Notes solving polynomial equations

14,650 views

Published on

Published in: Education
2 Comments
20 Likes
Statistics
Notes
No Downloads
Views
Total views
14,650
On SlideShare
0
From Embeds
0
Number of Embeds
9,079
Actions
Shares
0
Downloads
0
Comments
2
Likes
20
Embeds 0
No embeds

No notes for slide
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • \n
  • Notes solving polynomial equations

    1. 1. SolvingPolynomial Equations๏ Solve by Factoring๏ Identify solutions on Graph๏ Solve by Graphing
    2. 2. Solving Polynomial Equations★ Solving a polynomial equation is the same as solving a quadratic equation, except that the quadratic might be replaced by a different kind of polynomial (such as a cubic or a quartic).  
    3. 3. Solving Polynomial Equations★ Solving a polynomial equation is the same as solving a quadratic equation, except that the quadratic might be replaced by a different kind of polynomial (such as a cubic or a quartic).  ★ There are 3 ways to solve Polynomial Equations
    4. 4. Solving Polynomial Equations★ Solving a polynomial equation is the same as solving a quadratic equation, except that the quadratic might be replaced by a different kind of polynomial (such as a cubic or a quartic).  ★ There are 3 ways to solve Polynomial Equations (1) Using factoring and the zero product property
    5. 5. Solving Polynomial Equations★ Solving a polynomial equation is the same as solving a quadratic equation, except that the quadratic might be replaced by a different kind of polynomial (such as a cubic or a quartic).  ★ There are 3 ways to solve Polynomial Equations (1) Using factoring and the zero product property (2) Using the graphing calculator to graph
    6. 6. Solving Polynomial Equations★ Solving a polynomial equation is the same as solving a quadratic equation, except that the quadratic might be replaced by a different kind of polynomial (such as a cubic or a quartic).  ★ There are 3 ways to solve Polynomial Equations (1) Using factoring and the zero product property (2) Using the graphing calculator to graph (3) Using Synthetic Division (separate notes)
    7. 7. Total Number of Solutions★ The maximum number of solutions that a polynomial equation can have is equal to the degree of the polynomial.★ It is possible for a polynomial equation to have fewer solutions (or none at all).★ The degree of the polynomial gives you the maximum number of solutions that are theoretically possible.★ Some solutions may be irrational or even imaginary.
    8. 8. Total Number of Solutions★ The maximum number of solutions that a polynomial equation can have is equal to the degree of the polynomial. ( ) 3 f x = x + x − x +12★ It is possible for a polynomial equation to have fewer solutions (or none at all).★ The degree of the polynomial gives you the maximum number of solutions that are theoretically possible.★ Some solutions may be irrational or even imaginary.
    9. 9. Total Number of Solutions★ The maximum number of solutions that a polynomial equation can have is equal to the degree of the polynomial. ( ) 3 f x = x + x − x +12 Degree is 3. Therefore 3 or fewer solutions possible.
    10. 10. Total Number of Solutions★ The maximum number of solutions that a polynomial equation can have is equal to the degree of the polynomial. ( ) 3 f x = x + x − x +12 Degree is 3. Therefore 3 or fewer solutions possible.★ It is possible for a polynomial equation to have fewer solutions (or none at all).
    11. 11. Total Number of Solutions★ The maximum number of solutions that a polynomial equation can have is equal to the degree of the polynomial. ( ) 3 f x = x + x − x +12 Degree is 3. Therefore 3 or fewer solutions possible.★ It is possible for a polynomial equation to have fewer solutions (or none at all).★ The degree of the polynomial gives you the maximum number of solutions that are theoretically possible.
    12. 12. Total Number of Solutions★ The maximum number of solutions that a polynomial equation can have is equal to the degree of the polynomial. ( ) 3 f x = x + x − x +12 Degree is 3. Therefore 3 or fewer solutions possible.★ It is possible for a polynomial equation to have fewer solutions (or none at all).★ The degree of the polynomial gives you the maximum number of solutions that are theoretically possible.★ Some solutions may be irrational or even imaginary.
