Asymptote (geometry): Difference between revisions

Revision as of 17:11, 27 June 2012

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

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

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:

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

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 $y = 0$ .

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$ (Error compiling LaTeX. Unknown error_msg)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}$

Slanted Asymptotes

Slanted asymptotes are similar to horizontal asymptotes in that they describe the end-behavior of a function. For rational functions $\frac{P(x)}{Q(x)}$ , a slanted 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 slanted asymptotes, since again the slanted asymptotes just describe the behavior of the function as $x$ approaches $\pm \infty$ .

For rational functions, we can find the slant asymptote simply by long division.

Hyperbolas have two slant asymptotes. Given a hyperbola in the form of $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ , the equation of the asymptotes of the hyperbola are at $y - k = \pm \frac{b}{a}(x - h)$ (swap $a, b$ if the $y$ term is positive).

Problems

Introductory

Intermediate

Olympiad

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 == Horizontal Asymptotes ==
-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).
+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 <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>.
-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”.
+. 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>.
-Horizontal asymptotes also occur in the inverses of certain functions with vertical asymptotes, and can occur in rotated conics, namely [[hyperbola]]s. Then 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>. For example, the hyperbola <math>xy = 1 \Longrightarrow x = \frac{1}{y}</math> has a horizontal asymptote at <math>y = 0</math>.
+. If the degree of <math>Q(x)</math> is less than the degree of <math>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 <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>.
 ====Solution====
-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>.
+The numerator has the same degree as the denominator, so the horizontal asymptote is the quotient of the leading coefficients:
+</math>y= \frac {1} {-2}$
 == Slanted Asymptotes ==

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