r(x) = p(x) q(x) 4. r(x) = 2x2 1

Transcription

1 Chapter 4 Rational Functions 4. Introduction to Rational Functions If we add, subtract or multipl polnomial functions according to the function arithmetic rules defined in Section.5, we will produce another polnomial function. If, on the other hand, we divide two polnomial functions, the result ma not be a polnomial. In this chapter we stud rational functions - functions which are ratios of polnomials. Definition 4.. A rational function is a function which is the ratio of polnomial functions. Said differentl, r is a rational function if it is of the form r() = p() q(), where p and q are polnomial functions. a a According to this definition, all polnomial functions are also rational functions. (Take q() =). As we recall from Section.4, we have domain issues antime the denominator of a fraction is zero. In the eample below, we review this concept as well as some of the arithmetic of rational epressions. Eample 4... Find the domain of the following rational functions. Write them in the form p() q() for polnomial functions p and q and simplif.. f() = +. h() = Solution.. g() = + 4. r() =. To find the domain of f, we proceed as we did in Section.4: we find the zeros of the denominator and eclude them from the domain. Setting + = 0 results in =. Hence,

2 0 Rational Functions our domain is (, ) (, ). The epression f() is alread in the form requested and when we check for common factors among the numerator and denominator we find none, so we are done.. Proceeding as before, we determine the domain of g b solving + = 0. As before, we find the domain of g is (, ) (, ). To write g() in the form requested, we need to get a common denominator g() = + ( +) = + = + = + = ()( +) ()( +) + This formula is now completel simplified.. The denominators in the formula for h() areboth whose zeros are = ±. As a result, the domain of h is (, ) (, ) (, ). We now proceed to simplif h(). Since we have the same denominator in both terms, we subtract the numerators. We then factor the resulting numerator and denominator, and cancel out the common factor. h() = = + ( )( ) = ( +)( ) = + = ( ) ( ) = + = ( ) ( ) ( +) ( ) 4. To find the domain of r, it ma help to temporaril rewrite r() as r() = We need to set all of the denominators equal to zero which means we need to solve not onl = 0, but also = 0. We find = ± for the former and = for the latter. Our domain is (, ) (, ( ), ) (, ). We simplif r() b rewriting the division as multiplication b the reciprocal and then b canceling the common factor

3 4. Introduction to Rational Functions 0 r() = ( ) ( ) = ( ) ( ) = = = ( )( ) ( ) ( ) A few remarks about Eample 4.. are in order. Note that the epressions for f(), g() and h() work out to be the same. However, onl two of these functions are actuall equal. Recall that functions are ultimatel sets of ordered pairs, so for two functions to be equal, the need, among other things, to have the same domain. Since f() =g() andf and g have the same domain, the are equal functions. Even though the formula h() isthesameasf(), the domain of h is different than the domain of f, and thus the are different functions. We now turn our attention to the graphs of rational functions. Consider the function f() = + from Eample 4... Using a graphing calculator, we obtain Two behaviors of the graph are worth of further discussion. First, note that the graph appears to break at =. We know from our last eample that = is not in the domain of f which means f( ) is undefined. When we make a table of values to stud the behavior of f near = we see that we can get near = from two directions. We can choose values a little less than, for eample =., =.0, =.00, and so on. These values are said to approach fromtheleft. Similarl, the values = 0.9, = 0.99, = 0.999, etc., are said to approach fromtheright. If we make two tables, we find that the numerical results confirm what we see graphicall. f() (, f()). (., ).0 0 (.0, 0) (.00, 00) (.00, 000) f() (, f()) ( 0.9, 8) ( 0.99, 98) ( 0.999, 998) ( , 9998) As the values approach from the left, the function values become larger and larger positive numbers. We epress this smbolicall b stating as, f(). Similarl, using analogous notation, we conclude from the table that as +, f(). For this tpe of You should review Sections. and. if this statement caught ou off guard. We would need Calculus to confirm this analticall.

4 04 Rational Functions unbounded behavior, we sa the graph of = f() has a vertical asmptote of =. Roughl speaking, this means that near =, the graph looks ver much like the vertical line =. Theotherfeatureworthofnoteaboutthegraphof = f() is that it seems to level off on the left and right hand sides of the screen. This is a statement about the end behavior of the function. As we discussed in Section., the end behavior of a function is its behavior as as attains larger and larger negative values without bound,,andasbecomes large without bound,. Making tables of values, we find f() (, f()) 0. ( 0,.) ( 00,.00) ( 000,.000) ( 0000,.000) f() (, f()) 0.77 (0,.77) (00,.970) (000,.9970) (0000,.9997) From the tables, we see that as, f() + and as, f(). Here the + means from above and the means from below. In this case, we sa the graph of = f() has a horizontal asmptote of =. This means that the end behavior of f resembles the horizontal line =, which eplains the leveling off behavior we see in the calculator s graph. We formalize the concepts of vertical and horizontal asmptotes in the following definitions. Definition 4.. The line = c is called a vertical asmptote of the graph of a function = f() ifas c or as c +,eitherf() or f(). Definition 4.. The line = c is called a horizontal asmptote of the graph of a function = f() ifas or as, f() c. Note that in Definition 4., wewritef() c (not f() c + or f() c ) because we are unconcerned from which direction the values f() approach the value c, just as long as the do so. 4 In our discussion following Eample 4.., we determined that, despite the fact that the formula for h() reduced to the same formula as f(), the functions f and h are different, since = is in the domain of f, but = is not in the domain of h. Ifwegraphh() = using a graphing calculator, we are surprised to find that the graph looks identical to the graph of = f(). There is a vertical asmptote at =, but near =, everthing seem fine. Tables of values provide numerical evidence which supports the graphical observation. Here, the word larger means larger in absolute value. 4 As we shall see in the net section, the graphs of rational functions ma, in fact, cross their horizontal asmptotes. If this happens, however, it does so onl a finite number of times, and so for each choice of and, f() will approach c from either below (in the case f() c )orabove(inthecasef() c +.) We leave f() c generic in our definition, however, to allow this concept to appl to less tame specimens in the Precalculus zoo, such as Eercise 50 in Section 0.5.

