1. Splash Screen

2. Lesson Menu Five-Minute Check (over Lesson 2-4) Then/Now New Vocabulary Key Concept:Vertical and Horizontal Asymptotes Example 1:Find Vertical and Horizontal Asymptotes Key Concept:Graphs of Rational Functions Example 2:Graph Rational Functions: n < m and n > m Example 3:Graph a Rational Function: n = m Key Concept:Oblique Asymptotes Example 4:Graph a Rational Function: n = m + 1 Example 5:Graph a Rational Function with Common Factors Example 6:Solve a Rational Equation Example 7:Solve a Rational Equation with Extraneous Solutions Example 8:Real-World Example: Solve a Rational Equation

3. Over Lesson 2-4 5–Minute Check 1 List all possible rational zeros of f (x) = 2x 4 – x 3 – 3x 2 – 31x – 15. Then determine which, if any, are zeros. A. B. C. D.

4. Over Lesson 2-4 5–Minute Check 2 List all possible rational zeros of g (x) = x 4 – x 3 + 2x 2 – 4x – 8. Then determine which, if any, are zeros. A. B. C. D.

5. Over Lesson 2-4 5–Minute Check 3 Describe the possible real zeros of f (x) = 2x 5 – x 4 – 8x 3 + 8x 2 – 9x + 9. A.4 positive zeros; 1 negative zero B.4, 2, or 0 positive zeros; 1 negative zero C.3 or 1 positive zeros; 1 negative zero D.2 or 0 positive zeros; 2 or 0 negative zeros

6. Over Lesson 2-4 5–Minute Check 4 Describe the possible real zeros of g (x) = 3x 4 + 16x 3 – 7x 2 – 64x – 20. A.1 positive zero; 3 or 1 negative zeros B.1 positive zero; 3 negative zeros C.1 or 0 positive zeros; 3 or 1 negative zeros D.1 positive zero; 2 or 0 negative zeros

7. Over Lesson 2-4 5–Minute Check 5 Write a polynomial function of least degree with real coefficients in standard form that has 2, –5, and 3 + i as zeros. A.f (x) = x 4 + 3x 3 – x 2 + 27x – 90 B.f (x) = x 4 – 3x 3 – 20x 2 + 84x – 80 C.f (x) = x 4 – 9x 3 + 18x 2 + 30x – 100 D.f (x) = x 4 – 3x 3 – 18x 2 + 90x – 100

8. Then/Now You identified points of discontinuity and end behavior of graphs of functions using limits. (Lesson 1-3) Analyze and graph rational functions. Solve rational equations.

9. Vocabulary rational function asymptote vertical asymptote horizontal asymptote oblique asymptote holes

10. Key Concept 1

11. Example 1 Find Vertical and Horizontal Asymptotes Step 1Find the domain. The function is undefined at the real zero of the denominator b (x) = x – 1. The real zero of b (x) is 1. Therefore, the domain of f is all real numbers except x = 1. A. Find the domain of and the equations of the vertical or horizontal asymptotes, if any.

12. Example 1 Find Vertical and Horizontal Asymptotes Step 2Find the asymptotes, if any. Check for vertical asymptotes. Determine whether x = 1 is a point of infinite discontinuity. Find the limit as x approaches 1 from the left and the right. Because, you know that x = 1 is a vertical asymptote of f.

13. Example 1 Find Vertical and Horizontal Asymptotes Check for horizontal asymptotes. Use a table to examine the end behavior of f (x). The table suggests that 1. Therefore, you know that y = 1 is a horizontal asymptote of f.

14. Example 1 Find Vertical and Horizontal Asymptotes CHECKThe graph of shown supports each of these findings. Answer: D = {x | x ≠ 1, x }; vertical asymptote at x = 1; horizontal asymptote at y = 1

15. Example 1 Find Vertical and Horizontal Asymptotes Step 1The zeros of the denominator b (x) = 2x 2 + 1 are imaginary, so the domain of f is all real numbers. B. Find the domain of and the equations of the vertical or horizontal asymptotes, if any.

16. Example 1 Find Vertical and Horizontal Asymptotes Step 2Because the domain of f is all real numbers, the function has no vertical asymptotes. Using division, you can determine that. As the value of | x | increases, 2x 2 + 1 becomes an increasing large positive number and decreases, approaching 0. Therefore,

17. Example 1 Find Vertical and Horizontal Asymptotes CHECKYou can use a table of values to support this reasoning. The graph of shown also supports each of these findings. Answer: D = {x | x }; no vertical asymptotes; horizontal asymptote at y = 2

18. Example 1 Find the domain of and the equations of the vertical or horizontal asymptotes, if any. A.D = {x | x ≠ 4, x }; vertical asymptote at x = 4; horizontal asymptote at y = –10 B.D = {x | x ≠ 5, x }; vertical asymptote at x = 5; horizontal asymptote at y = 4 C.D = {x | x ≠ 4, x }; vertical asymptote at x = 4; horizontal asymptote at y = 5 D.D = {x | x ≠ 4, 4, x }; vertical asymptote at x = 4; horizontal asymptote at y = –2

