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Study Guide and Review - Chapter 9 State whether each sentence is true or false . If false , replace the underlined term to make a true sentence. 1. The axis of symmetry of a quadratic function can be found by using the equation x = . SOLUTION: The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation . The statement is true. 2. The vertex is the maximum or minimum point of a parabola. SOLUTION: The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true. 3. The graph of a quadratic function is a straight line. SOLUTION: 2 Quadratic functions are nonlinear and can be written in the form f (x) = ax + bx + c, where a ≠ 0. So, the statement is false. The graph of a quadratic function is a parabola . 2 4. The graph of a quadratic function has a maximum if the coefficient of the x -term is positive. SOLUTION: 2 When a > 0, the graph of y = ax + bx + c opens upward. The lowest point on the graph is the minimum. So, the 2 statement is false. The graph of a quadratic function has a minimum if the coefficient of the x -term is positive. 5. A quadratic equation with a graph that has two x-intercepts has one real root. SOLUTION: The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots. 2 6. The expression b − 4ac is called the discriminant. SOLUTION: 2 In the Quadratic Formula, the expression under the radical sign, b − 4ac, is called the discriminant. The statement is true. 7. The solutions of a quadratic equation are called roots. SOLUTION: The solutions of a quadratic equation are called roots. The statement is true. 2 8. The graph of the parent function is translated down to form the graph of f (x) = x + 5. SOLUTION: 2 If c > 0 in f (x) = x + c, the graph of the parent function is translated units up. So, the statement is false. The 2 graph of the parent function is translated up 5 units to form the graph of f (x) = x + 5. eSolutions Manual - Powered by Cognero State whether each sentence is true or false . If false , replace the underlined term to make a true sentence. Page 1 true. 7. The solutions of a quadratic equation are called roots. SOLUTION: Study Guide and Review - Chapter 9 The solutions of a quadratic equation are called roots. The statement is true. 2 8. The graph of the parent function is translated down to form the graph of f (x) = x + 5. SOLUTION: 2 If c > 0 in f (x) = x + c, the graph of the parent function is translated units up. So, the statement is false. The 2 graph of the parent function is translated up 5 units to form the graph of f (x) = x + 5. State whether each sentence is true or false . If false , replace the underlined term to make a true sentence. 9. The range of the greatest integer function is the set of all real numbers. SOLUTION: The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. State whether each sentence is true or false . If false , replace the underlined term to make a true sentence. 10. A function that is defined differently for different parts of its domain is called a piecewise-defined function. SOLUTION: true Consider each equation. a. Determine whether the function has a maximum or minimum value. b. State the maximum or minimum value. c. What are the domain and range of the function? 2 11. y = x − 4x + 4 SOLUTION: 2 a. For y = x – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a minimum value. b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is . The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y|y ≥ 0}. 2 12. y = −x + 3x SOLUTION: eSolutions Manual - Powered by Cognero 2 Page 2 a. For y = −x + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has a maximum value. The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y|y Study Guide and Review - Chapter 9 ≥ 0}. 2 12. y = −x + 3x SOLUTION: 2 a. For y = −x + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has a maximum value. b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is . The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate. The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y|y ≤ 2.25}. 2 13. y = x − 2x − 3 SOLUTION: 2 a. For y = x − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has a minimum value. b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is . The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate. eSolutions Manual - Powered by Cognero The minimum value is –4. Page 3 The maximum value is 2.25. c. Guide Study and Review The domain is all real- Chapter numbers. 9The range is all real numbers less than or equal to the minimum value, or {y|y ≤ 2.25}. 2 13. y = x − 2x − 3 SOLUTION: 2 a. For y = x − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has a minimum value. b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is . The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate. The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y|y ≥ –4}. 