Review of Rational FunctionsSolving f(x) = bMETHOD: 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
This equation simplifies to There is one obvious root, namely x = 1, so we can rewrite the equation as Since (x^{2} + 3) has not real roots, the only real solution is x = 1. Finding the DomainMETHOD: 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
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 xcoordinate b/a missing from the graph of f(x). EXAMPLE: Find all "holes" in the graph of
Solution: s(t) can be factored as
The terms 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
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
To find other asymptotes we divide top and bottom by x^{2} resulting in
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 xaxis. f(x): The denominator of f(x) is x^{2}  x + 2, which has no real roots, so f(x) has no vertical asymptotes. Dividing through by x^{2} gives
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 t^{3} and setting the appropriate terms to 0 will yield the constant 1. Thus the line These themes are put into a broader context in Stages 3 and 4.
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