Chapter 2 Linear and Quadratic Functions Section 2.1 7. y-intercept 1. From the equation y = 2 x − 3 , we see that the y- 8. −4; 3 intercept is −3 . Thus, the (0, −3) is on the graph. We can obtain a second point by choosing a value for x and finding the corresponding value 9. positive 10. True for y. Let x = 1 , then y = 2 (1) − 3 = −1 . Thus, 11. False. If x increases by 3, then y increases by 2. the point (1, −1) is also on the graph. Plotting the two points and connecting with a line yields the graph below. 12. False. The y-intercept is 8. The zero is −4 . 13. f ( x ) = 2x + 3 a. Slope = 2; y-intercept = 3 b. Plot the point (0, 3). Use the slope to find an additional point by moving 1 unit to the right and 2 units up. 2. m = y2 − y1 = 3 − 5 = −2 = 2 x2 − x1 −1 − 2 −3 3 3. To find the y-intercept, let x = 0: 02 + 2 y = 4 y = 2 So, the y-intercept is (0, 2). To find the x-intercept(s), let y = 0: x2 + 2(0) = 4 x2 = 4 x = ±2 14. c. average rate of change = 2 d. increasing g ( x ) = 5x − 4 a. = 5; y-intercept = −4 b. Plot the (0, −4) . Use the slope to find an additional point by moving 1 unit to the So, the x-intercepts are (−2, 0) 4. 60x − 900 = −15x + 2850 75x − 900 = 2850 75x = 3750 x = solution set is {50}. and (2, 0) . right and 5 units up. 5. f (−2) = (−2)2 − 4 = 4 − 4 = 0 h T 1)
20. G ( x ) = −2 23. f ( x ) = −5x + 10 a. Slope = 0; y-intercept = −2 b. Plot the (0, −2) and draw a horizontal line through it. a. −5x + 10 = 0 −5x = −10 x = 2 b. y-intercept = 10 21. c. average rate of change = 0 d. constant g ( x ) = 2 x − 8 24. f ( x ) = −6 x + 12 a. 2x − 8 = 0 2 x = 8 x = 4 a. −6 x + 12 = 0 −6 x = −12 x = 2 b. y-intercept = –8 b. y-intercept = 12 22. g ( x ) = 3x + 12 25. H ( x ) = − 1 x + 4 2 a. 3x + 12 = 0 3x = −12 x = −4 b. y-intercept = 12 a. − 1 x + 4 = 0 2 − 1 x = −4 2 x = 8 b. y-intercept = 4
26. G ( x ) = 1 x − 4 3 a. 1 x − 4 = 0 3 29. 1 x = 4 3 x = 12 b. y-intercept = –4 30. This is not a linear function, since the average rate of change is not constant. 28. This is a linear function with slope = –3, since the average rate of change is constant at –3. 31. is a linear function with slope = 4, since the average rate of change is constant at 4. 2 2 This is not a linear function, since the average rate of change is not constant. This is not a linear function since the average rate of change is not constant.
32. 35. f ( x ) = 4x −1; g ( x ) = −2 x + 5 a. f ( x ) = 0 4x −1 = 0 x = 1 4 b. f ( x ) > 0 4x −1 > 0 x > 1 4 The solution set is ⎧ x x > 1 ⎫ or ⎛ 1 , ∞ ⎞ . ⎨ 4 ⎬ ⎜ 4 ⎟ This is a linear function, since the average rate of ⎩ ⎭ ⎝ ⎠ c. f ( x ) = g ( x ) 33. is at 0.5 4 x = + 5 6 x = 6 x = 1 d. f ( x ) ≤ g ( x ) 4 x −1 ≤ −2 x + 5 6 x ≤ 6 x ≤ 1 The solution set is {x x ≤ 1} or (−∞, 1] . e. 34. This is a linear function with = 0, since the average rate of change is constant at 0. 36. f ( x ) = 3x + 5; g ( x ) = −2x + 15 a. f ( x ) = 0 3x + 5 = 0 5 This is not a linear function, since the average rate of change is not constant. x = − 3 b. f ( x ) < 0 3x + 5 < 0 x < − 5 3 The solution set is ⎧ x x 5 ⎫ or ⎛ − ,
than 40. Thus, the solution to {x x > 40} or (40, ∞) . f ( x) > 50 is Thus, x = 2 . b. f ( x ) ≤ g ( x ) when the graph of f is below e. The y-coordinates of the graph of y = f ( x) or intersects the of g. Thus, the are 80 when the x-coordinates are smaller than 88. Thus, the solution to f ( x) ≤ 80 is {x x ≤ 88} or (−∞, 88] . 41. a. solution is {x x ≤ 2} or (−∞, 2] . f ( x ) = g ( x ) when their graphs intersect. f. The y-coordinates of the graph of y = f ( x) Thus, x = −6 . are between 0 and 80 when the x-coordinates are between −40 and 88. Thus, the solution to 0 < f ( x) < 80 is {x −40 < x < 88} or (−40, 88) . b. g ( x ) ≤ f ( x ) < ( x ) when the graph of f is above or intersects the graph of g and below the graph of h. Thus, the solution is {x −6 ≤ x < 5} or [−6, 5) .
(300 million, 38.7 million ) are on the graph. T Thus, the points (7300, 730) , 40 (300, 38.7) the 20 (128, 0) (200, 16.2) p Team Payroll ($ million)
slope formula yields: m = Δy = 0 − 120000 = −120000 = −12000 x > 15000 = 7500 Δx 10 − 0 10 2 The company a profit if it sells more than 7500 units. The y-intercept is (0,120000) , so b = 120000 . Therefore, the linear function is V ( x ) = mx + b = −12, 000 x + 120, 000 .
