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- 1. First Degree Functions http://www.lahc.edu/math/precalculus/math_260a.html
- 2. Most mathematical functions used in the real world are “composed” with members from the following three groups of formulas. First Degree Functions
- 3. Most mathematical functions used in the real world are “composed” with members from the following three groups of formulas. * The algebraic family – these are polynomials, rational expressions and roots, etc.. First Degree Functions
- 4. Most mathematical functions used in the real world are “composed” with members from the following three groups of formulas. * The algebraic family – these are polynomials, rational expressions and roots, etc.. * The trigonometric family – these are sin(x), cos(x), .. etc that come from line measurements. First Degree Functions
- 5. Most mathematical functions used in the real world are “composed” with members from the following three groups of formulas. * The algebraic family – these are polynomials, rational expressions and roots, etc.. * The trigonometric family – these are sin(x), cos(x), .. etc that come from line measurements. * The exponential–log family – these are ex and ln(x) that come from exponential contexts. First Degree Functions
- 6. Most mathematical functions used in the real world are “composed” with members from the following three groups of formulas. * The algebraic family – these are polynomials, rational expressions and roots, etc.. * The trigonometric family – these are sin(x), cos(x), .. etc that come from line measurements. * The exponential–log family – these are ex and ln(x) that come from exponential contexts. Degree 1 or linear functions: f(x) = mx + b and degree 2 or quadratic functions: f(x) = ax2 + bx + c are especially important. First Degree Functions
- 7. Most mathematical functions used in the real world are “composed” with members from the following three groups of formulas. * The algebraic family – these are polynomials, rational expressions and roots, etc.. * The trigonometric family – these are sin(x), cos(x), .. etc that come from line measurements. * The exponential–log family – these are ex and ln(x) that come from exponential contexts. Degree 1 or linear functions: f(x) = mx + b and degree 2 or quadratic functions: f(x) = ax2 + bx + c are especially important. First Degree Functions We review below the basics of linear equations and linear functions.
- 8. The graphs of the equations Ax + By = C are straight lines. First Degree Functions
- 9. The graphs of the equations Ax + By = C are straight lines. Its easy to graph lines by graphing the x and y intercepts, First Degree Functions
- 10. a.2x – 3y = 12 The graphs of the equations Ax + By = C are straight lines. Its easy to graph lines by graphing the x and y intercepts, First Degree Functions
- 11. a.2x – 3y = 12 The graphs of the equations Ax + By = C are straight lines. Its easy to graph lines by graphing the x and y intercepts, i.e. set x = 0 to get the y-intercept, (0,-4) First Degree Functions
- 12. a.2x – 3y = 12 The graphs of the equations Ax + By = C are straight lines. Its easy to graph lines by graphing the x and y intercepts, i.e. set x = 0 to get the y-intercept, and set y = 0 for the x-intercept. (6,0) (0,-4) First Degree Functions
- 13. a.2x – 3y = 12 The graphs of the equations Ax + By = C are straight lines. Its easy to graph lines by graphing the x and y intercepts, i.e. set x = 0 to get the y-intercept, and set y = 0 for the x-intercept. (6,0) (0,-4) First Degree Functions
- 14. a.2x – 3y = 12 The graphs of the equations Ax + By = C are straight lines. Its easy to graph lines by graphing the x and y intercepts, i.e. set x = 0 to get the y-intercept, and set y = 0 for the x-intercept. If there is only one variable in the equation, we get a vertical or a horizontal line. (6,0) (0,-4) First Degree Functions
- 15. a.2x – 3y = 12 b. -3y = 12 The graphs of the equations Ax + By = C are straight lines. Its easy to graph lines by graphing the x and y intercepts, i.e. set x = 0 to get the y-intercept, and set y = 0 for the x-intercept. If there is only one variable in the equation, we get a vertical or a horizontal line. (6,0) (0,-4) First Degree Functions
- 16. a.2x – 3y = 12 b. -3y = 12 The graphs of the equations Ax + By = C are straight lines. Its easy to graph lines by graphing the x and y intercepts, i.e. set x = 0 to get the y-intercept, and set y = 0 for the x-intercept. If there is only one variable in the equation, we get a vertical or a horizontal line. (6,0) (0,-4) y = –4 First Degree Functions
- 17. a.2x – 3y = 12 b. -3y = 12 c. 2x = 12 The graphs of the equations Ax + By = C are straight lines. Its easy to graph lines by graphing the x and y intercepts, i.e. set x = 0 to get the y-intercept, and set y = 0 for the x-intercept. If there is only one variable in the equation, we get a vertical or a horizontal line. (6,0) (0,-4) y = –4 First Degree Functions
