Section III: Unit 5: Graphing Rational Functions

HabermanMTH 111Section III: Power, Polynomial, and Rational FunctionsUnit 5: Graphing Rational FunctionsIn the previous unit, we studied the long-run behavior of rational functions. Now we will studythe short-run behavior so that we can sketch the complete graphs of rational functions. Likepolynomial functions, the short-run behavior of rational functions includes roots and yintercepts. In addition, we must also pay attention to the values that are excluded from thedomains of rational functions.In the first two examples we will compare two simple rational functions, f ( x) g ( x) 1andx1. The reason to compare these functions is that they are very similar (both havex2horizontal asymptote y 0 and domain xx and x 0 ) and the only differencebetween them is that the factor of x in the denominator is squared in g but not in f. Comparingthese examples will help us determine how the degree of a factor in the denominator affectsthe shape of the graph of a rational function.EXAMPLE: Sketch a graph of the rational functionf ( x) 1.xSOLUTION:Let’s first establish the long-run behavior. Clearly, as x , f ( x ) gets closer andcloser to zero, so y 0 is the horizontal asymptote. When x , f ( x ) is positive, sothe graph is above the horizontal asymptote y 0 , but when x , f ( x ) is negative,so the graph is below the horizontal asymptote y 0 . This information leads to thefollowing graph of the long-run behavior:Figure 1: The long-run behaviorof f ( x) 1 .x

HabermanMTH 111Section III: Unit 52For the short-run behavior, we need to find the roots and the y-intercepts – but thisfunction has no roots and no y-intercept. (There is no y-intercept since 0 is not in thedomain of f, and there are no roots since 1 0 for all x.) So the only other thing to lookxfor are values that must be excluded from the domain of f. Obviously, 0 must beexcluded from the domain of f. Let’s investigate what happens to the graph of f as xgets closer and closer to 0. In Table 1 we’ll see what happens as x gets close to 0 fromthe right side of 0 (i.e., from above 0) and in Table 2 we’ll see what happens as x getsclose to 0 from the left side of 0 (i.e., from below 0).Table 1Table 2xf ( x)xf ( x)110110011000110000101 10–101001 100–10010001 1000–1000100001 10000–10000We can see in Table 1 that as x gets close to 0 from the right side of 0, f ( x ) gets largerand larger (so f increases without bound). From Table 2 we can see that as x gets closeto 0 from the left side of 0, f ( x ) gets smaller and smaller (so f decreases withoutbound). We can use this information to sketch the complete graph of f ; see Figure 2.Figure 2: The graph of f ( x ) 1 .xIn the example above, the line x 0 (i.e., the y-axis) is an important feature in the graph of f :the graph gets closer and closer to this line but never crosses it. This vertical line is anasymptote. Since it is a vertical line, it is called a vertical asymptote.DEFINITION: A vertical asymptote is a vertical line that the graph of a function nevercrosses. As the graph of a function approaches this vertical line, theoutputs either increase without bound or decrease without bound.

HabermanMTH 111Section III: Unit 5EXAMPLE: Sketch a graph of the rational function3g ( x) 1.x2SOLUTION:1 in the example above, g ( x ) gets closer and closer to 0 asxx , so y 0 is the horizontal asymptote. But unlike f, as x , g ( x ) is alwayspositive, so the graph is always above the line y 0 . The long-run behavior of g isJust as with f ( x) graphed in Figure 3.Figure 3: The long-run behaviorof g ( x ) 1.x21 , the function g ( x ) 1 has no roots and no y-intercept. And since g isn’txx2defined at 0 it probably has vertical asymptote x 0 just like f. Let’s study a couple ofLike f ( x) appropriate tables to make sure:Table 3Table 4xg ( x)xg ( x)1101001 101001100100001 100100001100010000001 100010000001100001000000001 10000100000000Tables 3 and 4 suggest that g ( x ) 1does in fact have a vertical asymptote at x 0 .x2