61. A linear function f ( x ) = mx + b will be odd 5. True provided f (− x ) = − f ( x ) . 6. True That is, provided m (− x ) + b = − (mx + b ) . −mx + b = −mx − b b = −b 7. Linear relation, m > 0 8. Nonlinear relation provided f (− x ) = f ( x ) . 13. a. 20 That is, provided m (− x ) + b = mx + b . −mx + b = mx + b −mxb = mx 0 = 2mx m = 0 0 10 0 So a linear function provided m = 0 . f ( x ) = mx + b will be even b. Answers will vary. We select (3, 4) and (9, 16). The slope of the line containing 62. The French word monter means “to climb”. Urban legend has it that the usage of m for slope resulted from monter. However, this is uncertain. Another theory is that the m comes from the term “modulus of the slope”. 63. The grade of a road is related to the slope of a line in that both terms describe “steepness”. A 4% grade means that for every 100 units of horizontal change in the road there is 4 units of vertical change. Answers will vary. 64. The pitch of a roof/staircase is related to the slope of a line in that both terms describe “steepness”. Answers will vary. Section 2.2 1. No, the relation is not a function because the x- coordinate 1 is paired with two different y- coordinates, 5 and 12. 2. Yes, the relation defined by y = 2x + 1 is a function because it is a line, which means each will be paired with a distinct y-coordinate. 3. scatter diagram 4. y = kx non-
these points is: m = 16 − 4 = 12 = 2 9 − 3 6 The equation of the line is: y − y1 = m( x − x1 ) y − 4 = 2( x − 3) y − 4 = 2x − 6 y = 2x − 2 c. 20 0 10 0 d. Using the LINear REGresssion program, the line of best fit is: y = 2.0357 x − 2.3571 e. 20 0 10 0
14. a. 12 0 15 –2 The equation of the line is: y − y1 = m( x − x1 ) y − (− 4) = 8 ( x − (− 2)) 3 y + 4 = 8 x + 16 3 3 y = 8 x + 4 b. Answers will vary. We select (5, 2) and (11, 9). The slope of the line containing 3 3 c. 6 points is: m = 9 − 2 = 7 11 − 5 6 The equation of the line is: y − y1 = m( x − x1 ) y − 2 = 7 ( x − 5) 6 –3 3 –6 y − 2 = 7 x − 35 6 6 y = 7 x − 23 6 6 d. Using the LINear REGresssion program, the line of best fit is: y = 2.2 x + 1.2 c. 12 e. 6 –3 0 15 –2 –6 d. Using the LINear REGression program, the line of best fit is: y = 1.1286 x − 3.8619 16. a. 8 e. 12 –3 3 0 15 –2 15. a. 6 –3 3 –6 –2 b. Answers will vary. We select (–2, 7) and (1, 2). The slope of the line containing these points is: m = 2 − 7 = − 5 = − 5 1 − (−2) 3 3
The equation of the line is: y − y1 = m( x − x1 ) y − 7 = − 5 ( x − (− 2)) 3 y − 7 = − 5 x − 10 b. Answers will vary. We select (–2,–4) and (1, 4). The slope of the line containing these points is: m = 4 − (− 4) = 8 . 1 − (− 2) 3 3 3 5 11 y =− 3 3
28 82 x Weight c. Answers will vary. We will use the points (42.3, 82) and (42.8, 93) . m = 93 − 82 = 11 = 22 42.8 − 42.3 0.5 y − 82 = 22 ( x − 42.3) y − 82 = 22 x − 930.6 y = 22 x − 848.6 d. y 94 92 90 b. Using the LINear REGression program, the line of best fit is: C ( H ) = 0.3734H + 7.3268 c. For each 1 inch increase in height, the circumference increases by 0.3734 inch. d. C (26) = 0.3734 (26) + 7.3268 ≈ 17.0 inches e. To find the height, we solve the following equation: 17.4 = 0.3734H + 7.3268 10.0732 = 0.3734H 26.98 ≈ H A child with a head circumference of 17.4 inches would have a height of about 26.98 inches. 88 28. a. 14 86 84 82 x Weight (grams) e. Let N represent the number of raisins in the box, and let w represent the weight (in grams) of the box of raisins. N ( w) = 22w − 848.6 f. N (42.5) = 22 ( 42.5) − 848.6 = 86.4 We predict that approximately 86 raisins will be in a box weighing 42.5 grams. 0 175 0 b. Using the LINear REGression program, the line of best fit is: L (G ) = + 7.8738 c. For each 1 day increase in Gestation period, the life expectancy increases by 0.0261 years (about 9.5 days). d. L (89) = 0.0261(89) + 7.8738 ≈ 10.2 years g. The slope of the line found is 22 raisins per gram. This means that if the weight is to be increased by one gram, then the number of raisins must be increased by 22 raisins.
29. a. The relation is not a function because 23 is paired with both 56 and 53. e. S ( A) = 2.0667 A + 292 .8869 b. f. Domain: {A A ≥ 0} g. S (25) = 2.0667 (25) + 292.8869 ≈ 344.5544 Sales are about $344,554. c. the LINear REGression program, the line of best fit is: D = −1.3355 p + 86.1974 d. As the price of the jeans increases by $1, the demand for the jeans decreases by about 1.34 pairs per day. 31. The data do not follow a linear pattern so it would not make sense to find the line of best fit. e. D ( p ) = −1.3355 p + 86.1974 f. Domain: { p 0 ≤ p ≤ 64} 32. The data do not follow a linear pattern so it would not make sense to find the line of best fit. Note that the p-intercept is roughly 64.54 and that the number of pairs cannot be negative. g. D (28) = −1.3355 (28) + 86.1974 ≈ 48.8034 Demand is about pairs. 30. a. The relation is not a function because 24 is paired with both 343 and 341. b. 33. Using the ordered pairs (1, 5) and (3, 8) , the line of best fit is y = 3 x + 7 2 2 or y = 1.5x + 3.5 . c. Using the LINear REGression program, the The correlation coefficient is r = 1 . This line of best fit is: S = 2.0667 A + 292 .8869 sense because two points completely determine a line.
35. If the student’s average in the class is directly proportional to the amount of time the student studies, then the relationship between the 11. f ( x ) = 0 x2 − 9 x = 0 average and time studying be linear. That x ( x − 9) = 0 is, we could express the relationship as A = kT where T is the time spent studying and A is the student’s average. x = 0 or x − 9 = 0 x = 9 The zeros of f ( x ) = x2 − 9 x are 0 and 9. The x- Section 2.3 12. intercepts of the of f are 0 and 9. f ( x ) = 0 1. x2 − 5x − 6 = ( x − 6) ( x + 2. 2 x2 − x − 3 = (2 x − 3) ( x + 1) x2 + 4x = 0 x ( x + 4) = 0 3. 2x − 3 = 11 x = 0 or x + 4 = 0 x = −4 2 x = 14 The zeros of f ( x ) = x2 + 4 x are −4 and 0. The x = 7 The solution set is {7}. 4. ( x − 3) (3x + 5) = 0 x − 3 = 0 or 3x + 5 = 0 13. of the graph of f are −4 and 0. g ( x ) = 0 x2 − 25 = 0 ( x + 5)( x − 5) = 0 x = 3 3x = −5 x 5 =− x + 5 = 0 or x = − 5 x − 5 = 0 x = 5 3 The solution set is ⎧− 5 ,3 . The zeros of g ( x ) = x2 − 25 are −5 and 5. The x-intercepts of the graph of g are −5 and 5. ⎨ 3 ⎬ ⎩ ⎭ 5. add; ⎛ 1 ⋅ 6 ⎞ = 9 14. G ( x ) = 0 x2 − 9 = 0 ⎜ 2 ⎟ ⎝ ⎠ ( x + 3)( x − 3) = 0
6. If f(4) = 10, then the point (4, 10) is on the graph of f. x + 3 = 0 or x = − 3 x − 3 = 0 x = 3 7. repeated; multiplicity 2 8. discriminant; negative 9. False. If the discriminant of a quadratic equation is 0, then the equation has one repeated real solution. If the discriminant is negative, then the 15. zeros of G ( x ) = x2 − 9 are −3 and 3. The x-intercepts of the graph of G are −3 and 3. F ( x ) = 0 x2 + x − 6 = 0 ( x + 3)( x − 2) = 0 equation has no real solutions. 10. True. x + 3 = 0 or x = − 3 x − 2 = 0 x = 2 The zeros of F ( x ) = x2 + x − 6 are −3 and 2. The x-intercepts of the graph of F are −3 and 2.
16. H ( x ) = 0 x2 + 7 x + 6 = 0 ( x + 6)( x + 1) = 0 20. H ( x ) = 0 2x2 − 50 = 0 2( x2 − 25) = 0 x + 6 = 0 or x + 1 = 0 2( x + 5)( x − 5) = 0 x = −6 x = −1 y + 5 = 0 or y − 5=0 The of H ( x ) = x2 + 7 x + 6 are −6 and −1 . y = − 5 y = 5 17. The of the graph of H are −6 and −1 . g ( x ) = 0 2 x2 − 5x − 3 = 0 (2 x + 1)( x − 3) = 0 21. The zeros of H ( x ) = 2 x2 − 50 are −5 and 5. The x-intercepts of the graph of H are −5 and 5. g ( x ) = 0 x ( x + 8) + 12 = 0 2x + 1 = 0 or x − 3 = 0 x2 + 8x + 12 = 0 1 x x = 3 ( x + 6) ( x + 2) = 0 x = −6 or x = −2 18. The zeros of g ( x ) = 2 x2 − 5x − 3 are − 1 2 The x-intercepts of the graph of g are − 1 2 f ( x ) = 0 3x2 + 2 = 0 and 3. and 3. 22. of g ( x ) = x ( x + 8) + 12 are −6 and −2 . The x-intercepts of the graph of g are −6 and −2 . f ( x ) = 0 x ( x − 4) −12 = 0 2 x − 4x −12 = 0 (3x + 2)( x + 1) = 0 ( x − 6) ( x + 2) = 0 3x + 2 = 0 or x + 1 = 0 x = −2 or x = 6 2 x x = −1 The zeros of f ( x ) = x ( x − 4) −12 are −2 and 6. The zeros of f ( x ) = 3x2 + 5x + 2 are −1 and
The x-intercepts of the of f are −2 and 6. − 2 . The x-intercepts of the graph of f are −1 3 and − 2 . 23. G ( x ) = 0 4x2 + 9 − 12 x = 0 2 3 4x −12x + 9 = 0 19. P ( x ) = 0 3 x2 − 48 = 0 (2 x − 3)(2 x − 3) = 0 2 x − 3 = 0 or 2 x − 3 = 0 3 3 3( x2 − 16) = 0 x = x = 2 2 3( x + 4)( x − 4) = 0 The only zero of G ( x ) = 4 x2 + 9 −12 x is 3 . 2 t + 4 = 0 or t = − 4 t − 4 = 0 t = 4 The only x-intercept of the graph of G is 3 . 2 The zeros of P ( x ) = 3 x2 − 48 are −4 and 4. The x-intercepts of the graph of P are −4 and 4.