5 4. Introduction to Rational Functions 05 h() (, h()) (0.9, 0.40) (0.99, 0.495) (0.999, 0.499) (0.9999, ) h() (, h()) (., 0.574) (.0, ) (.00, ) (.000, 0.500) We see that as, h() 0.5 and as +, h() In other words, the points on the graph of = h() are approaching (, 0.5), but since = is not in the domain of h, itwould be inaccurate to fill in a point at (, 0.5). As we ve done in past sections when something like this occurs, 5 we put an open circle (also called a hole in this case 6 )at(, 0.5). Below is a detailed graph of = h(), with the vertical and horizontal asmptotes as dashed lines Neither = nor = are in the domain of h, et the behavior of the graph of = h() is drasticall different near these -values. The reason for this lies in the second to last step when we simplified the formula for h() in Eample 4.., where we had h() = ( )( ) (+)( ). The reason = is not in the domain of h is because the factor ( + ) appears in the denominator of h(); similarl, = is not in the domain of h because of the factor ( ) in the denominator of h(). The major difference between these two factors is that ( ) cancels with a factor in the numerator whereas ( + ) does not. Loosel speaking, the trouble caused b ( ) in the denominator is canceled awa while the factor ( + ) remains to cause mischief. This is wh the graph of = h() has a vertical asmptote at = but onl a hole at =. These observations are generalized and summarized in the theorem below, whose proof is found in Calculus. 5 For instance, graphing piecewise defined functions in Section.6. 6 In Calculus, we will see how these holes can be plugged when embarking on a more advanced stud of continuit.

6 06 Rational Functions Theorem 4.. Location of Vertical Asmptotes and Holes: a Suppose r is a rational function which can be written as r() = p() q() where p and q have no common zeros.b Let c be a real number which is not in the domain of r. ( ) If q(c) 0, then the graph of = r() has a hole at. c, p(c) q(c) If q(c) = 0, then the line = c is a vertical asmptote of the graph of = r(). a Or, How to tell our asmptote from a hole in the graph. b In other words, r() is in lowest terms. In English, Theorem 4. sas that if = c is not in the domain of r but, when we simplif r(), it no longer makes the denominator 0, then we have a hole at = c. Otherwise, the line = c is a vertical asmptote of the graph of = r(). Eample 4... Find the vertical asmptotes of, and/or holes in, the graphs of the following rational functions. Verif our answers using a graphing calculator, and describe the behavior of the graph near them using proper notation.. f() =. h() = g() = r() = Solution.. To use Theorem 4., we first find all of the real numbers which aren t in the domain of f. To do so, we solve = 0 and get = ±. Since the epression f() is in lowest terms, there is no cancellation possible, and we conclude that the lines = and = are vertical asmptotes to the graph of = f(). The calculator verifies this claim, and from the graph, we see that as, f(),as +, f(),as, f(), and finall as +, f().. Solving 9=0gives = ±. In lowest terms g() = Since = continues to make trouble in the denominator, we know the line = isavertical asmptote of the graph of = g(). Since = no longer produces a 0 in the denominator, = ( )(+) ( )(+) = + we have a hole at =. To find the -coordinate of the hole, we substitute =into + + and find the hole is at (, 5 6). When we graph = g() using a calculator, we clearl see the vertical asmptote at =, but everthing seems calm near =. Hence, as, g(),as +, g(),as, g() 5 6,andas +, g()

7 4. Introduction to Rational Functions 07 The graph of = f() The graph of = g(). The domain of h isallrealnumbers,since + 9 = 0 has no real solutions. Accordingl, the graph of = h() is devoid of both vertical asmptotes and holes. 4. Setting = 0 gives us = as the onl real number of concern. Simplifing, we see r() = = ( )(+) = (+) +. Since = continues to produce a 0 in the denominator of the reduced function, we know = is a vertical asmptote to the graph. The calculator bears this out, and, moreover, we see that as, r() and as +, r(). The graph of = h() The graph of = r() Our net eample gives us a phsical interpretation of a vertical asmptote. This tpe of model arises from a famil of equations cheeril named doomsda equations. 7 Eample 4... A mathematical model for the population P, in thousands, of a certain species of bacteria, t das after it is introduced to an environment is given b P (t) = 00 (5 t),0 t<5.. Find and interpret P (0).. When will the population reach 00,000?. Determine the behavior of P as t 5. Interpret this result graphicall and within the contet of the problem. 7 These functions arise in Differential Equations. The unfortunate name will make sense shortl.

8 08 Rational Functions Solution.. Substituting t = 0 gives P (0) = 00 into the environment. (5 0) = 4, which means 4000 bacteria are initiall introduced. To find when the population reaches 00,000, we first need to remember that P (t) ismeasured in thousands. In other words, 00,000 bacteria corresponds to P (t) = 00. Substituting for P (t) gives the equation 00 (5 t) = 00. Clearing denominators and dividing b 00 gives (5 t) =, which, after etracting square roots, produces t =4ort = 6. Of these two solutions, onl t = 4 in in our domain, so this is the solution we keep. Hence, it takes 4 das for the population of bacteria to reach 00,000.. To determine the behavior of P as t 5, we can make a table t P (t) In other words, as t 5, P (t). Graphicall, the line t = 5 is a vertical asmptote of the graph of = P (t). Phsicall, this means that the population of bacteria is increasing without bound as we near 5 das, which cannot actuall happen. For this reason, t =5is called the doomsda for this population. There is no wa an environment can support infinitel man bacteria, so shortl before t = 5 the environment would collapse. Now that we have thoroughl investigated vertical asmptotes, we can turn our attention to horizontal asmptotes. The net theorem tells us when to epect horizontal asmptotes. Theorem 4.. Location of Horizontal Asmptotes: Suppose r is a rational function and r() = p() q(),wherepand q are polnomial functions with leading coefficients a and b, respectivel. Ifthedegreeofp() isthesameasthedegreeofq(), then = a b asmptote of the graph of = r(). is thea horizontal If the degree of p() is less than the degree of q(), then = 0 is the horizontal asmptote of the graph of = r(). If the degree of p() is greater than the degree of q(), then the graph of = r() has no horizontal asmptotes. a The use of the definite article will be justified momentaril. Like Theorem 4., Theorem4. is proved using Calculus. Nevertheless, we can understand the idea behind it using our eample f() = +. If we interpret f() as a division problem, ( ) (+),

9 4. Introduction to Rational Functions 09 we find that the quotient is with a remainder of. Using what we know about polnomial division, specificall Theorem.4, weget =( +). Dividing both sides b ( +) gives + = +. (You ma remember this as the formula for g() in Eample 4...) As becomes unbounded in either direction, the quantit + gets closer and closer to 0 so that the values of f() become closer and closer 8 to. In smbols, as ±, f(), and we have the result. 9 Notice that the graph gets close to the same value as or. This means that the graph can have onl one horizontal asmptote if it is going to have one at all. Thus we were justified in using the in the previous theorem. Alternativel, we can use what we know about end behavior of polnomials to help us understand this theorem. From Theorem., we know the end behavior of a polnomial is determined b its leading term. Appling this to the numerator and denominator of f(), we get that as ±, f() = + =. This last approach is useful in Calculus, and, indeed, is made rigorous there. (Keep this in mind for the remainder of this paragraph.) Appling this reasoning to the general case, suppose r() = p() q() where a is the leading coefficient of p() andb is the leading coefficient of q(). As ±, r() an b,wherenand m are the degrees of p() andq(), respectivel. m Ifthedegreeofp() andthedegreeofq() are the same, then n = m so that r() a b,which means = a b is the horizontal asmptote in this case. If the degree of p() is less than the degree of q(), then n<m,som nis a positive number, and hence, r() a 0as ±.If b m n the degree of p() is greater than the degree of q(), then n>m, and hence n m is a positive number and r() an m b, which becomes unbounded as ±. As we said before, if a rational function has a horizontal asmptote, then it will have onl one. (This is not true for other tpes of functions we shall see in later chapters.) Eample List the horizontal asmptotes, if an, of the graphs of the following functions. Verif our answers using a graphing calculator, and describe the behavior of the graph near them using proper notation.. f() = 5 + Solution.. g() = 4 +. h() = The numerator of f() is5, which has degree. The denominator of f() is +,which has degree. Appling Theorem 4., = 0 is the horizontal asmptote. Sure enough, we see from the graph that as, f() 0 and as, f() The numerator of g(), 4, has degree, but the degree of the denominator, +, has degree. B Theorem 4., there is no horizontal asmptote. From the graph, we see that the graph of = g() doesn t appear to level off to a constant value, so there is no horizontal asmptote. 0 8 As seen in the tables immediatel preceding Definition More specificall, as, f() +,andas, f(). 0 Sit tight! We ll revisit this function and its end behavior shortl.

