Review of Rational Functions

Solving f(x) = b

METHOD: Given f(x) = p(x)/q(x), set f(x) = b and solve the equation p(x) - b q(x) = 0 by whatever means is available.

EXAMPLE: Solve the equation f(x) = 3 where

		x³ + 2 x² + 3
f(x)	=
		x² - x + 2

Solution: We need to solve

x³ + 2 x² + 3 - 3(x² - x + 2) = 0.

This equation simplifies to

x³ - x² +3 x - 3 = 0.

There is one obvious root, namely x = 1, so we can rewrite the equation as

(x - 1)(x² + 3).

Since (x² + 3) has not real roots, the only real solution is x = 1.

Finding the Domain

METHOD: If f(x) = p(x)/q(x), to find the domain of f(x) we look for points at which q(x) is zero. The function f(x) is defined for all other real numbers.

Example: The rational function

		x + 3
a(x)	=
		3 x² + 4 x - 32

has denominator 3 x² + 4 x - 32 which factors as (3 x - 8)(x + 4). This is zero when x = 8/3 and x = -4, so a(x) is defined for all real numbers except for these two. m x + b.

Finding "Holes"

METHOD: To find holes in the graph of a rational function f(x), factor the numerator and denominator. If there is a factor (a x + b)ⁿ in the denominator with at least n powers of (a x + b) in the numerator, then the graph of f(x) is identical to the graph of the function obtained by cancelling all factors of (a x + b) except that there is a point with x-coordinate -b/a missing from the graph of f(x).

EXAMPLE: Find all "holes" in the graph of

		t³ + 2 t² - t - 2
s(t)	=
		t³ - t² -2 t + 2

Solution: s(t) can be factored as

		(t + 1)(t - 1) (t + 2)
s(t)	=
		(t - 1)(t - sqrt(2))(t + sqrt(2))

The terms (t - 1) in the numerator and denominator cancel, but s(t) is not defined there. The graph of s(t) therefore has a "hole" at t = 1. The graph does NOT have "holes" at t = ±sqrt(2) because these are not cancelled by terms in the numerator. The function becomes infinite as t nears these points.

Finding Asymptotes

METHOD: To find the vertical asymptotes of a rational function f(x), factor the numerator and denominator. If there is a factor (a x + b)ⁿ in the denominator with less than n powers of (a x + b) in the numerator, then the graph of f(x) has a vertical asymptote at x = -b/a.

Even if this proceedure results in a non-linear polynomial, the rational function is still asymptotic to the result!
To find the horizontal (and other) asymptotes of a rational function f(x), divide both the numerator and denominator by the largest power of x appearing in the denominator. Then set all the terms with negative exponents to zero. The rational function is then asymptotic to what is left.

Examples: We will find all asymptotes of the functions f(x), a(x), and s(t) which appear above.

a(x): The factored form of a(x) is

		x + 3
a(x)	=
		(3x - 8)(x + 4)

From this we see there are vertical asymptotes at x = 8/3 and x = -4..

To find other asymptotes we divide top and bottom by x² resulting in

		(1/x) + (3/x²)
a(x)	=
		3 + (4/x) - (32/x²)

All terms except the "3" get set to 0, leaving the polynomial 0. This means that a(x) has a horizontal asymptote at y = 0. In other words, it is asymptotic to the x-axis.

f(x): The denominator of f(x) is x² - x + 2, which has no real roots, so f(x) has no vertical asymptotes.

Dividing through by x² gives

		x + 2 + 3/x²
f(x)	=
		1 - 1/x + 2/x²

and setting all the terms with x's in the denominator to 0 gives 2 + x. Thus f(x) is asymptotic to the line 2 + x, which is an oblique asymptote.

s(t): As noted above, the denominator of s(t) has factors of (x + sqrt(2)) and (x - sqrt(2)) which are not in the numerator, so s(t) has vertical asymptotes at x = ±sqrt(2).

Dividing the numerator and deniminator by t³ and setting the appropriate terms to 0 will yield the constant 1. Thus the line t = 1 is a horizontal asymptote for s(t).

These themes are put into a broader context in Stages 3 and 4.

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