1. Copyright © Cengage Learning. All rights reserved. 7 Rational Functions

2. Copyright © Cengage Learning. All rights reserved. 7.1 Rational Functions and Variation

3. 3 Objectives  Identify a rational function.  Set up direct and inverse variation problems.  Find rational functions that model an application.  Find the domain of a rational function in an application.  Find the domain of a rational function.

4. 4 Rational Functions

5. 5 Rational functions are functions that contain fractions involving polynomials. These functions can be simple or very complex. Rational functions often result from combining two functions together using division.

6. 6 Rational Functions A function such as is a simple rational function because it has a variable in the denominator of the fraction. Any expression of the form where P(x) and Q(x) are polynomials and Q(x) ≠ 0 is called a rational expression. Notice that Q(x) cannot equal zero, or you would have division by zero, and the rational expression would be undefined. Division by zero is always a concern in working with rational expressions.

7. 7 Rational Functions

8. 8 Example 1 – Cost per student A group of students in the chess club wants to rent a bus to take them to the national chess competition. The bus is going to cost $1500 to rent and can hold up to 60 people. a. Write a model for the cost per student to rent the bus if s students take the bus and each student pays an equal share. b. How much would the cost per student be if 30 students take the bus?

9. 9 Example 1 – Cost per student c. How much would the cost per student be if 60 students take the bus? d. What would a reasonable domain and range be for this model? Explain. cont’d

10. 10 Example 1(a) – Solution Let C(s) be the cost per student in dollars for s students to take the bus to the national chess competition. Because each student is going to pay an equal amount, we might consider a few simple examples: If only one student takes the bus, that student would have to pay $1500. If two students take the bus, they will have to pay dollars each.

11. 11 Example 1(a) – Solution We are taking the total cost of $1500 and dividing it by the number of students taking the bus. This pattern would continue, and we would get the following function. cont’d

12. 12 Example 1(b) – Solution If 30 students take the bus, we can substitute 30 for s and calculate C(30). Therefore, if 30 students take the bus, it will cost $50 per person. cont’d

13. 13 Example 1(c) – Solution Substituting in s = 60, we get Therefore, if 60 students take the bus, it will cost $25 per person. cont’d

14. 14 Example 1(d) – Solution In an application problem, we will continue to avoid model breakdown when setting a domain. Because the bus can hold only up to 60 people, we must limit the domain to positive numbers up to 60. This means that we could have a possible domain of [1, 60]. With this domain, the range would be [25, 1500]. Of course, there are other possible domains and ranges, but these would be considered reasonable. cont’d

15. 15 Direct and Inverse Variation

16. 16 Direct and Inverse Variation The function which we found in part a of Example 1, is an example of inverse variation, and it could be stated that the cost, C(s), varies inversely with the number of students, s. That is, when one value increases, the other decreases. In Example 1, the more students who take the bus, the lower the per student cost will be.

17. 17 Direct and Inverse Variation Variation occurs when two or more variables are related to one another using multiplication or division. When two variables are related and both either increase together or decrease together, we call it direct variation. The equation D = 60t is an example of direct variation; when the value of t increases, so does the value of D.

18. 18 Direct and Inverse Variation A linear equation with a vertical intercept of (0, 0) is a simple representation of direct variation. The variation constant is the slope of the line.

19. 19 Example 2 – Cost for car repair labor The cost for labor at an auto repair shop is directly proportional to the time the mechanic spends working on the car. If a mechanic works on the car for five hours, the labor cost is $325. a. Write a model for the labor cost at this auto repair shop. b. What is the labor cost for two hours of work? c. If Sam were charged $292.50 for labor on a recent repair, how many hours did the mechanic work on the car?

20. 20 Example 2(a) – Solution Let C(h) be the labor cost in dollars for h hours of work done by the mechanic. Because the cost is directly proportional to the hours worked, the cost will be equal to a constant times the hours. C(h) = kh

21. 21 Example 2(a) – Solution We are told that for five hours of work, the cost was $325, so we substitute these values and solve for k. Now that we know that k = 65, we can write the model for the cost as cont’d Solve for k.

22. 22 Example 2(b) – Solution Substitute h = 2 into our model and calculate the cost. C(2) = 65(2) C(2) = 130 If a mechanic works on the car for two hours, the labor cost will be $130. cont’d

23. 23 Example 2(c) – Solution Sam was charged $292.50 for labor, so we substitute this value in for C(h) and solve for h. The mechanic worked on Sam’s car for 4.5 hours, resulting in a labor cost of $252.50. cont’d

24. 24 Example 3 – Illumination from a light source The illumination of a light source is inversely proportional to the square of the distance from the light source. A certain light has an illumination of 50 foot-candles at a distance of 5 feet from the light source. a. Write a model for the illumination of this light. b. What is the illumination of this light at a distance of 10 feet from the light source? c. What is the illumination of this light at a distance of 100 feet?