10 0 Rational Functions. The degrees of the numerator and denominator of h() are both three, so Theorem 4. tells us = 6 = is the horizontal asmptote. We see from the calculator s graph that as, h() +,andas, h(). The graph of = f() The graph of = g() The graph of = h() Our net eample of the section gives us a real-world application of a horizontal asmptote. Eample The number of students N at local college who have had the flu t months after the semester begins can be modeled b the formula N(t) = t for t 0.. Find and interpret N(0).. How long will it take until 00 students will have had the flu?. Determine the behavior of N as t. Interpret this result graphicall and within the contet of the problem. Solution.. N(0) = have had the flu. +(0) = 50. This means that at the beginning of the semester, 50 students. We set N(t) = 00 to get t = 00 and solve. Isolating the fraction gives +t = 00. Clearing denominators gives 450 = 00( + t). Finall, we get t = 5. This means it will take 5 months, or about das, for 00 students to have had the flu.. To determine the behavior of N as t, we can use a table. t N(t) The table suggests that as t, N(t) 500. (More specificall, 500.) This means as time goes b, onl a total of 500 students will have ever had the flu. Though the population below is more accuratel modeled with the functions in Chapter 6, we approimate it (using Calculus, of course!) using a rational function.

11 4. Introduction to Rational Functions We close this section with a discussion of the third (and final!) kind of asmptote which can be associated with the graphs of rational functions. Let us return to the function g() = 4 + in Eample Performing long division, we get g() = 4 + = +. Since the term + 0as ±, it stands to reason that as becomes unbounded, the function values g() = +. Geometricall, this means that the graph of = g() should resemble the line = as ±. We see this pla out both numericall and graphicall below. g() g() = g() and = = g() and = as as The wa we smbolize the relationship between the end behavior of = g() with that of the line = istowrite as ±, g(). In this case, we sa the line = isa slant asmptote to the graph of = g(). Informall, the graph of a rational function has a slant asmptote if, as or as, the graph resembles a non-horizontal, or slanted line. Formall, we define a slant asmptote as follows. Definition 4.4. The line = m + b where m 0 is called a slant asmptote of the graph of a function = f() ifas or as, f() m + b. A few remarks are in order. First, note that the stipulation m 0 in Definition 4.4 is what makes the slant asmptote slanted as opposed to the case when m = 0 in which case we d have a horizontal asmptote. Secondl, while we have motivated what me mean intuitivel b the notation f() m+b, like so man ideas in this section, the formal definition requires Calculus. Another wa to epress this sentiment, however, is to rephrase f() m + b as f() (m + b) 0. In other words, the graph of = f() has the slant asmptote = m + b if and onl if the graph of = f() (m + b) has a horizontal asmptote =0. See the remarks following Theorem 4.. Also called an oblique asmptote in some, ostensibl higher class (and more epensive), tets.

12 Rational Functions Our net task is to determine the conditions under which the graph of a rational function has a slant asmptote, and if it does, how to find it. In the case of g() = 4 +, the degree of the numerator 4is,whichiseactl one more than the degree if its denominator +which is. This results in a linear quotient polnomial, and it is this quotient polnomial which is the slant asmptote. Generalizing this situation gives us the following theorem. 4 Theorem 4.. Determination of Slant Asmptotes: Suppose r is a rational function and r() = p() q(), where the degree of p is eactl one more than the degree of q. Then the graph of = r() has the slant asmptote = L() wherel() is the quotient obtained b dividing p() bq(). In the same wa that Theorem 4. gives us an eas wa to see if the graph of a rational function r() = p() q() has a horizontal asmptote b comparing the degrees of the numerator and denominator, Theorem 4. gives us an eas wa to check for slant asmptotes. Unlike Theorem 4., which gives us a quick wa to find the horizontal asmptotes (if an eist), Theorem 4. givesusnosuch short-cut. If a slant asmptote eists, we have no recourse but to use long division to find it. 5 Eample Find the slant asmptotes of the graphs of the following functions if the eist. Verif our answers using a graphing calculator and describe the behavior of the graph near them using proper notation.. f() = 4 + Solution.. g() = 4. h() = + 4. The degree of the numerator is and the degree of the denominator is, so Theorem 4. guarantees us a slant asmptote. To find it, we divide = +into 4 + and get a quotient of +, so our slant asmptote is = +. We confirm this graphicall, and we see that as, the graph of = f() approaches the asmptote from below, and as, the graph of = f() approaches the asmptote from above. 6. As with the previous eample, the degree of the numerator g() = 4 of the denominator is, so Theorem 4. applies. In this case, is and the degree g() = 4 = ( +)( ) ( ) = ( +) ( ) = +, ( ) 4 Once again, this theorem is brought to ou courtes of Theorem.4 and Calculus. 5 That s OK, though. In the net section, we ll use long division to analze end behavior and it s worth the effort! 6 Note that we are purposefull avoiding notation like as, f() ( +) +. While it is possible to define these notions formall with Calculus, it is not standard to do so. Besides, with the introduction of the smbol in the net section, the authors feel we are in enough trouble alread.

13 4. Introduction to Rational Functions so we have that the slant asmptote = + is identical to the graph of = g() eceptat = (where the latter has a hole at (, 4).) The calculator supports this claim. 7. For h() = +, the degree of the numerator is and the degree of the denominator is so 4 again, we are guaranteed the eistence of a slant asmptote. The long division ( + ) ( 4 ) gives a quotient of just, so our slant asmptote is the line =. The calculator confirms this, and we find that as, the graph of = h() approaches the asmptote from below, and as, the graph of = h() approaches the asmptote from above. The graph of = f() The graph of = g() The graph of = h() The reader ma be a bit disappointed with the authors at this point owing to the fact that in Eamples 4.., 4..4, and4..6, weusedthecalculator to determine function behavior near asmptotes. We rectif that in the net section where we, in ecruciating detail, demonstrate the usefulness of number sense to reveal this behavior analticall. 7 While the word asmptote has the connotation of approaching but not equaling, Definitions 4. and 4.4 invite the same kind of pathologies we saw with Definitions. in Section.6.

