Difference between revisions of "Asymptote (geometry)"

(Horizontal Asymptotes)
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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]]
  
 
Linear asymptotes can be of three different kinds: horizontal, vertical or slanted (oblique).  
 
Linear asymptotes can be of three different kinds: horizontal, vertical or slanted (oblique).  
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1) To find the vertical asymptotes, let <math>x^2-5x=0</math>. Solving the equation:
 
1) To find the vertical asymptotes, let <math>x^2-5x=0</math>. Solving the equation:
  
<math>\begin{eqnarray*}x^2-5x&=&0\\x&=&\boxed{0,5}\end{eqnarray*}</math>
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<cmath>\begin{eqnarray*}x^2-5x&=&0\\x&=&\boxed{0,5}\end{eqnarray*}</cmath>
  
 
So the vertical asymptotes are <math>x=0,x=5</math>.
 
So the vertical asymptotes are <math>x=0,x=5</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:
 
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:
  
1. If the 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>.
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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.
  
 
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>.  
 
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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<math>y= \frac {1} {-2}</math>
 
<math>y= \frac {1} {-2}</math>
  
== Slanted Asymptotes ==
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== Oblique (Slanted) Asymptotes ==
Slanted asymptotes are similar to horizontal asymptotes in that they describe the end-behavior of a function. For rational functions <math>\frac{P(x)}{Q(x)}</math>, a slanted 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 slanted asymptotes, since again the slanted asymptotes just describe the behavior of the function as <math>x</math> approaches <math>\pm \infty</math>.
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For rational functions <math>\frac{P(x)}{Q(x)}</math>, an oblique 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 oblique asymptotes.
  
For rational functions, we can find the slant asymptote simply by long division.  
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For rational functions, we can find the oblique asymptote simply by long division, omitting the remainder and setting y=quotient.
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===Example Problem===
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Find the oblique 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>
  
Hyperbolas have two slant asymptotes. Given a hyperbola in the form of <math>\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1</math>, the equation of the asymptotes of the hyperbola are at <math>y - k = \pm \frac{b}{a}(x - h)</math> (swap <math>a, b</math> if the <math>y</math> term is positive).
 
  
==Problems==
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The oblique asymptote is <math>y=x+1</math>
===Introductory===
 
===Intermediate===
 
===Olympiad===
 
  
[[Category:Definition|Asymptote]]
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==External Links==
[[Category:Geometry|Asymptote]]
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3 minute asymptote lesson: [http://www.youtube.com/watch?v=4Cqm07W2teM&feature=plcp]

Latest revision as of 15:21, 11 October 2015

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 slanted (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: $y= \frac {1} {-2}$

Oblique (Slanted) Asymptotes

For rational functions $\frac{P(x)}{Q(x)}$, an oblique 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 oblique asymptotes.

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

Example Problem

Find the oblique 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 oblique asymptote is $y=x+1$

External Links

3 minute asymptote lesson: [1]

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