Math Chapter 2 Quadratic Functions Key

Chapter 2 NONLINEAR FUNCTIONS 6. The graph of y = (x+3)2 +2 is the graph of y = x2 translated 3 units to the left and 2 units upward. This is graph b. 7. The graph of y = ¡(3 ¡ x)2 + 2 is the same as the graph of y = ¡(x ¡ 3)2 + 2: This is the graph of y = x2 re‡ected in the x-axis, translated 3 units to the right, and translated 2 units upward. This is graph c. 8. The graph of y = ¡(x + 3)2 + 2 is the graph of y = x2 re‡ected in the x-axis, translated three units to the left, and two units upward. This is graph e. 9. y = x2 + 5x + 6 y = (x + 3)(x + 2) Set y = 0 to …nd the x-intercepts. 0 = (x + 3)(x + 2) x = ¡3; x = ¡2 The x-intercepts are ¡3 and ¡2: Set x = 0 to …nd the y-intercept. y = 02 + 5(0) + 6 y=6 The y-intercept is 6. The x-coordinate of the vertex is x= ¡5 5 ¡b = =¡ : 2a 2 2 Substitute to …nd the y-coordinate. 2.2 Quadratic Functions; Translation and Re‡ection 3. The graph of y = x2 ¡ 3 is the graph of y = x2 translated 3 units downward. This is graph d. 4. The graph of y = (x ¡ 3)2 is the graph of y = x2 translated 3 units to the right. This is graph f. 5. The graph of y = (x ¡ 3)2 + 2 is the graph of y = x2 translated 3 units to the right and 2 units upward. This is graph a. 5 25 25 1 5 y = (¡ )2 + 5(¡ ) + 6 = ¡ +6 =¡ 2 2 4 2 4 5 1 The vertex is (¡ ; ¡ ): 2 4 The axis is x = ¡ 52 ; the vertical line through the vertex. Section 2.2 Quadratic Functions; Translation and Re‡ection 10. y = x2 + 4x ¡ 5 = (x + 5)(x ¡ 1) The axis is x = ¡3; the vertical line through the vertex. Set y = 0 to …nd the x-intercepts. 0 = (x + 5)(x ¡ 1) x = ¡5 or x = 1 The x-intercepts are ¡5 and 1. Set x = 0 to …nd the y-intercept. y = 02 + 4(0) ¡ 5 y = ¡5 The y-intercept is ¡5. The x-coordinate of the vertex is x= ¡4 ¡b = = ¡2: 2a 2 12. y = ¡3x2 ¡ 6x + 4 Let y = 0: The equation Substitute to …nd the y-coordinate. y = (¡2)2 + 4(¡2) ¡ 5 =4¡8¡5 = ¡9 0 = ¡3x2 ¡ 6x + 4 cannot be solved by factoring. Use the quadratic formula. The vertex is (¡2; ¡9): The axis is x = ¡2; the vertical line through the vertex. 11. y = ¡2x2 ¡ 12x ¡ 16 = ¡2(x2 + 6x + 8) = ¡2(x + 4)(x + 2) Let y = 0: 2 ¡ 6x + 4 0 = ¡3xp 6 § 62 ¡ 4(¡3)(4) x= 2(¡3) p 6 § 36 + 48 = ¡6 p p 6 § 2 21 6 § 84 = = ¡6 ¡6 p 21 = ¡1 § 3 The x-intercepts are ¡1 + ¡1 ¡ p 21 3 p 21 3 ¼ 0:53 and ¼ ¡2:53 Let x = 0: 0 = ¡2(x + 4)(x + 2) x = ¡4; x = ¡2 ¡4 and ¡2 are the x-intercepts. Let x = 0: y = ¡3(0)2 ¡ 6(0) + 4 y=4 4 is the y-intercept. 