14 4 Rational Functions 4.. Eercises In Eercises - 8, for the given rational function f: Find the domain of f. Identif an vertical asmptotes of the graph of = f(). Identif an holes in the graph. Find the horizontal asmptote, if it eists. Find the slant asmptote, if it eists. Graph the function using a graphing utilit and describe the behavior near the asmptotes.. f() = 6 4. f() = + 7. f() = f() = 5 9. f() = f() =. f() = f() = + 5. f() = +7 ( +) 6. f() = + 8. f() = 4 4. f() = f() = f() = f() = + 6. f() = f() = f() = The cost C in dollars to remove p% of the invasive species of Ippizuti fish from Sasquatch Pond is given b C(p) = 770p 00 p, 0 p<00 (a) Find and interpret C(5) and C(95). (b) What does the vertical asmptote at = 00 mean within the contet of the problem? (c) What percentage of the Ippizuti fish can ou remove for $40000? 0. In Eercise 7 in Section.4, the population of Sasquatch in Portage Count was modeled b the function P (t) = 50t t +5, where t = 0 represents the ear 80. Find the horizontal asmptote of the graph of = P (t) and eplain what it means.

15 4. Introduction to Rational Functions 5. Recall from Eample.5. that the cost C (in dollars) to make dopi media plaers is C() = , 0. (a) Find a formula for the average cost C(). Recall: C() = C(). (b) Find and interpret C() and C(00). (c) How man dopis need to be produced so that the average cost per dopi is $00? (d) Interpret the behavior of C() as 0 +. (HINT: You ma want to find the fied cost C(0) to help in our interpretation.) (e) Interpret the behavior of C() as. (HINT: You ma want to find the variable cost (defined in Eample..5 in Section.) to help in our interpretation.). In Eercise 5 in Section., we fit a few polnomial models to the following electric circuit data. (The circuit was built with a variable resistor. For each of the following resistance values (measured in kilo-ohms, kω), the corresponding power to the load (measured in milliwatts, mw ) is given in the table below.) 8 Resistance: (kω) Power: (mw ) Using some fundamental laws of circuit analsis mied with a health dose of algebra, we can derive the actual formula relating power to resistance. For this circuit, it is P () = 5, (+.9) where is the resistance value, 0. (a) Graph the data along with the function = P () on our calculator. (b) Use our calculator to approimate the maimum power that can be delivered to the load. What is the corresponding resistance value? (c) Find and interpret the end behavior of P () as.. In his now famous 99 dissertation The Learning Curve Equation, Louis Leon Thurstone presents a rational function which models the number of words a person can tpe in four minutes as a function of the number of pages of practice one has completed. (This paper, which is now in the public domain and can be found here, isfromabgoneerawhenstudents at business schools took tping classes on manual tpewriters.) Using his original notation and original language, we have Y = L(X+P ) where L is the predicted practice limit in terms (X+P )+R of speed units, X is pages written, Y is writing speed in terms of words in four minutes, P is equivalent previous practice in terms of pages and R is the rate of learning. In Figure 5 of the paper, he graphs a scatter plot and the curve Y = 6(X+9) X+48. Discuss this equation with our classmates. How would ou update the notation? Eplain what the horizontal asmptote of the graph means. You should take some time to look at the original paper. Skip over the computations ou don t understand et and tr to get a sense of the time and place in which the stud was conducted. 8 The authors wish to thank Don Anthan and Ken White of Lakeland Communit College for devising this problem and generating the accompaning data set.

16 6 Rational Functions 4.. Answers. f() = 6 Domain: (, ) (, ) Vertical asmptote: = As,f() As +,f() No holes in the graph Horizontal asmptote: = As,f() As,f() +. f() = +7 5 Domain: (, 5 ) ( 5, ) Vertical asmptote: = 5 As 5,f() As 5 +,f() No holes in the graph Horizontal asmptote: = 7 As,f() 7 + As,f() 7. f() = + = ( +4)( ) Domain: (, 4) ( 4, ) (, ) Vertical asmptotes: = 4,= As 4,f() As 4 +,f() As,f() As +,f() No holes in the graph Horizontal asmptote: =0 As,f() 0 As,f() f() = + Domain: (, ) No vertical asmptotes No holes in the graph Horizontal asmptote: =0 As,f() 0 As,f() f() = +7 ( +) Domain: (, ) (, ) Vertical asmptote: = As,f() As +,f() No holes in the graph Horizontal asmptote: =0 9 As,f() 0 As,f() f() = + = + Domain: (, ) (, ) (, ) Vertical asmptote: = As,f() As +,f() Hole at (, ) Slant asmptote: = As, the graph is below = As, the graph is above = 9 This is hard to see on the calculator, but trust me, the graph is below the -ais to the left of = 7.

17 4. Introduction to Rational Functions 7 7. f() = 4 +4 Domain: (, ) No vertical asmptotes No holes in the graph Horizontal asmptote: =0 As,f() 0 As,f() f() = 4 4 = 4 ( +)( ) Domain: (, ) (, ) (, ) Vertical asmptotes: =,= As,f() As +,f() As,f() As +,f() No holes in the graph Horizontal asmptote: =0 As,f() 0 As,f() f() = + 6 = 4 Domain: (, ) (, ) (, ) Vertical asmptote: = As,f() As +,f() Hole at (, 7 ) 5 Horizontal asmptote: = As,f() + As,f() 0. f() = 5 ( +)( ) = 9 ( +)( ) Domain: (, ) (, ) (, ) Vertical asmptotes: =,= As,f() As +,f() As,f() As +,f() No holes in the graph Horizontal asmptote: = As,f() + As,f(). f() = + + ( +) = Domain: (, ) (, ) (, ) Vertical asmptote: = As,f() As +,f() Hole at (, 0) Slant asmptote: = + As, the graph is below = + As, the graph is above = +. f() = + + Domain: (, ) No vertical asmptotes No holes in the graph Slant asmptote: = As, the graph is above = As, the graph is below =

18 8 Rational Functions. f() = +5 + Domain: (, ( ), ) Vertical asmptote: = As,f() As +,f() No holes in the graph Slant asmptote: = + 9 As, the graph is above = As, the graph is below = f() = +4 = +4 9 ( )( +) Domain: (, ) (, ) (, ) Vertical asmptotes: =, = As,f() As +,f() As,f() As +,f() No holes in the graph Slant asmptote: = As, the graph is above = As, the graph is below = 5. f() = = ( ) Domain: (, ) (, ) Vertical asmptotes: = As,f() As +,f() No holes in the graph Slant asmptote: = 5 8 As, the graph is above = 5 8 As, the graph is below = f() = Domain: (, ) (, ) Vertical asmptote: = As,f() As +,f() No holes in the graph No horizontal or slant asmptote As, f() As, f() 8 7. f() = 9 = Domain: (, ) (, ) (, ) No vertical asmptotes Holes in the graph at (, ) and (, ) Horizontal asmptote = As ±, f() = 8. f() = = Domain: (, ) No vertical asmptotes No holes in the graph Slant asmptote: = 5 f() = 5 everwhere. 9. (a) C(5) = 590 means it costs $590 to remove 5% of the fish and and C(95) = 60 means it would cost $60 to remove 95% of the fish from the pond. (b) The vertical asmptote at = 00 means that as we tr to remove 00% of the fish from the pond, the cost increases without bound; i.e., it s impossible to remove all of the fish. (c) For $40000 ou could remove about 95.76% of the fish.