HabermanMTH 111Section III: Unit 541 , g increases without bound on both the left and right side of 0. Wexcan use this information to sketch a graph of g; see Figure 4.But unlike f ( x) Figure 4: The graph of g ( x ) 1.x2Before we move on, let’s observe what we can learn from these last two examples. Both1f ( x) 1 and g ( x ) 2 have vertical asymptotes at x 0 (this is because x is a factor inxxthe denominator). In function f, the factor x is a linear factor and the behavior at the verticalasymptote x 0 is different on the left and right sides of 0, while in function g, the factor x is aquadratic factor (i.e., it is squared) and the behavior is the same on both sides of 0.It turns out that this is always the case: if the factor is squared then the behavior is the sameon both sides of the asymptote but if the factor is linear, the behavior is different on both sidesof the asymptote. In fact, if the factor has any even power, the behaviors will be the same onboth sides of the asymptote, and if the factor has an odd power, the behaviors will be differenton both sides of the asymptote.

HabermanMTH 111Section III: Unit 552 x2 8EXAMPLE: Sketch the graph of the function h( x) .( x 1) 2SOLUTION:In Example 1 in Unit 4, we determined that the long-run behavior for h; see Figure 5.Figure 5: Thelong-runbehaviorofy h( x) . and its horizontalasymptote y 2 .To sketch the short-run behavior of h, we need to find the roots, y-intercept, and thevertical asymptotes. To find the roots, we need to find all x such that h( x ) 0 . Of course,the only way that h( x ) 0 is if the numerator of h is 0. So we need to solve 2 x2 8 0 :2x2 8 0 2( x 2 4) 0 2( x 2)( x 2) 0 x 2 or x 2Thus, 2 and –2 are the roots of h.To find the y-intercept, we need to evaluate h(0) :h(0) 2(0) 2 8(0 1) 2 81 8 So the y-intercept is (0, 8) .

HabermanMTH 111Section III: Unit 5To find the vertical asymptotes, we need to determine what values make the denominator0, so we need to solve ( x 1)2 0 . It’s easy to see that the only solution is 1. Therefore,the only vertical asymptote is the line x 1 . Since the factor in the denominator that givesus this vertical asymptote is a quadratic factor (i.e., it is squared), we know that the graphhas the same behavior on both sides of the vertical asymptote. When we plot all of theinformation we’ve determined thus far (see Figure 6), we can see that in order to avoidcontradicting what we’ve concluded about the roots and the horizontal asymptote, the onlypossibility is that the graph decreases without bound on both sides of the verticalasymptote.Figure 6: The long- and short-run behaviorof h( x) 2 x2 8.( x 1) 2Guided by the roots, y-intercept, vertical and horizontal asymptotes, we can sketch thecomplete graph of h; see Figure 7.2 x2 8,( x 1) 2along with asymptotes y 2and x 1 .Figure 7: The graph ofh( x ) 6

HabermanMTH 111Section III: Unit 57Be sure to convince yourself that based on what we discovered about the long-termbehavior and the roots and vertical asymptote, the graph of h( x) 2 x2 8MUST look( x 1) 2like the one pictured in Figure 7. You should also graph this function yourself on yourgraphing calculator for practice.Notice that we were forced to cross the horizontal asymptote at x 2.5 in order to havethe appropriate long-term behavior as well as the root x 2 .Key Point: It is perfectly reasonable for the graph of a rational function to cross itshorizontal asymptote when the x-values aren’t very large.EXAMPLE: Sketch the graph of the function k ( x) x.x 252SOLUTION:In Example 2 in Unit 4, we determined that the long-run behavior for k; see Figure 8.Figure 8: The long-run behavior of y k ( x ) .To sketch the short-run behavior of k, we need to find the roots, y-intercept, and the verticalasymptotes. To find the roots, we need to find all x such that k ( x ) 0 . Of course, theonly way that k ( x ) 0 is if the numerator of k is 0, which obviously occurs when x 0 ,so 0 is the only root.To find the y-intercept, we need to evaluate k (0) . Since 0 is a root, we know thatk (0) 0 , so the y-intercept is (0, 0) . (So the x-intercept and the y-intercept are the samepoint.)To find the vertical asymptotes, we need to determine what values make the denominator0, so we need to solve x2 25 0 :

