Slant (Oblique) Asymptotes, precalculus lesson

Far to the left and far to the right, the graph of

f(x) = \frac{2x^2 - 3x + 4}{x - 2}

lies almost on top of a tilted straight line. Its values grow without bound, so there is no horizontal asymptote, yet the graph still settles along a line. Polynomial division names that line and shows how fast the graph closes in on it.

Key terms

Polynomial division

Definition For polynomials $p$ and $q$ with $q$ not the zero polynomial, the division algorithm gives unique polynomials $s$ , the quotient, and $R$ , the remainder, with

\begin{gathered} p(x) = q(x) \, s(x) + R(x), \\[0.4em] R = 0 \ \text{or} \ \deg R < \deg q. \end{gathered}

Dividing by $q(x)$ gives, at every $x$ with $q(x) \ne 0$ ,

\frac{p(x)}{q(x)} = s(x) + \frac{R(x)}{q(x)}.

The quotient $s$ is called the polynomial part of the rational function. Cancelling a common factor of $p$ and $q$ before dividing does not change $s$ .

In plain words Division splits a top-heavy fraction into a polynomial plus a leftover fraction whose top has smaller degree than its bottom.

Example Divide $2x^2 - 3x + 4$ by $x - 2$ with synthetic division, using $c = 2$ and the coefficients $2, -3, 4$ . The bottom row is $2, 1, 6$ , so the quotient is $2x + 1$ and the remainder is $6$ . Multiplying back confirms it,

\begin{aligned} &(x - 2)(2x + 1) + 6 \\[0.4em] &\quad = (2x^2 - 3x - 2) + 6 \\[0.4em] &\quad = 2x^2 - 3x + 4. \end{aligned}

So for $x \ne 2$

f(x) = 2x + 1 + \frac{6}{x - 2}.

Remainder over divisor tends to zero

Definition If $R$ is not the zero polynomial and $\deg R < \deg q$ , then by the degree comparison for horizontal asymptotes

\begin{gathered} \lim_{x \to \infty} \frac{R(x)}{q(x)} = 0 \\[0.8em] \text{and} \\[0.4em] \lim_{x \to -\infty} \frac{R(x)}{q(x)} = 0. \end{gathered}

Hence $\frac{p(x)}{q(x)} - s(x) \to 0$ as $x \to \pm\infty$ . The sign of $\frac{R(x)}{q(x)}$ for large positive $x$ and for large negative $x$ tells on which side of the graph of $s$ the graph of $\frac{p}{q}$ lies at each end. The two graphs meet exactly at the real zeros of $R$ that lie in the domain.

In plain words The leftover fraction is the vertical gap between the curve and the polynomial part. Its bottom outgrows its top, so the gap shrinks toward $0$ at both ends.

Example For $f$ the gap is $\frac{6}{x - 2}$ .

$x$	$-998$	$-98$	$12$	$102$	$1002$
$f(x) - (2x + 1)$	$-0.006$	$-0.06$	$0.6$	$0.06$	$0.006$

Each time the distance $\lvert x - 2 \rvert$ grows tenfold, the gap shrinks tenfold in size. The gap is positive for $x > 2$ and negative for $x < 2$ , so the graph runs above the line on the right and below it on the left. The remainder $6$ is never $0$ , so the graph never meets the line.

Oblique asymptote

Definition The line $y = mx + b$ with $m \ne 0$ is an oblique asymptote, also called a slant asymptote, of the graph of $f$ if

\begin{gathered} \lim_{x \to \infty} \bigl(f(x) - (mx + b)\bigr) = 0 \\[0.4em] \text{or} \\[0.4em] \lim_{x \to -\infty} \bigl(f(x) - (mx + b)\bigr) = 0. \end{gathered}

The case $m = 0$ would be a horizontal asymptote. The gap in this definition is measured vertically, as the difference $f(x) - (mx + b)$ at the same input $x$ .

In plain words Far out, the graph hugs a tilted line, and the vertical distance between them shrinks to zero.

Example For $f$ the gap $f(x) - (2x + 1) = \frac{6}{x - 2}$ tends to $0$ in both directions, so $y = 2x + 1$ is an oblique asymptote at both ends. At $x = 102$ , for instance, the function gives $f(102) = 205.06$ and the line gives $2 \cdot 102 + 1 = 205$ .

Degree difference one

Definition Let $r = \frac{p}{q}$ , where $q$ has degree $d \ge 1$ and leading coefficient $B$ , and $p$ has degree $d + 1$ and leading coefficient $A$ . Then the quotient of $p$ by $q$ is a polynomial $s(x) = mx + b$ of degree $1$ with slope

m = \frac{A}{B},

and the line $y = mx + b$ is the oblique asymptote of the graph in both directions. If the degree of $p$ is at most $d$ , the graph has a horizontal asymptote instead, and if the degree of $p$ is $d + 2$ or more, it has neither a horizontal nor an oblique asymptote. If the remainder is $0$ , the graph lies on the line itself, apart from holes at the excluded values.