x −1 = ± 4 x −1 = ±2 x = 4 3 x = 0 x −1 = 2 or x − 1 = −2 The zeros of G ( x ) = (3x − 2)2 − 4 are 0 and 4 . x = 3 x = −1 3 The zeros of g ( x ) = ( x − 1)2 − 4 are −1 and 3. The x-intercepts of the graph of g are −1 and 3. The x-intercepts of the graph of G are 0 and 4 . 3
31. f ( x ) = 0 x2 + 4 x − 21 = 0 x2 + 4 x = 21 34. g ( x ) = 0 x2 + 2 x − 1 = 0 3 3 x2 + 4 x + 4 = 21 + 4 x2 + 2 x = 1 3 3 ( x + 2)2 = 25 x + 2 = ± 25 x2 + 2 x + 1 = 1 + 1 3 9 3 9 ⎛ 1 ⎞2 4 x + 2 = ±5 x = −2 ± 5 x = 3 or x = −7 ⎜ x + ⎟ ⎝ ⎠ x + 1 = ± = 9 4 = ± 2 The of f ( x ) = x2 + 4 x − 21 are −7 and 3. 3 9 3 1 2 =− ± The x-intercepts of the graph of f are −7 and 3. x 3 3 32. f ( x ) = 0 x2 − 6 x − 13 = 0 x2 − 6x + 9 = 13 + 9 x = 1 or x = −1 3 The zeros of g ( x ) = x2 + 2 x − 1 are −1 and 1 . 3 3 3 1 ( x − 3)2 = 22 x − 3 = ± 22 x = 3 ± 22 35. x-intercepts of the graph of g are −1 and . 3 F ( x ) = 0 The zeros of f ( x ) = x2 − 6 x − 13 are 3 − 22 3x2 + x − 1 = 0 2 and 3 + 22 . The x-intercepts of the graph of f x2 + 1 x − 1 = 0 are 3 − 22 and 3 + 22 . 3 6 x2 + 1 x = 1 33. g ( x ) = 0 3 6 1 1 1 1 x2 − 1 x − 3 = 0 2 16 x2 + x + = + 3 36 6 36 2 x2 − 1 x = 3 ⎛ x + 1 ⎞ = 7
2 16 ⎜ 6 ⎟ 36 x2 − 1 x + 1 = 3 + 1 ⎝ ⎠ 1 7 7 2 16 16 16 x + = ± 6 = ± 36 6 x − ⎝ 1 ⎞2 1 ⎟ = ⎠ 4 x = −1 ± 7 6 x − 1 = ± 1 = ± 1 The zeros of F ( x ) = 3x2 + x − 1 are −1 − 7 and 4 4 2 2 6 x = 1 ± 1 4 2 −1 + 7 . The x-intercepts of the graph of F are 6 x = 3 or x = − 1 −1 − 7 6 and −1 + 7 . 6 4 4 The zeros of g ( x ) = x2 − 1 x − 3 2 16 are − 1 4 and 3 . 4 The x-intercepts of the graph of g are − 1 4 and 3 . 4
36. G ( x ) = 0 2x2 − 3x −1 = 0 39. g ( x ) = 0 x2 − 4x −1 = 0 x2 − 3 x − 1 = 0 a = 1, b = − 4, c = −1 2 2 −(− 4) ± (− 4)2 − 4(1)(−1) 4 ± 16 + 4 x2 − 3 x = 1 2 2 x = = 2(1) 2 x2 − 3 x + 9 = 1 + 9 2 16 2 16 = 4 ± 20 = 4 ± 2 5 = 2 ± 5 2 2 x − 3 = 17 The of g ( x ) = x2 − 4x − 1 are 2 − 5 and ⎝ 4 ⎠ 16 2 + 5 . The x-intercepts of the graph of g are x − 3 = ± 17 = ± 17 2 − 5 and 2 + 5 . 4 16 4 x = 3 ± 17 4 40. g ( x ) = 0 x2 + 6x + 1 = 0 The zeros of G ( x ) = 2 x2 − 3x −1 are 3 − 17 and a = 1, b = 6, c = 1 4 3 + 17 . The x-intercepts of the graph of G are 4 − 6 ± x = 62 − 4(1)(1) 2(1) = − 6 ± 36 − 4 2 37. 3 − 17 4 and 3 + 17 . 4 f ( x ) = 0 = − 6 ± 32 = − 6 ± 4 2 = −3 ± 2 2 2 2 The zeros of g ( x ) = x2 + 6x + 1 are −3 − 2 2 x2 − 4x + 2 = 0 and −3 + 2 2 . The x-intercepts of the graph of a = 1, b = − 4, c = 2 g are −3 − 2 2 and −3 + 2 2 . −(−4) ± (− 4)2 − 4(1)(2) 4 ± 16 − 8 x = = 2(1) 2 41. F ( x ) = 0 2x2 − 5x + 3 = 0 = 4 ± 8 = 4 ± 2 2 = 2 ± 2 a = 2, b = − 5,
c = 3 2 2 −(− 5) ± (− 5)2 − 4(2)(3) 5 ± 25 − 24 The zeros of f ( x ) = x2 − 4 x + 2 are 2 − 2 and x = = 2(2) 4 2 + 2 . The x-intercepts of the graph of f are = 5 ± 1 = 3 or 1 2 − 2 and 2 + 2 . 4 2 The zeros of F ( x ) = 2 x2 − 5x + 3 are 1 and 3 . 38. f ( x ) = 0 x2 + 4 x + 2 = 0 2 The x-intercepts of the graph of F are 1 and 3 . 2 a = 1, b = 4, c = 2 42. g ( x ) = 0 − 4 ± 42 − 4(1)(2) − 4 ± 16 − 8 x = = 2x2 + 5x + 3 = 0 2(1) 2 a = 2, b = 5, c = 3 = − 4 ± 8 = − 4 ± 2 2 = − 2 ± 2 −5 ± 52 − 4(2)(3) −5 ± 25 − 24 2 2 x = = The zeros of f ( x ) = x2 + 4 x + 2 are − 2 − 2 2(2) 4 = −5 ± 1 = −1 or − 3 and − 2 + 2 . The x-intercepts of the graph of f 4 2 are − 2 − 2 and − 2 + 2 . The zeros of g ( x ) = 2 x2 + 5x + 3 are − 3 2 The x-intercepts of the graph of g are − 3 2 and −1 . and −1 .
