Homework on Rational Functions - Solutions

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1 Homework on Rational Functions - Solutions Fall, 2 Philippe B. Laval Name 1. 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 3x +3 (a) r 1 (x) = x 3 Domain. r 1 (x) is defined whenever its denominator is not. Thus, its domain is the set of real number except 3. x-intercepts. The x-intercepts are the points at which the graph of r 1 (x) intersect with the x-axis. Therefore, they are the points for which r 1 (x) =. r 1 (x) =whenever its numerator is. Thus, the x-intercepts of r 1 (x) (of any rational function in fact) are the zeros of its numerator. In tis case, there is one x-intercept, x = 1. y-intercept. The y-intercept is the point at which the graph of r 1 (x) intersect with the y-axis. Thus, to find it, we set x =,inotherwords, the y-intercept is given by y = r 1 () = 1. Vertical asymptote. 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 x =3. Behavior near the vertical asymptote. We know that as x 3 and as x 3 +, r 1 (x) go to infinity. We wish to know if it is or. We recall that a fraction whose numerator approaches a non-zero number, and whose denominator approaches will go to infinity. It will go to, if the fraction is positive. Otherwise, it will go to. To simulate x 3, we try to plug in r 1 (x) values of x close to 3 and less than 3. Such values are 2.9, 2.99,... We see that r 1 (2.9) = 117., r 1 (2.99) = So, as x 3, r 1 (x). Similarly, to simulate, we can try 3.1, 3.1,... We see that r 1 (3.1) = 123., r 1 (3.1) = Thus, as x 3 +, r 1 (x). Horizontal asymptote. The horizontal asymptote of a rational function are found by studying the behavior of the function as x ±. We recall that as x ±, a rational function behaves like the quotient of the terms of highest degree. In this case, as x ±, r 1 (x) behaves like 3x =3. Therefore, the line y =3is a horizontal asymptote. x End behavior. Here, we want to study the behavior of the function as x ±, like for the horizontal asymptote. When a rational function

2 has a horizontal asymptote, then its end behavior is determined by the horizontal asymptote. In this case, we simply say that as x ±, r 1 (x) behaves like the line y =3. 3x +3 Graph of y = x 3 2 y x (b) r 2 (x) = x2 2 x +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 explanations. Domain. All reals except 1. x-intercepts. There are two, x = ± 2 y-intercept. y = 2 Vertical asymptotes. The vertical line x = 1 Behavior near the vertical asymptote. As x 1 +, x 2 2 <, x+1 and is positive (you can see this by trying x =.99). Therefore, r 2 (x). As x 1, x 2 2 <, x +1 and is negative (you can see this by trying x = 1.1). Therefore, r 2 (x).asx 1 +. Horizontal asymptote. As x ±, r 2 (x) behaves like x2 = x, thus x goes to infinity. It follows that r 2 (x) has no horizontal asymptote. End behavior. Since r 2 (x) has no horizontal asymptote, we know that r 2 (x) goes to ± as x ±. Wewishtoknowhow?Tofind this, we perform long division and find that x 2 2 x +1 = x 1 1 x +1 1 As x ±, x +1. It follows that as x ±, r 2 (x) will behave like x 1. Thiscanbeverified on the graph below. 2

3 Graph of y = x2 2 x y x (c) r 3 (x) = x2 x 6 x 2 +3x Domain. We have x 2 +3x = x (x +3)= x =or x = 3 Thus, the domain is all reals except and 3. x-intercepts. We have x 2 x 6=(x 3) (x +2) Thus, the zeros of the numerator are 2 and 3. It follows that the x- intercepts are x = 2 and x =3 y-intercept. None, since x cannot be. Vertical asymptotes. They are the vertical lines x =and x = 3. Behavior near the vertical asymptotes. First, we need to find the behavior of r 3 (x) as x and as x +.Forthefirst, we compute r 3 (.1) = and conclude that as x, r 3 (x). For the second, we compute r 3 (.1) = and conclude that as x +, r 3 (x). Next,weneedtostudythebehaviorofr 3 as x 3 and as x 3 +. For the first, we compute r 3 ( 3.1) = and conclude that as x 3, r 3 (x). For the second, we compute r 3 ( 2.99) = and conclude that as x 3 +, r 3 (x). Horizontal asymptote. As x ±, r 3 (x) behaves like x2 x 2 the horizontal asymptote is y =1. =1. Thus, 3

4 End behavior. Since r 3 has a horizontal asymptote, its end behavior is determined by its horizontal asymptote. Thus, as x ±, r 3 behaves like y =1. Graph of y = x2 x 6 x 2 +3x 2 y x (d) r 4 (x) = 6x4 x 2 3 Domain. Since x 2 3=when x = ± 3, the domain of r 4 is all reals except 3 and 3. x-intercepts. One x-intercept, x =. y-intercept. Since r 4 () =, they-interceptisy =. Vertical asymptotes. There are two vertical asymptotes. They are the vertical lines x = 3 and x = 3. Behavior near the vertical asymptotes. As x 3, r 4 (x). As x 3 +, r 4 (x). As x 3, r 4 (x). As x 3 +, r 4 (x). Horizontal asymptote. As x ±, r 4 (x) behaves like 6x4 =6x 2. So, x 2 as x ±, r 4 (x).thus,r 4 has no horizontal asymptote. End behavior. Above, we saw that as x ±, r 4 (x). We wish to know how r 4 approaches. For this, we perform long division. We obtain 6x 4 x 2 3 =6x x As x ±, x 2 3. Thus, r 4 (x) will behave like 6x as x ±. Graph of y = 6x4 x 2 3 4