    13. 13. Solving by Factoring★ When using factoring, remember to always look for a GCF (Greatest Common Factor) first. This will help make the remaining factoring easier.★ Factor completely!★ Set each factor equal to 0 and solve.
    14. 14. Solving by Factoring★ When using factoring, remember to always look for a GCF (Greatest Common Factor) first. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0★ Factor completely!★ Set each factor equal to 0 and solve.
    15. 15. Solving by Factoring★ When using factoring, remember to always look for a GCF (Greatest Common Factor) first. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0★ Factor completely!
    16. 16. Solving by Factoring★ When using factoring, remember to always look for a GCF (Greatest Common Factor) first. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0★ Factor completely! Keep going here.
    17. 17. Solving by Factoring★ When using factoring, remember to always look for a GCF (Greatest Common Factor) first. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0★ Factor completely! Keep going here. 3x ( x − 2 ) ( x + 2 ) = 0
    18. 18. Solving by Factoring★ When using factoring, remember to always look for a GCF (Greatest Common Factor) first. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0★ Factor completely! Keep going here. 3x ( x − 2 ) ( x + 2 ) = 0★ Set each factor equal to 0 and solve.
    19. 19. Solving by Factoring★ When using factoring, remember to always look for a GCF (Greatest Common Factor) first. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0★ Factor completely! Keep going here. 3x ( x − 2 ) ( x + 2 ) = 0★ Set each factor equal to 0 and solve. 3x = 0 ( x − 2) = 0 ( x + 2) = 0
    20. 20. Solving by Factoring★ When using factoring, remember to always look for a GCF (Greatest Common Factor) first. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0★ Factor completely! Keep going here. 3x ( x − 2 ) ( x + 2 ) = 0★ Set each factor equal to 0 and solve. 3x = 0 ( x − 2) = 0 ( x + 2) = 0 x=0 x=2 x = −2
    21. 21. Example: Solve by Factoring 4 2 x − 6x = 27
    22. 22. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up to 4 solutions.
    23. 23. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0
    24. 24. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 )
    25. 25. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2x −9=0
    26. 26. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2x −9=0 x +3= 0
    27. 27. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2x −9=0 x +3= 0 2 x =9
    28. 28. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2x −9=0 x +3= 0 2 x =9 x=± 9
    29. 29. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2x −9=0 x +3= 0 2 x =9 x=± 9 x = ±3
    30. 30. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2x −9=0 x +3= 0 2 2 x =9 x = −3 x=± 9 x = ±3
    31. 31. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2x −9=0 x +3= 0 2 2 x =9 x = −3 x=± 9 x = ± −3 x = ±3
    32. 32. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2x −9=0 x +3= 0 2 2 x =9 x = −3 x=± 9 x = ± −3 x = ±3 x = ±i 3
    33. 33. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) Solutions: 2x −9=0 2 x +3= 0 {±3, ±i 3} 2 2 x =9 x = −3 x=± 9 x = ± −3 x = ±3 x = ±i 3
    34. 34. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0
    35. 35. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions.
    36. 36. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2
    37. 37. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2
    38. 38. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 ( ) x − 1 ( x + 3) = 0 2
    39. 39. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 ( ) x − 1 ( x + 3) = 0 2 2 x −1= 0
    40. 40. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 ( ) x − 1 ( x + 3) = 0 2 2 x −1= 0 x+3= 0
    41. 41. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 ( ) x − 1 ( x + 3) = 0 2 2 x −1= 0 x+3= 0 x = −3
    42. 42. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 ( ) x − 1 ( x + 3) = 0 2 2 x −1= 0 x+3= 0 2 x =1 x = −3
    43. 43. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 ( ) x − 1 ( x + 3) = 0 2 2 x −1= 0 x+3= 0 2 x =1 x = −3 x = ±1
    44. 44. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 Solutions: ( ) x − 1 ( x + 3) = 0 2 {1, −1, −3} 2 x −1= 0 x+3= 0 2 x =1 x = −3 x = ±1
    45. 45. Try this: Solve by Factoring 3 2 x + x − 4x = 0
    46. 46. Try this: Solve by Factoring 3 2 x + x − 4x = 0 Degree is 3 so can up to 3 solutions.