43. P ( x ) = 0 4x2 − x + 2 = 0 The zeros of f ( x ) = 2 x2 −1 + 2x are −1 − 3 a = 4, b = −1, c = 2 and −1 + 3 . The x-intercepts of the of f 2 −(−1) ± (−1)2 − 4(4)(2) 1 ± 1 − 32 are −1 − 3 and −1 + 3 . x = = 2(4) 8 = 1 ± −31 = not real 8 47. 2 2 G ( x ) = 0 4 x2 − 9 x − 2 = 0 The function P ( x ) = 4x2 − x + 2 has no real a = 4, b = −9, c = −2 44. and the graph of P has no x-intercepts. H ( x ) = 0 4x2 + x + 1 = 0 − (−9 ) ± (−9 )2 − 4(4)(−2) 9 ± 81 + 32 x = = 2(4) 8 = 9 ± 113 a = 4, b = 1, c = 1 8 G x = 4 x2 − 9 x − 2 are 9 − 113 −1 ± 12 − 4(4)(1) t = = −1 ± 1 − 16 The zeros of ( ) 2(4) 8 and 9 + 113 . The x-intercepts of the graph of G 8 = −1 ± −15 = not real 8 The function H ( x ) = 4x2 + x + 1 has no real are 9 − 113 8 and 9 + 113 . 8 45. zeros, and the graph of H has no x-intercepts. f ( x ) = 0 48. F ( x ) = 0 5x − 4 x2 − 5 = 0 0 = 4 x2 − 5x + 5 4 x2 − 1 + 2 x = 0 a = 4, b = −5, c = 5 4 x2 + 2 x − 1 = 0 − (−5) ± (−5)2 − 4(4)(5) 5 ± 25 − 80 a = 4, b = 2, c = −1 x = = 2(4) 8 − 2 ± 22 − 4(4)(−1) − 2 ± 4 + 16 5 ± −55
x = = 2(4) 8 = − 2 ± 20 = − 2 ± 2 5 = −1 ± 5 = = not real 8 The function ( ) F x = 5x − 4x2 − 5 has no real 8 8 4 zeros, and the graph of F has no x-intercepts. The zeros of f ( x ) = 4 x2 −1 + 2x are −1 − 5 49. f ( x ) = 0 and −1 + 5 . The x-intercepts of the of f 4 x2 − 5 = 0 x2 = 5 ⇒ x = ± 5 are −1 − 5 4 and −1 + 5 . 4 The zeros of f ( x ) = x2 − 5 are − 5 and 5 . 46. f ( x ) = 0 2 x2 − 1 + 2 x = 0 2 x2 + 2 x − 1 = 0 50. The x-intercepts of the of f are − 5 . f ( x ) = 0 5 and a = 2, b = 2, c = −1 x2 − 6 = 0 − 2 ± 22 − 4(2)(−1) − 2 ± 4 + 8 x2 = 6 ⇒ x = ± 6 x = = 2(2) 4 The zeros of f ( x ) = x2 − 6 are − 6 and 6 . = − 2 ± 12 = − 2 ± 2 3 = −1 ± 3 4 4 2 The x-intercepts of the of f are − 6 . 6 and
51. g ( x ) = 0 16 x2 − 8x + 1 = 0 (4 x − 1)2 = 0 4 x − 1 = 0 ⇒ x = 1 4 55. P ( x ) = 0 6 x2 − x − 2 = 0 (3x − 2) (2x + 1) = 0 3x − 2 = 0 or 2 x + 1 = 0 2 1 =− The only real zero of g ( x ) = 16 x2 − 8x + 1 is 1 . 4 x = x 3 2 The only x-intercept of the graph of g is 1 . 4 The zeros of P ( x ) = 6 x2 − x − 2 are − 1 2 and 2 . 3 52. F ( x ) = 0 4 x2 − 12 x + 9 = 0 (2 x − 3)2 = 0 2x − 3 = 0 ⇒ x = 3 2 The only real zero of F ( x ) = 4x2 − 12 x + 9 is 3 . 56. The of the graph of P are − 1 2 H ( x ) = 0 6 x2 + x − 2 = 0 (3x + 2) (2 x − 1) = 0 3x + 2 = 0 or 2 x − 1 = 0 2 1 and 2 . 3 2 The only x-intercept of the graph of F is 3 . x = − x = 3 2 2 2 1 2 The zeros of H ( x ) = 6 x + x − 2 are − and . 3 2 53. G ( x ) = 0 10x2 − 19 x −15 = 0 (5x + 3) (2x − 5) = 0 57. The x-intercepts of the of H are − 2 3 G ( x ) = 0 and 1 . 2 5x + 3 = 0 or 2 x − 5 = 0 x2 + 2 x − 1 = 3 x =− x = 5 ⎛ 1 ⎞ 5 2 The zeros of G ( x ) = 10 x2 − 19 x −15 are − 3 and 2 ⎜ x2 + ⎝ 2 2 x − ⎟ = (0) (2) ⎠ 5 2 x + 2 2x − 1 = 0 5 . The x-intercepts of the graph of G are − 3 a = 2, b = 2 2 , c = −1 2 and 5 . 5 ( ) x = −(2 2 ) ± 2
60. g ( x ) = 0 1 x x = −1 x2 + x −1 = 0 The x-coordinates of the points of intersection a = 1, b = 1, c = −1 are −1 and − 1 . The y-coordinates are 2 −(1) ± (1) 2 − 4 (1)(−1) −1 ± 5 g (−1) = 3 (−1) + 2 = −3 + 2 = −1 and 2(1) 2 g ⎛ − 1 ⎞ = 3 ⎛ − 1 ⎞ + 2 = − 3 + 2 = 1 . The of g ( x ) = x2 + x − 1 are −1 − 5 and ⎝ 2 ⎠ ⎝ 2 ⎠ 2 2 2 −1 + 5 . The x-intercepts of the graph of g are The graphs of the f and g intersect at the points (−1, −1) and ⎛ − 1 , 1 ⎞ . 2 ⎜ 2 2 ⎟ −1 − 5 2 and ⎝ ⎠ −1 + 5 . 2
− 7 and −1 . The of the graph of f 2 x3 + 8 = 0 or x3 = −8 x3 −1 = 0 x3 = 1 77. are − 7 2 and −1 . f ( x ) = 0 x = −2 x = 1 (3x + 4)2 − 6 (3x + 4) + 9 = 0 74. The zeros of g ( x ) = x6 + 7 x3 − 8 are −2 and 1. The x-intercepts of the graph of g are −2 and 1. g ( x ) = 0 x6 − 7 x3 − 8 = 0 ( x3 − 8)( x3 + 1) = 0 Let u = 3x + 4 → u 2 u 2 − 6u + 9 = 0 (u − 3)2 = 0 u − 3 = 0 u = 3 3x + 4 = 3 = (3x + 4)2 x3 − 8 = 0 or x3 = 8 x = 2 x3 + 1 = 0 x3 = −1 x = −1 x = − 1 3 The only zero of f ( x ) = (3x + 4)2 − 6 (3x + 4) + 9 The of g ( x ) = x6 − 7 x3 − 8 are −1 and 2. is − 1 . The x-intercept of the graph of f is − 1 . The x-intercepts of the graph of g are −1 and 2. 3 3
78. H ( x ) = 0 (2 − x )2 + (2 − x ) − 20 = 0 The zeros of H ( x ) = 3 (1 − x )2 + 5 (1 − x ) + 2 are 5 Let u = 2 − x → u 2 = (2 − x )2 and 2. The x-intercepts of the graph of H are 3 5 u 2 + u − 20 = 0 (u + 5) (u − 4) = 0 and 2. 3 u + 5 = 0 or u − 4 = 0 81. G ( x ) = 0 u = −5 u = 4 x − 4 x = 0 2 − x = −5 2 − x = 4 Let u = x → u 2 = x x = 7 x = −2 u 2 − 4u = 0 79. The zeros of H ( x ) = (2 − x )2 + (2 − x ) − 20 are −2 and 7. The x-intercepts of the graph of H are −2 and 7. P ( x ) = 0 u (u − 4) = 0 u = 0 or x = 0 u − 4 = 0 u = 4 x = 4 2 2 2 ( x + 1)2 − 5 ( x + 1) − 3 = 0 x = 0 = 0 x = 4 = 16 Let u = x + 1 → u2 = ( x + 1)2 2u 2 − 5u − 3 = 0 (2u + 1) (u − 3) = 0 G (0) = 0 − 4 0 = 0 G (16) = 16 − 4 16 = 16 − 16 = 0 2u + 1 = 0 or u − 3 = 0 The zeros of G ( x ) = x − 4 x are 0 and 16. The 1 u u = 3 x + 1 = 3 x-intercepts of the graph of G are 0 and 16. x + 1 = − 1 x = 2 82. f ( x ) = 0 2 x + 8 x 3
− = 160 n2 − 3n − 160 = 0 a = 1, b = −3, c = −160 = −20 ± −1560 −9.8 There is no real solution. The object never −(−3) ± t = (−3)2 − 4(1)(−160) 2(1) reaches a height of 100 meters. = 3 ± 649 2 Since the solutions are not integers, a polygon with 80 diagonals is not possible.
