15.1: Asymptotes
- Page ID
- 181512
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| Title | Level of Approach | Type | Length |
|---|---|---|---|
| A Review of Rational Functions | College Algebra (Algebra 2.5) | Review | 16:30 |
| Identifying the Vertical Asymptotes of Rational Functions | College Algebra (Algebra 2.5) | Lecture | 7:44 |
| Identifying the Horizontal Asymptotes of Rational Functions | College Algebra (Algebra 2.5) | Lecture | 16:19 |
| Identifying the Slant Asymptotes of Rational Functions | College Algebra (Algebra 2.5) | Lecture | 12:44 |
| Asymptotic Behavior of Rational Functions | Precalculus (Algebra 3) | Lecture | 56:49 |
Definitions and Theorems
An asymptote of a curve is a straight line such that the distance between the curve and the line approaches zero as one or both of the coordinates increase without bound.
The line \(x=a\) is a vertical asymptote of the graph of \(f\) if the values of \(f(x)\) increase or decrease without bound as \(x\) approaches \(a\) from the left, from the right, or from both sides. We write\[ \begin{array}{rcll}
f(x) \to \infty & \text{ as } & x \to a^- & \text{ if the graph increases without bound as }x\text{ approaches }a\text{ from the left,} \\[6pt]
f(x) \to -\infty & \text{ as } & x \to a^- & \text{ if the graph decreases without bound as }x\text{ approaches }a\text{ from the left,} \\[6pt]
f(x) \to \infty & \text{ as } & x \to a^+ & \text{ if the graph increases without bound as }x\text{ approaches }a\text{ from the right,} \\[6pt]
& \text{ and } & & \\[6pt]
f(x) \to -\infty & \text{ as } & x \to a^+ & \text{ if the graph decreases without bound as }x\text{ approaches }a\text{ from the right.} \\[6pt]
\end{array} \nonumber \]
A rational function is a function of the form \(f(x)=\dfrac{p(x)}{q(x)}\), where \(p(x)\) and \(q(x)\) are polynomials and \(q(x)\) is not the zero polynomial. The domain of \(f\) is the set of all real numbers except the real zeros of \(q(x)\).
A single point where the graph of a function is not defined, indicated by an open circle on the graph, is called a removable discontinuity.
A removable discontinuity for the graph of a rational function occurs at \(x=a\) if
- \(a\) is a zero for a factor in the denominator that is common with a factor in the numerator, and
- after simplification, \( a \) is no longer a zero of the denominator.
- Proof
- The most elegant proof of this theorem occurs in Calculus.
Let \(f(x)=\dfrac{p(x)}{q(x)}\) be a rational function written in lowest terms, so that \(p(x)\) and \(q(x)\) share no common factors. Then the line \(x=a\) is a vertical asymptote of the graph of \(f\) for each real number \(a\) satisfying \(q(a)=0\).
The theorem requires lowest terms. If a factor is common to both \(p(x)\) and \(q(x)\), that factor produces a hole (a removable discontinuity), not a vertical asymptote. Always divide out common factors first; only the zeros of the reduced denominator give vertical asymptotes.
The end behavior of a graph is a description, usually as a function, of what the function values of the graph tend to approach as the inputs increase or decrease without bound.
The line \(y=c\) is a horizontal asymptote of the graph of \(f\) if the values of \(f(x)\) approach \(b\) as \(x\) increases without bound, as \(x\) decreases without bound, or both. We write\[ \begin{array}{rcll}
f(x) \to b & \text{ as } & x \to \infty & \text{ if the graph approaches the line }y = b\text{ as }x\text{ increases without bound} \\[6pt]
& \text{ and } & & \\[6pt]
f(x) \to b & \text{ as } & x \to -\infty & \text{ if the graph approaches the line }y = b\text{ as }x\text{ decreases without bound.} \\[6pt]
\end{array} \nonumber \]
The line \(y=mx+b\), with \(m\neq 0\), is an oblique asymptote (also called a slant asymptote) of the graph of \(f\) if the distance between \(f(x)\) and the line approaches zero as \(x\) increases or decreases without bound.
Let \( f(x) = \frac{N(x)}{D(x)} \) be a rational function. Then
- \( f(x) \) has a horizontal asymptote of \( y = 0 \) if \( \deg(D(x)) > \deg(N(x)) \)
- \( f(x) \) has a horizontal asymptote of \( y = \frac{a}{b} \) if \( \deg(D(x)) = \deg(N(x)) \), where \( a \) is the lead coefficient of \( N(x) \) and \( b \) is the lead coefficient of \( D(x) \)
- \( f(x) \) has a slant or oblique asymptote if \( \deg(D(x)) < \deg(N(x)) \). The asymptote is a slant asymptote if \( \deg(N(x)) = \deg(D(x)) + 1 \); otherwise, it is an oblique asymptote. In either case, the equation of the asymptote is the quotient after dividing \( N(x) \) by \( D(x) \).
Let \(f(x)=\dfrac{p(x)}{q(x)}\) be a rational function in which \(p(x)\) has degree \(m\) and \(q(x)\) has degree \(n\).
- If \(m\lt n\), then the graph of \(f\) has the horizontal asymptote \(y=0\).
