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Difference between revisions of "Asymptote (geometry)"

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:''For the vector graphics language, see [[Asymptote (Vector Graphics Language)]].''
 
:''For the vector graphics language, see [[Asymptote (Vector Graphics Language)]].''
  
 
An '''asymptote''' is a [[line]] or [[curve]] that a certain [[function]] approaches.
 
An '''asymptote''' is a [[line]] or [[curve]] that a certain [[function]] approaches.
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[[File:Asymptote_graph.png|thumb|300px|The function <math>y=\tfrac{2x}{x-2}</math> has a vertical asymptote at x=2 and a horizontal asymptote at y=2]]
  
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Asymptotes can be of three different kinds: horizontal, vertical or slanted (oblique).  
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Linear asymptotes can be of three different kinds: horizontal, vertical or slant (oblique).  
  
  
 
== Vertical Asymptotes ==
 
== Vertical Asymptotes ==
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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.
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The vertical asymptote can be found by finding values of <math>x</math> that make the function undefined. Generally, it is found by setting the denominator of a rational function to zero.
  
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===Example Problem===
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If the numerator and denominator of a rational function share a factor, this factor is not a vertical asymptote. Instead, it appears as a hole in the graph.
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Find the vertical asymptotes of <math>\frac{1}{x^{2}}</math>.
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A rational function may have more than one vertical asymptote.
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===Example Problems===
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Find the vertical asymptotes of 1) <math>y = \frac{1}{x^2-5x}</math> 2) <math>\tan 3x</math>.
  
 
====Solution====
 
====Solution====
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To find the vertical asymptotes, <math>x^2</math> must equal zero. Solving the equation:
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1) To find the vertical asymptotes, let <math>x^2-5x=0</math>. Solving the equation:
  
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<math>\begin{eqnarray*}x^2&=&0\\x&=&\boxed{0}\end{eqnarray*}</math>
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<cmath>\begin{eqnarray*}x^2-5x&=&0\\x&=&\boxed{0,5}\end{eqnarray*}</cmath>
  
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So the vertical asymptote is <math>x=0</math>, or just the y-axis
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So the vertical asymptotes are <math>x=0,x=5</math>.
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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} + \frac{n\pi}{3}</math>.
  
 
== Horizontal Asymptotes ==
 
== Horizontal Asymptotes ==
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The horizontal asymptote can be found in the same method as vertical asymptotes, but in relation to <math>y</math> instead of <math>x</math>.
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For rational functions in the form of <math>\frac{P(x)}{Q(x)}</math> where <math>P(x), Q(x)</math> are both [[polynomial]]s:
  
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In general, to find a horizontal asymptote, take the <math>\lim_{x \rightarrow \infty} f(x)</math> and <math>\lim_{x \rightarrow -\infty} f(x)</math> to find the end behavior of the function. For rational functions in the form of <math>\frac{P(x)}{Q(x)}</math> where <math>P(x), Q(x)</math> are both [[polynomial]]s, if the degree of the <math>Q(x)</math> is greater than that of the degree of <math>P(x)</math>, then the horizontal asymptote is at <math>y = 0</math>. If the degree of <math>Q(x)</math> is equal to that of the degree of <math>P(x)</math>, then the horizontal asymptote is at the quotient of the leading coefficient of <math>P(x)</math> over the leading coefficient of <math>Q(x)</math>. (If the degree of <math>Q(x)</math> is less than that of <math>P(x)</math>, then you get a slant asymptote, explained in the next section).
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1. If the [[Degree of a polynomial | degree ]] of <math>Q(x)</math> is greater than that of the degree of <math>P(x)</math>, then the horizontal asymptote is at <math>y = 0</math>. This can be seen by noting that as <math>x</math> increases, <math>Q(x)</math> increases much faster than <math>P(x)</math> does. Since the denominator increases faster than the numerator, as x approaches infinity, y gets smaller until it approaches zero. A similar trend can be seen as x decreases.
  
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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”.  
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2. If the degree of <math>Q(x)</math> is equal to that of the degree of <math>P(x)</math>, then the horizontal asymptote is at the quotient of the leading coefficient of <math>P(x)</math> over the leading coefficient of <math>Q(x)</math>.  
  
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Horizontal asymptotes also occur in the inverses of certain functions with vertical asymptotes, and can occur in rotated conics, namely [[hyperbola]]s. For example, the hyperbola <math>xy = 1</math> has a horizontal asymptote at <math>y = 0</math>.
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3. If the degree of <math>Q(x)</math> is less than the degree of <math>P(x)</math>, see below (slanted asymptotes)
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A function may not have more than one horizontal asymptote. Functions with a "middle section" may cross the horizontal asymptote at one point. To find this point, set y=horizontal asymptote and solve.
  
 
===Example Problem===
 
===Example Problem===
 
Find the horizontal asymptote of <math>f(x) = \frac{x^2 - 3x + 2}{-2x^2 + 15x + 10000}</math>.
 