2 14. y = −x + 2 SOLUTION: 2 a. For y = −x + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a maximum value. b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is . The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate. eSolutions Manual - Powered by Cognero The maximum value is 2. Page 4 The minimum value is –4. c. Guide Study and Review The domain is all real- Chapter numbers. 9The range is all real numbers greater than or equal to the minimum value, or {y|y ≥ –4}. 2 14. y = −x + 2 SOLUTION: 2 a. For y = −x + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a maximum value. b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is . The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate. The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y|y ≤ 2}. 2 15. ROCKET A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t + 32t gives the height of the rocket t seconds after it is launched. a. Determine whether the function has a maximum or minimum value. b. State the maximum or minimum value. c. State a reasonable domain and range of this situation. SOLUTION: 2 a. For h = −16t + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function has a maximum value. b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is . The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate. eSolutions Manual - Powered by Cognero Page 5 The maximum value is 2. c. Guide Study and Review The domain is all real- Chapter numbers. 9The range is all real numbers less than or equal to the maximum value, or {y|y ≤ 2}. 2 15. ROCKET A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t + 32t gives the height of the rocket t seconds after it is launched. a. Determine whether the function has a maximum or minimum value. b. State the maximum or minimum value. c. State a reasonable domain and range of this situation. SOLUTION: 2 a. For h = −16t + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function has a maximum value. b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is . The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate. The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}. Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth. 2 16. x − 3x − 4 = 0 SOLUTION: 2 Graph the related function f (x) = x − 3x − 4. The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4. CHECK: Check each solution in the original equation. eSolutions Manual - Powered by Cognero Page 6 The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then itand willReview return to the ground. Study Guide - Chapter 9 Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}. Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth. 2 16. x − 3x − 4 = 0 SOLUTION: 2 Graph the related function f (x) = x − 3x − 4. The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4. CHECK: Check each solution in the original equation. Therefore, the solutions are – 1 and 4. 2 17. −x + 6x − 9 = 0 SOLUTION: 2 Graph the related function f (x) = −x + 6x − 9. The x-intercept of the graph appears to be only at 3, so the only solution is 3. CHECK: Check the solution in the original equation. eSolutions Manual - Powered by Cognero Page 7 Study Guide and 9 4. Therefore, theReview solutions- Chapter are – 1 and 2 17. −x + 6x − 9 = 0 SOLUTION: 2 Graph the related function f (x) = −x + 6x − 9. The x-intercept of the graph appears to be only at 3, so the only solution is 3. CHECK: Check the solution in the original equation. Therefore, the solution is 3. 2 18. x − x − 12 = 0 SOLUTION: 2 Graph the related function f (x) = x − x − 12. The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4. CHECK: Check each solution in the original equation. eSolutions Manual - Powered by Cognero Therefore, the solutions are – 3 and 4. Page 8 Study Guide and Review - Chapter 9 Therefore, the solution is 3. 2 18. x − x − 12 = 0 SOLUTION: 2 Graph the related function f (x) = x − x − 12. The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4. CHECK: Check each solution in the original equation. Therefore, the solutions are – 3 and 4. 2 19. x + 4x − 3 = 0 SOLUTION: 2 Graph the related function f (x) = x + 4x − 3. The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function. –4.9 –4.8 –4.7 1.41 0.84 0.29 –4.4 –4.3 –4.2 x eSolutions Manual - Powered by Cognero –1.24 –1.71 –2.16 y 0.1 0.2 0.3 x x y –4.6 –0.24 –4.1 –2.59 0.4 Page 9 Guide and Review - Chapter 9 Study Therefore, the solutions are – 3 and 4. 2 19. x + 4x − 3 = 0 SOLUTION: 2 Graph the related function f (x) = x + 4x − 3. The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function. x y x y x y x y –4.9 1.41 –4.4 –1.24 0.1 –2.59 0.6 –0.24 –4.8 0.84 –4.3 –1.71 0.2 –2.16 0.7 0.29 –4.7 0.29 –4.2 –2.16 0.3 –1.71 0.8 0.84 –4.6 –0.24 –4.1 –2.59 0.4 –1.24 0.9 1.41 For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6. 2 20. x − 10x = −21 SOLUTION: Rewrite the equation in standard form. 2 Graph the related function f (x) = x − 10x + 21. eSolutions Manual - Powered by Cognero The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7. CHECK: Check each solution in the original equation. Page 10 x y 0.6 –0.24 0.7 0.29 0.8 0.84 0.9 1.41 Study and Review - Chapter 9 is closest to zero when the sign changes is –0.24. Thus, the solutions are ForGuide each table, the function value that approximately –4.6 and 0.6. 