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# Introducing Rational Functions

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Section 7.1 Introducing Rational Functions 603Version: Fall 20077.1 Introducing Rational FunctionsIn the previous chapter, we studied polynomials, functions having equation formp(x) = a0+ a1x + a2x2+ ··· + anxn. (1)Even though this polynomial is presented in ascending powers of x, the leading termof the polynomial is still anxn, the term with the highest power of x. The degree ofthe polynomial is the highest power of x present, so in this case, the degree of thepolynomial is n.In this section, our study will lead us to the rational functions. Note the root word“ratio” in the term “rational.” Does it remind you of the word “fraction”? It should,as rational functions are functions in a very specific fractional form.Definition 2. A rational function is a function that can be written as a quotientof two polynomial functions. In symbols, the functionf(x) =a0+ a1x + a2x2+ ··· + anxnb0+ b1x + b2x2+ ··· + bmxm(3)is called a rational function.For example,f(x) =1 + xx + 2, g(x) =x2− 2x − 3x + 4, and h(x) =3 − 2x − x2x3+ 2x2− 3x − 5(4)are rational functions, whilef(x) =1 +√xx2+ 1, g(x) =x2+ 2x − 31 + x1/2− 3x2, and h(x) =rx2− 2x − 3x2+ 4x − 12(5)are not rational functions.Each of the functions in equation (4) are rational functions, because in each case,the numerator and denominator of the given expression is a valid polynomial.However, in equation (5), the numerator of f(x) is not a polynomial (polynomialsdo not allow the square root of the independent variable). Therefore, f is not a rationalfunction.Similarly, the denominator of g(x) in equation (5) is not a polynomial. Fractionsare not allowed as exponents in polynomials. Thus, g is not a rational function.Finally, in the case of function h in equation (5), although the radicand (theexpression inside the radical) is a rational function, the square root prevents h frombeing a rational function.Copyrighted material. See: http://msenux.redwoods.edu/IntAlgText/1604 Chapter 7 Rational FunctionsVersion: Fall 2007An important skill to develop is the ability to draw the graph of a rational function.Let’s begin by drawing the graph of one of the simplest (but most fundamental) rationalfunctions.The Graph of y = 1/xIn all new situations, when we are presented with an equation whose graph we’ve notconsidered or do not recognize, we begin the process of drawing the graph by creatinga table of points that satisfy the equation. It’s important to remember that the graphof an equation is the set of all points that satisfy the equation. We note that zero isnot in the domain of y = 1/x (division by zero makes no sense and is not defined), andcreate a table of points satisfying the equation shown in Figure 1.x10y10x y = 1/x−3 −1/3−2 −1/2−1 −11 12 1/23 1/3Figure 1. At the right is a table of points satisfying the equation y =1/x. These points are plotted as solid dots on the graph at the left.At this point (see Figure 1), it’s pretty clear what the graph is doing betweenx = −3 and x = −1. Likewise, it’s clear what is happening between x = 1 and x = 3.However, there are some open areas of concern.1. What happens to the graph as x increases without bound? That is, what happensto the graph as x moves toward ∞?2. What happens to the graph as x decreases without bound? That is, what happensto the graph as x moves toward −∞?3. What happens to the graph as x approaches zero from the right?4. What happens to the graph as x approaches zero from the left?Let’s answer each of these questions in turn. We’ll begin by discussing the “end-behavior” of the rational function defined by y = 1/x. First, the right end. Whathappens as x increases without bound? That is, what happens as x increases toward∞? In Table 1(a), we computed y = 1/x for x equalling 100, 1 000, and 10 000. Notehow the y-values in Table 1(a) are all positive and approach zero.Students in calculus use the following notation for this idea.limx→∞y = limx→∞1x= 0 (6)Section 7.1 Introducing Rational Functions 605Version: Fall 2007They say “the limit of y as x approaches infinity is zero.” That is, as x approachesinfinity, y approaches zero.x y = 1/x100 0.011 000 0.00110 000 0.0001x y = 1/x−100 −0.01−1 000 −0.001−10 000 −0.0001(a) (b)Table 1. Examining the end-behavior of y = 1/x.A completely similar event happens at the left end. As x decreases without bound,that is, as x decreases toward −∞, note that the y-values in Table 1(b) are all negativeand approach zero. Calculus students have a similar notation for this idea.limx→−∞y = limx→−∞1x= 0. (7)They say “the limit of y as x approaches negative infinity is zero.” That is, as xapproaches negative infinity, y approaches zero.These numbers in Tables 1(a) and 1(b), and the ideas described above, predict thecorrect end-behavior of the graph of y = 1/x. At each end of the x-axis, the y-valuesmust approach zero. This means that the graph of y = 1/x must approach the x-axisfor x-values at the far right- and left-ends of the graph. In this case, we say that thex-axis acts as a horizontal asymptote for the graph of y = 1/x. As x approaches eitherpositive or negative infinity, the graph of y = 1/x approaches the x-axis. This behavioris shown inFigure 2.x10y10Figure 2. The graph of 1/x ap-proaches the x-axis as x increases ordecreases without bound.Our last investigation will be on the interval from x = −1 to x = 1. Readers areagain reminded that the function y = 1/x is undefined at x = 0. Consequently, we willbreak this region in half, first investigating what happens on the region between x = 0606 Chapter 7 Rational FunctionsVersion: Fall 2007and x = 1. We evaluate y = 1/x at x = 1/2, x = 1/4, and x = 1/8, as shown in thetable inFigure 3, then plot the resulting points.x10y10x y = 1/x1/2 21/4 41/8 8Figure 3. At the right is a table of points satisfying the equation y =1/x. These points are plotted as solid dots on the graph at the left.Note that the x-values in the table in Figure 3 approach zero from the right, thennote that the corresponding y-values are getting larger and larger. We could continuein this vein, adding points. For example, if x = 1/16, then y = 16. If x = 1/32, theny = 32. If x = 1/64, then y = 64. Each time we halve our value of x, the resultingvalue of x is closer to zero, and the corresponding y-value doubles in size. Calculusstudents describe this behavior with the notationlimx→0+y = limx→0+1x= ∞. (8)That is,