Rational Functions:
Tricks of the trade

Zero Quotient Rule. The quotient a/b equals 0 if two conditions are met: (1) a=0 and (2) b≠0. This zero quotient rule applies as well to quotients of any two functions: the quotient p(x)/q(x) equals 0 if two conditions are met: (1) p(x)=0 and (2) q(x)≠0. A typical application involves finding values of x that meet these two conditions.

Rational Functions Not Expressible as a Polynomial. Does a rational expression like (x–1)2/(x–1) reduce, through cancellation, to the polynomial x–1? The answer is no! It reduces to the term x–1 only with the extra condition: x≠1. The cancellation is called "improper" if you exclude the condition x≠1. Because of this extra condition, it does not qualify to be a "full-fledged" polynomial. True, it is "almost" a polynomial, but that is not the same as "being" a polynomial. In fact, unlike the polynomial function y=x–1, the graph of y=(x–1)2/(x–1) has a hole in it, to signify that the function is not defined at x=1. Here is the graph with the hole in it.

Continuity and "Smoothness". Rational functions are continuous at every point in their domain. The domain excludes all denominator zeros (values of x that make the denominator equal to zero). Denominator zeros give rise to breaks in the curve, either holes or vertical asymptotes. As for smoothness, rational functions never have corners.

Vertical Asymptotes and Holes. Whenever the denominator is zero at a particular value, say x=c, there is either a hole or vertical asymptote at x=c. If (xc) is a common factor that can be eliminated from the denominator by improper cancellation, it gives rise to a hole; otherwise, there is a vertical asymptote at x=c. Either way, the function is not defined at x=c. It is for this reason that the curve can never intersect a vertical asymptote (it has no y-value, which is required to have a point of intersection).

Maximum Number of x-Intercepts. The numerator of a rational function is a polynomial and x-intercepts can only occur at zeros of the numerator (values of x that make the numerator equal to zero). So, as in the case of polynomial functions, the maximum number of x-intercepts equals the degree of the polynomial in the numerator.

Rational Function End Behaviors. Every rational function can be expressed as the sum of a polynomial quotient plus a fractional term in which the numerator has degree less than that of the denominator. For example,

(x2+1)/(x–1) = (x+1) + 2/(x–1)

Here, the polynomial quotient is (x+1) and in the fractional term, the numerator, 2, has degree 0, which is less than that of the denominator, x–1, which has degree 1. The end behaviors of a rational function follow those of its polynomial quotient, so in this example the end behaviors follow those of y=x+1: down on the left and up on the right. Also, we say that the rational function is asymptotic (in its end behavior) to the polynomial y=x+1.

What are Asymptotes? Asymptotes are dashed lines or dashed polynomial curves that are displayed with the graph of rational and other nonpolynomial functions to help clarify certain types of limiting behaviors. Strictly speaking, asymptotes (whether vertical, horizontal, oblique, or some other polynomial curve) are not part of the function and it is for this reason that they are traditionally displayed as dashed lines, to distinguish them from the solid lines that represent the function. This manner of displaying asymptotes is standard for mathematicians and mathematics textbooks. Less sophisticated graphing calculators display the asymptotes either as a solid line or as no line at all, either of which can cause confusion with standard notation. On WebGraphing.com, asymptotes are always displayed as dashed.

                         

Straight Line Asymptotes. A rational function may, but need not, have a vertical asymptote. Further, a rational function may be asymptotic to a nonvertical straight line in two ways. (1) If the degree of the numerator polynomial is less than or equal to the degree of the denominator polynomial, the rational function has a horizontal asymptote. (2) If the degree of the numerator polynomial exceeds by 1 the degree of the denominator polynomial, the rational function has a slant or oblique asymptote. These "rules" are quick ways to determine the existence of horizontal or slant asymptotes. To determine an asymptote explicitly, you need to compute the polynomial quotient.

Below is an example of a rational function with a horizontal asymptote. Note that the curve intersects the horizontal asymptote at (0,0). While it is impossible for a rational function to ever intersect any of its vertical asymptotes, this example illustrates how it is possible for a rational function to intersect its end behavior polynomial asymptote.


Graphing an Elementary Rational Function Given in Factored Form:
Example: y=(x1)/(x+2).

First, plot the y-intercept (set x=0 and solve for y) and plot the x-intercept (set y=0 and solve for x). The y-intercept is y=(01)/(0+2)=–(1/2). The x-intercept(s) is (are) determined by solving for x in the quotient: (x–1)/(x+2)=0. According to the zero quotient rule, the quotient is zero only when the numerator is zero and the denominator is not zero. Here, setting the numerator equal to 0, x–1=0, and solving for x we get x=1. Note that the denominator is not zero when x=1, so the zero quotient rule is satisfied and there is one x-intercept at x=1. To determine if there are any vertical asymptotes, set the denominator equal to zero and solve for x: x+2=0. Consequently, x=2 is a vertical asymptote. Lastly, if you divide the numerator by the denominator, the polynomial quotient is 1 and y=1 represents the polynomial asymptotic end behavior. In this case, y=1 is a horizontal asymptote. Now, you can plot the x- and y-intercepts together with the vertical and horizontal asymptotes.

 

For rational functions, the x-intercepts and vertical asymptotes are the only places where the y-value may, but need not, change sign. So, with one x-intercept at x=1 and one vertical asymptote at x=2, the x-axis is split into three open intervals, (∞,2), (2,1), and (1,∞), on each of which the function is either always positive (the graph is in the upper half-plane) or always negative (the graph is in the lower half-plane). Selecting convenient "test values" on each interval, say x=3, x=0, and x=2, and substituting into the formula for y, we construct a table to determine the sign of y on each interval. The sign tells us whether the curve lies above or below the x-axis on each interval.

Test value

y=(x1)/(x+2)

Sign of y

3

 4

+

0

1/2

2

1/4

+


So far, there are no points plotted to the left of x=2 but we can plot the test values from the table to remedy this and gain further insight. Also, using the information about the sign enables us to begin to sketch the graph nearby the plotted points:

To the left of x=2, we can complete the graph by extending the curve so it is asymptotic to the vertical asymptote, x=2, and the horizontal asymptote y=2. Note that on this half-plane, since the curve cannot cross the x-axis to the left of x=2 (otherwise, the crossing would be another zero), there is only one way to approach the vertical asymptote: up (downward would mean crossing the x-axis). To the right of x=2, we can complete the graph by extending the curve so it is asymptotic to the vertical asymptote, x=2, and the horizontal asymptote y=2. Note that on this half-plane, since the curve cannot cross the x-axis again, there is only one way to approach the vertical asymptote: down (upward from (0,–1/2) would mean crossing the x-axis).

Question: Does the curve intersect the horizontal asymptote y=1?

Answer: Suppose there is a value of x for which y=1. "If" there truly is one, we can find it by solving the equation

1=(x–1)/(x+2)

Multiplying both sides by x+2 we get

x+2=x–1

Subtracting x from both sides, we get

2=–1

This is impossible. That is, the equation cannot be solved for x; so the curve cannot intersect the horizontal asymptote.

 

 

       

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