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## What is a rational function?

Just as in the case of rational numbers, when speaking of rational functions, the word ‘rational’ refers to a ratio. In the case of rational functions, it refers to the ratio of two polynomials.

A rational function is defined by a ratio of two polynomials.

For example, let $$f ( x ) = \dfrac{1}{x}$$. The numerator is a zero degree polynomial and the denominator is a first degree polynomial.

## Exercise 2.6.1

Sketch the graph of $$f ( x ) = \dfrac{1}{x}$$ for values of $$x$$ between $$- 2$$ and + 2, exclusive of $$0$$.

See Solution

## A vertical asymptote

Notice that as one picks input values $$x$$ closer to $$0$$, the output values $$y$$ get further from $$0$$ in size. For example, if $$x=0.1$$ then $$y=10$$ and if $$x=0.01$$ then $$y=100$$, etc. For this particular function, the graph ‘approaches’ the $$y$$-axis as $$x$$ approaches $$0$$ from either side. The $$y$$-axis is a vertical asymptote of the graph. If $$y$$ is a function of $$x$$ and if the value of $$y$$ increases (or decreases) without bound as $$x$$ approaches $$a$$ from either the left ($$x\to a^-$$) or from the right ($$x\to a^+$$) then the vertical line $$x=a$$ is a vertical asymptote of the graph of the function.

## Exercise 2.6.2

You just sketched the graph of the function $$y=\dfrac{1}{x}$$ between $$-2$$ and $$+2$$ exclusive of $$0$$. Now sketch the graph for values of $$x$$ between $$2$$ and $$10$$ and between $$-10$$ and $$-2$$.

See Solution

## Horizontal and vertical asymptotes

Notice that as one picks input values further from $$0$$, the output values approach $$0$$ in size. The $$x$$-axis is a horizontal asymptote of the graph. If $$y$$ is a function of $$x$$ and if it is true that the further $$x$$ gets from $$0$$ (in either or both directions) $$y$$ approaches some constant value $$b$$, then the horizontal line $$y=b$$ is a horizontal asymptote of the graph.

Rational functions may have vertical and horizontal asymptotes.

Assuming that the numerator and denominator of a rational function have no common factors, the vertical asymptotes occur at the zeros of the denominator.

In the example $$y=\dfrac{1}{x}$$ above, 0 is the zero of the denominator, so the vertical asymptote occurs at $$x = 0$$. In fact, $$x = 0$$ is the equation of the vertical asymptote. [recall that vertical lines have equations of the form $$x = c$$, where $$(c,0)$$ is the point where the vertical line crosses the $$x$$-axis.]

## Exercise 2.6.3

Write the equations of the vertical asymptotes of $$f ( x ) = \dfrac{1}{x^2 - 1}$$.

See Solution

## Horizontal and quotient asymptotes

In order to find the horizontal asymptote of a rational function (there can be at most one), it is necessary to find the quotient of the ratio. For example, if $$f ( x ) = 1 / x$$, the quotient is $$0$$ and the remainder is $$1$$. This means that $$x$$ divides into $$1$$ zero times with a remainder of $$1$$. Thus $$y = 0$$ (that is, the $$x$$-axis) is the horizontal asymptote. More generally, it is the quotient asymptote. But in this case, the equation of $$y =$$ quotient is the equation of a horizontal line, so it is called a horizontal asymptote.

If the degree of the numerator is greater than the degree of the denominator, then the quotient asymptote will not be a horizontal line. For example, consider the rational function $$y=\dfrac{x^2-x}{x-2}$$. If we divide the denominator into the numerator using either long division or synthetic division, we find that the quotient is $$q(x)=x+1$$. The graph of $$y=q(x)$$ is the quotient asymptote of the graph of $$y=\dfrac{x^2-x}{x-2}$$, but it is not a horizontal line. It is a line, but not a horizontal line. Whenever the numerator is exactly one degree higher than the degree of the numerator, the quotient asymptote will be a non-horizontal line. Such a quotient asymptote is sometimes called a slant asymptote or an oblique asymptote.

## Exercise 2.6.4

Find the equations and sketch the graphs of of the vertical and horizontal asymptotes of the polynomial function $$f ( x ) = \dfrac{2x^2 - 8}{ x^2 - 1}$$.

See Solution

## Intercepts and asymptotes

If the numerator and denominator of a rational function contain no common factors, then the $$x$$-intercepts occur at the zeros of the numerator. The $$x$$-intercepts will be either transitive or intransitive according to the same principle which applies to polynomial functions: If $$c$$ is a zero of the numerator and is of odd multiplicity, then the point $$( c, 0 )$$ is a transitive $$x$$-intercept. If $$c$$ is a zero of the numerator and is of even multiplicity, then the point $$( c,0 )$$ is an intransitive $$x$$-intercept. [Recall that the graph crosses the $$x$$-axis at a transitive $$x$$-intercept, and intersects but does not cross the $$x$$-axis at an intransitive $$x$$-intercept.]

The graph may or may not cross the horizontal (quotient) asymptote. The points of intersection of a graph and its horizontal (quotient) asymptote are called the horizontal (quotient) intercepts.

The quotient intercepts of a polynomial function occur at the zeros of the remainder.

The quotient and remainder are both found by dividing the denominator into the numerator using the process of long division of polynomials. The quotient intercepts may be either transitive or intransitive depending upon whether the zeros of the remainder are of odd or even multiplicity.

## Example 2.6.a

Consider the function mentioned above: $$y=\dfrac{x^2-x}{x-2}$$

We can re-write this equation in two different ways to obtain information about its graph.

1. In factored form $$y=\dfrac{x(x-1)}{x-2}$$ we see that it has transitive $$x$$-intercepts at $$x=0$$ and at $$x=1$$ and a transitive vertical asymptote at $$y=2$$. Yes, vertical asymptotes can be transitive or intransitive also, just like $$x$$-intercepts. For transitive vertical asymptotes, the graph approaches the top half of the asymptote on one side and the bottom half on the other. For intransitive vertical asymptotes, the graph approaches the same half of the vertical asymptote on both sides.
2. In quotient-remainder form, the equation of the function is $$y=x+1+\dfrac{2}{x-2}$$. So we see that there is a quotient asymptote $$y=q(x)=x+1$$ which is an oblique asymptote. And since the remainder is $$2$$ the remainder never equals $$0$$, so the graph of the function never intersects its quotient asymptote.

Since the graph must cross the $$x$$-axis at $$0$$ and $$1$$ and must never intersect either its vertical or quotient asymptote, the graph must appear as pictured below:

## Exercise 2.6.5

Let $$f ( x ) = \dfrac{2 x^2 + x}{ x^2 + 1 }$$.

Find all vertical and horizontal asymptotes and sketch their graphs. Find all $$x$$ and $$y$$ and quotient intercepts. Sketch the graph of $$y=f(x)$$.

See Solution

## Exercise 2.6.6

Sketch the graph of the polynomial function. Indicate all asymptotes and intercepts. Plot additional points only when necessary to resolve ambiguities in the graph.

$$f ( x ) = \dfrac{x^2 - 3}{x^2 - x - 2}$$

See Solution