2 y = ¡2(0) + 12(0) ¡ 16 ¡16 is the y-intercept. Vertex: x = 12 ¡b = = ¡3 2a ¡4 y = ¡2(¡3)2 ¡ 12(¡3) ¡ 16 = ¡18 + 36 ¡ 16 = 2 The vertex is (¡3; 2): Vertex: x = 6 6 ¡b = = = ¡1 2a 2(¡3) ¡6 y = ¡3(¡1)2 ¡ 6(¡1) + 4 = ¡3 + 6 + 4 =7 The vertex is (¡1; 7): Chapter 2 The axis is x = ¡1; the vertical line through the vertex. NONLINEAR FUNCTIONS 14. f(x) = ¡x2 + 6x ¡ 6 Let f (x) = 0: 0 = ¡x2 + 6x ¡ 6 p 62 ¡ 4(¡1)(¡6) x= 2(¡1) p ¡6 § 36 ¡ 24 = ¡2 p ¡6 § 12 = ¡2 p ¡6 § 2 3 = ¡2 p =3§ 3 p p The x-intercepts are 3 + 3 ¼ 4:73 and 3 ¡ 3 ¼ 1:27: ¡6 § 13. y = 2x2 + 8x ¡ 8 Let y = 0: 2x2 + 8x ¡ 8 = 0 x2 + 4x ¡ 4 = 0 p ¡4 § 42 ¡ 4(1)(¡4) x= p p 2(1) ¡4 § 4 2 ¡4 § 32 = = 2 2 p = ¡2 § 2 2 p The x-intercepts are ¡2 § 2 2 ¼ 0:83 or ¡4:83: Let x = 0: y = 2(0)2 + 8(0) ¡ 8 = ¡8 The y-intercept is ¡8: Let x = 0: y = ¡02 + 6(0) ¡ 6 ¡6 is the y-intercept. ¡6 ¡6 ¡b = = =3 Vertex: x = 2a 2(¡1) ¡2 y = ¡32 + 6(3) ¡ 6 = ¡9 + 18 ¡ 6 =3 The vertex is (3; 3): The axis is x = 3: The x-coordinate of the vertex is x= 8 ¡b = ¡ = ¡2: 2a 4 If x = ¡2; y = 2(¡2)2 + 8(¡2) ¡ 8 = 8 ¡ 16 ¡ 8 = ¡16: The vertex is (¡2; ¡16): The axis is x = ¡2: 15. f(x) = 2x2 ¡ 4x + 5 Let f (x) = 0: 0 = 2x2 ¡ 4x + 5 x= ¡(¡4) § p (¡4)2 ¡ 4(2)(5) 2(2) p 16 ¡ 40 = 4 p 4 § ¡24 = 4 4§ Since the radicand is negative, there are no xintercepts. Section 2.2 Quadratic Functions; Translation and Re‡ection Let x = 0: Vertex: x = y = 2(0)2 ¡ 4(0) + 5 y=5 5 is the y-intercept. Vertex: x = ¡(¡4) 4 ¡b = = =1 2a 2(2) 4 ¡6 ¡6 ¡b = ¡1¢ = = ¡6 2a 1 2 2 1 y = (¡6)2 + 6(¡6) + 24 2 = 18 ¡ 36 + 24 = 6 The vertex is (¡6; 6): The axis x = ¡6: y = 2(1)2 ¡ 4(1) + 5 = 2 ¡ 4 + 5 = 3 The vertex is (1; 3): The axis is x = 1: 17. f(x) = ¡2x2 + 16x ¡ 21 Let f(x) = 0 Use the quadratic formula. 1 16. f (x) = x2 + 6x + 24 2 Let f(x) = 0: 1 0 = x2 + 6x + 24 2 Multiply by 2 to clear fractions. 0 = x2 + 12x + 48 Use the quadratic formula. p ¡12 § 122 ¡ 4(1)(48) x= 2(1) p ¡12 § 144 ¡ 192 = 2 p ¡12 § ¡48 = 2 Since the radicand is negative, there are no xintercepts. Let x = 0. 1 y = (0)2 + 6(0) + 24 2 y = 24 24 is the y-intercept. p 162 ¡ 4(¡2)(¡21) x= 2(¡2) p ¡16 § 88 = ¡4 p ¡16 § 2 22 = ¡4 p 22 =4§ 2 ¡16 § The x-intercepts are 4+ 1:65: p 22 2 ¼ 6:35 and 4¡ Let x = 0: y = ¡2(0)2 + 16(0) ¡ 21 y = ¡21 ¡21 is the y-intercept. Vertex: x = ¡16 ¡16 ¡b = = =4 2a 2 (¡2) ¡4 y = ¡2(4)2 + 16(4) ¡ 21 = ¡32 + 64 ¡ 21 = 11 The vertex is (4; 11): p 22 2 ¼ Chapter 2 The axis is x = 4: NONLINEAR FUNCTIONS 1 1 8 19. y = x2 ¡ x + 3 3 3 Let y = 0: 1 8 1 0 = x2 ¡ x + 3 3 3 Multiply by 3. 0 = x2 ¡ 8x + 1 p (¡8)2 ¡ 4(1)(1) x= 2(1) p p 8 § 60 8 § 64 ¡ 4 = = 2 2 p p 8 § 2 15 = 4 § 15 = 2 3 18. f (x) = x2 ¡ x ¡ 4 2 ¡(¡8) § Let f(x) = 0: 3 0 = x2 ¡ x ¡ 4 2 Multiply by 2 to clear fractions, then solve. 0 = 3x2 ¡ 2x ¡ 8 = (3x + 4)(x ¡ 2) 4 x = ¡ ;2 3 p p The x-intercepts are 4+ 15 ¼ 7:87 and 4¡ 15 ¼ 0:13: Let x = 0: The x-intercepts are ¡ 43 and 2. 1 8 1 y = (0)2 ¡ (0) + 3 3 3 Let x = 0: 3 y = (0)2 ¡ 0 ¡ 4 2 y = ¡4 The y-intercept is ¡4: ¡(¡1) ¡b 1 = ¡3¢ = 2a 3 2 2 μ ¶2 1 2 24 25 1 3 1 =¡ ¡ ¡4 = ¡ ¡ y= 2 3 3 6 6 6 6 ¡ 1 25 ¢ The vertex is 3 ; ¡ 6 : Vertex: x = The axis x = 1 3: 1 3 is the y-intercept. ¡ ¢ ¡ ¡ 83 ¡b ¡ ¢ = = Vertex: x = 2a 2 13 8 3 2 3 =4 1 1 8 y = (4)2 ¡ (4) + 3 3 3 15 16 32 1 ¡ + = ¡ = ¡5 = 3 3 3 3 The vertex is (4; ¡5): The axis is x = 4: Section 2.2 Quadratic Functions; Translation and Re‡ection 1 7 20. y = ¡ x2 ¡ x ¡ 2 2 Let y = 0: 1 7 0 = ¡ x2 ¡ x ¡ 2 2 Multiply by ¡2 to clear fractions. 