b2 − 4ac 4ac 4k 2 = 1 2c (−b + b2 − 4ac ) 2c (−b + b2 − 4ac ) k 2 = = 2a = 1 k =± 1 4 and −b + b2 − 4ac x2 k = 1 or k = − 1 −b + b − 4 (c ) (a ) x * −b + b − 4ac 2 2 96. In to have one repeated real zero, we need 2 2 = = 2 2c 2c −b + b 2 − 4ac −b − b 2 − 4ac = ⋅ the discriminant to be 0. 2c −b − b2 − 4ac b2 − 4ac = 0 (−k )2 − 4 (1) (4) = 0 b2 − (b2 − 4ac ) 4ac ( ) ( ) k 2 − 16 = 0 2c −b − b2 − 4ac 2c −b − b2 − 4ac (k − 4) (k + 4) = 0 = 2a = 1 k = 4 or k = −4 −b − b2 − 4ac x1
2. To find the y-intercept, let x = 0 : y = 2 (0)2 + 7 (0) − 4 = −4 . To find the x-intercept(s), let 2x2 + 7 x − 4 = 0 (2 x − 1) ( x + 4) = 0 y = 0 :
20. f ( x) = 2 x2 Using the graph of factor of 2. y = x2 , stretch vertically by a 23. f ( x) = 1 x2 + 2 4 Using the graph of y = x2 , compress vertically by a factor of 1 , then shift up 2 units. 4 21. f ( x) = 1 x2 − 2 4 Using the graph of y = x2 , compress vertically 24. f ( x) = 2 x2 + 4 by a factor of 2, then shift down 2 units. the graph of y = x2 , stretch vertically by a factor of 2, then shift up 4 units. 22. f ( x) = 2 x2 − 3 Using the graph of y = x2 , stretch vertically by a factor of 2, then shift down 3 units. 25. f ( x) = 1 x2 + 1 4 the graph of y = x2 , compress vertically by a factor of 1 , then up 1 unit. 4
26. f ( x) = −2 x2 − 2 Using the graph of y = x2 , stretch vertically by a 29. f ( x) = 2 x2 − 4 x + 1 = 2 ( x2 − 2x ) + 1 factor of 2, reflect across the x-axis, then shift down 2 units. = 2( 2 x + 1) + 1 − 2 = 2( x − 1)2 −1 Using the graph of y = x2 , shift right 1 unit, stretch vertically by a factor of 2, then shift down 1 unit. 27. f ( x) = x2 + 4 x + 2 = ( x2 + 4x + 4) + 2 − 4 = ( x + 2)2 − 2 Using the graph of shift down 2 units. y = x2 , shift left 2 units, then 30. f ( x) = + 6 x = 3 ( x2 + 2 x ) = 3( x2 + 2 x + 1) − 3 = 3( x + 1)2 − 3 Using the graph of y = x2 , shift left 1 unit, stretch vertically by a factor of 3, then shift down 3 units. 28. f ( x) = x2 − 6 x − 1 = ( x2 − 6 x + 9) −1 − 9 = ( x − 3)2 − 10 Using the graph of y = x2 , shift right 3 units, then down 10 units. 31. f ( x) = − x2 − 2 x = − ( x2 + 2x ) = −( x2 + 2 x + 1) + 1 = −( x + 1)2 + 1 Using the graph of y = x2 , shift left 1 unit, reflect across the x-axis, then shift up 1 unit.
32. f ( x) = −2 x2 + 6 x + 2 = −2 ( x2 − 3x ) + 2 34. f ( x) = 2 x2 + 4 3 3 x − 1 = 2 x2 + 2 x − 1 = −2 ⎛ x2 − 3x + 9 ⎞ + 2 + 9 3 ( ) ⎜ 4 ⎟ 2 ⎝ ⎠ 2 2 = −2 ⎛ x − 3 + 13 = 3 ( x + 2 x + 1) −1 − 3 ⎝ 2 ⎠ 2 = 2 ( x + 1)2 − 5 Using the graph of y = x2 , right 3 units, 3 3 2 reflect about the x-axis, stretch vertically by a Using the graph of y = x2 , shift left 1 unit, 2 factor of 2, then shift up 13 units. 2 compress vertically by a factor of , then shift 3 down 5 3 unit. 33. f ( x) = 1 x2 + x −1 2 35. a. For f ( x) = x2 + 2x , a = 1 , b = 2 , c = 0. = 1 ( x2 + 2 x ) − 1 Since a = 1 > 0 , the graph opens up. The x-coordinate of the vertex is = 1 ( x2 + 2 x + 1) −1 − 1 −b −(2) −2 2 2 x = = = = −1 . 2a 2(1) 2 = 1 ( x + 1)2 − 3 2 2 The y-coordinate of the vertex is Using the graph of y = x2 , left 1 unit, f ⎛ −b ⎞ = f (−1) = (−1)2 + 2(−1) = 1 − 2 = −1. ⎝ ⎠ compress vertically by a factor of 1 , then shift 2 down 3 units. 2
f (0) = 0 . The y-intercept is f (0) = 0 . b. The domain is (−∞, ∞) . The range is (−∞, 9] . c. Increasing on (−∞, − 3) ; decreasing on (−3, ∞) .
3)2 = 0 x = −3. The x-intercept is −3 . The y-intercept is f (0) = 2 . The y-intercept is f (0) = 9 . b. The domain is (−∞, ∞) . The range is ⎡15 , ∞ ⎞ . ⎢⎣ 8 ⎟ c. Increasing on ⎛ 1 , ∞ ⎞ ; decreasing on ⎛ −∞, 1 ⎞ . ⎜ 4 ⎟ ⎜ 4 ⎟ ⎝ ⎠ ⎝ ⎠
and −1 + 3 . 3 (2 x + 3)( x + 1) = 0 x = − 3 or x = −1. The y-intercept is f (0) = 2 . 2 The x-intercepts are − 3 and − 1 . 2 The y-intercept is f (0) = 3 .
= 6 ± 68 = 6 ± 2 17 = 3 ± 17 = = = 6 6 3 −8 −8 −4 4 + 10 4 − 10 The x-intercepts are −3 + 17 4 and −3 − 17 . 4 The x-intercepts are 3 and . 3 The y-intercept is f (0) = 2 . The y-intercept is f (0) = 2 .