19. Key Concept 2

20. Example 2 Graph Rational Functions: n m A. For, determine any vertical and horizontal asymptotes and intercepts. Then graph the function and state its domain. Step 2There is a vertical asymptote at x = –5. The degree of the polynomial in the numerator is 0, and the degree of the polynomial in the denominator is 1. Because 0 < 1, the graph of k has a horizontal asymptote at y = 0. Step 1The function is undefined at b (x) = 0, so the domain is {x | x ≠ –5, x  }

21. Example 2 Graph Rational Functions: n m Step 3The function in the numerator has no real zeros, so k has no x-intercepts. Because k(0) = 1.4, the y-intercept is 1.4. Step 4Graph the asymptotes and intercepts. Then choose x-values that fall in the test intervals determined by the vertical asymptote to find additional points to plot on the graph. Use smooth curves to complete the graph.

22. Example 2 Graph Rational Functions: n m Answer: vertical asymptote at x = –5; horizontal asymptote at y = 0; y-intercept: 1.4; D = {x | x≠ –5, x };

23. Example 2 Graph Rational Functions: n m B. For, determine any vertical and horizontal asymptotes and intercepts. Then graph the function and state its domain. Factoring the denominator yields. Notice that the numerator and denominator have no common factors, so the expression is in simplest form.

24. Example 2 Graph Rational Functions: n m Step 2There are vertical asymptotes at x = 2 and x = –2. Compare the degrees of the numerator and denominator. Because 1 < 2, there is a horizontal asymptote at y = 0. Step 3The numerator has a zero at x = –1, so the x-intercept is –1. f(0) =  0.25, so The y-intercept is –0.25. Step 1The function is undefined at b(x) = 0, so the domain is {x | x ≠ 2, –2, x  }.

25. Example 2 Graph Rational Functions: n m Step 4Graph the asymptotes and intercepts. Then find and plot points in the test intervals determined by the intercepts and vertical asymptotes: (–∞, –2), (–2, –1), (–1, 2), (2, ∞). Use smooth curves to complete the graph.

26. Example 2 Graph Rational Functions: n m Answer: vertical asymptotes at x = 2 and x = –2; horizontal asymptote at y = 0. x-intercept: –1; y-intercept: –0.25; D = {x | x ≠ 2, –2, x }

27. Example 2 A.vertical asymptotes x = –4 and x = 3; horizontal asymptote y = 0; y-intercept: –0.0833 B.vertical asymptotes x = –4 and x = 3; horizontal asymptote y = 1; intercept: 0 C.vertical asymptotes x = 4 and x = 3; horizontal asymptote y = 0; intercept: 0 D.vertical asymptotes x = 4 and x = –3; horizontal asymptote y = 1; y-intercept: –0.0833 Determine any vertical and horizontal asymptotes and intercepts for.

28. Example 3 Graph a Rational Function: n = m Determine any vertical and horizontal asymptotes and intercepts for. Then graph the function, and state its domain. Factoring both numerator and denominator yields with no common factors. Step 1The function is undefined at b (x) = 0, so the domain is {x | x ≠ –2, 2, x  }.

29. Example 3 Graph a Rational Function: n = m Step 2There are vertical asymptotes at x = –2 and x = 2. There is a horizontal asymptote at or y = 0.5, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. Step 3The x-intercepts are –3 and 4, the zeros of the numerator. The y-intercept is 1.5 because f(0) = 1.5.

30. Example 3 Graph a Rational Function: n = m Step 4Graph the asymptotes and intercepts. Then find and plot points in the test intervals (–∞, –3), (–3, –2), (–2, 2), (2, 4), (4, ∞).

31. Example 3 Graph a Rational Function: n = m Answer: vertical asymptotes at x = –2 and x = 2; horizontal asymptote at y = 0.5; x-intercepts: 4 and –3; y-intercept: 1.5;

32. Example 3 A.vertical asymptote x = 2; horizontal asymptote y = 6; x-intercept: –0.833; y-intercept: –2.5 B.vertical asymptote x = 2; horizontal asymptote y = 6; x-intercept: –2.5; y-intercept: –0.833 C.vertical asymptote x = 6; horizontal asymptote y = 2; x-intercepts: –3 and 0; y-intercept: 0 D.vertical asymptote x = 6, horizontal asymptote y = 2; x-intercept: –2.5; y-intercept: –0.833 Determine any vertical and horizontal asymptotes and intercepts for.

33. Key Concept 3

34. Example 4 Graph a Rational Function: n = m + 1 Determine any asymptotes and intercepts for. Then graph the function, and state its domain. Step 2There is a vertical asymptote at x = –3. The degree of the numerator is greater than the degree of the denominator, so there is no horizontal asymptote. Step 1The function is undefined at b (x) = 0, so the domain is D = {x | x ≠ –3, x  }.

35. Example 4 Graph a Rational Function: n = m + 1 Because the degree of the numerator is exactly one more than the degree of the denominator, f has an oblique asymptote. Using polynomial division, you can write the following. f(x) = Therefore, the equation of the oblique asymptote is y = x – 2.