0 = x2 + 2x + 7 Use the quadratic formula. p ¡2 § 4 ¡ 4(1)(7) x= 2 p ¡2 § ¡24 = 2 Since the radicand is negative, there are no xintercepts. Let x = 0: 7 7 1 y = ¡ (0)2 ¡ 0 ¡ = ¡ 2 2 2 ¡ 72 is the y-intercept. Vertex: ¡(¡1) 1 ¡b = ¡ 1¢ = = ¡1 2a ¡1 2 ¡2 1 7 y = ¡ (¡1)2 ¡ (¡1) ¡ 2 2 7 1 =¡ +1¡ 2 2 p 22. The graph of y = x ¡ 2 ¡ 4 is the graph of y = p x translated 2 units to the right and 4 units downward. This is graph a. p 23. The p graph of y = ¡x + 2 ¡ 4 is the graph of p y = ¡(x ¡ 2) ¡ 4; which is the graph of y = x re‡ected in the y-axis, translated 2 units to the right, and translated 4 units downward. This is graph c. p 24. The graph of p y = ¡x ¡ 2 ¡ 4 is the same as the graph of y = ¡(x + 2) ¡ 4: This is the graph of p y = x re‡ected in the y-axis, translated 2 units to the left, and translated 4 units downward. This is graph b. p 25. The graph of y = ¡ x + 2 ¡ 4 is the graph of p y = x re‡ected in the x-axis, translated 2 units to the left, and translated 4 units downward. This is graph e. p 26. The graph of y = ¡ x ¡ 2 ¡ 4 is the graph of p y = x re‡ected in the x-axis, translated 2 units to the right, and translated 4 units downward. This is graph f. 27. The graph of y = ¡f (x) is the graph of y = f (x) re‡ected in the x-axis. = ¡3 The vertex is (¡1; ¡3): The axis is x = ¡1: 28. The graph of y = f (x ¡ 2) + 2 is the graph of y = f(x) translated 2 units to the right and 2 units upward. p 21. The graph of y = x + 2 ¡ 4 is the graph of p y = x translated 2 units to the left and 4 units downward. This is graph d. Chapter 2 29. The graph of y = f (¡x) is the graph of y = f (x) re‡ected in the y-axis. NONLINEAR FUNCTIONS p 33. f (x) = ¡p2 ¡ x ¡ 2 = ¡ ¡(x ¡ 2) ¡ 2 Re‡ect the graph of f(x) vertically and horizontally. Translate the graph 2 units right and 2 units down. 30. y = f (2 ¡ x) + 2 y = f [¡(x ¡ 2)] + 2 This is the graph of y = f(x) re‡ected in the yaxis, translated 2 units to the right, and translated 2 units upward. p p 34. f (x) = ¡ 2 ¡ x + 2 = ¡ ¡(x ¡ 2) + 2 Re‡ect the graph of f(x) vertically and horizontally. Translate the graph 2 units right and 2 units up. 31. f (x) = p x¡2+2 Translate the graph of f(x) = and 2 units up. 32. f (x) = p p x 2 units right 35. If 0 < a < 1; the graph of f (ax) will be ‡atter and wider than the graph of f(x): Multiplying x by a fraction makes the y-values less than the original y-values. x+2¡3 Translate the graph of f (x) = 3 units down. p x 2 units left and Section 2.2 Quadratic Functions; Translation and Re‡ection 36. If 1 < a; the graph of f (ax) will be taller and thinner than the graph of f(x): Multiplying x by a constant greater than 1 pairs x-values of smaller absolute value with y-values of points for which the x-values have larger absolute value. 