5 2 15 y x Same question with f (x) = 3x2 3x 6. What do you notice, what did this teach you? Domain. All reals except 2. x-intercepts. Since 3x 2 3x 6=3(x +1)(x 2) It follows that the zeros of f are 1 and 2. Therefore, the x-intercepts should be x = 1 and x =2. However, 2 is not in the domain. The problem here is that f is not simplified. Its numerator and denominator have a common factor, 3(x +1)(x 2). So, f (x) = =3(x +1) when x 6= 2. This means that f is a linear function with a hole at x =2. y-intercept. The y-intercept is y = f () = 3. Vertical asymptotes. None since f is linear (with a hole). One can also try to study the behavior of f as x 2 + and as x 2 by computing f (x) for x close to 2. 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 (x) ± as x ± End behavior. f behaves like 3(x +1). Graph of y = 3x2 3x 6 5

6 x There should be a hole at x =2,itsimplydoesnotshowhere. 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 example, though once simplified the function became 3(x +1), the original function was 3x2 3x 6 which is not defined at 2. Eventhough 2 is a zero of the denominator of this function, x =2 is not a vertical asymptote. This can be seen because as x 2 or as x 2 +, 3x 2 3x 6 does not approach ±, which must happen in order to have an asymptote. 3. Find a rational function which has vertical asymptotes x = 1 and x =2, and horizontal asymptote y =1. The vertical asymptotes are the zeros of the denominator. Hence, the denominator of the rational function must have () and (x +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) x 2 + Bx + C where D can be any constant. B and C can be any constant D () (x +1) as long as and x +1are not factors of the numerator. A specific example x 2 +1 would be () (x +1). x 4 (x 2 +1)() (x +1) The next four problems are from pages 275,276 6

7 6. (a) 1 x + 1 y 1 y = 1 F = 1 F 1 x = x F xf So, y = xf x F When F =55,weget y = 55x x 55 (b) If the object moves far away from the lens, then x. Thusthisquestion is equivalent to finding the horizontal asymptotes of this function. As x, 55x 55x behaves like =55. Thus, as x, y 55 (which is F, the focal x 55 x length). In conclusion, as the object moves far from the lens, the focusing distance becomes the focal length of the lens. (c) As x, y. 61. P (t) = 3t t +1 (a) The domain is all reals except 1. However, since t, P (t) will always be defined. The x-intercepts is t =. The y-intercept is P =. There is a vertical asymptote at t = 1, but we only study P when t. Ast, P (t) 3, thus the horizontal asymptote is P = 3. Therefore, the graph of P is: t 7

8 (b) This equation is equivalent to asking what happens when t,inotherwords, what is the horizontal asymptote. We have already established that the horizontal asymptote was y =3. Thus, the population approaches C (t) = 3t t 2 +2 (a) The domain is all reals. However, we restrict the study to t. We start measuring the drug concentration after the drug is injected. The x-intercepts is t =. The y-intercept is C =. There is no vertical asymptote. As t, C (t), thus the horizontal asymptote is C =. Therefore, the graph of C is: t (b) This question is equivalent to asking what is C (t) approaching as t,inother words, what is the horizontal asymptote of C (t). We have already seen that as t, C (t). Thus, after a long period of time, the drug concentration goes to. (c) This question amounts to finding the solution of C (t) =.3. Looking at the graph, we see this will happen twice. However, the first time, the concentration is rising. So, the correct answer is the largest of the two solutions. In any case, first, we need to solve 3t t 2 +2 =.3 3t t = 3t t t2 +.6 = t t.3t 2.6 = t 2 +2 Afractionis when its numerator is. Thus, we solve 3t.3t 2.6 = 8

9 The solutions are: t = and t = As we said earlier, forthe first solution, C is still rising. So, C will drop below.3 when t> Note: Whenever working with an actual problem, the student should make sure that his/her answers make sense in relation with the current problem. Think about what should happen when drug is injected into a body, and its concentration is measured. First, it will take a while for the drug to spread to the entire body. Thus, at the beginning, the concentration starts at (the drug has not spread yet), then increases (as the drug starts to spread). Since a finite amount of drug is injected, the concentration should stop increasing when the drug is spread evenly in the entire body. Finally, as the body start using the drug, its concentration should start to decrease. As the body will use all the drug, the concentration should drop to. This behavior is shown on the graph. µ µ s P (v) =P.Inourcase,P = 44, s = 332.Thus,P (v) =44 = s v 332 v v (a) The domain is all reals except 332. There are no x-intercepts. The y-intercept is P = 44. There is a vertical asymptote at v =332. As v 332, P (v). As v, P (v), thus the horizontal asymptote is P =. Therefore, the graph of P is: v (b) When the speed of the train approaches the speed of sound, we see that the pitch of the whistle approaches infinity. Haven t you heard the sound a plane makes as it approaches, then exceeds the speed of sound? 9