2 20. x − 10x = −21 SOLUTION: Rewrite the equation in standard form. 2 Graph the related function f (x) = x − 10x + 21. The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7. CHECK: Check each solution in the original equation. Therefore, the solutions are 3 and 7. 2 21. 6x − 13x = 15 SOLUTION: Rewrite the equation in standard form. 2 Graph the related function f (x) = 6x − 13x – 15. eSolutions - Powered The Manual x-intercepts of by theCognero graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation. Page 11 Guide and Review - Chapter 9 Study Therefore, the solutions are 3 and 7. 2 21. 6x − 13x = 15 SOLUTION: Rewrite the equation in standard form. 2 Graph the related function f (x) = 6x − 13x – 15. The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation. To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function. x y x y x x –0.9 1.56 –0.6 –5.04 –0.3 –10.56 –0.8 –0.76 –0.5 –7 –0.2 –12.16 –0.7 –2.96 –0.4 –8.84 –0.1 –13.64 The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately – 0.8. So, the solutions are about –0.8 and 3. 22. NUMBER THEORY Find two numbers that have a sum of 2 and a product of −15. SOLUTION: From the given information, write two equations. Let x and y represent the numbers. x +y = 2 x • y = –15 eSolutions Manual - Powered by Cognero Solve the first equation for y. Page 12 x x –0.3 –10.56 –0.2 –12.16 –0.1 –13.64 Study Guide andvalue Review The function that-isChapter closest to9 zero when the sign changes is –0.76. Thus, the second root is approximately – 0.8. So, the solutions are about –0.8 and 3. 22. NUMBER THEORY Find two numbers that have a sum of 2 and a product of −15. SOLUTION: From the given information, write two equations. Let x and y represent the numbers. x +y = 2 x • y = –15 Solve the first equation for y. Substitute this value for y into the second equation. 2 Graph the related function f (x) = x − 2x – 15. The x-intercepts of the graph appear to be at –3 and 5, so the solutions for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5. 2 Describe how the graph of each function is related to the graph of f (x) = x . 2 23. f (x) = x + 8 SOLUTION: 2 The graph of f (x) = x + c represents a translation up or down of the parent graph. Since c = 8, the translation is up. So, the graph is shifted up 8 units from the parent function. 2 24. f (x) = x − 3 SOLUTION: eSolutions Manual - Powered by Cognero 2 Page 13 The graph of f (x) = x + c represents a translation up or down of the parent graph. Since c = –3, the translation is down. So, the graph is shifted down 3 units from the parent function. SOLUTION: 2 The graph of f (x) = x + c represents a translation up or down of the parent graph. Since c = 8, the translation is up. Study andisReview Chapter 9 the parent function. So,Guide the graph shifted -up 8 units from 2 24. f (x) = x − 3 SOLUTION: 2 The graph of f (x) = x + c represents a translation up or down of the parent graph. Since c = –3, the translation is down. So, the graph is shifted down 3 units from the parent function. 25. f (x) = 2x 2 SOLUTION: 2 2 The graph of f (x) = ax stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x is the 2 graph of y = x vertically stretched. 2 26. f (x) = 4x − 18 SOLUTION: 2 The function can be written f (x) = ax + c, where a = 4 and c = –18. Since –18 < 0 and > 1, the graph of y = 4x 2 2 − 18 is the graph of y = x vertically stretched and shifted down 18 units. 27. f (x) = x 2 SOLUTION: 2 The graph of f (x) = ax stretches or compresses the parent graph vertically. Since a = , the graph of y = 2 x is 2 the graph of y = x vertically compressed. 28. f (x) = x 2 SOLUTION: 2 The graph of f (x) = ax stretches or compresses the parent graph vertically. ince a = , the graph of y = 2 x is 2 the graph of y = x vertically compressed. 29. Write an equation for the function shown in the graph. SOLUTION: Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have 2 the form y = ax + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this point to find the value of a. eSolutions Manual - Powered by Cognero Page 14 SOLUTION: 2 The graph of f (x) = ax stretches or compresses the parent graph vertically. ince a = Study Guide and Review - Chapter 9 2 the graph of y = x vertically compressed. 2 x is , the graph of y = 29. Write an equation for the function shown in the graph. SOLUTION: Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have 2 the form y = ax + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this point to find the value of a. 2 Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x – 3. 2 30. PHYSICS A ball is dropped off a cliff that is 100 feet high. The function h = −16t + 100 models the height h of the 2 ball after t seconds. Compare the graph of this function to the graph of h = t . SOLUTION: 2 Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at + c, 2 where a = –16 and c = 100. Since 100 > 0 and > 1, the graph of h = −16t + 100 is the graph of y = x vertically stretched and shifted up 100 units from the parent function. 