37. If ¡1 < a < 0; the graph of f (ax) will be re‡ected horizontally, since a is negative. It will be ‡atter because multiplying x by a fraction decreases the corresponding y-values. 39. If 0 < a < 1; the graph of af (x) will be ‡atter and wider than the graph of f (x): Each y-value is only a fraction of the height of the original y-values. 40. If 1 < a (or a > 1); the graph of af(x) will be taller than the graph of f(x): The absolute value of the y-value will be larger than the original yvalues, while the x-values will remain the same. 41. If ¡1 < a < 0; the graph will be re‡ected vertically, since a will be negative. Also, because a is a fraction, the graph will be ‡atter because each yvalue will only be a fraction of its original height. 38. If a < ¡1; the graph of f (ax) will be re‡ected horizontally, since a is negative. It will also be taller and thinner. Chapter 2 42. If a < ¡1; the graph of af (x) will be re‡ected vertically. It will also be taller than the graph of f (x) since the absolute value of each y-value will be larger than the original y-values, while the xvalues stay the same. NONLINEAR FUNCTIONS 45. (a) (b) Break-even quantities are values of x = number of widgets for which revenue and cost are equal. Set R(x) = C(x) and solve for x: ¡x2 + 8x = 2x + 5 x ¡ 6x + 5 = 0 (x ¡ 5)(x ¡ 1) = 0 x ¡ 5 = 0 or x ¡ 1 = 0 x = 5 or x=1 2 43. (a) Since the graph of y = f(x) is re‡ected vertically to obtain the graph of y = ¡f (x); the xintercept is unchanged. The x-intercept of the graph of y = f (x) is r: (b) Since the graph of y = f (x) is re‡ected horizontally to obtain the graph of y = f (¡x); the x-intercept of the graph of y = f (¡x) is ¡r: (c) Since the graph of y = f (x) is re‡ected both horizontally and vertically to obtain the graph of y = ¡f (¡x); the x-intercept of the graph of y = ¡f (¡x) is ¡r: So, the break-even quantities are 1 and 5. The minimum break-even quantity is x = 1: (c) The maximum revenue occurs at the vertex of R: Since R(x) = ¡x2 + 8x; then the x-coordinate of the vertex is x=¡ 8 b =¡ = 4: 2a 2(¡1) So, the maximum revenue is R(4) = ¡42 + 8(4) = 16: (d) The maximum pro…t is the maximum di¤erence R(x) ¡ C(x): Since 44. (a) Since the graph of y = f (x) is re‡ected vertically to obtain the graph of y = ¡f (x); the yintercept of the graph of y = ¡f(x) is ¡b: (b) Since the graph of y = f (x) is re‡ected horizontally to obtain the graph of y = f (¡x); the y-intercept is unchanged. The y-intercept of the graph of y = f (¡x) is b: P (x) = R(x) ¡ C(x) = ¡x2 + 8x ¡ (2x + 5) = ¡x2 + 6x ¡ 5 is a quadratic function, we can …nd the maximum pro…t by …nding the vertex of P: This occurs at x=¡ ¡6 b = = 3: 2a 2(¡1) Therefore, the maximum pro…t is (c) Since the graph of y = f (x) is re‡ected both horizontally and vertically to obtain the graph of y = ¡f(¡x); the