Homework on Rational Functions - Solutions

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1 Homework on Rational Functions - Solutions Group Work 1 Consider f () = 1. Answer the questions below, eplain your answers. 1. What is the domain of f? f is a fraction, hence to be defined, its denominator cannot be. Therefore, its domain is: R\{}. With the help of a table of values, determine the value f approaches when approaches. An equivalent way of asking this is: find what f approaches when. Here, we want to evaluate f for larger and larger values of. We use a table of values f () It appears that as, f (). 3. Same question when Here, we want to evaluate f for smaller and smaller values of. We use a table of values f () It appears that as, f (). 4. Obviously, f is not defined at. We wish to study the behavior of f as approaches. can either approach from the right ( is larger than and approaches ), in this case we write +. can also approach from the left ( is smaller than and approaches ), in this case we write. Using a table of values, determine what f approaches when +,when First,westudythebehavioroff as +. We use a table of values, with values of approaching and larger than f () It appears that as +, f (). Net, we study the behavior of f as. We use a table of values, with values of approaching and smaller than f () It appears that as, f ().. Sketch the graph of f and its asymptotes. See your book on page. Group Work. Consider g () = 1. Answer the questions below, eplain your answers.

2 1. What is the domain of g? g is a fraction, hence to be defined, its denominator cannot be. Therefore, its domain is: R\{}. With the help of a table of values, determine the value g approaches when approaches. An equivalent way of asking this is: find what g approaches when. Here, we want to evaluate g for larger and larger values of. We use a table of values g () It appears that as, g (). 3. Same question when Here, we want to evaluate g for smaller and smaller values of. We use a table of values g () It appears that as, g (). 4. Determine what g approaches when +,when First, we study the behavior of g as +. We use a table of values, with values of approaching and larger than g () It appears that as +, g (). Net, we study the behavior of g as. We use a table of values, with values of approaching and smaller than g () It appears that as g, g ().. Sketch the graph of g and of its asymptotes. See answer below. 6. Questions 1 - can also be answered by considering g as a transformation of f (see.4). Show how you would do this by first determining which transformation would produce g from f. g can be obtained from f replacing by. Thus,g is obtained by translating f units to the right. Problems, p 9 For each function below, do the following: find the domain find the intercepts find the asymptotes find the end behavior sketch the graph and the asymptotes

3 3 +3 (a) r 1 () = 3 Domain. r 1 () is defined whenever its denominator is not. Thus, its domain is the set of real number ecept 3. -intercepts. The -intercepts are the points at which the graph of r 1 () intersect with the -ais. Therefore, they are the points for which r 1 () =. r 1 () =whenever its numerator is. Thus, the -intercepts of r 1 () (of any rational function in fact) are the zeros of its numerator. In tis case, there is one -intercept, = 1. y-intercept. The y-intercept is the point at which the graph of r 1 () intersect with the y-ais. Thus, to find it, we set =,inotherwords, the y-intercept is given by y = r 1 () = 1. Vertical asymptotes. The vertical asymptotes of a rational fraction are at the zeros of its denominator. In this case, there is one, it is the vertical line =3. Behavior near the vertical asymptote. We know that as 3 and as 3 +, r 1 () approaches infinity. We wish to know if it is or. For this, we look at a table of value. To simulate 3,we trytopluginr 1 () values of close to 3 and less than 3. Such values are.9,.99,... We obtain. So, we see r 1 () that as 3, r 1 (). Similarly, to simulate 3 +,wecantry , 3.1,... We obtain r 1 () Thus, as 3+, r 1 (). Horizontal asymptote. The horizontal asymptote of a rational function are found by studying the behavior of the function as ±. We recall that as ±, a rational function behaves like the quotient of the terms of highest degree. In this case, as ±, r 1 () behaves like 3 =3. Therefore, the line y =3is a horizontal asymptote. End behavior. Here, we want to study the behavior of the function as ±, like for the horizontal asymptote. When a rational function has a horizontal asymptote, then its end behavior is determined by the horizontal asymptote. In this case, we simply say that as ±, r 1 () behaves like the line y = Graph of y = 3 3

4 (b) r () = +1 For the remaining problems, only the answers will be provided. The answers are derived using the same techniques as in the previous problem. Refer to it for eplanations. Domain. All reals ecept 1. -intercepts. There are two, = ± y-intercept. y = Vertical asymptotes. The vertical line = 1 Behavior near the vertical asymptote. Using a table of value, we can see that r ().As 1 and r ().As 1 +. Horizontal asymptote. As ±, r () behaves like =, thus goes to infinity. It follows that r () has no horizontal asymptote. End behavior. Since r () has no horizontal asymptote, we know that r () goes to ± as ±.Wewishtoknowhow?Tofind this, we perform long division and find that +1 = As ±, +1. It follows that as ±, r () will behave like 1. Thiscanbeverified on the graph below. Graph of y = +1 4

