus marks if you can describe the sequence of transformations from the third function to this. When $x$ gets really, really large, it follows that $qx+r$ gets really, really large and $\displaystyle \frac{p}{qx+r}$ equals $p$ divided by a reaaaaaaaally huge number, which gives us a reaaaaaaaally small number, which approaches $0$ . Thus, overall, $\displaystyle y=ax+b+\frac{p}{qx+r}$ tends towards $\displaystyle y=ax+b$ since the $\displaystyle \frac{p}{qx+r}$ tends toward $0$ . Therefore $\displaystyle y=ax+b$ is our oblique asymptote.

How can we make $y$ go to infinity? We do this by letting the denominator equal $0$ . In this case, $qx+r=0$ and $\displaystyle x=-\frac{r}{q}$ . Hence, that is our vertical asymptote.

In sum, to find the asymptotes, we ask the two questions

What happens when $x$ gets really, really large? (to obtain horizontal or oblique asymptotes)
What happens when the denominator equals zero? (to obtain vertical asymptotes)

*Footnote: The last function assumes $q \neq 0$ as division by zero is undefined. Also, if $q = 0$ , we actually get a linear function $\displaystyle y=ax+b+\frac{p}{r}$ , now assuming that $r \neq 0$ as division by zero is undefined. Note also that if $a = 0$ then we get the initial examples without oblique asyptotes.