1. Copyright © Cengage Learning. All rights reserved. 4 Rational Functions and Conics

2. 4.2 GRAPHS OF RATIONAL FUNCTIONS Copyright © Cengage Learning. All rights reserved.

3. 3 Analyze and sketch graphs of rational functions. Sketch graphs of rational functions that have slant asymptotes. Use graphs of rational functions to model and solve real-life problems. What You Should Learn

4. 4 Analyzing Graphs of Rational Functions

5. 5 To sketch the graph of a rational function, use the following guidelines.

6. 6 Analyzing Graphs of Rational Functions You may also want to test for symmetry when graphing rational functions, especially for simple rational functions. Recall that the graph of f (x) = 1/x is symmetric with respect to the origin.

7. 7 Example 1 – Sketching the Graph of a Rational Function Sketch the graph of and state its domain. Solution: y-intercept: because g(0) = x-intercept: None, because 3 ≠ 0 Vertical asymptote: x = 2, zero of denominator Horizontal asymptote: y = 0, because degree of N(x) < degree of D(x)

8. 8 Example 1 – Solution Additional points: By plotting the intercepts, asymptotes, and a few additional points, you can obtain the graph shown in Figure 4.8. The domain of g is all real numbers except x = 2. cont’d Figure 4.8

9. 9 Analyzing Graphs of Rational Functions The graph of g in Example 1 is a vertical stretch and a right shift of the graph of because

10. 10 Slant Asymptotes

11. 11 Slant Asymptotes Consider a rational function whose denominator is of degree 1 or greater. If the degree of the numerator is exactly one more than the degree of the denominator, the graph of the function has a slant (or oblique) asymptote.

12. 12 Slant Asymptotes For example, the graph of has a slant asymptote, as shown in Figure 4.12. To find the equation of a slant asymptote, use long division. Figure 4.12

13. 13 Slant Asymptotes For instance, by dividing x + 1 into x 2 – x, you obtain As x increases or decreases without bound, the remainder term 2/(x + 1) approaches 0, so the graph of f approaches the line y = x – 2, as shown in Figure 4.12. Slant asymptote (y = x – 2)

14. 14 Example 5 – A Rational Function with a Slant Asymptote Sketch the graph of f (x) = Solution: First write f (x) in two different ways. Factoring the numerator allows you to recognize the x-intercepts.

15. 15 Example 5 – Solution Long division allows you to recognize that the line y = x is a slant asymptote of the graph. y-intercept: (0, 2), because f (0) = 2 x-intercepts: (–1, 0) and (2, 0), because f (–1) = 0 and f (2) = 0 Vertical asymptote: x = 1, zero of denominator cont’d

16. 16 Example 5 – Solution Slant asymptote: y = x Additional points: The graph is shown in Figure 4.13. cont’d Figure 4.13

17. 17 Application

18. 18 Example 6 – Finding a Minimum Area A rectangular page is designed to contain 48 square inches of print. The margins at the top and bottom of the page are each 1 inch deep. The margins on each side are inches wide. What should the dimensions of the page be so that the least amount of paper is used?

19. 19 Example 6 – Solution Let A be the area to be minimized. From Figure 4.14, you can write A = (x + 3)(y + 2). Figure 4.14

20. 20 Example 6 – Solution The printed area inside the margins is modeled by 48 = xy or y = 48/x. To find the minimum area, rewrite the equation for A in terms of just one variable by substituting 48/x for y. Use the table feature of a graphing utility to create a table of values for the function beginning at x = 1. cont’d

21. 21 Example 6 – Solution From the table, you can see that the minimum value of y 1 occurs when x is somewhere between 8 and 9, as shown in Figure 4.16. To approximate the minimum value of y 1 to one decimal place, change the table so that it starts at x = 8 and increases by 0.1. cont’d Figure 4.16

22. 22 Example 6 – Solution The minimum value of y 1 occurs when x ≈ 8.5, as shown in Figure 4.17. The corresponding value of y is 48/8.5 ≈ 5.6 inches. So, the dimensions should be x + 3 ≈ 11.5 inches by y + 2 ≈ 7.6 inches. cont’d Figure 4.17

23. 23 Application If you go on to take a course in calculus, you will learn an analytic technique for finding the exact value of x that produces a minimum area. In this case, that value is x = ≈ 8.485.