5 = a (−2)2 + 1 5 = 4a + 1 4 = 4a 1 = a The quadratic function is f ( x ) = ( x − 2)2 + 1 = x2 − 4 x + 5 . have h = 1 and k = −3 . The also passes through the point ( x, y ) = (3, 5) . Substituting these values for x, y, h, and k, we can solve for a: 5 = a (3 − 1)2 + (−3) 5 = a (2)2 − 3 5 = 4a − 3 8 = 4a 2 = a The quadratic function is f ( x ) = 2 ( x − 1)2 − 3 = 2 x2 − 4 x − 1 .
f ( x) = 4 x2 − 4 x , a = 4, b = −4, c = 0. occurs at x = −b = −12 = −12 = 3. 2a 2(−2) −4 The maximum value is Since a = 4 > 0, the graph opens up, so the vertex is a minimum point. The minimum occurs at x = −b = −(− 4) = 4 = 1 . The f (3) = −2(3)2 + 12(3) = −18 + 36 = 18 . 2a minimum value is 2(4) 8 2 61. For f ( x) = 2 x2 + 12 x − 3, a = 2, b = 12, ⎛ 1 ⎞ ⎛ 1 ⎞2 f ⎜ ⎟ = 4 ⎜ ⎟ − 4 ⎛ 1 ⎞ = 1 − 2 = −1 . c = −3. Since a = 2 > 0, the graph opens up, so ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ the vertex is a minimum point. The minimum f x = a x − h 2 + k . occurs at x = −b = −12 = −12 = −3. The 67. Use the form ( ) ( ) 2a 2(2) 4 The vertex is (0, 2) , so h = 0 and k = 2. minimum value is f x = a x − 0 2 + 2 = ax2 + 2 . f (−3) = 2(−3)2 + 12(−3) − 3 = 18 − 36 − 3 = −21 . ( ) ( ) Since the graph passes through (1, 8) , f (1) = 8 . 62. For f ( x) = 4 x2 − 8x + 3, a = 4, b = −8, c = 3. f ( x ) = ax2 + 2 Since a = 4 > 0, the graph opens up, so the vertex is a minimum point. The minimum 8 = a (1)2 + 2 8 = a + 2 occurs at x = −b = −(−8) = 8 = 1. The 2a minimum value is 2(4) 8 6 = a f ( x ) = 6 x2 + 2 . f (1) = 4(1)2 − 8(1) + 3 = 4 − 8 + 3 = −1 . a = 6, b = 0, c = 2
b. f ( x ) = g ( x ) − x2 + 4 = −2 x + 1 0 = x2 − 2x − 3 0 = ( x + 1) ( x − 3) 73. a and d. x + 1 = 0 or x = −1 x − 3 = 0 x = 3 c. f (1) = − (−1)2 + 4 = −1 + 4 = 3 g (1) = −2 (−1) + 1 = 2 + 1 = 3 f (3) = − (3)2 + 4 = −9 + 4 = −5 g (3) = −2 (3) + 1 = −6 + 1 = −5 b. f ( x ) = g ( x ) 2 2 Thus, the graphs of f and g intersect at the points (−1, 3) and (3, −5) . − x + 5x = x 0 = 2 x + 3x − 4 2 − 2 x − 4 72. a and d. 0 = x2 − x − 2 0 = ( x + 1) ( x − 2) x + 1 = 0 or x = −1 x − 2 = 0 x = 2 c. f (−1) = − (−1)2 + 5 (−1) = −1 − 5 = −6 g (−1) = (−1)2 + 3 (−1) − 4 = 1 − 3 − 4 = −6 f (2) = − (2)2 + 5 (2) = −4 + 10 = 6 g (2) = 22 + 3(2) − 4 = 4 + 6 − 4 = 6 Thus, the graphs of f and g intersect at the points (−1, −6) and (2, 6) . b. f ( x ) = g ( x ) − x2 + 9 = 2 x + 1 0 = x2 + 2x − 8 0 = ( x + 4) ( x − 2) 74. a and d. x + 4 = 0 or x = −4 x − 2 = 0 x = 2 c. f (−4) = − (−4)2 + 9 = −16 + 9 = −7 g (−4) = 2 (−4) + 1 = −8 + 1 = −7 f (2) = − (2)2 + 9 = −4 + 9 = 5
b. f ( x ) = g ( x ) 2 2 g (2) = 2 (2) + 1 = 4 + 1 = 5 Thus, the graphs of f and g intersect at the − x + 7 x − 6 = x 0 = 2 x + x − 6 2 − 6 x points (−4, −7 ) and (2, 5) . 0 = 2 x ( x − 3) 2 x = 0 or x = 0 x − 3 = 0 x = 3
The zeros of f are −7 and 3. Likewise, the x-intercepts of f are −7 and 3. c. f ( x ) = −21 x2 + 4 x − 21 = −21 x2 + 4 x = 0 x ( x + 4) = 0 f ( x) = 1( x − (−5))( x − 3) = ( x + 5)( x − 3) = x2 + 2x −15 x = 0 or x + 4 = 0 x = −4 For a = 2 : f ( x) = 2( x − (−5))( x − 3) The solutions f ( x ) = −21 are −4 and 0. = 2( x + 5)( x − 3) = 2( x2 + 2 x −15) = 2 x2 + 4 x − 30 Thus, the points (−4, −21) and (0, −21) are on the graph of f.
7. x2 − 3x −10 < 0 f ( x) = x2 − 3x − 10 x2 − 3x −10 = 0 ( x − 5)( x + 2) = 0 x = 5, x =− 2 are the zeros of f . 3. a. f ( x) > 0 when the graph of f is above the x- axis. Thus, {x x < −2 or x > 2} or, using interval notation, (−∞, −2) ∪ (2, ∞ ) . The solution set is { x − 2 < x < 5 } or, using b. f ( x) ≤ 0 when the graph of f is below or intersects the x-axis. Thus, {x − 2 ≤ x ≤ 2} or, using interval notation, [−2, 2] . interval notation, (−2, 5) .
8. x2 + 3x −10 > 0 f ( x ) = x2 + 3x −10 x2 + 3x −10 = 0 ( x + 5)( x − 2) = 0 x = −5, x = 2 are the of f . 11. x2 − 9 < 0 f ( x) = x2 − 9 x2 − 9 = 0 ( x − 3)( x + 3) = 0 x = −3, x = 3 are the zeros of f . Interval (−∞, − 5) (−5, 2) (2, ∞) Test Number −6 0 6 Value of f 8 −10 8 Conclusion Positive Negative Positive The solution set is { x x < −5 or x > 2 } or, The solution set is { x − 3 < x < 3 } or, using using interval notation, (−∞, −5) ∪ (2, ∞ ) . 9. x2 − 4 x > 0 f ( x) = x2 − 4 x x2 − 4 x = 0 x( x − 4) = 0 x = 0, x = 4 are the zeros of f . 12. notation, (−3, 3) . x2 − 1 < 0 f ( x) = x2 −1 x2 − 1 = 0 ( x − 1)( x + 1) = 0 x = −1, x = 1 are the zeros of f . Interval (−∞, 0) (0, 4) (4, ∞) Test Number −1 1 5 Value of f 5 −3 5 Conclusion Positive Negative Positive The solution set is { x x < 0 or x > 4 } or, The solution set is { x −1 < x < 1 } or, using 10. using (−∞, 0) ∪ (4, ∞ ) . x2 + 8x > 0 f ( x) = x2 + 8x x2 + 8x = 0 x( x + 8) = 0 x = – 8, x = 0 are the zeros of f . 13. (−1, 1) . x2 + x > 12 x2 + x −12 > 0 f ( x) = x2 + x −12 x2 + x − 12 = 0 ( x + 4)( x − 3) = 0 x = −4, x = 3 are the of f no
x = −5, x = 4 are the zeros of f . x 2 , x = 3 3 2 are the The solution set is { x x < −5 or x > 4} or, using interval notation, (−∞, −5) ∪ (4, ∞ ) .