19 4. Introduction to Rational Functions 9 0. The horizontal asmptote of the graph of P (t) = 50t t+5 is = 50 and it means that the model predicts the population of Sasquatch in Portage Count will never eceed 50.. (a) C() = , >0. (b) C() = 00 and C(00) = 0. When just dopi is produced, the cost per dopi is $00, but when 00 dopis are produced, the cost per dopi is $0. (c) C() = 00 when = 0. So to get the cost per dopi to $00, 0 dopis need to be produced. (d) As 0 +, C(). This means that as fewer and fewer dopis are produced, the cost per dopi becomes unbounded. In this situation, there is a fied cost of $000 (C(0) = 000), we are tring to spread that $000 over fewer and fewer dopis. (e) As, C() This means that as more and more dopis are produced, the cost per dopi approaches $00, but is alwas a little more than $00. Since $00 is the variable cost per dopi (C() = ), it means that no matter how man dopis are produced, the average cost per dopi will alwas be a bit higher than the variable cost to produce a dopi. As before, we can attribute this to the $000 fied cost, which factors into the average cost per dopi no matter how man dopis are produced.. (a) (b) The maimum power is approimatel.60 mw which corresponds to.9 kω. (c) As, P() 0 + which means as the resistance increases without bound, the power diminishes to zero.

20 0 Rational Functions 4. Graphs of Rational Functions In this section, we take a closer look at graphing rational functions. In Section 4., we learned that the graphs of rational functions ma have holes in them and could have vertical, horizontal and slant asmptotes. Theorems 4., 4. and 4. 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.4, 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 4. 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 4.., 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 (+) ( ) 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.

21 4. 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. 4. 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. 4. 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. 5. Analze the end behavior of r. Find the horizontal or slant asmptote, if one eists. 6. Use a sign diagram and plot additional points, as needed, to sketch the graph of = r().

22 Rational Functions Eample 4... Sketch a detailed graph of f() = 4. Solution. We follow the si step procedure outlined above.. As usual, we set the denominator equal to zero to get 4 = 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. 4. 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 4., = 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, 4 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 4., 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 6, the quantit ( ) would be ver close to 4, 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() = ( )( +) 6 ( 4) (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. 4 The sign diagram in step 6 will also determine the behavior near the vertical asmptotes.

23 4. Graphs of Rational Functions The term ver big ( ) means a number with a large absolute value which is negative. 5 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() 6 ( 4) (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() 6 (ver small ( )) (4) = (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 6 5. 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 4. 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() = 4. 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 5 The actual retail value of f(.00000) is approimatel,500, We have deliberatel left off the labels on the -ais because we know onl the behavior near = ±, not the actual function values.

24 4 Rational Functions the denominator 4 would be ( billion) 4, 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 6. 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.

25 4. Graphs of Rational Functions 5 ( ) (+) 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 4... Sketch a detailed graph of g() = 5 6. Solution.. Setting 6=0gives = and =. Our domain is (, ) (, ) (, ).. Factoring g() givesg() = ( 5)(+) ( )(+). 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 ( 5)( + ) = 0. We obtain = 5 and =. Since both of these numbers are in the domain of g, wehavetwo-intercepts, ( 5, 0) and (, 0). To find the -intercept, we set = 0 and find = g(0) = 5 6,soour-intercept is ( 0, 5 6). 4. Since g() was given to us in lowest terms, we have, once again b Theorem 4. vertical asmptotes = and =. Keeping in mind g() = ( 5)(+) ( )(+), 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)( ) ( 5)(ver small ( )) 9 ver big (+) ver small (+)

26 6 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() ()(4) (versmall( ))(5) 4 ver big ( ) ver small ( ) Hence, as, g().as +,weget g() 4 ver big (+) ver small (+) so g(). Graphicall, we have (again, without labels on the -ais) 4 5. Since the degrees of the numerator and denominator of g() are the same, we know from Theorem 4. 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 4., we use the result of the long division 5 ) ( 6 ) to rewrite g() = 5 as g() = 7. We focus our 6 6 attention on the term 7 6. That is, if ou use a calculator to graph. Once again, Calculus is the ultimate graphing power tool.

27 4. Graphs of Rational Functions 7 The behavior of = g() as : If imagine substituting = billion into 7, we estimate ver small ( ). Hence, billion billion g() = 7 ver small ( ) =+versmall(+) 6 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 6 substituting = billion and, going through the usual mental routine, find 7 ver small (+) 6 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) 6. Finall we construct our sign diagram. We place an above = and =,anda 0 above = 5 and =. Choosing test values in the test intervals gives us f() is(+)on the intervals (, ), (, 5 ( ) and (, ), and ( ) ontheintervals(, ) and 5, ).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 6 6 =0. This gives 7 = 0, or = 7. Note that 7 is the remainder when 5 is divided b 6, 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 (,.96). 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 6. It s eas to see wh the 6 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.

28 8 Rational Functions (+) ( ) 0 (+) 0 ( ) (+) Our net eample gives us an opportunit to more thoroughl analze a slant asmptote. Eample 4... 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, ). 4. From Theorem 4., 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 4. and get 0. Hence, we have a hole on Bet ou never thought ou d never see that stuff again before the Final Eam!

29 4. 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 5. For end behavior, we note that the degree of the numerator of h(), , is and the degree of the denominator, + +, is so b Theorem 4., 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 4., 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.

30 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. 4 4 But rest assured, some graphs do!

31 4. 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 4 +. 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. 4. This step doesn t appl to r, since its domain is all real numbers. 5. For end behavior, we note that since the degree of the numerator is eactl two more than the degree of the denominator, neither Theorems 4. nor 4. appl. 5 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 5 This won t stop us from giving it the old communit college tr, however!

32 Rational Functions a graph. 6 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. 6 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.