Find the horizontal asymptote of <math>f(x) = \frac{x^2 - 3x + 2}{-2x^2 + 15x + 10000}</math>.
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====Solution====
 
====Solution====
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If we take <math>\lim_{x \rightarrow \pm\infty} f(x)</math>, notice that the <math>x^2</math> term grows at a faster rate than the rest of the terms; hence our answer is <math>-\frac{1}{2}</math>.
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The numerator has the same degree as the denominator, so the horizontal asymptote is the quotient of the leading coefficients:
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<math>y= \frac {1} {-2}</math>
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== Slant Asymptotes ==
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[[File:Slantasymptote.png|thumb|500px|The function <math>y=\tfrac{x^2+2x+4} {x+1}</math> has a slant asymptote at <math>y=x+1</math> ]]
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For rational functions <math>\frac{P(x)}{Q(x)}</math>, a slant asymptote occurs when the degree of <math>P(x)</math> is one greater than the degree of <math>Q(x)</math>. If the degree of <math>P(x)</math> is two or more greater than the degree of <math>Q(x)</math>, then we get a curved asymptote. Again, like horizontal asymptotes, it is possible to get crossing points of slant asymptotes.
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For rational functions, we can find the slant asymptote simply by long division, omitting the remainder and setting y=quotient.
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===Example Problem===
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Find the slant asymptote of <math>y= \frac{x^2+2x+4} {x+1}</math>
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'''Solution'''
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<math>\frac{x^2+2x+4}{x+1}= x+1+\frac{3} {x+1}</math>
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The slant asymptote is <math>y=x+1</math>
  
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== Slanted Asymptotes ==
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==External Links==
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3 minute asymptote lesson: [http://www.youtube.com/watch?v=4Cqm07W2teM&feature=plcp]
  
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{{stub}}
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[[Category:Mathematics]]
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[[Category:Definition|Asymptote]]
 
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[[Category:Geometry|Asymptote]]
 

Latest revision as of 10:46, 28 September 2024

For the vector graphics language, see Asymptote (Vector Graphics Language).

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

The function $y=\tfrac{2x}{x-2}$ has a vertical asymptote at x=2 and a horizontal asymptote at y=2

Linear asymptotes can be of three different kinds: horizontal, vertical or slant (oblique).


Vertical Asymptotes

The vertical asymptote can be found by finding values of $x$ that make the function undefined. Generally, it is found by setting the denominator of a rational function to zero.

If the numerator and denominator of a rational function share a factor, this factor is not a vertical asymptote. Instead, it appears as a hole in the graph.

A rational function may have more than one vertical asymptote.

Example Problems

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

Solution

1) To find the vertical asymptotes, let $x^2-5x=0$. Solving the equation:

\begin{eqnarray*}x^2-5x&=&0\\x&=&\boxed{0,5}\end{eqnarray*}

So the vertical asymptotes are $x=0,x=5$.

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 $x=\frac{\pi}{6} + \frac{n\pi}{3}$.

Horizontal Asymptotes

For rational functions in the form of $\frac{P(x)}{Q(x)}$ where $P(x), Q(x)$ are both polynomials:

1. If the degree of $Q(x)$ is greater than that of the degree of $P(x)$, then the horizontal asymptote is at $y = 0$. This can be seen by noting that as $x$ increases, $Q(x)$ increases much faster than $P(x)$ does. Since the denominator increases faster than the numerator, as x approaches infinity, y gets smaller until it approaches zero. A similar trend can be seen as x decreases.

2. 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 $Q(x)$.

3. If the degree of $Q(x)$ is less than the degree of $P(x)$, see below (slanted asymptotes)

A function may not have more than one horizontal asymptote. Functions with a "middle section" may cross the horizontal asymptote at one point. To find this point, set y=horizontal asymptote and solve.

Example Problem

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

Solution

The numerator has the same degree as the denominator, so the horizontal asymptote is the quotient of the leading coefficients: $y= \frac {1} {-2}$

Slant Asymptotes

The function $y=\tfrac{x^2+2x+4} {x+1}$ has a slant asymptote at $y=x+1$

For rational functions $\frac{P(x)}{Q(x)}$, a slant asymptote occurs when the degree of $P(x)$ is one greater than the degree of $Q(x)$. If the degree of $P(x)$ is two or more greater than the degree of $Q(x)$, then we get a curved asymptote. Again, like horizontal asymptotes, it is possible to get crossing points of slant asymptotes.

For rational functions, we can find the slant asymptote simply by long division, omitting the remainder and setting y=quotient.

Example Problem

Find the slant asymptote of $y= \frac{x^2+2x+4} {x+1}$

Solution

$\frac{x^2+2x+4}{x+1}= x+1+\frac{3} {x+1}$


The slant asymptote is $y=x+1$

External Links

3 minute asymptote lesson: [1]

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