    47. 47. Try this: Solve by Factoring 3 2 x + x − 4x = 0 Degree is 3 so can up to 3 solutions. ( 2 x x +x−4 =0 )
    48. 48. Try this: Solve by Factoring 3 2 x + x − 4x = 0 Degree is 3 so can up to 3 solutions. ( x x +x−4 =02 )x=0
    49. 49. Try this: Solve by Factoring 3 x + x − 4x = 0 2 Degree is 3 so can up to 3 solutions. ( x x +x−4 =02 ) 2x=0 x +x−4=0
    50. 50. Try this: Solve by Factoring 3 x + x − 4x = 0 2 Degree is 3 so can up to 3 solutions. ( x x +x−4 =02 ) 2 Not factorable so usex=0 x +x−4=0 the quadratic formula
    51. 51. Try this: Solve by Factoring 3 x + x − 4x = 0 2 Degree is 3 so can up to 3 solutions. ( x x +x−4 =02 ) 2 Not factorable so usex=0 x +x−4=0 the quadratic formula a = 1; b = 1; c = −4 −1 ± 12 − 4 (1) ( −4 ) x= 2 (1)
    52. 52. Try this: Solve by Factoring 3 x + x − 4x = 0 2 Degree is 3 so can up to 3 solutions. ( x x +x−4 =02 ) 2 Not factorable so usex=0 x +x−4=0 the quadratic formula a = 1; b = 1; c = −4 −1 ± 12 − 4 (1) ( −4 ) x= 2 (1) −1 ± 17 x= 2
    53. 53. Try this: Solve by Factoring 3 x + x − 4x = 0 2 Degree is 3 so can up to 3 solutions. ( x x +x−4 =02 ) 2 Not factorable so usex=0 x +x−4=0 the quadratic formula a = 1; b = 1; c = −4 −1 ± 12 − 4 (1) ( −4 ) x= 2 (1) Solutions: −1 ± 17  −1 ± 17    x= 0,  2   2  
    54. 54. Try this: Solve by Factoring f ( x ) = x + 64 3
    55. 55. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions.
    56. 56. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + (4) 3
    57. 57. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + (4) 3 Sum of cubes. Apply the formula.
    58. 58. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 30 = ( x + 4 ) ( x − 4x + 16 ) 2
    59. 59. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 30 = ( x + 4 ) ( x − 4x + 16 ) 2x+4=0
    60. 60. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 30 = ( x + 4 ) ( x − 4x + 16 ) 2 2x+4=0 x − 4x + 16 = 0
    61. 61. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 2 x+4=0 x − 4x + 16 = 0−4 = x
    62. 62. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 Not factorable so 2 x+4=0 x − 4x + 16 = 0 use completing the square.−4 = x
    63. 63. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 Not factorable so 2 x+4=0 x − 4x + 16 = 0 use completing the square. 2−4 = x x − 4x + ( −2 ) = −16 + 4 2
    64. 64. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 Not factorable so 2 x+4=0 x − 4x + 16 = 0 use completing the square. 2−4 = x x − 4x + ( −2 ) = −16 + 4 2 2 ( x − 2 ) = −12
    65. 65. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 Not factorable so 2 x+4=0 x − 4x + 16 = 0 use completing the square. 2−4 = x x − 4x + ( −2 ) = −16 + 4 2 2 ( x − 2 ) = −12 x − 2 = ± −12
    66. 66. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 Not factorable so 2 x+4=0 x − 4x + 16 = 0 use completing the square. 2−4 = x x − 4x + ( −2 ) = −16 + 4 2 2 ( x − 2 ) = −12 x − 2 = ± −12 x = 2 ± 2i 3
    67. 67. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 Not factorable so 2 x+4=0 x − 4x + 16 = 0 use completing the square. 2−4 = x x − 4x + ( −2 ) = −16 + 4 2 2 ( x − 2 ) = −12 Solutions: x − 2 = ± −12 {−4, 2 ± 2i 3} x = 2 ± 2i 3
    68. 68. Practice Time!★ Follow this link to practice solving polynomial equations using Factoring.