y-intercept of the graph of y = ¡f (¡x) is ¡b: P (3) = ¡(3)2 + 6(3) ¡ 5 = 4: Section 2.2 Quadratic Functions; Translation and Re‡ection 46. (a) 47. (a) (b) Break-even quantities are values of x = number of widgets for which revenue and cost are equal. Set R(x) = C(x) and solve for x: ¡ 4 ¡ x2 + 10x = 2x + 15 5 3 x2 + 5x = x + 3 2 2 ¡x2 + 10x = 3x + 6 x2 ¡ 7x + 6 = 0 (x ¡ 1)(x ¡ 6) = 0 x ¡ 1 = 0 or x ¡ 6 = 0 x = 1 or x=6 So, the break-even quantities are 1 and 6. The minimum break-even quantity is x = 1: (c) The maximum revenue occurs at the vertex of 2 R: Since R(x) = ¡ x2 + 5x; then the x-coordinate of the vertex is ¡5 ¡b = ¡ 1 ¢ = 5: x= 2a 2 ¡2 So, the maximum revenue is R(8) = ¡ 52 + 5(5) = 12:5: 2 (d) The maximum pro…t is the maximum di¤erence R(x) ¡ C(x): Since μ ¶ 3 x2 x+3 P (x) = R(x) ¡ C(x) = ¡ + 5x ¡ 2 2 x2 7 =¡ + x¡3 2 2 is a quadratic function, we can …nd the maximum pro…t by …nding the vertex of P: This occurs at x=¡ (b) Break-even quantities are values of x = number of widgets for which revenue equals cost. Set R(x) = C(x) and solve for x: ¡7 7 b = ¡ 21 ¢ = : 2a 2 2 ¡2 Therefore, the maximum pro…t is ¡ ¢2 μ ¶ μ ¶ ¡ 72 7 7 7 + P = ¡ 3 = 3:125: 2 2 2 2 4 2 x ¡ 8x + 15 = 0 5 4x2 ¡ 40x + 75 = 0 4x ¡ 10x ¡ 30x + 75 = 0 2x(2x ¡ 5) ¡ 15(2x ¡ 5) = 0 (2x ¡ 5)(2x ¡ 15) = 0 2x ¡ 5 = 0 or 2x ¡ 15 = 0 x = 2:5 or x = 7:5 2 So, the break-even quantities are 2.5 and 7.5 with x = 2:5 the minimum break-even quantity. (c) The maximum revenue occurs at the vertex of R: Since R(x) = ¡ 45 x2 + 10x; then the xcoordinate of the vertex is x=¡ 10 b = ¡ ¡ ¡4 ¢ = 6:25: 2a 2 5 So, the maximum revenue is R(6:25) = 31:25: (d) The maximum pro…t is the maximum di¤erence R(x) ¡ C(x): Since P (x) = R(x) ¡ C(x) 4 = ¡ x2 + 10x ¡ (2x + 15) 5 4 = ¡ x2 + 8x ¡ 15 5 is a quadratic function, we can …nd the maximum pro…t by …nding the vertex of P: This occurs at x=¡ 8 b = ¡ ¡ ¡4 ¢ = 5: 2a 2 5 Therefore, the maximum pro…t is 4 P (5) = ¡ 52 + 8(5) ¡ 15 = 5: 5 Chapter 2 NONLINEAR FUNCTIONS 49. R(x) = 8000 + 70x ¡ x2 = ¡x2 + 70x + 8000 48. (a) The maximum revenue occurs at the vertex. ¡70 ¡b = = 35 x= 2a 2(¡1) y = 8000 + 70(35) ¡ (35)2 = 8000 + 2450 ¡ 1225 = 9225 (b) Break-even quantities are values of x for which revenue equals cost. Set R(x) = C(x) and solve for x: ¡4x2 + 36x = 16x + 24 4x ¡ 20x + 24 = 0 4(x ¡ 2)(x ¡ 3) = 0 x ¡ 2 = 0 or x ¡ 3 = 0 x = 2 or x=3 2 So, the break-even quantites are 2 and 3. The minimum break-even quantity is x = 2: (c) The maximum revenue occurs at the vertex of R: Since R(x) = ¡4x2 + 36x; then x-coordinate of the vertex is ¡36 9 ¡b = = : x= 2a 2(¡4) 2 So, the maximum revenue is R(5) = ¡4 μ ¶2 μ ¶ 9 9 + 36 = 81: 2 2 (d) The maximum pro…t is the maximum di¤erence R(x) ¡ C(x): Since P (x) = R(x) ¡ C(x) = ¡4x2 + 36x ¡ (16x + 24) = ¡4x2 + 20x ¡ 24 The vertex is (35; 9225): The maximum revenue of $9225 is realized when 35 seats are left unsold. 