HabermanMTH 111Section III: Unit 58x 2 25 0 ( x 5)( x 5) 0 x 5 or x 5Therefore, the lines x 5 and x 5 are the vertical asymptotes of k. Let’s plot all ofthe graphical behavior we have determined thus far in Figure 9:Figure 9: The long-run behavior of y k ( x ) ,and its vertical asymptotes x 5and x 5 , and the point (0, 0).Since the factors in the denominator that gives us the vertical asymptote are linear factors(i.e., both ( x 5) and ( x 5) have an (invisible) exponent of 1), we know that the graphhas the different behavior on both sides of the vertical asymptote. (So if the graphincreases without bound on one side of x 5 then it will decrease without bound on theother side of x 5 , and the same is true about x 5 .) Using what we’ve plotted inFigure 9, we can finish the graph of k by making sure that all of the behavior we’vedetermined is satisfied and avoiding any contradictions; see Figure 10.Figure 10: The graph of k ( x) alongx,x 252with its the verticalasymptotes x 5 and x 5.

HabermanMTH 111Section III: Unit 59Be sure to convince yourself that based on what we discovered about the long-runbehavior and the roots and vertical asymptote, the graph of k ( x) xMUST lookx 252like the one pictured in Figure 10. You should also graph this function yourself on yourgraphing calculator for practice.x3 16 xEXAMPLE: Sketch a graph of the function m( x) 2.x 64SOLUTION:In Example 3 in Unit 4, we determined that the long-run behavior for m; see Figure 11.Figure 11: The long-run behavior of y m( x )and its oblique asymptote y x .To sketch the short-run behavior of m, we need to find the roots, y-intercept, and thevertical asymptotes. To find the roots, we need to find all x such that m( x ) 0 . So weneed to solve x3 16 x 0 :x3 16 x 0 x( x 2 16) 0 x( x 4)( x 4) 0x 0 or x 4 or x 4Thus, 0, 4 and –4 are the roots of m.To find the y-intercept, we need to evaluate m(0) .m(0) 0 , so the y-intercept is (0, 0) .Since 0 is a root, we know that

HabermanMTH 111Section III: Unit 5To find the vertical asymptotes, we need to determine what values make the denominator0, so we need to solve x2 64 0 :x 2 64 0 ( x 8)( x 8) 0 x 8 or x 8Therefore, the lines x 8 and x 8 are the vertical asymptotes of m. Let’s plot all of thegraphical behavior we have determined thus far:y m( x) , itsasymptotes y x , x 8 and x 8 ,and the points (0, 0) , (0, 4) , and (0, 4) .Figure 12: The long-run behavior ofUsing what we’ve plotted in Figure 12, we can finish the graph of m by making sure that allof the behavior we’ve determined is satisfied and avoiding any contradictions; see Fig. 13.x3 16 x,x 2 64along with its asymptote y x ,x 8 and x 8 .Figure 13: The graph ofm( x ) 10

HabermanMTH 111Section III: Unit 511Be sure to convince yourself that based on what we discovered about the long-runbehavior and the roots and vertical asymptote, the graph of m( x) x3 16 xMUST lookx 2 64like the one pictured in Figure 13. You should also graph this function yourself on yourgraphing calculator for practice.EXAMPLE: Find an algebraic rule for the functiong graphed in Figure 14.Figure 14: The graph of y g ( x ) .SOLUTION:Since the graph has vertical asymptotes, we know that g is a rational function. Since thevertical asymptotes are x 4 and x 4 , we know that the factors ( x 4) and ( x 4)must appear in the denominator. Since g has roots 3 and 5 , its rule must have factors( x 2) and ( x 6) in the numerator. This tells us that the rule for g looks likeg ( x) a ( x 2)( x 6).( x 4)( x 4)Now we need to find a . Let’s use the horizontal asymptote y 4 to find a . Sincey 4 is the horizontal asymptote, we know that for extreme x values (i.e., as x ),g ( x) 4 :g ( x) Therefore, g ( x ) a ( x 2)( x 6) a x 2 2 a 4 .( x 4)( x 4)x 4( x 2)( x 6).( x 4)( x 4)