    69. 69. Solutions by Observing the Graph★ The degree of the function tells you the maximum number of solutions possible.
    70. 70. Solutions by Observing the Graph★ The degree of the function tells you the maximum number of solutions possible.★ The real solutions are where the function crosses or touches the x-axis.
    71. 71. Solutions by Observing the Graph★ The degree of the function tells you the maximum number of solutions possible.★ The real solutions are where the function crosses or touches the x-axis.★ The graph below has 4 solutions because it crosses the x-axis in 4 places. Notice 2 are positive real numbers and 2 are negative real numbers.
    72. 72. Example: Find all real zeros on the graph.
    73. 73. Example: Find all real zeros on the graph.
    74. 74. Example: Find all real zeros on the graph.
    75. 75. Example: Find all real zeros on the graph.
    76. 76. Example: Find all real zeros on the graph.
    77. 77. Example: Find all real zeros on the graph.★ The real zeros for the graph below are {−3, −1,1, 2}
    78. 78. You try: Find all real zeros on the graph.
    79. 79. You try: Find all real zeros on the graph.
    80. 80. You try: Find all real zeros on the graph.
    81. 81. You try: Find all real zeros on the graph.
    82. 82. You try: Find all real zeros on the graph.★ The real zeros for the graph below are {−1, 2, 5}
    83. 83. Practice Time!★ Follow this link to practice solving polynomial equations using Factoring.
    84. 84. Solving by Graphing in Calculator★ Graph the left side of the equation in Y1.
    85. 85. Solving by Graphing in Calculator★ Graph the left side of the equation in Y1.★ Graph the right side of the equation in Y2.
    86. 86. Solving by Graphing in Calculator★ Graph the left side of the equation in Y1.★ Graph the right side of the equation in Y2.★ Find all the points the two graphs intersect. The x- coordinate is the solution.
    87. 87. Solving by Graphing in Calculator★ Graph the left side of the equation in Y1.★ Graph the right side of the equation in Y2.★ Find all the points the two graphs intersect. The x- coordinate is the solution.★ If you are given a function such as f(x) = x2 - 1, use zero for f(x). So Y1 = 0 and Y2 = x2 - 1. The find all the intersections.
    88. 88. Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1)
    89. 89. Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1)Y1 = 0 2Y 2 = 0.25 ( x + 2 ) ( x − 1)
    90. 90. Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1)Y1 = 0 2Y 2 = 0.25 ( x + 2 ) ( x − 1)
    91. 91. Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1)Y1 = 0 2Y 2 = 0.25 ( x + 2 ) ( x − 1)
    92. 92. Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1)Y1 = 0 2Y 2 = 0.25 ( x + 2 ) ( x − 1)
    93. 93. Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1)Y1 = 0 2Y 2 = 0.25 ( x + 2 ) ( x − 1) Solutions: {−2,1}
    94. 94. Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1)Y1 = 0 2Y 2 = 0.25 ( x + 2 ) ( x − 1) Solutions: {−2,1} Notice if you used the zero product property, x = 1 would have occurred twice. We say 1 has multiplicity of 2.
    95. 95. Example: Solve by Graphing 3 2 4x − 8x = x − 2
    96. 96. Example: Solve by Graphing 3 2 4x − 8x = x − 2 3 2Y1 = 4x − 8xY2 = x − 2
    97. 97. Example: Solve by Graphing 3 2 4x − 8x = x − 2 3 2Y1 = 4x − 8xY2 = x − 2
    98. 98. Example: Solve by Graphing 3 2 4x − 8x = x − 2 3 2Y1 = 4x − 8xY2 = x − 2
    99. 99. Example: Solve by Graphing 3 2 4x − 8x = x − 2 3 2Y1 = 4x − 8xY2 = x − 2
    100. 100. Example: Solve by Graphing 3 2 4x − 8x = x − 2 3 2Y1 = 4x − 8xY2 = x − 2
    101. 101. Example: Solve by Graphing 3 2 4x − 8x = x − 2 3 2Y1 = 4x − 8xY2 = x − 2 Solutions:  1 1  − , , 2   2 2 
    102. 102. The end.

    ×