33 4. Graphs of Rational Functions 4.. Eercises In Eercises - 6, use the si-step procedure to graph the rational function. Be sure to draw an asmptotes as dashed lines.. f() = 4 +. f() = 5 6. f() = 4. f() = 5. f() = f() = f() = f() = f() = + 6. f() = 6 +. f() = f() = f() = 5 9. f() = 4. 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() = 4+ is not a polnomial even though its formula reduced to 4 + 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, 0.5]. What do ou see?

34 4 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 4..4 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 4-7. Compare and contrast their features. Which features can the si-step process reveal and which features cannot be detected b it? 4. f() = + 5. f() = + 6. f() = + 7. f() = +

35 4. Graphs of Rational Functions Answers. f() = 4 + Domain: (, ) (, ) No -intercepts -intercept: (0, ) Vertical asmptote: = As,f() As +,f() Horizontal asmptote: =0 As,f() 0 As,f() f() = 5 6 Domain: (, ) (, ) -intercept: (0, 0) -intercept: (0, 0) Vertical asmptote: = As,f() As +,f() Horizontal asmptote: = 5 As,f() 5 + 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()

36 6 Rational Functions 4. f() = + = ( )( +4) Domain: (, 4) ( 4, ) (, ) No -intercepts -intercept: (0, ) Vertical asmptotes: = 4 and = As 4,f() As 4 +,f() As,f() As +,f() Horizontal asmptote: =0 As,f() 0 + As,f() f() = 5 + = ( )( +) 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() f() = + = ( )( +4) Domain: (, 4) ( 4, ) (, ) -intercept: (0, 0) -intercept: (0, 0) Vertical asmptotes: = 4 and = As 4,f() As 4 +,f() As,f() As +,f() Horizontal asmptote: =0 As,f() 0 As,f()

37 4. Graphs of Rational Functions 7 7. f() = 4 +4 Domain: (, ) -intercept: (0, 0) -intercept: (0, 0) No vertical asmptotes No holes in the graph Horizontal asmptote: =0 As,f() 0 As,f() f() = 4 4 = 4 ( +)( ) 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() = + 6 = 4 Domain: (, ) (, ) (, ) -intercept: (4, 0) -intercept: (0, ) Vertical asmptote: = As,f() As +,f() Hole at (, 7 ) 5 Horizontal asmptote: = As,f() + As,f()

38 8 Rational Functions 0. f() = 5 ( +)( ) = 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() = 6 ( )( +) = + + Domain: (, ) (, ) -intercepts: (, 0), (, 0) -intercept: (0, 6) 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 =

39 4. 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() = +4 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 = f() = + + Domain: (, ) -intercept: (0, 0) -intercept: (0, 0) Slant asmptote: = As, the graph is below = As, the graph is above = 4 4

40 40 Rational Functions 6. 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 g() = Verticall stretch the graph of = b a factor of. Reflect the graph of = about the -ais. Shift the graph of = up unit

41 4. Graphs of Rational Functions 4 9. h() = + = + Shift the graph of = down units j() = 7 = 7 Shift the graph of = to the right units. Reflect the graph of = about the -ais. Shift the graph of = up units

42 4 Rational Functions 4. Rational Inequalities and Applications In this section, we solve equations and inequalities involving rational functions and eplore associated application problems. Our first eample showcases the critical difference in procedure between solving a rational equation and a rational inequalit. Eample Solve + =.. Solve +. Use our calculator to graphicall check our answers to and. Solution.. To solve the equation, we clear denominators. + = ( ) ( ) + ( ) = ( ) 4 + = + epand = 0 ( +)( ) = 0 factor =, 0, Since we cleared denominators, we need to check for etraneous solutions. Sure enough, we see that = does not satisf the original equation and must be discarded. Our solutions are = and =0.. To solve the inequalit, it ma be tempting to begin as we did with the equation namel b multipling both sides b the quantit ( ). The problem is that, depending on, ( ) ma be positive (which doesn t affect the inequalit) or ( ) could be negative (which would reverse the inequalit). Instead of working b cases, we collect all of the terms on one side of the inequalit with 0 on the other and make a sign diagram using the technique given on page in Section ( + ) ( ) + (( )) 0 get a common denominator ( ) 0 epand

43 4. Rational Inequalities and Applications 4 Viewing the left hand side as a rational function r() we make a sign diagram. The onl value ecluded from the domain of r is = which is the solution to = 0. The zeros of r are the solutions to = 0, which we have alread found to be =0, = and =, the latter was discounted as a zero because it is not in the domain. Choosing test values in each test interval, we construct the sign diagram below. (+) 0 ( ) 0 (+) (+) 0 We are interested in where r() 0. We find r() > 0, or (+), on the intervals (, ), (0, ( ) and (, ). We add to these intervals the zeros of r, and 0, to get our final solution:, ] [0, ) (, ).. Geometricall, if we set f() = + and g() =, the solutions to f() =g() are the -coordinates of the points where the graphs of = f() and = g() intersect. The solution to f() g() represents not onl where the graphs meet, but the intervals over which the graph of = f() isabove(>) the graph of g(). We obtain the graphs below. The Intersect command confirms that the graphs cross when = and =0. Itisclear from the calculator that the graph of = f() isabovethegraphof = g() on (, ) as well as on (0, ). According to the calculator, our solution is then (, ] [0, ) which almost matches the answer we found analticall. We have to remember that f is not defined at =, and, even though it isn t shown on the calculator, there is a hole in the graph of = f() when =whichiswh = is not part of our final answer. Net, we eplore how rational equations can be used to solve some classic problems involving rates. Eample 4... Carl decides to eplore the Meander River, the location of several recent Sasquatch sightings. From camp, he canoes downstream five miles to check out a purported Sasquatch nest. Finding nothing, he immediatel turns around, retraces his route (this time traveling upstream), There is no asmptote at = since the graph is well behaved near =. According to Theorem 4., there must be a hole there.

44 44 Rational Functions and returns to camp hours after he left. If Carl canoes at a rate of 6 miles per hour in still water, how fast was the Meander River flowing on that da? Solution. We are given information about distances, rates (speeds) and times. The basic principle relating these quantities is: distance = rate time The first observation to make, however, is that the distance, rate and time given to us aren t compatible : the distance given is the distance for onl part of the trip, the rate given is the speed Carl can canoe in still water, not in a flowing river, and the time given is the duration of the entire trip. Ultimatel, we are after the speed of the river, so let s call that R measured in miles per hour to be consistent with the other rate given to us. To get started, let s divide the trip into its two parts: the initial trip downstream and the return trip upstream. For the downstream trip, all we know is that the distance traveled is 5 miles. distance downstream = rate traveling downstream time traveling downstream 5 miles = rate traveling downstream time traveling downstream Since the return trip upstream followed the same route as the trip downstream, we know that the distance traveled upstream is also 5 miles. distance upstream = rate traveling upstream time traveling upstream 5 miles = rate traveling upstream time traveling upstream We are told Carl can canoe at a rate of 6 miles per hour in still water. How does this figure into the rates traveling upstream and downstream? The speed the canoe travels in the river is a combination of the speed at which Carl can propel the canoe in still water, 6 miles per hour, and the speed of the river, which we re calling R. When traveling downstream, the river is helping Carl along, so we add these two speeds: rate traveling downstream = rate Carl propels the canoe + speed of the river = 6 miles hour + R miles hour So our downstream speed is (6 + R) miles hour. Substituting this into our distance-rate-time equation for the downstream part of the trip, we get: 5 miles = rate traveling downstream time traveling downstream 5 miles = (6 + R) miles hour time traveling downstream When traveling upstream, Carl works against the current. Since the canoe manages to travel upstream, the speed Carl can canoe in still water is greater than the river s speed, so we subtract the river s speed from Carl s canoing speed to get: rate traveling upstream = rate Carl propels the canoe river speed = 6 miles hour R miles hour Proceeding as before, we get

