4.2 Graphs of Rational Functions

Transcription

1 0 Rational Functions. Graphs of Rational Functions In this section, we take a closer look at graphing rational functions. In Section., we learned that the graphs of rational functions ma have holes in them and could have vertical, horizontal and slant asmptotes. Theorems.,. and. tell us eactl when and where these behaviors will occur, and if we combine these results with what we alread know about graphing functions, we will quickl be able to generate reasonable graphs of rational functions. One of the standard tools we will use is the sign diagram which was first introduced in Section., and then revisited in Section.. In those sections, we operated under the belief that a function couldn t change its sign without its graph crossing through the -ais. The major theorem we used to justif this belief was the Intermediate Value Theorem, Theorem.. It turns out the Intermediate Value Theorem applies to all continuous functions, not just polnomials. Although rational functions are continuous on their domains, Theorem. tells us that vertical asmptotes and holes occur at the values ecluded from their domains. In other words, rational functions aren t continuous at these ecluded values which leaves open the possibilit that the function could change sign without crossing through the -ais. Consider the graph of = h() from Eample.., recorded below for convenience. We have added its -intercept at (, 0) for the discussion that follows. Suppose we wish to construct a sign diagram for h(). Recall that the intervals where h() > 0, or (+), correspond to the -values where the graph of = h() isabove the -ais; the intervals on which h() < 0, or ( ) correspond to where the graph is below the -ais. 8 7 (+) ( ) 0 (+) (+) As we eamine the graph of = h(), reading from left to right, we note that from (, ), the graph is above the -ais, so h() is (+) there. At =, we have a vertical asmptote, at which point the graph jumps across the -ais. On the interval (, ), the graph is below the Recall that, for our purposes, this means the graphs are devoid of an breaks, jumps or holes Another result from Calculus.

2 . Graphs of Rational Functions -ais, so h() is( ) there. The graph crosses through the -ais at (, 0) and remains above the -ais until =, where we have a hole in the graph. Since h() is undefined, there is no sign here. So we have h() as (+) on the interval (, ). Continuing, we see that on (, ), the graph of = h() isabovethe-ais, so we mark (+) there. To construct a sign diagram from this information, we not onl need to denote the zero of h, but also the places not in the domain of h. Asisourcustom,wewrite 0 above on the sign diagram to remind us that it is a zero of h. We need a different notation for and, and we have chosen to use - a nonstandard smbol called the interrobang. We use this smbol to conve a sense of surprise, caution and wonderment - an appropriate attitude to take when approaching these points. The moral of the stor is that when constructing sign diagrams for rational functions, we include the zeros as well as the values ecluded from the domain. Steps for Constructing a Sign Diagram for a Rational Function Suppose r is a rational function.. Place an values ecluded from the domain of r on the number line with an abovethem.. Find the zeros of r and place them on the number line with the number 0 above them.. Choose a test value in each of the intervals determined in steps and.. Determine the sign of r() for each test value in step, and write that sign above the corresponding interval. We now present our procedure for graphing rational functions and appl it to a few ehaustive eamples. Please note that we decrease the amount of detail given in the eplanations as we move through the eamples. The reader should be able to fill in an details in those steps which we have abbreviated. Suppose r is a rational function.. Find the domain of r. Steps for Graphing Rational Functions. Reduce r() to lowest terms, if applicable.. Find the - and-intercepts of the graph of = r(), if the eist.. Determine the location of an vertical asmptotes or holes in the graph, if the eist. Analze the behavior of r on either side of the vertical asmptotes, if applicable.. Analze the end behavior of r. Find the horizontal or slant asmptote, if one eists.. Use a sign diagram and plot additional points, as needed, to sketch the graph of = r().