In plain words A slanted asymptote appears exactly when the top has degree one more than the bottom. Its slope is the ratio of the leading coefficients, and the division supplies the intercept.

Example For

r(x) = \frac{3x^3 - 2x + 1}{x^2 + 1}

the degrees are $3$ and $2$ , so expect a slant asymptote with slope $\frac{3}{1} = 3$ . Long division gives

\begin{aligned} 3x^3 - 2x + 1 &= (x^2 + 1)(3x) \\[0.4em] &\quad + (1 - 5x), \end{aligned}

r(x) = 3x + \frac{1 - 5x}{x^2 + 1},

and the asymptote is $y = 3x$ , with intercept $0$ . At $x = 10$ the function is about $29.51$ while the line is at $30$ .

Polynomial asymptotes of higher degree

Definition The same reasoning works for any degree difference. If $\deg p \ge \deg q + 2$ and $s$ is the quotient of $p$ by $q$ , then $s$ has degree $\deg p - \deg q \ge 2$ and

\begin{gathered} \frac{p(x)}{q(x)} - s(x) = \frac{R(x)}{q(x)} \to 0 \\[0.8em] \text{as } x \to \pm\infty. \end{gathered}

So the graph of $\frac{p}{q}$ approaches the curve $y = s(x)$ at both ends. This curve is called a curvilinear or polynomial asymptote, and a parabolic asymptote when $s$ has degree $2$ . Since $s$ has degree at least $2$ , for every line $y = mx + b$ the size of $s(x) - (mx + b)$ grows without bound far out, so the graph has no horizontal or oblique asymptote.

In plain words When the top outgrows the bottom by two or more degrees, the graph follows a polynomial curve far out instead of a line. Division finds that curve just as it finds a slant line.

Example Synthetic division of $x^3 + x^2 + 2$ by $x + 1$ , with $c = -1$ and the coefficients $1, 1, 0, 2$ , gives the bottom row $1, 0, 0, 2$ . So for $x \ne -1$

h(x) = \frac{x^3 + x^2 + 2}{x + 1} = x^2 + \frac{2}{x + 1}.

Far out the graph of $h$ follows the parabola $y = x^2$ . At $x = 9$ the gap is $\frac{2}{10} = 0.2$ , and at $x = 99$ it is $\frac{2}{100} = 0.02$ .

Check yourself

Find the slant asymptote of the function below.

r(x) = \frac{x^2 + 4x - 1}{x + 2}

Show the answer

Synthetic division with $c = -2$ on the coefficients $1, 4, -1$ gives the bottom row $1, 2, -5$ . So for $x \ne -2$

r(x) = x + 2 - \frac{5}{x + 2},

and the slant asymptote is $y = x + 2$ .

Which of the following functions has a slant asymptote?

\begin{gathered} f(x) = \frac{x^2 + 1}{x^2 - 1}, \\[0.8em] g(x) = \frac{x^3}{x^2 - 4}, \\[0.8em] h(x) = \frac{x^4}{x^2 + 1} \end{gathered}

Show the answer

Only $g$ does. The degrees in $f$ are equal, so $f$ has the horizontal asymptote $y = 1$ . In $g$ the degree difference is one, and $x^3 = (x^2 - 4)x + 4x$ gives

g(x) = x + \frac{4x}{x^2 - 4},

so $y = x$ is the slant asymptote. In $h$ the difference is two, and $x^4 = (x^2 + 1)(x^2 - 1) + 1$ shows that the graph follows the parabola $y = x^2 - 1$ instead.

Find the slant asymptote of the function below and the point where the graph crosses it.

r(x) = \frac{2x^3 + x^2 - 1}{x^2 + 1}

Show the answer

Long division gives

\begin{aligned} 2x^3 + x^2 - 1 &= (x^2 + 1)(2x + 1) \\[0.4em] &\quad - 2x - 2, \end{aligned}

r(x) = 2x + 1 - \frac{2x + 2}{x^2 + 1}.

The asymptote is $y = 2x + 1$ . The leftover fraction is $0$ only at $x = -1$ , so the graph crosses the line at $(-1, -1)$ . As a check, $r(-1) = \frac{-2 + 1 - 1}{2} = -1$ , and the line also gives $2(-1) + 1 = -1$ .

Slant (Oblique) Asymptotes

Key terms

Polynomial division

Remainder over divisor tends to zero

Oblique asymptote

Degree difference one

Polynomial asymptotes of higher degree

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus