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5–Minute Check 3

Evaluate | x – 2 y | – |2 x – y | – xy if x = –2 and y = 7. A. –9 B. 9 C. 19 D. 41. 5–Minute Check 3. Evaluate | x – 2 y | – |2 x – y | – xy if x = –2 and y = 7. A. –9 B. 9 C. 19 D. 41. 5–Minute Check 3. Factor 8 xy 2 – 4 xy. A. 2 x (4 xy 2 – y )

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5–Minute Check 3

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  1. Evaluate |x – 2y| – |2x – y| – xy if x = –2 and y = 7. A.–9 B.9 C.19 D.41 5–Minute Check 3

  2. Evaluate |x – 2y| – |2x – y| – xy if x = –2 and y = 7. A.–9 B.9 C.19 D.41 5–Minute Check 3

  3. Factor 8xy2 – 4xy. A.2x(4xy2– y) B.4xy(2y – 1) C.4xy(y2– 1) D.4y2(2x – 1) 5–Minute Check 4

  4. Factor 8xy2 – 4xy. A.2x(4xy2– y) B.4xy(2y – 1) C.4xy(y2– 1) D.4y2(2x – 1) 5–Minute Check 4

  5. A. B. C. D. 5–Minute Check 5

  6. A. B. C. D. 5–Minute Check 5

  7. interval notation • function • function notation • independent variable • dependent variable • piecewise-defined function • relevant domain Vocabulary

  8. Key Concept 1

  9. Use Interval Notation A. Write –2 ≤x ≤12 using interval notation. The set includes all real numbers greater than or equal to –2 and less than or equal to 12. Answer: Example 2

  10. Use Interval Notation A. Write –2 ≤x ≤12 using interval notation. The set includes all real numbers greater than or equal to –2 and less than or equal to 12. Answer:[–2, 12] Example 2

  11. Use Interval Notation B. Write x > –4 using interval notation. The set includes all real numbers greater than –4. Answer: Example 2

  12. Answer: (–4, ) Use Interval Notation B. Write x > –4 using interval notation. The set includes all real numbers greater than –4. Example 2

  13. Use Interval Notation C. Write x < 3 or x ≥ 54 using interval notation. The set includes all real numbers less than 3 and all real numbers greater than or equal to 54. Answer: Example 2

  14. Answer: Use Interval Notation C. Write x < 3 or x ≥ 54 using interval notation. The set includes all real numbers less than 3 and all real numbers greater than or equal to 54. Example 2

  15. A. B. C. (–1, 5) D. Write x > 5 or x < –1 using interval notation. Example 2

  16. A. B. C. (–1, 5) D. Write x > 5 or x < –1 using interval notation. Example 2

  17. Key Concept 3

  18. Key Concept 3a

  19. Identify Relations that are Functions B. Determine whether the table represents y as a function of x. Answer: Example 3

  20. Identify Relations that are Functions B. Determine whether the table represents y as a function of x. Answer:No; there is more than one y-value for an x-value. Example 3

  21. Identify Relations that are Functions C. Determine whether the graph represents y as a function of x. Answer: Example 3

  22. Identify Relations that are Functions C. Determine whether the graph represents y as a function of x. Answer:Yes; there is exactly one y-value for each x-value. Any vertical line will intersect the graph at only one point. Therefore, the graph represents y as a function of x. Example 3

  23. Divide each side by 3. Take the square root of each side. Identify Relations that are Functions D. Determine whether x = 3y2 represents y as a function of x. To determine whether this equation represents y as a function of x, solve the equation for y. x = 3y2 Original equation Example 3

  24. Identify Relations that are Functions This equation does not represent y as a function of x because there will be two corresponding y-values, one positive and one negative, for any x-value greater than 0. Let x = 12. Answer: Example 3

  25. Identify Relations that are Functions This equation does not represent y as a function of x because there will be two corresponding y-values, one positive and one negative, for any x-value greater than 0. Let x = 12. Answer:No; there is more than one y-value for an x-value. Example 3

  26. Determine whether 12x2 + 4y = 8 represents y as a function of x. A. Yes; there is exactly one y-value for each x-value. B. No; there is more than one y-value for an x-value. Example 3

  27. Determine whether 12x2 + 4y = 8 represents y as a function of x. A. Yes; there is exactly one y-value for each x-value. B. No; there is more than one y-value for an x-value. Example 3

  28. Find Function Values A. If f(x) = x2 – 2x – 8, find f(3). To find f(3), replace x with 3 in f(x) =x2 –2x –8. f(x) =x2 –2x –8 Original function f(3) =32–2(3) –8 Substitute 3 for x. = 9 – 6 – 8 Simplify. = –5 Subtract. Answer: Example 4

  29. Find Function Values A. If f(x) = x2 – 2x – 8, find f(3). To find f(3), replace x with 3 in f(x) =x2 –2x –8. f(x) =x2 –2x –8 Original function f(3) =32–2(3) –8 Substitute 3 for x. = 9 – 6 – 8 Simplify. = –5 Subtract. Answer:–5 Example 4