3 Rational Functions Eample... Sketch a detailed graph of f() =. Solution. We follow the si step procedure outlined above.. As usual, we set the denominator equal to zero to get = 0. We find = ±, so our domain is (, ) (, ) (, ).. To reduce f() to lowest terms, we factor the numerator and denominator which ields f() = ( )(+). There are no common factors which means f() is alread in lowest terms.. To find the -intercepts of the graph of = f(), we set = f() = 0. Solving ( )(+) =0 results in =0. Since = 0 is in our domain, (0, 0) is the -intercept. To find the -intercept, we set = 0 and find = f(0) = 0, so that (0, 0) is our -intercept as well.. The two numbers ecluded from the domain of f are = and =. Sincef() didn t reduce at all, both of these values of still cause trouble in the denominator. Thus b Theorem., = and = are vertical asmptotes of the graph. We can actuall go a step further at this point and determine eactl how the graph approaches the asmptote near each of these values. Though not absolutel necessar, it is good practice for those heading off to Calculus. For the discussion that follows, it is best to use the factored form of f() = ( )(+). The behavior of = f() as : Suppose. If we were to build a table of values, we d use -values a little less than, sa.,.0 and.00. While there is no harm in actuall building a table like we did in Section., we want to develop a number sense here. Let s think about each factor in the formula of f() asweimagine substituting a number like = into f(). The quantit would be ver close to, the quantit ( ) would be ver close to, and the factor ( +) would be ver close to 0. More specificall, ( + ) would be a little less than 0, in this case, We will call such a number a ver small ( ), ver small meaning close to zero in absolute value. So, mentall, as, we estimate f() = ( )( +) ( ) (ver small ( )) = (ver small ( )) Now, the closer gets to, the smaller ( + ) will become, so even though we are multipling our ver small ( ) b, the denominator will continue to get smaller and smaller, and remain negative. The result is a fraction whose numerator is positive, but whose denominator is ver small and negative. Mentall, f() (ver small ( )) ver big ( ) ver small ( ) As we mentioned at least once earlier, since functions can have at most one -intercept, once we find that (0, 0) is on the graph, we know it is the -intercept. The sign diagram in step will also determine the behavior near the vertical asmptotes.

4 . Graphs of Rational Functions The term ver big ( ) means a number with a large absolute value which is negative. What all of this means is that as, f(). Now suppose we wanted to determine the behavior of f() as +. If we imagine substituting something a little larger than infor, sa , we mentall estimate f() ( ) (ver small (+)) = (ver small (+)) ver big (+) ver small (+) We conclude that as +, f(). The behavior of = f() as : Consider. We imagine substituting = Approimating f() as we did above, we get f() (ver small ( )) () = (ver small ( )) ver big ( ) ver small ( ) We conclude that as, f(). Similarl, as +, we imagine substituting = to get f() ver small (+) ver big (+). So as +,f(). Graphicall, we have that near = and =thegraphof = f() looks like. Net, we determine the end behavior of the graph of = f(). Since the degree of the numerator is, and the degree of the denominator is, Theorem. tells us that =0 is the horizontal asmptote. As with the vertical asmptotes, we can glean more detailed information using number sense. For the discussion below, we use the formula f() =. The behavior of = f() as : If we were to make a table of values to discuss the behavior of f as, we would substitute ver large negative numbers in for, sa for eample, = billion. The numerator would then be billion, whereas The actual retail value of f(.00000) is approimatel,00,000. We have deliberatel left off the labels on the -ais because we know onl the behavior near = ±, not the actual function values.

5 Rational Functions the denominator would be ( billion), which is prett much the same as (billion). Hence, f ( billion) billion (billion) ver small ( ) billion Notice that if we substituted in = trillion, essentiall the same kind of cancellation would happen, and we would be left with an even smaller negative number. This not onl confirms the fact that as, f() 0, it tells us that f() 0.Inother words, the graph of = f() is a little bit below the -ais as we move to the far left. The behavior of = f() as : On the flip side, we can imagine substituting ver large positive numbers in for and looking at the behavior of f(). For eample, let = billion. Proceeding as before, we get f ( billion) billion (billion) ver small (+) billion The larger the number we put in, the smaller the positive number we would get out. In other words, as, f() 0 +, so the graph of = f() isalittlebitabove the -ais as we look toward the far right. Graphicall, we have 7. Lastl, we construct a sign diagram for f(). The -values ecluded from the domain of f are = ±, and the onl zero of f is = 0. Displaing these appropriatel on the number line gives us four test intervals, and we choose the test values 8 =, =, = and =. We find f( ) is ( ), f( ) is (+), f() is ( ) andf() is (+). Combining this with our previous work, we get the graph of = f() below. 7 As with the vertical asmptotes in the previous step, we know onl the behavior of the graph as ±.For that reason, we provide no -ais labels. 8 In this particular case, we can eschew test values, since our analsis of the behavior of f near the vertical asmptotes and our end behavior analsis have given us the signs on each of the test intervals. In general, however, this won t alwas be the case, so for demonstration purposes, we continue with our usual construction.

