Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Asymptote (geometry)"

(→‎Horizontal Asymptotes: expand; example not very suitable)
(→‎Vertical Asymptotes: expand)
Line 9: Line 9:
  
 
== Vertical Asymptotes ==
 
== Vertical Asymptotes ==
−
The vertical asymptote can be found by finding values of <math>x</math> that make the function undefined. One of the common ways is to have the function divided by zero, which is undefined. This can be shown by example.
+
The vertical asymptote can be found by finding values of <math>x</math> that make the function undefined, generally because it results in a division by zero, which is undefined.  
 +
 
 +
Be careful to distinguish this from a removable discontinuity (hole) in the graph; in a hole the factor that causes the division by zero can be canceled with another term.  
  
 
===Example Problem===
 
===Example Problem===
−
Find the vertical asymptotes of <math>\frac{1}{x^{2}}</math>.
+
Find the vertical asymptotes of 1) <math>y = \frac{1}{x^2-5x}</math> 2) <math>\tan 3x</math>.
  
 
====Solution====
 
====Solution====
−
To find the vertical asymptotes, <math>x^2</math> must equal zero. Solving the equation:
+
1) To find the vertical asymptotes, <math>x^2-5x</math> must equal zero. Solving the equation:
 +
 
 +
<math>\begin{eqnarray*}x^2-5x&=&0\\x&=&\boxed{0,5}\end{eqnarray*}</math>
  
−
<math>\begin{eqnarray*}x^2&=&0\\x&=&\boxed{0}\end{eqnarray*}</math>
+
So the vertical asymptote is <math>x=0,x=5</math>.
  
−
So the vertical asymptote is <math>x=0</math>, or just the y-axis
+
2) Since <math>\tan 3x = \frac{\sin 3x}{\cos 3x}</math>, we need to find where <math>\cos 3x = 0</math>. The cosine function is zero at <math>\frac{\pi}{2} + n\pi</math> for all [[integer]]s <math>n</math>; thus the functions is undefined at <math>x=\frac{\pi}{6} + n\frac{\pi}{3}</math>.
  
 
== Horizontal Asymptotes ==
 
== Horizontal Asymptotes ==

Revision as of 20:43, 9 November 2007

This is an AoPSWiki Word of the Week for Nov 8-14
For the vector graphics language, see Asymptote (Vector Graphics Language).

An asymptote is a line or curve that a certain function approaches.

Asymptotes can be of three different kinds: horizontal, vertical or slanted (oblique).


Vertical Asymptotes

The vertical asymptote can be found by finding values of $x$ that make the function undefined, generally because it results in a division by zero, which is undefined.

Be careful to distinguish this from a removable discontinuity (hole) in the graph; in a hole the factor that causes the division by zero can be canceled with another term.

Example Problem

Find the vertical asymptotes of 1) $y = \frac{1}{x^2-5x}$ 2) $\tan 3x$.

Solution

1) To find the vertical asymptotes, $x^2-5x$ must equal zero. Solving the equation:

$\begin{eqnarray*}x^2-5x&=&0\\x&=&\boxed{0,5}\end{eqnarray*}$ (Error compiling LaTeX. Unknown error_msg)

So the vertical asymptote is $x=0,x=5$.

2) Since $\tan 3x = \frac{\sin 3x}{\cos 3x}$, we need to find where $\cos 3x = 0$. The cosine function is zero at $\frac{\pi}{2} + n\pi$ for all integers $n$; thus the functions is undefined at $x=\frac{\pi}{6} + n\frac{\pi}{3}$.

Horizontal Asymptotes

The horizontal asymptote can be found in the same method as vertical asymptotes, but in relation to $y$ instead of $x$.

In general, to find a horizontal asymptote, take the $\lim_{x \rightarrow \infty} f(x)$ and $\lim_{x \rightarrow -\infty} f(x)$ to find the end behavior of the function. For rational functions in the form of $\frac{P(x)}{Q(x)}$ where $P(x), Q(x)$ are both polynomials, if the degree of the $Q(x)$ is greater than that of the degree of $P(x)$, then the horizontal asymptote is at $y = 0$. If the degree of $Q(x)$ is equal to that of the degree of $P(x)$, then the horizontal asymptote is at the quotient of the leading coefficient of $P(x)$ over the leading coefficient of $Q(x)$. (If the degree of $Q(x)$ is less than that of $P(x)$, then you get a slant asymptote, explained in the next section).

Note a crucial difference between horizontal asymptotes and vertical asymptotes: a function can never be defined at a vertical asymptote, but it can be defined at a horizontal asymptote. This is because the function is undefined (division by zero) at vertical asymptotes. However, a horizontal asymptote only gives the values for the ends of the function, but doesn’t have anything to do with the behavior of the function in the “middle”.

Horizontal asymptotes also occur in the inverses of certain functions with vertical asymptotes, and can occur in rotated conics, namely hyperbolas. For example, the hyperbola $xy = 1$ has a horizontal asymptote at $y = 0$.

Example Problem

Find the horizontal asymptote of $f(x) = \frac{x^2 - 3x + 2}{-2x^2 + 15x + 10000}$.

Solution

If we take $\lim_{x \rightarrow \pm\infty} f(x)$, notice that the $x^2$ term grows at a faster rate than the rest of the terms; hence our answer is $-\frac{1}{2}$.

Slanted Asymptotes

This article is a stub. Help us out by expanding it.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Asymptote_(geometry)&oldid=19279"