45 4. Rational Inequalities and Applications 45 5 miles = rate traveling upstream time traveling upstream 5 miles = (6 R) miles hour time traveling upstream The last piece of information given to us is that the total trip lasted hours. If we let t down denote the time of the downstream trip and t up the time of the upstream trip, we have: t down +t up = hours. Substituting t down and t up into the distance-rate-time equations, we get (suppressing the units) three equations in three unknowns: E (6+R) t down = 5 E (6 R) t up = 5 E t down + t up = Since we are ultimatel after R, we need to use these three equations to get at least one equation involving onl R. To that end, we solve E fort down b dividing both sides b the quantit (6+R) to get t down = 5 6+R. Similarl, we solve E fort up and get t up = 5 6 R. Substituting these into E, we get: R R =. Clearing denominators, we get 5(6 R)+5(6+R) =(6+R)(6 R) which reduces to R = 6. We find R = ±4, and since R represents the speed of the river, we choose R = 4. On the da in question, the Meander River is flowing at a rate of 4 miles per hour. One of the important lessons to learn from Eample 4.. is that speeds, and more generall, rates, are additive. As we see in our net eample, the concept of rate and its associated principles can be applied to a wide variet of problems - not just distance-rate-time scenarios. Eample 4... Working alone, Talor can weed the garden in 4 hours. If Carl helps, the can weed the garden in hours. How long would it take for Carl to weed the garden on his own? Solution. The ke relationship between work and time which we use in this problem is: amount of work done = rate of work time spent working We are told that, working alone, Talor can weed the garden in 4 hours. In Talor s case then: amount of work Talor does = rate of Talor working time Talor spent working garden = (rate of Talor working) (4 hours) So we have that the rate Talor works is garden 4 hours = garden 4 hour. We are also told that when working together, Talor and Carl can weed the garden in just hours. We have: This is called a sstem of equations. No doubt, ou ve had eperience with these things before, and we will stud sstems in greater detail in Chapter 8. While we usuall discourage dividing both sides of an equation b a variable epression, we know (6 + R) 0 since otherwise we couldn t possibl multipl it b t down and get 5. 4 The reader is encouraged to verif that the units in this equation are the same on both sides. To get ou started, the units on the is hours.

46 46 Rational Functions amount of work done together = rate of working together time spent working together garden = (rate of working together) ( hours) From this, we find that the rate of Talor and Carl working together is garden hours = garden hour.weare asked to find out how long it would take for Carl to weed the garden on his own. Let us call this unknown t, measured in hours to be consistent with the other times given to us in the problem. Then: amount of work Carl does = rate of Carl working time Carl spent working garden = (rate of Carl working) (t hours) In order to find t, we need to find the rate of Carl working, so let s call this quantit R, with units. Using the fact that rates are additive, we have: garden hour so that R = rate working together = rate of Talor working + rate of Carl working garden hour = garden 4 hour + R garden hour garden hour. Substituting this into our work-rate-time equation for Carl, we get: garden = (rate of Carl working) (t hours) ) (t hours) garden = ( garden hour Solving = t,wegett =, so it takes Carl hours to weed the garden on his own.5 As is common with word problems like Eamples 4.. and 4.., there is no short-cut to the answer. We encourage the reader to carefull think through and appl the basic principles of rate to each (potentiall different!) situation. It is time well spent. We also encourage the tracking of units, especiall in the earl stages of the problem. Not onl does this promote uniformit in the units, it also serves as a quick means to check if an equation makes sense. 6 Our net eample deals with the average cost function, first introduced on page 8, as applied to PortaBo Game sstems from Eample..5 in Section.. Eample Given a cost function C(), which returns the total cost of producing items, recall that the average cost function, C() = C() computes the cost per item when items are produced. Suppose the cost C, indollars,toproduceportabo game sstems for a local retailer is C() = , 0.. Find an epression for the average cost function C().. Solve C() < 00 and interpret. 5 Carl would much rather spend his time writing open-source Mathematics tets than gardening anwa. 6 In other words, make sure ou don t tr to add apples to oranges!

47 4. Rational Inequalities and Applications 47. Determine the behavior of C() as and interpret. Solution.. From C() = C() 80+50, we obtain C() =. The domain of C is 0, but since =0 causes problems for C(), we get our domain to be >0, or (0, ).. Solving C() < 00 means we solve < 00. We proceed as in the previous eample < < < 0 common denominator 50 0 < 0 If we take the left hand side to be a rational function r(), we need to keep in mind that the applied domain of the problem is >0. This means we consider onl the positive half of the number line for our sign diagram. On (0, ), r is defined everwhere so we need onl look for zeros of r. Setting r() = 0 gives 50 0 =0,sothat = 5 =7.5. The test intervals on our domain are (0, 7.5) and (7.5, ). We find r() < 0on(7.5, ). 0 (+) 0 ( ) 7.5 In the contet of the problem, represents the number of PortaBo games sstems produced and C() is the average cost to produce each sstem. Solving C() < 00 means we are tring to find how man sstems we need to produce so that the average cost is less than $00 per sstem. Our solution, (7.5, ) tells us that we need to produce more than 7.5 sstems to achieve this. Since it doesn t make sense to produce half a sstem, our final answer is [8, ).. When we appl Theorem 4. to C() we find that = 80 is a horizontal asmptote to the graph of = C(). To more precisel determine the behavior of C() as,we first use long division 7 and rewrite C() = As, 0+, which means C() 80 + ver small (+). Thus the average cost per sstem is getting closer to $80 per sstem. If we set C() = 80, we get 50 = 0, which is impossible, so we conclude that C() > 80 for all >0. This means that the average cost per sstem is alwas greater than $80 per sstem, but the average cost is approaching this amount as more and more sstems are produced. Looking back at Eample..5, we realize $80 is the variable cost per sstem 7 In this case, long division amounts to term-b-term division.