25. f ( x ) = x2 −1; g ( x ) = 3x + 3 a. f ( x ) = 0 x2 − 1 = 0 ( x −1) ( x + 1) = 0 x = 1; x = −1 Solution set: { − 1,1} . b. g ( x ) = 0 3x + 3 = 0 3x = −3 x = −1 Solution set: { −1} . The solution set is { x x < −1 or x > 4 } or, using interval notation, (−∞, −1) ∪ (4, ∞ ) . g. f ( x ) ≥ 1 x2 −1 > 1 x2 − 2 > 0 c. f ( x ) = g ( x ) ( x − 2 )( x + 2 ) > 0 x2 −1 = 3x + 3 The zeros are x = – 2 and x = 2 . ( ) 2 x2 − 3x − 4 = 0 ( x − 4) ( x + 1) = 0 x = 4; x = −1 Solution set: { −1, 4} . d. f ( x ) > 0 x2 − 1 > 0 ( x −1) ( x + 1) > 0 p x = x − 2 The zeros are x = –1 and x = 1 . The solution set is { x x <− 2 or x > 2 } or, using interval notation, (−∞, − 2 ) ∪ ( 2 , ∞ ) . 26. f ( x ) = x2 −1; g ( x ) = −3x + 3 The solution set is { x x < −1 or x > 1 } or, using interval notation, (−∞, −1) ∪ (1, ∞ ) . e. g ( x ) ≤ 0
3x + 3 ≤ 0 3x ≤ −3 x ≤ −1 The solution set is { x x ≤ −1} or, interval notation, (−∞, −1] . f. f ( x ) > g ( x ) x2 −1 > 3x + 3 x2 − 3x − 4 > 0 ( x − 4) ( x + 1) > 0 The zeros are x = –1 and x = 4 . p ( x ) = x2 − 3x − 4 a. f ( x ) = 0 x2 − 1 = 0 ( x −1) ( x + 1) = 0 x = 1; x = −1 Solution set: {−1, 1} . b. g ( x ) = 0 −3x + 3 = 0 −3x = −3 x = 1 Solution set: {1} . c. f ( x ) = g ( x ) x2 −1 = −3x + 3 x2 + 3x − 4 = 0 ( x + 4) ( x − 1) = 0 x = −4; x = 1 Solution set: {−4, 1} .
d. f ( x ) > 0 x2 − 1 > 0 ( x −1) ( x + 1) > 0 The are x = –1 and x = 1 . { } The solution set is x x <− 2 or x > 2 or, using interval notation, The solution set is { x x < −1 or x > 1 } or, (−∞, − 2 ) ∪ ( 2 , ∞ ) . interval notation, (−∞, −1) ∪ (1, ∞ ) . 27. ( ) ( ) e. g ( x ) ≤ 0 f x = − x2 + 1; a. f ( x ) = 0 g x = 4 x + 1 −3x + 3 ≤ 0 − x2 + 1 = 0 2 −3x ≤ −3 1 − x = 0 x ≥ 1 The solution set is { x x ≥ 1} or, using interval notation, [1, ∞ ) . f. f ( x ) > g ( x ) x2 −1 > −3x + 3 x2 + 3x − 4 > 0 (1 − x ) (1 + x ) = 0 x = 1; x = −1 Solution set: {−1, 1} . b. g ( x ) = 0 4 x + 1 = 0 4 x = −1 1 =− ( x + 4) ( x − 1) > 0 The zeros are x = –4 and x = 1 . x 4 ⎧ 1 ⎫ p ( x ) = x2 + 3x − 4 Solution set: ⎨− ⎬ . ⎩ 4 ⎭ The solution set is { x x < −4 or x > 1 } or, using interval notation, (−∞, −4) ∪ (1, ∞ ) . g. f ( x ) ≥ 1
x2 −1 > 1 x2 − 2 > 0 c. f ( x ) = g ( x ) − x2 + 1 = 4 x + 1 0 = x2 + 4x 0 = x ( x + 4) x = −4; x = 0 Solution set: {−4, 0} . d. f ( x ) > 0 − x2 + 1 > 0 1 − x2 > 0 (1 − x ) (1 + x ) > 0 x = 1; x = −1 ( x − 2 )( x + 2 ) > 0 The are x = –1 and x = 1 . The are x = – 2 and x = 2 . p ( x ) = x2 − 2 The solution set is { x −1 < x < 1 } or, interval notation, (−1, 1) .
e. g ( x ) ≤ 0 4x + 1 ≤ 0 4x ≤ −1 x 1 ≤− 4 b. g ( x ) = 0 − x − 2 = 0 − x = 2 x = −2 Solution set: {−2} . The solution set is ⎧ x x 1 ⎫ or, using c. f ( x ) = g ( x ) ⎨ ≤− ⎬ ⎩ 4 ⎭ interval notation, ⎛ −∞, − 1 ⎤ . − x2 + 4 = − x − 2 0 = x2 − x − 6 ⎜ 4 ⎥ ⎝ ⎦ f. f ( x ) > g ( x ) − x2 + 1 > 4x + 1 0 > x2 + 4x 0 > x ( x + 4) The zeros are x = –4 and x = 0 . p ( x ) = x ( x + 4) 0 = ( x − 3) ( x + 2) x = 3; x = −2 Solution set: {−2, 3} . d. f ( x ) > 0 − x2 + 4 > 0 4 − x2 > 0 (2 − x ) (2 + x ) > 0 The are x = –2 and x = 2 . The solution set is { x − 4 < x < 0} or, using The solution set is { x − 2 < x < 2 } or, interval notation, (−4, 0) . g. f ( x ) ≥ 1 − x2 + 1 ≥ 1 − x2 ≥ 0 x2 ≤ 0 using interval notation, (−2, 2) . e. g ( x ) ≤ 0 − x − 2 ≤ 0 − x ≤ 2 x ≥ −2 The zero is x = 0 . p ( x ) = x2 The set is { x x ≥ −2} or, using interval notation, [−2, ∞ ) . f. f ( x ) > g ( x ) − x2 + 4 > − x − 2 0 > x2 − x − 6 The solution set is {0} . 0 > ( x − 3) ( x + 2) The zeros are x = –2 and x = 3 . 28. f ( x ) = − x2 + 4; g ( x ) = − x − 2 ( ) ( )( )
g. f ( x ) ≥ 1 − x2 + 4 > 1 0 ≥ x2 − 3 0 ≥ ( x − 3 )( x + 3 ) The are x = – 3 and x = 3 . p ( x ) = ( x − 3 )( x + 3 ) The set is { x x < −2 or x > 2 } or, using interval notation, (−∞, −2) ∪ (2, ∞ ) . e. g ( x ) ≤ 0 − x2 + 4 ≤ 0 4 − x2 ≤ 0 (2 − x ) (2 + x ) ≤ 0 The solution set is { x − 3 ≤ x ≤ 3} or, The zeros are x = –2 and x = 2 . using interval notation, 3, 3 ⎤⎦ . 29. f ( x ) = x2 − 4; g ( x ) = − x2 + 4 a. f ( x ) = 0 x2 − 4 = 0 ( x − 2) ( x + 2) = 0 x = 2; x = −2 Solution set: {−2, 2} . b. g ( x ) = 0 The solution set is { x x ≤ −2 or x ≥ 2 } or, using interval notation, (−∞, −2] ∪ [2, ∞ ) . f. f ( x ) > g ( x ) x2 − 4 > − x2 + 4 2 − x2 + 4 = 0 4 − x2 = 0 (2 − x ) (2 + x ) = 0 x = 2; x = −2 Solution set: {−2, 2} . c. f ( x ) = g ( x ) x2 − 4 = − x2 + 4 2x2 − 8 = 0 2 ( x2 − 4) = 0 2 ( x − 2) ( x + 2) = 0 x = 2; x = −2 Solution set: {−2, 2} . d. f ( x ) > 0 x2 − 4 > 0 ( x − 2) ( x + 2) > 0 The zeros are x = –2 and x = 2 . 2x − 8 > 0
2 ( x2 − 4) > 0 2 ( x − 2) ( x + 2) > 0 The zeros are x = –2 and x = 2 . p ( x ) = 2 ( x − 2) ( x + 2) The solution set is { x x < −2 or x > 2 } or, using interval notation, (−∞, −2) ∪ (2, ∞ ) .