50. (a) The revenue is R(x) = (Price per ticket) ¢ (Number of people ‡ying). Number of people ‡ying = 100 ¡ x Price per ticket = 200 + 4x R(x) = (200 + 4x)(100 ¡ x) = 20,000 + 200x ¡ 4x2 (b) R(x) = ¡4x2 + 200x+ 20,000 x-intercepts: 0 = (200 + 4x)(100 ¡ x) x = ¡50 or x = 100 y-intercept: y = ¡4(0)2 + 200(0) + 20,000 = 20,000 Vertex: x = ¡200 ¡b = = 25 2a ¡8 y = ¡4(25)2 + 200(25) + 20,000 = 22,500 This is a parabola which opens downward. The vertex is at (25; 22,500): is a quadratic function, we can …nd the maximum pro…t by …nding the vertex of P: This occurs at x=¡ 20 5 b =¡ = = 2:5: 2a 2(¡4) 2 Therefore, the maximum pro…t is P μ ¶ μ ¶ μ ¶2 5 5 5 + 20 = ¡4 ¡ 24 = 1: 2 2 2 (c) The maximum revenue occurs at the vertex, (25, 22,500). This will happen when x = 25; or there are 25 unsold seats. (d) The maximum revenue is $22,500, as seen from the graph. Section 2.2 Quadratic Functions; Translation and Re‡ection 51. p = 500 ¡ x (d) Find the vertex. (a) The revenue is R(x) = px = (500 ¡ x)(x) = 500x ¡ x2 : (b) x= 0 ¡b = =0 2a ¡20 y = 8000 ¡ 20(0)2 = 8000 The vertex is (0; 8000): To produce maximum revenue, wait 0 weeks. Pick the peaches now. (e) R(0) = 8000 ¡ 20(0)2 = 8000 or the maximum revenue is 8000 cents per tree or $80.00 per tree. 53. Let x = the number of $25 increases. (a) Rent per apartment: 800 + 25x (c) From the graph, the vertex is halfway between x = 0 and x = 500; so x = 250 units corresponds to maximum revenue. Then the price is p = 500 ¡ x = 500 ¡ 250 = $250: Note that price, p; cannot be read directly from the graph of R(x) = 500x ¡ x2 : (d) R(x) = 500x ¡ x2 = ¡x2 + 500x Find the vertex. (b) Number of apartments rented: 80 ¡ x (c) Revenue: R(x) = (number of apartments rented) £ (rent per apartment) = (80 ¡ x)(800 + 25x) = ¡25x2 + 1200x + 64,000 (d) Find the vertex: x= ¡1200 ¡b = = 24 2a 2(¡25) y = ¡25(24)2 + 1200(24) + 64,000 = 78,400 ¡500 ¡b = = 250 2a 2(¡1) The vertex is (24; 78,400). The maximum revenue occurs when x = 24: y = ¡(250)2 + 500(250) = 62,500 (e) The maximum revenue is the y-coordinate of the vertex, or $78,400. x= The vertex is (250; 62,500): The maximum revenue is $62,500. 52. Let x = the number of weeks to wait. (a) Income per pound (in cents): 80 ¡ 4x (b) Yield in pounds per tree: 100 + 5x (c) Revenue per tree (in cents): R(x) = (100 + 5x)(80 ¡ 4x) R(x) = 8000 ¡ 20x2 1 54. S(x) = ¡ (x ¡ 10)2 + 40 for 0 · x · 10 4 1 (a) S(0) = ¡ (0 ¡ 10)2 + 40 = ¡25 + 40 = 15 4 The increase in sales is $15,000. (b) For $10,000, x = 10: 1 S(10) = ¡ (10 ¡ 10)2 + 40 4 S(10) = 40 The increase in sales is $40,000. Chapter 2 (c) The graph of S(x) = ¡ 14 (x ¡ 10)2 + 40 is the graph of y = ¡ 14 x2 translated 10 units to the right and 40 units upward. NONLINEAR FUNCTIONS 56. (a) When t = 14; L = 2:024, so the length of the crown is 2.024 mm at 14 weeks. When t = 24; L = 6:104, so the length is 6.104 mm at 24 weeks. (b) The vertex occurs at t= ¡0:788 ¡b = 2a 2(¡0:01) ¼ 39:4: When t = 39:4; L ¼ 8:48, so the maximum length is 8.48 mm and occurs 39.4 weeks after conception. 