5 (c) r 3 () = 6 +3 Domain. Wehave +3 = ( +3)= =or = 3 Thus, the domain is all reals ecept and 3. -intercepts. Wehave 6=( 3) ( +) Thus, the zeros of the numerator are and 3. It follows that the - intercepts are = and =3 y-intercept. None, since cannot be. Vertical asymptotes. They are the vertical lines =and = 3. Behavior near the vertical asymptotes. First, we need to find the behavior of r 3 () near =. Using a table of values, we can check that as, r 3 () and as +, r 3 (). Net, we need to study the behavior of r 3 () near =3. Again, using a table of values, we can see that as 3, r 3 () and as 3 +, r 3 (). Horizontal asymptote. As ±, r 3 () behaves like =1.Thus, the horizontal asymptote is y =1. End behavior. Sincer 3 has a horizontal asymptote, its end behavior is determined by its horizontal asymptote. Thus, as ±, r 3 behaves like y =1. Graph of y = 6 +3

6 (d) r 4 () = 64 3 Domain. Since 3=when = ± 3, the domain of r 4 is all reals ecept 3 and 3. -intercepts. One -intercept, =. y-intercept. Since r 4 () =, the y-intercept is y =. Vertical asymptotes. There are two vertical asymptotes. They are the vertical lines = 3 and = 3. Behavior near the vertical asymptotes. As 3, r 4 (). As 3 +, r 4 ().As 3, r 4 ().As 3 +, r 4 (). Horizontal asymptote. As ±, r 4 () behaves like 64 =6. So, as ±, r 4 ().Thus,r 4 has no horizontal asymptote. End behavior. Above, we saw that as ±, r 4 ().Wewish to know how r 4 approaches. For this, we perform long division. We obtain = As ±, 3. Thus, r 4 () will behave like as ±. Graph of y =

7 E.11p1 Samequestionwithf () = What do you notice, what did this teach you? Domain. All reals ecept. -intercepts. Since 3 3 6=3( +1)( ) It follows that the zeros of f are 1 and. Therefore, the -intercepts should be = 1 and =. However, is not in the domain. The problem here is that f is not simplified. Its numerator and denominator have a common factor, 3( +1)( ). So, f () = =3( +1) when. This means that f is a linear function with a hole at =. y-intercept. The y-intercept is y = f () = 3. Vertical asymptotes. Nonesincef is linear (with a hole). One can also try to study the behavior of f as + and as by computing f () for close to. It will be cleat that f does not go to infinity, which is what must happen to have a vertical asymptote. Behavior near the vertical asymptote. There are no vertical asymptotes. Horizontal asymptote. None,f () ± as ± End behavior. f behaves like 3( +1). Graph of y =

8 There should be a hole at =, it simply does not show here. We learned the following. All the theory developed for rational function assumed that the numerator and denominator had no common factors. So, it is important to find the common factors if any, and cancel them. However, it is also important to keep in mind what the domain is. In this eample, though once simplified the function became 3( +1), the original function was which is not defined at. Even though is a zero of the denominator of this function, = is not a vertical asymptote. This can be seen because as or as +, does not approach ±, which must happen in order to have an asymptote. E. 1 p 1 Find a rational function which has vertical asymptotes = 1 and =, and horizontal asymptote y =1. The vertical asymptotes are the zeros of the denominator. Hence, the denominator of the rational function must have ( ) and ( +1) as factors. Since the horizontal asymptote is y =1, the numerator and the denominator must have the same degree. They must also have the same leading term. finally, the numerator and the denominator must not have common factors. Below, we list several possibilities. (a) (b) + B + C where D can be any constant. B and C can be any constant D ( ) ( +1) as long as and +1are not factors of the numerator. A specific eample +1 would be ( ) ( +1). 4 ( +1)( ) ( +1) #47 p 38. (a) To find the horizontal asymptote, we study the behavior of N (t) as t.from the theorem in the handout, we see that as t, N (t).8 =.16. 8

9 (b) This means that the concentration will settle to.16 parts per million. Thus, the medication never completely dissolves from the body, a trace will always remain. #49 p 38. Given that P (t) = t t +9 (a) P () =., P (1) = 4.4, P (3) =.6, P (8)= (b) To find the horizontal asymptote, we study the behavior of P (t) as t.from the theorem in the handout, we see that as t, P (t). (c) The population will become etinct, no one will live in Lordsburg in the very distant future. 9