- If \(m=n\), then the graph of \(f\) has a horizontal asymptote whose value is the ratio of the leading coefficient of \(p(x)\) to the leading coefficient of \(q(x)\).
- If \(m=n+1\), then the graph of \(f\) has an oblique asymptote equal to the quotient obtained when \(p(x)\) is divided by \(q(x)\).
- If \(m\gt n+1\), then the graph of \(f\) has no horizontal or oblique asymptote.
A rational function has at most one horizontal or oblique asymptote—never both—but it may have several vertical asymptotes. When the degree of the numerator exceeds the degree of the denominator by more than one, the graph still follows a curved guide at its ends, but that guide is not a straight line and so is not an asymptote in the sense used here.
Examples
Find the vertical asymptotes of \(f(x)=\dfrac{x+4}{x^2-x-6}\).
- Solution
-
Factor the denominator to locate its zeros.
\[x^2-x-6=(x-3)(x+2).\nonumber\]The denominator is zero at \(x=3\) and \(x=-2\). Evaluating the numerator at these values gives \(3+4=7\) and \(-2+4=2\), both nonzero, so the fraction is already in lowest terms and neither zero cancels. Therefore the graph has two vertical asymptotes:
\[x=-2 \quad \text{and} \quad x=3.\nonumber\]Because the degree of the numerator is less than the degree of the denominator, the graph also has the horizontal asymptote \(y=0\), shown in the figure below.
Figure \(\PageIndex{1}\): The graph of \(f(x)=\dfrac{x+4}{x^2-x-6}\), with the vertical asymptotes \(x=-2\) and \(x=3\) and the horizontal asymptote \(y=0\) drawn as dashed lines.
Find the horizontal asymptote of \(f(x)=\dfrac{3x-2}{x^2+1}\).
- Solution
-
The numerator has degree \(1\) and the denominator has degree \(2\), so the degree of the numerator is less than the degree of the denominator. By the degree comparison, the horizontal asymptote is
\[y=0.\nonumber\]The denominator \(x^2+1\) has no real zeros, so this graph has no vertical asymptote.
Find the horizontal asymptote of \(f(x)=\dfrac{5x^2+1}{x^2+3}\).
- Solution
-
The numerator and denominator both have degree \(2\). When the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. The leading coefficient of the numerator is \(5\) and the leading coefficient of the denominator is \(1\), so
\[y=\dfrac{5}{1}=5.\nonumber\]As with the previous example, the denominator \(x^2+3\) has no real zeros, so there is no vertical asymptote.
Find all asymptotes of \(f(x)=\dfrac{x^2+1}{x-1}\).
- Solution
-
The denominator is zero at \(x=1\), and the numerator there equals \(1^2+1=2\neq 0\), so the graph has the vertical asymptote \(x=1\).
The numerator has degree \(2\) and the denominator has degree \(1\), so the degree of the numerator is exactly one greater than the degree of the denominator. The graph therefore has an oblique asymptote given by the quotient of \(x^2+1\) divided by \(x-1\):
\[x^2+1=(x-1)(x+1)+2, \quad \text{so} \quad f(x)=x+1+\dfrac{2}{x-1}.\nonumber\]The remainder term \(\dfrac{2}{x-1}\) approaches \(0\) as \(x\) increases or decreases without bound, so the oblique asymptote is
\[y=x+1.\nonumber\]
Figure \(\PageIndex{2}\): The graph of \(f(x)=\dfrac{x^2+1}{x-1}\), with the vertical asymptote \(x=1\) and the oblique asymptote \(y=x+1\) drawn as dashed lines.
Find all asymptotes and any holes of \(f(x)=\dfrac{x-2}{x^2-4}\).
- Solution
-
Factor the denominator and reduce.
\[f(x)=\dfrac{x-2}{(x-2)(x+2)}=\dfrac{1}{x+2}, \quad x\neq 2.\nonumber\]The factor \(x-2\) is common to the numerator and denominator, so it produces a hole rather than a vertical asymptote. The hole occurs at \(x=2\); substituting into the reduced form gives the point \(\left(2,\dfrac{1}{4}\right)\).
The reduced denominator \(x+2\) is zero at \(x=-2\), where the reduced numerator is nonzero, so the graph has the single vertical asymptote
\[x=-2.\nonumber\]Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y=0\).
Determine the asymptotes of \(f(x)=\dfrac{2x^3-1}{x-1}\).
- Solution
-
The denominator is zero at \(x=1\), and the numerator there equals \(2(1)^3-1=1\neq 0\), so the graph has the vertical asymptote \(x=1\).
The numerator has degree \(3\) and the denominator has degree \(1\), so the degree of the numerator exceeds the degree of the denominator by \(2\), which is more than one. By the degree comparison, the graph has no horizontal or oblique asymptote. Dividing produces the quotient \(2x^2+2x+2\), a quadratic rather than a line, so the graph follows a curved guide at its ends—but this is not a linear asymptote.
Sources
Several parts of this text use modifications from the following source:
- Wikipedia article: "Asymptote"
This source is released under the Creative Commons Attribution-Share-Alike License 4.0.