6 . Graphs of Rational Functions ( ) (+) 0 ( ) (+) 0 A couple of notes are in order. First, the graph of = f() certainl seems to possess smmetr with respect to the origin. In fact, we can check f( ) = f() to see that f is an odd function. In some tetbooks, checking for smmetr is part of the standard procedure for graphing rational functions; but since it happens comparativel rarel 9 we ll just point it out when we see it. Also note that while = 0 is the horizontal asmptote, the graph of f actuall crosses the -ais at (0, 0). The mth that graphs of rational functions can t cross their horizontal asmptotes is completel false, 0 as we shall see again in our net eample. Eample... Sketch a detailed graph of g() =. Solution.. Setting =0gives = and =. Our domain is (, ) (, ) (, ).. Factoring g() givesg() = ( )(+) ( )(+). There is no cancellation, so g() isinlowestterms.. To find the -intercept we set = g() = 0. Using the factored form of g() above, we find the zeros to be the solutions of ( )( + ) = 0. We obtain = and =. Since both of these numbers are in the domain of g, wehavetwo-intercepts, (, 0) and (, 0). To find the -intercept, we set = 0 and find = g(0) =,soour-intercept is ( 0, ).. Since g() was given to us in lowest terms, we have, once again b Theorem. vertical asmptotes = and =. Keeping in mind g() = ( )(+) ( )(+), we proceed to our analsis near each of these values. The behavior of = g() as : As, we imagine substituting a number a little bit less than. We have g() 9 And Jeff doesn t think much of it to begin with... 0 That s wh we called it a MYTH! ( 9)( ) ( )(ver small ( )) 9 ver big (+) ver small (+)

7 Rational Functions so as, g(). On the flip side, as +,weget g() 9 ver big ( ) ver small ( ) so g(). The behavior of = g() as : As, we imagine plugging in a number just sh of. We have g() ()() (versmall( ))() ver big ( ) ver small ( ) Hence, as, g().as +,weget g() ver big (+) ver small (+) so g(). Graphicall, we have (again, without labels on the -ais). Since the degrees of the numerator and denominator of g() are the same, we know from Theorem. that we can find the horizontal asmptote of the graph of g b taking the ratio of the leading terms coefficients, = =. However, if we take the time to do a more detailed analsis, we will be able to reveal some hidden behavior which would be lost ( otherwise. As in the discussion following Theorem., we use the result of the long division ) ( ) to rewrite g() = as g() = 7. We focus our attention on the term 7. That is, if ou use a calculator to graph. Once again, Calculus is the ultimate graphing power tool.

8 . Graphs of Rational Functions 7 The behavior of = g() as : If imagine substituting = billion into 7, we estimate ver small ( ). Hence, 7 billion billion g() = 7 ver small ( ) =+versmall(+) In other words, as, the graph of = g() is a little bit above the line =. 7 The behavior of = g() as. To consider as, we imagine substituting = billion and, going through the usual mental routine, find 7 ver small (+) Hence, g() ver small (+), in other words, the graph of = g() is just below the line =as. On = g(), we have (again, without labels on the -ais). Finall we construct our sign diagram. We place an above = and =,anda 0 above = and =. Choosing test values in the test intervals gives us f() is(+)on the intervals (, ), (, ( ) and (, ), and ( ) ontheintervals(, ) and, ).As we piece together all of the information, we note that the graph must cross the horizontal asmptote at some point after = in order for it to approach = from underneath. This is the subtlet that we would have missed had we skipped the long division and subsequent end behavior analsis. We can, in fact, find eactl when the graph crosses =. As a result of the long division, we have g() = 7 7. For g() =, we would need =0. This gives 7 = 0, or = 7. Note that 7 is the remainder when is divided b, so it makes sense that for g() to equal the quotient, the remainder from the division must be 0. Sure enough, we find g(7) =. Moreover, it stands to reason that g must attain a relative minimum at some point past = 7. Calculus verifies that at =, we have such a minimum at eactl (,.9). The reader is challenged to find calculator windows which show the graph crossing its horizontal asmptote on one window, and the relative minimum in the other. In the denominator, we would have (billion) billion. It s eas to see wh the is insignificant, but to ignore the billion seems criminal. However, compared to ( billion), it s on the insignificant side; it s 0 8 versus 0 9. We are once again using the fact that for polnomials, end behavior is determined b the leading term, so in the denominator, the term wins out over the term.

9 8 Rational Functions (+) ( ) 0 (+) 0 ( ) (+) Our net eample gives us an opportunit to more thoroughl analze a slant asmptote. Eample... Sketch a detailed graph of h() = Solution.. For domain, ou know the drill. Solving + + = 0 gives = and =. Our answer is (, ) (, ) (, ).. To reduce h(), we need to factor the numerator and denominator. To factor the numerator, we use the techniques set forth in Section. and we get h() = = ( +)( +) ( +)( +) = ( +)( +) ( +) ( +) = ( +)( +) + We will use this reduced formula for h() as long as we re not substituting =. To make this eclusion specific, we write h() = (+)(+) +,.. To find the -intercepts, as usual, we set h() = 0 and solve. Solving (+)(+) + = 0 ields = and =. The latter isn t in the domain of h, so we eclude it. Our onl - intercept is (, 0). To find the -intercept, we set =0. Since0, we can use the reduced formula for h() and we get h(0) = for a -intercept of ( 0, ).. From Theorem., we know that since = still poses a threat in the denominator of the reduced function, we have a vertical asmptote there. As for =, the factor ( +) was canceled from the denominator when we reduced h(), so it no longer causes trouble there. This means that we get a hole when =. To find the -coordinate of the hole, we substitute = into (+)(+) +, per Theorem. and get 0. Hence, we have a hole on Bet ou never thought ou d never see that stuff again before the Final Eam!

10 . Graphs of Rational Functions 9 the -ais at (, 0). It should make ou uncomfortable plugging = intothereduced formula for h(), especiall since we ve made such a big deal concerning the stipulation about not letting = for that formula. What we are reall doing is taking a Calculus short-cut to the more detailed kind of analsis near = which we will show below. Speaking of which, for the discussion that follows, we will use the formula h() = (+)(+) +,. The behavior of = h() as : As, we imagine substituting a number ( )( ) a little bit less than. We have h() (ver small ( )) (ver small ( )) ver big ( ) thus as, h(). On the other side of, as +, we find that h() ver small (+) ver big (+), so h(). The behavior of = h() as. As, we imagine plugging in a number ( )(ver small ( )) a bit less than =. We have h() = ver small (+) Hence, as, h() 0 +. This means that as, the graph is a bit above the point (, 0). As + ( )(ver small (+)),wegeth() =versmall( ). This gives us that as +, h() 0, so the graph is a little bit lower than (, 0) here. Graphicall, we have. For end behavior, we note that the degree of the numerator of h(), + + +, is and the degree of the denominator, + +, is so b Theorem., the graph of = h() has a slant asmptote. For ±, we are far enough awa from = tousethe reduced formula, h() = (+)(+) +,. To perform long division, we multipl out the numerator and get h() = ++ +,, and rewrite h() = + +,. B Theorem., the slant asmptote is = =, and to better see how the graph approaches the asmptote, we focus our attention on the term generated from the remainder, +. The behavior of = h() as : Substituting = billion into +,wegetthe estimate billion ver small ( ). Hence, h() = + + +ver small ( ). This means the graph of = h() is a little bit below the line = as.

11 0 Rational Functions The behavior of = h() as : If,then + ver small (+). This means h() + ver small (+), or that the graph of = h() is a little bit above the line = as. Graphicall we have. To make our sign diagram, we place an above = and = and a 0 above =. On our four test intervals, we find h() is (+) on (, ) and (, ) and h() is( ) on (, ) and (, ). Putting all of our work together ields the graph below. ( ) (+) ( ) 0 (+) We could ask whether the graph of = h() crosses its slant asmptote. From the formula h() = + +,, we see that if h() =, we would have + = 0. Since this will never happen, we conclude the graph never crosses its slant asmptote. But rest assured, some graphs do!