36. Example 4 Graph a Rational Function: n = m + 1 Step 4Graph the asymptotes and intercepts. Then find and plot points in the test intervals (–∞, –3.37), (–3.37, –3), (–3, 2.37), (2.37, ∞). Step 3The x-intercepts are the zeros of the numerator, and, or about 2.37 and –3.37. The y-intercept is about –2.67 because f(0) ≈

37. Example 4 Graph a Rational Function: n = m + 1

38. Example 4 Graph a Rational Function: n = m + 1 Answer:vertical asymptote at x = –3; oblique asymptote at y = x – 2; x-intercepts: and ; y-intercept: ;

39. Example 4 Determine any asymptotes and intercepts for. A.vertical asymptote at x = –2; oblique asymptote at y = x; x-intercepts: 2.5 and 0.5; y-intercept: 0.5 B.vertical asymptote at x = –2; oblique asymptote at y = x – 5; x-intercepts at ; y-intercept: 0.5 C.vertical asymptote at x = 2; oblique asymptote at y = x – 5; x-intercepts: ; y-intercept: 0 D.vertical asymptote at x = –2; oblique asymptote at y = x 2 – 5x + 11; x-intercepts: 0 and 3; y-intercept: 0

40. Example 5 Graph a Rational Function with Common Factors Determine any vertical and horizontal asymptotes, holes, and intercepts for. Then graph the function and state its domain. Factoring both the numerator and denominator yields h(x) = Step 1The function is undefined at b (x) = 0, so the domain is D = {x | x ≠ –2, 3, x  }.

41. Example 5 Graph a Rational Function with Common Factors Step 2There is a vertical asymptote at the real zero of the simplified denominator x = –2. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. Step 3The x-intercept is –3, the zero of the simplified numerator. The y-intercept is because h(0) =.

42. Example 5 Graph a Rational Function with Common Factors Step 4Graph the asymptotes and intercepts. Then find and plot points in the test intervals There is a hole at because the original function is undefined when x = 3.

43. Example 5 Graph a Rational Function with Common Factors Answer:vertical asymptote at x = –2; horizontal asymptote at y = 1; x-intercept: –3 and y-intercept: ; hole: ; ; –4 –22 4

44. Example 5 A.vertical asymptote at x = –2, horizontal asymptote at y = –2; no holes B.vertical asymptotes at x = –5 and x = –2; horizontal asymptote at y = 1; hole at (–5, 3) C.vertical asymptotes at x = –5 and x = –2; horizontal asymptote at y = 1; hole at (–5, 0) D.vertical asymptote at x = –2; horizontal asymptote at y = 1; hole at (–5, 3) Determine the vertical and horizontal asymptotes and holes of the graph of.

45. Example 6 Original Equation Solve a Rational Equation Solve. Multiply by the LCD, x – 6. Simplify. Quadratic Formula Simplify.

46. Example 6 Solve a Rational Equation Answer: CHECKBecause the zeros of the graph of the related function appear to be at about x = 6.6 and x = –0.6, this solution is reasonable.

47. Example 6 A.–22 B.–2 C.2 D.8 Solve.

48. Example 7 Solve a Rational Equation with Extraneous Solutions The LCD of the expressions is x – 1. Solve. x 2 – x + x= 3x – 2Simplify. x 2 – 3x + 2= 0Subtract 3x – 2 from each side. Multiply by the LCD. Original Equation

49. Example 7 Solve a Rational Equation with Extraneous Solutions (x – 2)(x – 1)= 0Factor. x = 2 or x = 1Solve. Because the original equation is not defined when x = 1, you can eliminate this extraneous solution. So, the only solution is 2. Answer: 2

50. Example 7 A.–2, 1 B.1 C.–2 D.–2, 5 Solve.

51. Example 8 Solve a Rational Equation WATER CURRENT The rate of the water current in a river is 4 miles per hour. In 2 hours, a boat travels 6 miles with the current to one end of the river and 6 miles back. If r is the rate of the boat in still water, r + 4 is its rate with the current, r – 4 is its rate against the current, and, find r.

52. Example 8 Solve a Rational Equation 6r + 24 + 6r – 24= 2r 2 – 32Simplify. 12r= 2r 2 – 32Combine like terms. 0= 2r 2 – 12r – 32Subtract 12r from each side. 0= r 2 – 6r – 16Divide each side by 2. Multiply by the LCD. Original Equation

53. Example 8 Answer:8 Solve a Rational Equation 0= (r – 8)(r + 2)Factor. r = 8 or r = –2Solve. Because r is the rate of the boat, r cannot be negative. Therefore, r is 8 miles per hour.

54. Example 8 A.1.7 or 8.3 seconds B.2 or 7 seconds C.4.7 seconds D.12 seconds ELECTRONICS Suppose the current I, in amps, in an electric circuit is given by the formula, where t is time in seconds. At what time is the current 2 amps?

55. End of the Lesson