  30. Find Function Values B. If f(x) = x2 – 2x – 8, find f(–3d). To find f(–3d), replace x with –3d in f(x) = x2 – 2x – 8. f(x) = x2 – 2x – 8 Original function f(–3d) = (–3d)2– 2(–3d) – 8 Substitute –3d for x. = 9d2 + 6d – 8 Simplify. Answer: Example 4

  31. Find Function Values B. If f(x) = x2 – 2x – 8, find f(–3d). To find f(–3d), replace x with –3d in f(x) = x2 – 2x – 8. f(x) = x2 – 2x – 8 Original function f(–3d) = (–3d)2– 2(–3d) – 8 Substitute –3d for x. = 9d2 + 6d – 8 Simplify. Answer:9d2 + 6d – 8 Example 4

  32. Find Function Values C. If f(x) = x2 – 2x – 8, find f(2a – 1). To find f(2a – 1), replace x with 2a – 1 in f(x) = x2 – 2x – 8. f(x) = x2 – 2x – 8 Original function f(2a – 1) = (2a – 1)2– 2(2a – 1) – 8 Substitute 2a – 1 for x. = 4a2 – 4a + 1 – 4a + 2 – 8 Expand (2a – 1)2 and 2(2a – 1). = 4a2 – 8a – 5 Simplify. Answer: Example 4

  33. Find Function Values C. If f(x) = x2 – 2x – 8, find f(2a – 1). To find f(2a – 1), replace x with 2a – 1 in f(x) = x2 – 2x – 8. f(x) = x2 – 2x – 8 Original function f(2a – 1) = (2a – 1)2– 2(2a – 1) – 8 Substitute 2a – 1 for x. = 4a2 – 4a + 1 – 4a + 2 – 8 Expand (2a – 1)2 and 2(2a – 1). = 4a2 – 8a – 5 Simplify. Answer:4a2 – 8a – 5 Example 4

  34. If , find f(6). A. B. C. D. Example 4

  35. If , find f(6). A. B. C. D. Example 4

  36. A. State the domain of the function . Because the square root of a negative number cannot be real, 4x – 1 ≥ 0. Therefore, the domain of g(x) is all real numbers x such that x ≥ , or . Find Domains Algebraically Answer: Example 5

  37. A. State the domain of the function . Because the square root of a negative number cannot be real, 4x – 1 ≥ 0. Therefore, the domain of g(x) is all real numbers x such that x ≥ , or . Answer:all real numbers x such that x ≥ , or Find Domains Algebraically Example 5

  38. B. State the domain of the function . When the denominator of is zero, the expression is undefined. Solving t2 – 1 = 0, the excluded values in the domain of this function are t = 1 and t = –1. The domain of this function is all real numbers except t = 1 and t = –1, or . Find Domains Algebraically Answer: Example 5

  39. B. State the domain of the function . When the denominator of is zero, the expression is undefined. Solving t2 – 1 = 0, the excluded values in the domain of this function are t = 1 and t = –1. The domain of this function is all real numbers except t = 1 and t = –1, or . Answer: Find Domains Algebraically Example 5

  40. C. State the domain of the function . This function is defined only when 2x – 3 > 0. Therefore, the domain of f(x) is or . Find Domains Algebraically Answer: Example 5

  41. C. State the domain of the function . This function is defined only when 2x – 3 > 0. Therefore, the domain of f(x) is or . Answer: or Find Domains Algebraically Example 5

  42. State the domain of g(x) = . A. or [4, ∞) B. or [–4, 4] C. or (− , −4] D. Example 5

  43. State the domain of g(x) = . A. or [4, ∞) B. or [–4, 4] C. or (− , −4] D. Example 5

  44. Evaluate a Piecewise-Defined Function A. FINANCE Realtors in a metropolitan area studied the average home price per square foot as a function of total square footage. Their evaluation yielded the following piecewise-defined function. Find the average price per square foot for a home with the square footage of 1400 square feet. Example 6

  45. Because 1400 is between 1000 and 2600, use to find p(1400). Function for 1000 ≤ a < 2600 Substitute 1400 for a. Subtract. Evaluate a Piecewise-Defined Function = 85 Simplify. Example 6

  46. Evaluate a Piecewise-Defined Function According to this model, the average price per square foot for a home with a square footage of 1400 square feet is $85. Answer: Example 6

  47. Evaluate a Piecewise-Defined Function According to this model, the average price per square foot for a home with a square footage of 1400 square feet is $85. Answer:$85 per square foot Example 6

  48. Evaluate a Piecewise-Defined Function B. FINANCE Realtors in a metropolitan area studied the average home price per square foot as a function of total square footage. Their evaluation yielded the following piecewise-defined function. Find the average price per square foot for a home with the square footage of 3200 square feet. Example 6

  49. Because 3200 is between 2600 and 4000, use to find p(3200). Function for 2600 ≤ a < 4000. Substitute 3200 for a. Simplify. Evaluate a Piecewise-Defined Function Example 6

  50. Evaluate a Piecewise-Defined Function According to this model, the average price per square foot for a home with a square footage of 3200 square feet is $104. Answer: Example 6

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