12 . Graphs of Rational Functions We end this section with an eample that shows it s not all pathological weirdness when it comes to rational functions and technolog still has a role to pla in studing their graphs at this level. Eample... Sketch the graph of r() = + +. Solution.. The denominator + is never zero so the domain is (, ).. With no real zeros in the denominator, + is an irreducible quadratic. Our onl hope of reducing r() isif + is a factor of +. Performing long division gives us + + = + + The remainder is not zero so r() is alread reduced.. To find the -intercept, we d set r() = 0. Since there are no real solutions to + + =0,we have no -intercepts. Since r(0) =, we get (0, ) as the -intercept.. This step doesn t appl to r, since its domain is all real numbers.. For end behavior, we note that since the degree of the numerator is eactl two more than the degree of the denominator, neither Theorems. nor. appl. We know from our attempt to reduce r() thatwecanrewriter() = +, so we focus our attention + on the term corresponding to the remainder, It should be clear that as ±, + + ver small (+), which means r() + ver small (+). So the graph = r() is a little bit above the graph of the parabola = as ±. Graphicall,. There isn t much work to do for a sign diagram for r(), since its domain is all real numbers and it has no zeros. Our sole test interval is (, ), and since we know r(0) =, we conclude r() is (+) for all real numbers. At this point, we don t have much to go on for This won t stop us from giving it the old communit college tr, however!

13 Rational Functions a graph. Below is a comparison of what we have determined analticall versus what the calculator shows us. We have no wa to detect the relative etrema analticall 7 apart from brute force plotting of points, which is done more efficientl b the calculator. As usual, the authors offer no apologies for what ma be construed as pedantr in this section. We feel that the detail presented in this section is necessar to obtain a firm grasp of the concepts presented here and it also serves as an introduction to the methods emploed in Calculus. As we have said man times in the past, our instructor will decide how much, if an, of the kinds of details presented here are mission critical to our understanding of Precalculus. Without further dela, we present ou with this section s Eercises. So even Jeff at this point ma check for smmetr! We leave it to the reader to show r( ) =r() sor is even, and, hence, its graph is smmetric about the -ais. 7 Without appealing to Calculus, of course.

14 . Graphs of Rational Functions.. Eercises In Eercises -, use the si-step procedure to graph the rational function. Be sure to draw an asmptotes as dashed lines.. f() = +. f() =. f() =. f() =. f() = + 7. f() = +. f() = f() = 9. f() = +. f() = +. f() = + +. f() = f() = 9. f() =. f() = f() = + + In Eercises 7-0, graph the rational function b appling transformations to the graph of =. 7. f() = 9. h() = + 8. g() = (Hint: Divide) 0. j() = 7 (Hint: Divide). Discuss with our classmates how ou would graph f() = a + b. What restrictions must c + d be placed on a, b, c and d so that the graph is indeed a transformation of =?. In Eample.. in Section. we showed that p() = + is not a polnomial even though its formula reduced to + for 0. However, it is a rational function similar to those studied in the section. With the help of our classmates, graph p(). 8 Once ou ve done the si-step procedure, use our calculator to graph this function on the viewing window [0, ] [0, 0.]. What do ou see?

15 Rational Functions. Let g() = With the help of our classmates, find the - and intercepts of the graph of g. Find the intervals on which the function is increasing, the intervals on which it is decreasing and the local etrema. Find all of the asmptotes of the graph of g and an holes in the graph, if the eist. Be sure to show all of our work including an polnomial or snthetic division. Sketch the graph of g, using more than one picture if necessar to show all of the important features of the graph. Eample.. showed us that the si-step procedure cannot tell us everthing of importance about the graph of a rational function. Without Calculus, we need to use our graphing calculators to reveal the hidden msteries of rational function behavior. Working with our classmates, use a graphing calculator to eamine the graphs of the rational functions given in Eercises - 7. Compare and contrast their features. Which features can the si-step process reveal and which features cannot be detected b it?. f() = +. f() = +. f() = + 7. f() = +