g. f ( x ) ≥ 1 x2 − 4 ≥ 1 x2 − 5 ≥ 0 The solution set is { x x < 1 or x > 1 } or, using interval notation, (−∞, 1) ∪ (1, ∞ ) . ( x − 5 )( x + 5 ) ≥ 0 e. g ( x ) ≤ 0 2 The zeros are x = – 5 and x = 5 . − x + 1 ≤ 0 1 − x2 ≤ 0 (1 − x ) (1 + x ) ≤ 0 p ( x ) = ( x − 5 )( x + 5 ) The zeros are x = –1 and x = 1 . Interval (−∞, − 5 ) (− 5, 5 ) ( 5, ∞ ) Test Number −3 0 3 of p 4 −5 4 Conclusion Positive Negative Positive The solution set is { x x ≤− 5 or x ≥ using interval notation, (−∞, − ⎤ ⎡ 5 } or, 5, ∞ ) . The solution set is { x x ≤ −1 or x ≥ 1 } or, using interval notation, (−∞, −1] ∪ [1, ∞ ) . f. f ( x ) > g ( x ) 30. f ( x ) = x2 − 2 x + 1; g ( x ) = − x2 + 1 x2 − 2 x + 1 = − x2 + 1 a. f ( x ) = 0 x2 − 2 x + 1 = 0 ( x −1) ( x − 1) = 0 x = 1 Solution set: {1} . b. g ( x ) = 0 − x2 + 1 = 0 1 − x2 = 0 (1 − x ) (1 + x ) = 0 x = 1; x = −1 Solution set: {−1, 1} . c. f ( x ) = g ( x ) x2 − 2 x + 1 = − x2 + 1 2x2 − 2 x = 0 2x ( x − 1) = 0 x = 0, x = 1 Solution set: {0, 1} . d. f ( x ) > 0 x2 − 2 x + 1 = 0 ( x −1) ( x − 1) = 0 The zero is x = 1 . 2x2 − 2 x = 0 2x ( x − 1) = 0
x = 0, x = 1 The are x = 0 and x = 1 . p ( x ) = 2x ( x − 1) The solution set is { x x < 0 or x > 1 } or, using interval notation, (−∞, 0) ∪ (1, ∞ ) . g. f ( x ) ≥ 1 x2 − 2x + 1 ≥ 1 x2 − 2x ≥ 0 x ( x − 2) ≥ 0 The zeros are x = 0 and x = 2 . p ( x ) = x ( x − 2) The solution set is { x x ≤ 0 or x ≥ 2} or, using interval notation, (−∞, 0] ∪ [2, ∞ ) .
31. f ( x ) = x2 − x − 2; g ( x ) = x2 + x − 2 f. f ( x ) > g ( x ) a. f ( x ) = 0 x2 − x − 2 = 0 ( x − 2) ( x + 1) = 0 x = 2, x = −1 Solution set: {−1, 2} . b. g ( x ) = 0 x2 + x − 2 = 0 ( x + 2) ( x − 1) = 0 x2 − x − 2 > x2 + x − 2 −2 x > 0 x < 0 The solution set is {x x < 0} or, using interval notation, (−∞, 0) . g. f ( x ) ≥ 1 x2 − x − 2 ≥ 1 x2 − x − 3 ≥ 0 x = −2; x = 1 Solution set: {−2, 1} . − (−1) ± x = (−1)2 − 4 (1) (−3) 2 (1) c. f ( x) = g ( x) x2 − x − 2 = x2 + x − 2 −2 x = 0 = 1 ± 1 + 12 = 1 ± 13 2 2 The zeros are x = 0 Solution set: {0} . x = 1 − 13 ≈ −1.30 and x = 1 + 13 2 2 ≈ 2.30 d. f ( x ) > 0 x2 − x − 2 > 0 ( x − 2) ( x + 1) > 0 The zeros are x = −1 and x = 2 . p ( x ) = x2 − x − 3 −∞, 2 2 , 2 2 , ∞ The solution set is ⎧⎪ x x ≤ 1 − 13 or x ≥ 1 + 13 ⎫⎪ or, using ⎩⎪ 2 2 ⎭⎪ The solution set is {x x < −1 or x > 2} or, interval notation, (−∞, −1) ∪ (2, ∞ ) . interval notation, ⎛ 1 − 13 ⎤ ⎡1 + 13 ⎞ ⎜ −∞, 2 ⎥ ∪ ⎢ 2 , ∞ ⎟ . e. g ( x ) ≤ 0 ⎝ ⎦⎥ ⎣⎢ ⎠ x2 + x − 2 ≤ 0 32. f ( x ) = − x2 − x + 1; g ( x ) = − x2 + x + 6
( x + 2) ( x − 1) ≤ 0 a. f ( x ) = 0 2 The zeros are x = –2 and x = 1 . − x − x + 1 = 0 x2 + x −1 = 0 2 ( ) ( ) ()( ) x = − 1 ± 1 − 4 1 −1 2 (1) = −1 ± 1 + 4 = −1 ± 5 2 2 The solution set is { x − 2 ≤ x ≤ 1} or, ⎧⎪ Solution set: ⎨ −1 − 5 , −1 + 5 ⎫⎪⎬ . 2 2 interval notation, [−2, 1] . ⎪⎩ ⎭⎪
b. g ( x ) = 0 − x2 + x + 6 = 0 x2 − x − 6 = 0 ( x − 3) ( x + 2) = 0 x = 3; x = −2 Solution set: {−2, 3} . c. f ( x ) = g ( x ) − x2 − x + 1 = − x2 + x + 6 −2 x − 5 = 0 −2x = 5 The solution set is { x x ≤ −2 or x ≥ 3 } or, using interval notation, (−∞, 2] ∪ [3, ∞) . f. f ( x ) > g ( x ) − x2 − x + 1 > − x2 + x + 6 5 x −2x > Solution set: ⎧− 5 ⎫ . x <− 2 ⎨ 2 ⎬ ⎩ ⎭ d. f ( x ) > 0 − x2 − x + 1 > 0 x2 + x −1 < 0 The solution set is { x x <− 5 } or, interval notation, (−∞, − 5 ) . g. f ( x ) ≥ 1 − (1) ± x = (1)2 − 4 (1) (−1) − − x + 1 ≥ 1 − x2 − x ≥ 2 (1) = −1 ± 1 + 4 = −1 ± 5 0 − x ( x + 1) ≥ 0 2 2 The zeros are x = −1 − 5 ≈ −1.62 and x = −1 + 5 ≈ 0.62 2 2 p ( x ) = x2 + x − 1 The are x = −1 and x = 0 . p ( x ) = − x ( x + 1) −∞, 2 2 , 2 2 , ∞ The solution set is { x −1 ≤ x ≤ 0 } or, using interval notation, [−1, 0] . The solution set is x −1 − 5 < x < −1 + 5 33. a. The ball strikes the ground when 2 ⎪⎨ 2 2 ⎪⎬ s (t ) = 80t − 16t = 0 . ⎛ −1 − 5 −1 + 5 ⎞ or, using interval notation, ⎜ , ⎟ . 80t − 16t 2 = 0 16t (5 − t ) = 0 e. g ( x ) ≤ 0
− x2 + x + 6 ≤ 0 ⎝ 2 2 ⎠ t = 0, t = 5 The ball strikes the ground after 5 seconds. b. Find the values of t for which 2 x2 − x − 6 ≥ 0 80t − 16t 2 > 96 ( x − 3) ( x + 2) ≥ 0 −16t + 80t − 96 > 0 The zeros are x = –2 and x = 3 . p ( x ) = x2 − x − 6 16t 2 − 80t + 96 < 0 16(t 2 − 5t + 6) < 0 16(t − 2)(t − 3) < 0 The zeros are t = 2 and t = 3 . s(t ) = 16t 2 − 80t + 96
p = 0, p = 1000 p2 −1900 p + 1200000 < 0 2 Thus, the revenue equals zero when p = $0 and when p = $1000. b. Find the values of p for R ( p ) = − 4 p2 + 4000 p > 800000 1 ( p2 − 3800 p + 2400000) < 0 2 1 ( p − 800) ( p − 3000) < 0 2 − 4 p2 + 4000 p > 800000 The zeros are p = 800 and p = 3000 . − 4 p2 + 4000 p − 800000 > 0 s( p) 1 p2 + 1900 p − 1200000 2