Asymptote (geometry)

Revision as of 17:13, 27 June 2012 by Jellymoop (talk | contribs) (Horizontal Asymptotes)
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) $\tan 3x$.


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. ! Missing \endgroup inserted.)

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 $f(x) = \frac{x^2 - 3x + 2}{-2x^2 + 15x + 10000}$.


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).





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