57. (a) The vertex of the quadratic function y = 0:057x ¡ 0:001x2 is at x=¡ 2 55. S(x) = 1 ¡ 0:058x ¡ 0:076x (a) 0:50 = 1 ¡ 0:058x ¡ 0:076x2 2 0:076x + 0:058x ¡ 0:50 = 0 76x2 + 58x ¡ 500 = 0 38x2 + 29x ¡ 250 = 0 p ¡29 § (29)2 ¡ 4(38)(¡250) x= 2(38) p ¡29 § 38,841 = 76 p ¡29¡ 38,841 76 ¼ ¡2:97 and p ¡29+ 38;841 76 ¼ 2:21 We ignore the negative value. The value x = 2:2 represents 2.2 decades or 22 years, and 22 years after 65 is 87. The median length of life is 87 years. (b) If nobody lives, S(x) = 0: 1 ¡ 0:058x ¡ 0:076x2 = 0 76x2 + 58x ¡ 1000 = 0 38x2 + 29x ¡ 500 = 0 p (29)2 ¡ 4(38)(¡500) x= 2(38) p ¡29 § 76,841 = 76 ¡29 § p ¡29¡ 76;841 76 ¼ ¡4:03 and p ¡29+ 76;841 76 ¼ 3:27 We ignore the negative value. The value x = 3:3 represents 3.3 decades or 33 years, and 33 years after 65 is 98. Virtually nobody lives beyond 98 years. 0:057 b =¡ = 28:5: 2a 2(¡0:001) Since the coe¢cient of the leading term, ¡0:001, is negative, then the graph of the function opens downward, so a maximum is reached at 28.5 weeks of gestation. (b) The maximum splenic artery resistance reached at the vertex is y = 0:057(28:5) ¡ 0:001(28:5)2 ¼ 0:81: (c) The splenic artery resistance equals 0, when y = 0: 0:057x ¡ 0:001x2 = 0 Substitute in the expression in x for y: x(0:057 ¡ 0:001x) = 0 Factor. x = 0 or 0:057 ¡ 0:001x = 0 Set each factor equal to 0. 0:057 = 57 x= 0:001 So, the splenic artery resistance equals 0 at 0 weeks or 57 weeks of gestation. No, this is not reasonable because at x = 0 or 57 weeks, the fetus does not exist. 58. The vertex of the quadratic function y = 10:5x2 ¡ 125x + 17,186 is at x= ¡(¡125) ¡b = ¼ 66: 2a 2(10:5) Since the coe¢cient of the leading term, 10.5, is positive, the graph of the function opens upward, so a minimum of …rst year enrollments is reached when x ¼ 6; which is the year 1999. For 2004, x = 11. The domain of f(x) is [0, 11]. Section 2.2 Quadratic Functions; Translation and Re‡ection 59. (a) (c) (c) y = 0:002726x2 ¡ 0:3113x + 29:33 61. f(x) = 60:0 ¡ 2:28x + 0:0232x2 ¡2:28 ¡b =¡ ¼ 49:1 2a 2(0:0232) The minimum value occurs when x ¼ 49:1: The age at which the accident rate is a minimum is 49 years. The minimum rate is (d) Given that (h; k) = (60; 20:3); the equation has the form y = a(x ¡ 60)2 + 20:3: Since (100; 25:1) is also on the curve. 25:1 = a(100 ¡ 60)2 + 20:3 4:8 = 1600a a = 0:003 (e) f(49:1) = 60:0 ¡ 2:28(49:1) + 0:0232(49:1)2 = 60:0 ¡ 111:948 + 55:930792 ¼ 3:98: 62. h = 32t ¡ 16t2 = ¡16t2 + 32t (a) Find the vertex. x= ¡32 ¡b = =1 2a ¡32 y = ¡16(1)2 + 32(1) = 16 The vertex is (1; 16); so the maximum height is 16 ft. (b) When the object hits the ground, h = 0; so 32t ¡ 16t2 = 0 16t(2 ¡ t) = 0 t = 0 or t = 2: The two graphs are very close. 