25. 25 Example 3(a) – Solution Let I(d) be the illumination of the light in foot-candles and let d be the distance from the light source in feet. Because we are told that the illumination, I(d), is inversely proportional to the square of the distance, d 2, from the light source, the illumination will be equal to a constant divided by distance squared.

26. 26 Example 3(a) – Solution We still need to find the variation constant k. Because we are told that at 5 feet from the light source, the illumination is 60 foot-candles, we can substitute these values in and find k. Substitute the given values for I and d. Multiply both sides by 25 to solve for k.

27. 27 Example 3(a) – Solution Now that we know that k = 1250, we can write the model for illumination as cont’d

28. 28 Example 3(b) – Solution We are given the distance of 10 feet, so we can substitute d = 10 and calculate the illumination. Therefore, the illumination at 10 feet from the light source is 12.5 foot-candles. cont’d

29. 29 Example 3(c) – Solution We are given the distance of 100 feet, so we can substitute d = 100 and calculate the illumination. Therefore, the illumination at 100 feet from the light source is 0.125 foot-candle. cont’d

30. 30 Domain of a Rational Function

31. 31 Domain of a Rational Function When considering the domain of a rational function, we will mainly be concerned with excluding values from the domain that would result in the denominator being zero. The easiest way to determine the domain of a rational function is to set the denominator equal to zero and solve. The domain then becomes all real numbers except those values that make the denominator equal zero.

32. 32 Domain of a Rational Function Any place where the denominator is zero would result in a vertical asymptote or a hole with a missing value. Basic logarithmic functions had the y-axis as a vertical asymptote. Vertical asymptotes are also similar to the horizontal asymptotes that we can see in the graphs of exponential functions.

33. 33 Domain of a Rational Function The graph of a function will not touch a vertical asymptote but instead will get as close as possible and then jump over it and continue on the other side. Whenever an input value makes the numerator and denominator both equal to zero, a hole in the graph will occur instead of a vertical asymptote.

34. 34 Domain of a Rational Function Consider the two graphs below to see when a vertical asymptote occurs and when a hole occurs.

35. 35 Domain of a Rational Function A vertical asymptote occurs when an input value makes the denominator equal zero but the numerator does not equal zero. A hole occurs in a graph when an input value makes both the numerator and denominator equal zero.

36. 36 Example 6 – Finding the domain of a rational function Find the domain of the following rational functions. Determine whether the excluded values represent where a vertical asymptote or a hole appear in the graph. a. b. c. d.

37. 37 Example 6 – Finding the domain of a rational function e. cont’d

38. 38 Example 6(a) – Solution Because the denominator of the function would be zero when x = 0, we have a domain of all real numbers except zero. This can also be written simply as x ≠ 0. When x = 0,

39. 39 Example 6(a) – Solution The denominator is zero but the numerator is not, so a vertical asymptote occurs when x = 0. Looking at the graph of f (x), we see that the function jumps over the input value x = 0, and there is a vertical asymptote in its place. cont’d

40. 40 Example 6(b) – Solution The denominator of the function would be zero when so its domain is all real numbers such that When The denominator is zero but the numerator is –4, so there is a vertical asymptote at x = –9. cont’d

41. 41 Example 6(b) – Solution Looking at this graph again, we see a vertical asymptote. Pay attention to the way in which this function must be entered into the calculator with parentheses around the numerator and another set of parentheses around the denominator of the fraction. cont’d

42. 42 Example 6(c) – Solution If you set the denominator of the function equal to zero, you get (x + 4)(x – 7) = 0 x + 4 = 0 x – 7 = 0 x = –4 x = 7 Therefore, the domain is all real numbers except x = –4 or 7. When x = –4, cont’d

43. 43 Example 6(c) – Solution Both the numerator and denominator equal zero, so a hole occurs in the graph when x = –4. When x = 7, The denominator equals zero but the numerator equals 11, so there is a vertical asymptote at x = 7. cont’d

44. 44 Example 6(c) – Solution This graph is shown in two parts so that you can see the hole that appears at x = –4 and then the asymptote at x = 7. Without setting up two windows, it is almost impossible to see the hole. cont’d

45. 45 Example 6(d) – Solution Set the denominator of the function equal to zero. x 2 + 5x + 6 = 0 (x + 3)(x + 2) = 0 x + 3 = 0 x + 2 = 0 x = –3 x = –2 cont’d

46. 46 Example 6(d) – Solution Therefore, the domain is all real numbers except x = –3 or –2. For both x = –3 and –2, the denominator equals zero but the numerator does not. Therefore, there are vertical asymptotes at x = –3 and x = –2. This graph has an interesting shape, but it does have two vertical asymptotes. Again the numerator and denominator of the fraction need parentheses around them to create the graph correctly. cont’d

47. 47 Example 6(e) – Solution The graph of this function shows a vertical asymptote at about x = 3, so the domain should be all real numbers except x = 3. We cannot see any holes in the given graph. cont’d