48 48 Rational Functions the cost per sstem above and beond the fied initial cost of $50. Another wa to interpret our answer is that infinitel man sstems would need to be produced to effectivel zero out the fied cost. Our net eample is another classic bo with no top problem. Eample A bo with a square base and no top is to be constructed so that it has a volume of 000 cubic centimeters. Let denote the width of the bo, in centimeters as seen below. height depth width,. Epress the height h in centimeters as a function of the width and state the applied domain.. Solve h() and interpret.. Find and interpret the behavior of h() as 0 + and as. 4. Epress the surface area S of the bo as a function of and state the applied domain. 5. Use a calculator to approimate (to two decimal places) the dimensions of the bo which minimize the surface area. Solution.. We are told that the volume of the bo is 000 cubic centimeters and that represents the width, in centimeters. From geometr, we know Volume = width height depth. Since the base of the bo is a square, the width and the depth are both centimeters. Using h for the height, we have 000 = h,sothath = 000. Using function notation, 8 h() = 000 As for the applied domain, in order for there to be a bo at all, >0, and since ever such choice of will return a positive number for the height h we have no other restrictions and conclude our domain is (0, ).. To solve h(), we proceed as before and collect all nonzero terms on one side of the inequalit in order to use a sign diagram. 8 That is, h() means h of, not h times here.

49 4. Rational Inequalities and Applications 49 h() common denominator We consider the left hand side of the inequalit as our rational function r(). We see r is undefined at = 0, but, as in the previous eample, the applied domain of the problem is >0, so we are considering onl the behavior of r on (0, ). The sole zero of r comes when 000 =0,whichis = 0. Choosing test values in the intervals (0, 0) and (0, ) gives the following diagram. (+) 0 0 ( ) 0 We see r() > 0on(0, 0), and since r() =0at = 0, our solution is (0, 0]. In the contet of the problem, h represents the height of the bo while represents the width (and depth) of the bo. Solving h() is tantamount to finding the values of which result in a bo where the height is at least as big as the width (and, in this case, depth.) Our answer tells us the width of the bo can be at most 0 centimeters for this to happen.. As 0 +, h() = 000. This means that the smaller the width (and, in this case, depth), the larger the height h has to be in order to maintain a volume of 000 cubic centimeters. As, we find h() 0 +, which means that in order to maintain a volume of 000 cubic centimeters, the width and depth must get bigger as the height becomes smaller. 4. Since the bo has no top, the surface area can be found b adding the area of each of the sides to the area of the base. The base is a square of dimensions b, and each side has dimensions b h. We get the surface area, S = +4h. To get S as a function of, we substitute h = 000 to obtain S = +4 ( ) 000. Hence, as a function of, S() = The domain of S is the same as h, namel (0, ), for the same reasons as above. 5. A first attempt at the graph of = S() on the calculator ma lead to frustration. Chances are good that the first window chosen to view the graph will suggest = S() has the -ais as a horizontal asmptote. From the formula S() = , however, we get S() as,sos(). Readjusting the window, we find S does possess a relative minimum at.60. As far as we can tell, 9 this is the onl relative etremum, so it is the absolute minimum as well. This means that the width and depth of the bo should each measure 9 without Calculus, that is...

50 50 Rational Functions approimatel.60 centimeters. To determine the height, we find h(.60) 6.0, so the height of the bo should be approimatel 6.0 centimeters. 4.. Variation In man instances in the sciences, rational functions are encountered as a result of fundamental natural laws which are tpicall a result of assuming certain basic relationships between variables. These basic relationships are summarized in the definition below. Definition 4.5. Suppose, and z are variable quantities. We sa varies directl with (or is directl proportional to) if there is a constant k such that = k. varies inversel with (or is inversel proportional to) if there is a constant k such that = k. z varies jointl with (or is jointl proportional to) and if there is a constant k such that z = k. The constant k in the above definitions is called the constant of proportionalit. Eample Translate the following into mathematical equations using Definition Hooke s Law: TheforceF eerted on a spring is directl proportional the etension of the spring.. Bole s Law: At a constant temperature, the pressure P of an ideal gas is inversel proportional to its volume V.. The volume V of a right circular cone varies jointl with the height h of the cone and the square of the radius r of the base. 4. Ohm s Law: The current I through a conductor between two points is directl proportional to the voltage V between the two points and inversel proportional to the resistance R between the two points.

51 4. Rational Inequalities and Applications 5 5. Newton s Law of Universal Gravitation: Suppose two objects, one of mass m and one of mass M, are positioned so that the distance between their centers of mass is r. The gravitational force F eerted on the two objects varies directl with the product of the two masses and inversel with the square of the distance between their centers of mass. Solution.. Appling the definition of direct variation, we get F = k for some constant k.. Since P and V are inversel proportional, we write P = k V.. There is a bit of ambiguit here. It s clear that the volume and the height of the cone are represented b the quantities V and h, respectivel, but does r represent the radius of the base or the square of the radius of the base? It is the former. Usuall, if an algebraic operation is specified (like squaring), it is meant to be epressed in the formula. We appl Definition 4.5 to get V = khr. 4. Even though the problem doesn t use the phrase varies jointl, it is implied b the fact that the current I is related to two different quantities. Since I varies directl with V but inversel with R, wewritei = kv R. 5. We write the product of the masses mm and the square of the distance as r.wehavethat F varies directl with mm and inversel with r,sof = kmm r. In man of the formulas in the previous eample, more than two varing quantities are related. In practice, however, usuall all but two quantities are held constant in an eperiment and the data collected is used to relate just two of the variables. Comparing just two varing quantities allows us to view the relationship between them as functional, as the net eample illustrates. Eample According to this website the actual data relating the volume V of a gas and its pressure P used b Bole and his assistant in 66 to verif the gas law that bears his name is given below. V P V P Use our calculator to generate a scatter diagram for these data using V as the independent variable and P as the dependent variable. Does it appear from the graph that P is inversel proportional to V? Eplain.. Assuming that P and V do var inversel, use the data to approimate the constant of proportionalit.

52 5 Rational Functions. Use our calculator to determine a Power Regression for this data 0 and use it verif our results in and. Solution.. If P reall does var inversel with V,thenP = k V the points do seem to lie along a curve like = k. for some constant k. From the data plot,. To determine the constant of proportionalit, we note that from P = k V,wegetk = PV. Multipling each of the volume numbers times each of the pressure numbers, we produce a number which is alwas approimatel 400. We suspect that P = V. Graphing = along with the data gives us good reason to believe our hpotheses that P and V are, in fact, inversel related. The graph of the data The data with = 400. After performing a Power Regression, the calculator fits the data to the curve = a b where a 400 and b with a correlation coefficient which is darned near perfect. In other words, = 400 or = 400, as we guessed. 0 We will talk more about this in the coming chapters. You can use tell the calculator to do this arithmetic on the lists and save ourself some time. We will revisit this eample once we have developed logarithms in Chapter 6 to see how we can actuall linearize this data and do a linear regression to obtain the same result.