Linear fractional functions: asymptotes, monotonicity, and symmetry

Content updated: October 22, 2026

A nondegenerate quotient of two linear expressions has a two-branch graph. Check the denominator coefficient and determinant before using asymptote formulas; one-sided infinite-limit signs depend on the coefficients.

Objectives: determine the domain, two asymptotes, derivative signs, and center of symmetry; handle degenerate forms separately.

Prerequisites: the quotient rule, one-sided limits, and rational functions.

On this page: Definition · Method · Examples · Practice · Further learning

1. Conditions for the nondegenerate form

Consider $f(x)=(ax+b)/(cx+d)$ with $c\ne0$ and $A=ad-bc\ne0$. Set $h=-d/c$ and $k=a/c$. The domain is $\mathbb R\setminus\{h\}$. Division yields the form

The condition $c\ne0$ ensures a linear denominator. If $c=0,d\ne0$, the function is affine or constant; if $c=d=0$, the expression is undefined. If $c\ne0$ but $ad-bc=0$, the function equals the constant $a/c$ except at the excluded input; that point is a hole, rather than a vertical asymptote.

2. Features of the two branches

Since $K/(x-h)\to0$ in both directions at infinity, the horizontal asymptote is $y=k$. As $x\to h^+$, the limit is $+\infty$ for $K>0$ and $-\infty$ for $K<0$; as $x\to h^-$, the signs reverse. Thus $x=h$ is a vertical asymptote, but there is no single sign rule independent of the coefficients.

The derivative's denominator is positive on the domain. Thus the function is strictly increasing on each of $(-\infty;h)$ and $(h;+\infty)$ for $A>0$, and strictly decreasing on each for $A<0$. Do not combine these into one monotonicity assertion on the disconnected domain. The derivative never vanishes, so there are no local extrema on the domain.

From the standard form, for $t\ne0$, we have $f(h+t)=k+K/t$ and $f(h-t)=k-K/t$. Their sum is $2k$, so the asymptotes' intersection $I=(h;k)$ is a center of symmetry, not a point on the graph.

For graphing, first draw both asymptotes as dashed lines, then plot intercepts when present. The sign of $K/(x-h)$ determines whether a branch is above or below the horizontal asymptote. Neither branch touches the excluded vertical line.

3. Worked example

The function y = (4x + 1)/(2x − 1)

We have $c=2\ne0$ and $ad-bc=4(-1)-2=-6\ne0$. The domain is $\mathbb R\setminus\{1/2\}$. The standard form is $f(x)=2+3/(2x-1)=2+(3/2)/(x-1/2)$.

The vertical asymptote is $x=1/2$ and the horizontal asymptote $y=2$. Since $K=3/2>0$, the left-hand limit at $1/2$ is $-\infty$ and the right-hand limit $+\infty$. At both infinities the function tends to 2: from below on the left and from above on the right.

The function decreases on each of $(-\infty;1/2)$ and $(1/2;+\infty)$ and has no local extrema. Its center of symmetry is $(1/2;2)$. The horizontal-axis intercept has x-coordinate $-1/4$; the vertical-axis intercept is $(0;-1)$. Both lie on the left branch.

The left branch runs from values near 2 as $x\to-\infty$ down to $-\infty$ as $x\to(1/2)^-$. The right branch runs from $+\infty$ down toward 2 as $x\to+\infty$. Draw them separately; do not join across the vertical asymptote.

4. Practice

1. For $f(x)=(x+1)/(x-1)$, find both asymptotes and the derivative sign.
2. Does $(2x+2)/(x+1)$ have a vertical asymptote at $x=-1$?
3. Does $(3x+1)/2$ belong to the two-branch form under consideration?

5. Common mistakes

Do not use $-d/c$ when $c=0$. Do not assign the same infinite-limit signs to all coefficient sets. A hole in a constant expression is not a vertical asymptote. State each domain interval when discussing increase or decrease.

6. Further learning

Graphs of cubic functions (English lesson) · Graphs of even quartic functions (English lesson) · Limits of functions (English lesson) · Derivatives (English lesson).

Original exercise material on Scribd.

GeoGebra graphing tool from the original lesson.