16 . Graphs of Rational Functions.. Answers. f() = + Domain: (, ) (, ) No -intercepts -intercept: (0, ) Vertical asmptote: = As,f() As +,f() Horizontal asmptote: =0 As,f() 0 As,f() f() = Domain: (, ) (, ) -intercept: (0, 0) -intercept: (0, 0) Vertical asmptote: = As,f() As +,f() Horizontal asmptote: = As,f() + As,f() f() = Domain: (, 0) (0, ) No -intercepts No -intercepts Vertical asmptote: =0 As 0,f() As 0 +,f() Horizontal asmptote: =0 As,f() 0 + As,f() 0 +

17 Rational Functions. f() = + = ( )( +) Domain: (, ) (, ) (, ) No -intercepts -intercept: (0, ) Vertical asmptotes: = and = As,f() As +,f() As,f() As +,f() Horizontal asmptote: =0 As,f() 0 + As,f() 0 +. f() = + = ( )( +) Domain: (, ) (, ) (, ) No -intercepts -intercept: (0, ) f() = +, Hole in the graph at (, 7 ) Vertical asmptote: = As,f() As +,f() Horizontal asmptote: =0 As,f() 0 + As,f() 0 7. f() = + = ( )( +) Domain: (, ) (, ) (, ) -intercept: (0, 0) -intercept: (0, 0) Vertical asmptotes: = and = As,f() As +,f() As,f() As +,f() Horizontal asmptote: =0 As,f() 0 As,f() 0 +

18 . Graphs of Rational Functions 7 7. f() = + Domain: (, ) -intercept: (0, 0) -intercept: (0, 0) No vertical asmptotes No holes in the graph Horizontal asmptote: =0 As,f() 0 As,f() f() = = ( +)( ) Domain: (, ) (, ) (, ) -intercept: (0, 0) -intercept: (0, 0) Vertical asmptotes: =,= As,f() As +,f() As,f() As +,f() No holes in the graph Horizontal asmptote: =0 As,f() 0 As,f() f() = + = Domain: (, ) (, ) (, ) -intercept: (, 0) -intercept: (0, ) Vertical asmptote: = As,f() As +,f() Hole at (, 7 ) Horizontal asmptote: = As,f() + As,f()

19 8 Rational Functions 0. f() = ( +)( ) = 9 ( +)( ) Domain: (, ) (, ) (, ) -intercepts: (, 0),(, 0) -intercept: ( 0, ) 9 Vertical asmptotes: =,= As,f() As +,f() As,f() As +,f() No holes in the graph Horizontal asmptote: = As,f() + As,f() f() = ( )( +) = + + Domain: (, ) (, ) -intercepts: (, 0), (, 0) -intercept: (0, ) Vertical asmptote: = As,f() As +,f() Slant asmptote: = As, the graph is above = As, the graph is below = f() = = ( ) Domain: (, ) (, ) -intercepts: (0, 0), (, 0) -intercept: (0, 0) Vertical asmptote: = As,f() As +,f() Slant asmptote: = As, the graph is above = As, the graph is below =

20 . Graphs of Rational Functions 9. f() = + + ( +) = Domain: (, ) (, ) (, ) -intercept: (0, 0) -intercept: (0, 0) Vertical asmptote: = As,f() As +,f() Hole at (, 0) Slant asmptote: = + As, the graph is below = + As, the graph is above = f() = + 9 Domain: (, ) (, ) (, ) -intercepts: (, 0), (0, 0), (, 0) -intercept: (0, 0) Vertical asmptotes: =,= As,f() As +,f() As,f() As +,f() Slant asmptote: = As, the graph is above = As, the graph is below = 7 7. f() = + + Domain: (, ) -intercept: (0, 0) -intercept: (0, 0) Slant asmptote: = As, the graph is below = As, the graph is above =

21 0 Rational Functions. f() = + + Domain: (, ) (, 0) (0, ) (, ) f() = ( +), No -intercepts No -intercepts Vertical asmptotes: = and =0 As,f() As +,f() As 0,f() As 0 +,f() Hole in the graph at (, 0) Horizontal asmptote: =0 As,f() 0 As,f() f() = Shift the graph of = to the right units. 8. g() = Verticall stretch the graph of = b a factor of. Reflect the graph of = about the -ais. Shift the graph of = up unit. 7

22 . Graphs of Rational Functions 9. h() = + = + Shift the graph of = down units. 0. j() = 7 = 7 Shift the graph of = to the right units. Reflect the graph of = about the -ais. Shift the graph of = up units.