1. A. 4 positive zeros; 1 negative zero
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
5–Minute Check 3

2. A. 4 positive zeros; 1 negative zero
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
5–Minute Check 3

3. Analyze and graph rational functions. Solve rational equations.
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.
Then/Now

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

5. Key Concept 1

6. Find Vertical and Horizontal Asymptotes
A. Find the domain of and the equations of the vertical or horizontal asymptotes, if any.
Step 1 Find 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.
Example 1

7. Step 2 Find the asymptotes, if any. Check for vertical asymptotes.
Find Vertical and Horizontal Asymptotes
Step 2 Find 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.
Example 1

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

9. CHECK The graph of shown supports each of these findings.
Find Vertical and Horizontal Asymptotes
CHECK The graph of shown supports each of these findings.
Answer:
Example 1

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

11. Find Vertical and Horizontal Asymptotes
B. Find the domain of and the equations of the vertical or horizontal asymptotes, if any.
Step 1 The zeros of the denominator b (x) = 2x2 + 1 are imaginary, so the domain of f is all real numbers.
Example 1

12. Find Vertical and Horizontal Asymptotes
Step 2 Because 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 becomes an increasing large positive number and decreases, approaching 0. Therefore,
Example 1

13. Find Vertical and Horizontal Asymptotes
CHECK You can use a table of values to support this reasoning. The graph of shown also supports each of these findings.
Answer:
Example 1

14. Find Vertical and Horizontal Asymptotes
CHECK You 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
Example 1

15. 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
Example 1

16. 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
Example 1

17. Key Concept 2

18. Step 2 There is a vertical asymptote at x = –5.
Graph Rational Functions: n < m and n > m
A. For , determine any vertical and horizontal asymptotes and intercepts. Then graph the function and state its domain.
Step 1 The function is undefined at b (x) = 0, so the domain is {x | x ≠ –5, x  }
Step 2 There 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.
Example 2

19. Graph Rational Functions: n < m and n > m
Step 3 The 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 4 Graph 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.
Example 2

20. Graph Rational Functions: n < m and n > m
Answer:
Example 2

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

22. Graph Rational Functions: n < m and 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.
Example 2

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

24. Graph Rational Functions: n < m and n > m
Step 4 Graph 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.
Example 2

25. Graph Rational Functions: n < m and 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 }
Example 2

26. Determine any vertical and horizontal asymptotes and intercepts for .
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
Example 2

27. Determine any vertical and horizontal asymptotes and intercepts for .
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
Example 2

28. 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 1 The function is undefined at b (x) = 0, so the domain is {x | x ≠ –2, 2, x  }.
Example 3

29. Graph a Rational Function: n = m
Step 2 There 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 3 The x-intercepts are –3 and 4, the zeros of the numerator. The y-intercept is 1.5 because f(0) = 1.5.
Example 3

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

31. 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;
Example 3

32. Determine any vertical and horizontal asymptotes and intercepts for .
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
Example 3

33. Determine any vertical and horizontal asymptotes and intercepts for .
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
Example 3

34. Key Concept 3

35. Step 2 There is a vertical asymptote at x = –3.
Graph a Rational Function: n = m + 1
Determine any asymptotes and intercepts for Then graph the function, and state its domain.
Step 1 The function is undefined at b (x) = 0, so the domain is D = {x | x ≠ –3, x  }.
Step 2 There 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.
Example 4

36. Therefore, the equation of the oblique asymptote is y = x – 2.
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.
Example 4

37. Graph a Rational Function: n = m + 1
Step 3 The 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) ≈
Step 4 Graph the asymptotes and intercepts. Then find and plot points in the test intervals (–∞, –3.37), (–3.37, –3), (–3, 2.37), (2.37, ∞).
Example 4

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

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

40. 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 = x2– 5x + 11; x-intercepts: 0 and 3; y-intercept: 0
Example 4

41. 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 = x2– 5x + 11; x-intercepts: 0 and 3; y-intercept: 0
Example 4

42. Factoring both the numerator and denominator yields h(x) =
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 1 The function is undefined at b (x) = 0, so the domain is D = {x | x ≠ –2, 3, x  }.
Example 5

43. Graph a Rational Function with Common Factors
Step 2 There 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 3 The x-intercept is –3, the zero of the simplified numerator. The y-intercept is because h(0) = .
Example 5

44. Graph a Rational Function with Common Factors
Step 4 Graph 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.
Example 5

45. Answer: Graph a Rational Function with Common Factors –4 –2 2 4
Example 5

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

47. Determine the vertical and horizontal asymptotes and holes of the graph of .
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)
Example 5

48. Determine the vertical and horizontal asymptotes and holes of the graph of .
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)
Example 5

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

50. Solve a Rational Equation
CHECK Because 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.
Answer:
Example 6

51. Solve a Rational Equation
CHECK Because 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.
Answer:
Example 6

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

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

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

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

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

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

58. 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.
Example 8

59. 12r = 2r 2 – 32 Combine like terms.
Solve a Rational Equation
Original Equation
Multiply by the LCD.
6r r – 24 = 2r 2 – 32 Simplify.
12r = 2r 2 – 32 Combine like terms.
0 = 2r 2 – 12r – 32 Subtract 12r from each side.
0 = r 2 – 6r – 16 Divide each side by 2.
Example 8

60. 0 = (r – 8)(r + 2) Factor. r = 8 or r = –2 Solve.
Solve a Rational Equation
0 = (r – 8)(r + 2) Factor.
r = 8 or r = –2 Solve.
Because r is the rate of the boat, r cannot be negative. Therefore, r is 8 miles per hour.
Answer: 8
Example 8

61. 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?
A. 1.7 or 8.3 seconds
B. 2 or 7 seconds
C. 4.7 seconds
D. 12 seconds
Example 8

62. 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?
A. 1.7 or 8.3 seconds
B. 2 or 7 seconds
C. 4.7 seconds
D. 12 seconds
Example 8

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