2 Solve each equation by completing the square. Round to the nearest tenth if necessary. 2 31. x + 6x + 9 = 16 SOLUTION: The solutions are 1 and –7. eSolutions Manual - Powered by Cognero 2 32. −a − 10a + 25 = 25 Page 15 SOLUTION: 2 Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at + c, 2 where a =and –16Review and c =-100. Since 9100 > 0 and > 1, the graph of h = −16t + 100 is the graph of y = x Study Guide Chapter vertically stretched and shifted up 100 units from the parent function. 2 Solve each equation by completing the square. Round to the nearest tenth if necessary. 2 31. x + 6x + 9 = 16 SOLUTION: The solutions are 1 and –7. 2 32. −a − 10a + 25 = 25 SOLUTION: The solutions are 0 and –10. 2 33. y − 8y + 16 = 36 SOLUTION: eSolutions Manual - Powered by Cognero Page 16 Study Guide and Review - Chapter 9 The solutions are 0 and –10. 2 33. y − 8y + 16 = 36 SOLUTION: The solutions are 10 and –2. 2 34. y − 6y + 2 = 0 SOLUTION: Use a calculator to approximate each value of y. The solutions are approximately 5.6 and 0.4. 2 35. n − 7n = 5 SOLUTION: eSolutions Manual - Powered by Cognero Page 17 Guide and Review - Chapter 9 Study The solutions are approximately 5.6 and 0.4. 2 35. n − 7n = 5 SOLUTION: Use a calculator to approximate each value of n. The solutions are approximately 7.7 and –0.7. 2 36. −3x + 4 = 0 SOLUTION: Use a calculator to approximate each value of x. The solutions are approximately 1.2 and –1.2. 37. NUMBER THEORY Find two numbers that have a sum of −2 and a product of −48. SOLUTION: From the given information, write two equations. Let x and y represent the numbers. x + yManual = –2 - Powered by Cognero eSolutions x • y = –48 Page 18 Guide and Review - Chapter 9 Study The solutions are approximately 1.2 and –1.2. 37. NUMBER THEORY Find two numbers that have a sum of −2 and a product of −48. SOLUTION: From the given information, write two equations. Let x and y represent the numbers. x + y = –2 x • y = –48 Solve the first equation for y. Substitute this value for y into the second equation. Now solve by completing the square. The solutions are 6 and –8. Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary. 2 38. x − 8x = 20 SOLUTION: Rewrite the equation in standard form. For this equation, a = 1, b = –8, and c = –20. eSolutions Manual - Powered by Cognero Page 19 Guide and Review - Chapter 9 Study The solutions are 6 and –8. Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary. 2 38. x − 8x = 20 SOLUTION: Rewrite the equation in standard form. For this equation, a = 1, b = –8, and c = –20. The solutions are 10 and –2. 2 39. 21x + 5x − 7 = 0 SOLUTION: For this equation, a = 21, b = 5, and c = –7. The solutions are approximately 0.5 and –0.7. eSolutions Manual - Powered by Cognero 2 40. d − 5d + 6 = 0 SOLUTION: Page 20 Study Guide and are Review - Chapter 9 The solutions 10 and –2. 2 39. 21x + 5x − 7 = 0 SOLUTION: For this equation, a = 21, b = 5, and c = –7. The solutions are approximately 0.5 and –0.7. 2 40. d − 5d + 6 = 0 SOLUTION: For this equation, a = 1, b = –5, and c = 6. The solutions are 3 and 2. 2 41. 2f + 7f − 15 = 0 SOLUTION: For this equation, a = 2, b = 7, and c = –15. eSolutions Manual - Powered by Cognero Page 21 Study Guide and Review - Chapter 9 The solutions are 3 and 2. 2 41. 2f + 7f − 15 = 0 SOLUTION: For this equation, a = 2, b = 7, and c = –15. The solutions are 1.5 and –5. 2 42. 2h + 8h + 3 = 3 SOLUTION: Rewrite the equation in standard form. For this equation, a = 2, b = 8, and c = 0. The solutions are 0 and –4. 2 43. 4x + 4x = 15 SOLUTION: eSolutions Manual byin Cognero Rewrite the- Powered equation standard form. Page 22 Study Guide and Review - Chapter 9 The solutions are 0 and –4. 2 43. 4x + 4x = 15 SOLUTION: Rewrite the equation in standard form. For this equation, a = 4, b = 4, and c = –15. The solutions are 1.5 and –2.5. 44. GEOMETRY The area of a square can be quadrupled by increasing the side length and width by 4 inches. What is the side length? SOLUTION: 2 If x = the original side length, then the area of the square = x . If x is increased by 4, the area of the square is quadrupled. Write an equation to represent the new area of the square. For this equation, a = –3, b = 8, and c = 16. eSolutions Manual - Powered by Cognero Page 23 Study Guide and Review - Chapter 9 The solutions are 1.5 and –2.5. 44. GEOMETRY The area of a square can be quadrupled by increasing the side length and width by 4 inches. What is the side length? SOLUTION: 2 If x = the original side length, then the area of the square = x . If x is increased by 4, the area of the square is quadrupled. Write an equation to represent the new area of the square. For this equation, a = –3, b = 8, and c = 16. Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches. Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data. 45. SOLUTION: First differences: Second differences: Since the second differences are equal, a quadratic function models the data. 2 Write an equation for the function that models the data. The equation has the form y = ax . Use the ordered pairPage (1, 24 3) to find the value of a. eSolutions Manual - Powered by Cognero Study Guide and Review - Chapter 9 Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches. Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data. 45. SOLUTION: First differences: Second differences: Since the second differences are equal, a quadratic function models the data. 2 Write an equation for the function that models the data. The equation has the form y = ax . Use the ordered pair (1, 3) to find the value of a. 2 An equation that models the data is y = 3x . 46. SOLUTION: Calculate the first differences. Calculate the second differences. Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios. eSolutions Manual - Powered by Cognero The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential Page 25 2 Study andthat Review - Chapter AnGuide equation models the data is9y = 3x . 46. SOLUTION: Calculate the first differences. Calculate the second differences. Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios. The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential x function. The equation has the form y = ab . The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the value of a. x x An equation that models the data is y = 1 • 2 or y = 2 . 47. SOLUTION: Calculate the first differences. Calculate the second differences. eSolutions Manual - Powered by Cognero Page 26 Since the second differences are equal, a quadratic function models the data. 2 Write an equation for the function that models the data. The equation has the form y = ax . Use the ordered pair (1, Study Guide and Review - Chapter 9 x x An equation that models the data is y = 1 • 2 or y = 2 . 47. SOLUTION: Calculate the first differences. Calculate the second differences. Since the second differences are equal, a quadratic function models the data. 2 Write an equation for the function that models the data. The equation has the form y = ax . Use the ordered pair (1, –1) to find the value of a. 2 An equation that models the data is y = –x . Graph each function. State the domain and range. 48. SOLUTION: Make a table of values. x 0 0.5 1 1.5 2 2.5 3 f (x) 0 0 1 1 2 2 3 Because the dots and eSolutions Manual - Powered by circles Cognero overlap, The range is all integers. the graph will cover all possible values of x, so the domain is all real numbers. Page 27 Study Guide and Review - Chapter 9 2 An equation that models the data is y = –x . Graph each function. State the domain and range. 48. SOLUTION: Make a table of values. x 0 0.5 1 1.5 2 2.5 3 f (x) 0 0 1 1 2 2 3 Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers. 49. f (x) = SOLUTION: Make a table of values. x 0 0.25 0.5 1 1.25 1.5 2 2x 0 0.5 1 2 2.5 3 4 f (x) 0 0 1 2 2 3 4 eSolutions Manual - Powered by Cognero Page 28 Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers. Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. Study Guide - Chapter 9 The rangeand is allReview integers. 49. f (x) = SOLUTION: Make a table of values. x 0 0.25 0.5 1 1.25 1.5 2 2x 0 0.5 1 2 2.5 3 4 f (x) 0 0 1 2 2 3 4 Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers. 50. f (x) = |x| SOLUTION: Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0. Make a table of values. Be sure to include the domain values for which the function changes. x f (x) −2 2 −1 1 0 0 1 1 2 2 The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8, so the range is {y | y 0}. 51. f (x) = |2x − 2| SOLUTION: eSolutions Manual - Powered by Cognero Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0. Page 29 The graphand willReview cover all- possible of x, so the domain is all real numbers. The graph will go no lower than y = 8, Study Guide Chaptervalues 9 so the range is {y | y 0}. 51. f (x) = |2x − 2| SOLUTION: Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0. Make a table of values. Be sure to include the domain values for which the function changes. x f (x) −1 4 0 2 1 0 2 2 3 4 The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0, so the range is {y | y 0}. 52. SOLUTION: This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes. x 0 1 2 −2 −1 f (x) 3 6 −4 −3 −2 Notice that both functions are linear. The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between −1 and 3. Thus, the range is {y | y < −1 or y 3}. 53. eSolutions Manual - Powered by Cognero Page 30 SOLUTION: This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the The graphand willReview cover all- possible of x, so the domain is all real numbers. The graph excluded y-values between Study Guide Chaptervalues 9 −1 and 3. Thus, the range is {y | y < −1 or y 3}. 53. SOLUTION: This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes. x 0 1 2 3 −1 f (x) 1 4 −5 −3 −1 Notice that both functions are linear. The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values between 1 and 3. Thus, the range is {y | y 1 or y > 3}. eSolutions Manual - Powered by Cognero Page 31