60. (a) The vertex of the quadratic function y = 0:016391x2 ¡ 1:2375x + 90:901 is at x= ¡(¡1:2375) ¡b = ¼ 38: 2a 2(0:016391) This corresponds to the year 1988. (b) When x = 38; y = 0:016391(38)2 ¡ 1:2375(38) + 90:901 ¼ 67:54: When t = 0, the object is thrown upward. When t = 2; the object hits the ground; that is, after 2 sec. 63. y = 0:056057x2 + 1:06657x (a) If x = 25 mph, y = 0:056057(25)2 + 1:06657(25) y ¼ 61:7: At 25 mph, the stopping distance is approximately 61.7 ft. Chapter 2 0:056057x2 + 1:06657x = 150 0:056057x2 + 1:06657x ¡ 150 = 0 p ¡1:06657 § (1:06657)2 ¡ 4(0:056057)(¡150) x= 2(0:056057) (b) x ¼ 43:08 or x ¼ ¡62:11 We ignore the negative value. To stop within 150 ft, the fastest speed you can drive is 43 mph. 64. Let x = the width. Then 380 ¡ 2x = the length. Area = x(380 ¡ 2x) = ¡2x2 + 380x Find the vertex: x= ¡380 ¡b = = 95 2a ¡4 y = ¡2(95)2 + 380(95) = 18,050 The graph of the area function is a parabola with vertex (95; 18,050): The maximum area of 18,050 sq ft occurs when the width is 95 ft and the length is Find the vertex: ¡190 ¡b = = 95 2a ¡2 y = ¡(95)2 + 190(95) = 9025 This is a parabola with vertex (95; 9025) that opens downward. The maximum area is the value of A at the vertex, or 9025 sq ft. 66. Draw a sketch of the arch with the vertex at the origin. Since the arch is a parabola that opens downward, the equation of the parabola is the form y = a(x ¡ h)2 + k; where the vertex (h; k) = (0; 0) and a < 0: That is, the equation is of the form y = ax2 : Since the arch is 30 meters wide at the base and 15 meters high, the points (15; ¡15) and (¡15; ¡15) are on the parabola. Use (15; ¡15) as one point on the parabola. ¡15 = a(15)2 a= ¡15 1 =¡ 152 15 So, the equation is 380 ¡ 2x = 380 ¡ 2(95) = 190 ft. y=¡ 65. Let x = the length of the lot and y = the width of the lot. The perimeter is given by P = 2x + 2y 380 = 2x + 2y 190 = x + y 190 ¡ x = y: Area = xy (quantity to be maximized) A = x(190 ¡ x) = 190x ¡ x2 = ¡x2 + 190x NONLINEAR FUNCTIONS 1 2 x : 15 Ten feet from the ground (the base) is at y = ¡5: Substitute ¡5 for y and solve for x: ¡5 = ¡ 1 2 x 15 x2 = ¡5(¡15) = 75 p p x = § 75 = §5 3 The width of the arch ten feet from the ground is then p p p 5 3 ¡ (¡5 3) = 10 3 meters ¼ 17:32 meters. Section 2.3 Polynomial and Rational Functions 67. Sketch the culvert on the xy-axes as a parabola that opens upward with vertex at (0; 0): The equation is of the form y = ax2 : Since the culvert is 18 ft wide at 12 ft from its vertex, the points (9; 12) and (¡9; 12) are on the parabola. Use (9; 12) as one point on the parabola. 12 = a(9)2 12 = 81a 12 =a 81 4 =a 27 Thus, y= 4 2 x : 27 To …nd the width 8 feet from the top, …nd the points with y-value = 12 ¡ 8 = 4: Thus, 4 2 x 27 108 = 4x2 27 = x2 x2 = 27p 27 x=§ p x = §3 3: 4= ¯ p ¯ p Thepwidth of the culvert is 3 3 + ¯¡3 3¯ = 6 3 ft ¼ 10:39 ft.