Asymptote (geometry)

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.

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

Horizontal Asymptotes

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. Then the horizontal asymptote can be found in the same method as vertical asymptotes, but in relation to $y$ instead of $x$ . For example, the hyperbola $xy = 1 \Longrightarrow x = \frac{1}{y}$ 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

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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Asymptote (geometry)

Contents

Vertical Asymptotes

Example Problem

Solution

Horizontal Asymptotes

Example Problem

Solution

Slanted Asymptotes

Problems

Introductory

Intermediate

Olympiad