- 18. a.2x – 3y = 12 b. -3y = 12 c. 2x = 12 The graphs of the equations Ax + By = C are straight lines. Its easy to graph lines by graphing the x and y intercepts, i.e. set x = 0 to get the y-intercept, and set y = 0 for the x-intercept. If there is only one variable in the equation, we get a vertical or a horizontal line. (6,0) (0,-4) y = –4 x = 6 First Degree Functions
- 19. a.2x – 3y = 12 b. -3y = 12 c. 2x = 12 If both x and y are present, we get a tilted line. The graphs of the equations Ax + By = C are straight lines. Its easy to graph lines by graphing the x and y intercepts, i.e. set x = 0 to get the y-intercept, and set y = 0 for the x-intercept. If there is only one variable in the equation, we get a vertical or a horizontal line. (6,0) (0,-4) y = –4 x = 6 First Degree Functions
- 20. a.2x – 3y = 12 b. -3y = 12 c. 2x = 12 If both x and y are present, we get a tilted line. If the equation is y = c, we get a horizontal line. The graphs of the equations Ax + By = C are straight lines. Its easy to graph lines by graphing the x and y intercepts, i.e. set x = 0 to get the y-intercept, and set y = 0 for the x-intercept. If there is only one variable in the equation, we get a vertical or a horizontal line. (6,0) (0,-4) y = –4 x = 6 First Degree Functions
- 21. a.2x – 3y = 12 b. -3y = 12 c. 2x = 12 If both x and y are present, we get a tilted line. If the equation is y = c, we get a horizontal line. The graphs of the equations Ax + By = C are straight lines. Its easy to graph lines by graphing the x and y intercepts, i.e. set x = 0 to get the y-intercept, and set y = 0 for the x-intercept. If there is only one variable in the equation, we get a vertical or a horizontal line. (6,0) (0,-4) y = –4 x = 6 If the equation is x = c, we get a vertical line. First Degree Functions
- 22. First Degree Functions Given Ax + By = C with B = 0,
- 23. First Degree Functions Given Ax + By = C with B = 0, treating x as the input and y as the output, we may solve for y and put the equation in a function form: y = f(x) = mx + b,
- 24. First Degree Functions Given Ax + By = C with B = 0, treating x as the input and y as the output, we may solve for y and put the equation in a function form: y = f(x) = mx + b, m is called the slope and b is the y intercept,
- 25. First Degree Functions Given Ax + By = C with B = 0, treating x as the input and y as the output, we may solve for y and put the equation in a function form: y = f(x) = mx + b, m is called the slope and b is the y intercept, and the form is called the slope-intercept form.
- 26. First Degree Functions Given Ax + By = C with B = 0, treating x as the input and y as the output, we may solve for y and put the equation in a function form: y = f(x) = mx + b, m is called the slope and b is the y intercept, and the form is called the slope-intercept form. From the examples above, a. 2x – 3y = 12
- 27. First Degree Functions Given Ax + By = C with B = 0, treating x as the input and y as the output, we may solve for y and put the equation in a function form: y = f(x) = mx + b, m is called the slope and b is the y intercept, and the form is called the slope-intercept form. 2 3 2 3 From the examples above, a. 2x – 3y = 12 y = x – 4, so the slope =
- 28. First Degree Functions Given Ax + By = C with B = 0, treating x as the input and y as the output, we may solve for y and put the equation in a function form: y = f(x) = mx + b, m is called the slope and b is the y intercept, and the form is called the slope-intercept form. 2 3 2 3 From the examples above, a. 2x – 3y = 12 y = x – 4, so the slope = b. -3y = 12
- 29. First Degree Functions Given Ax + By = C with B = 0, treating x as the input and y as the output, we may solve for y and put the equation in a function form: y = f(x) = mx + b, m is called the slope and b is the y intercept, and the form is called the slope-intercept form. 2 3 2 3 From the examples above, a. 2x – 3y = 12 y = x – 4, so the slope = b. -3y = 12 y = 0x – 4, so the slope = 0
- 30. First Degree Functions Given Ax + By = C with B = 0, treating x as the input and y as the output, we may solve for y and put the equation in a function form: y = f(x) = mx + b, m is called the slope and b is the y intercept, and the form is called the slope-intercept form. 2 3 2 3 From the examples above, a. 2x – 3y = 12 y = x – 4, so the slope = b. -3y = 12 y = 0x – 4, so the slope = 0 c. 2x = 12,
- 31. First Degree Functions Given Ax + By = C with B = 0, treating x as the input and y as the output, we may solve for y and put the equation in a function form: y = f(x) = mx + b, m is called the slope and b is the y intercept, and the form is called the slope-intercept form. 2 3 2 3 From the examples above, a. 2x – 3y = 12 y = x – 4, so the slope = b. -3y = 12 y = 0x – 4, so the slope = 0 c. 2x = 12, the slope is undefined since we can't solve for y.
- 32. First Degree Functions Given Ax + By = C with B = 0, treating x as the input and y as the output, we may solve for y and put the equation in a function form: y = f(x) = mx + b, m is called the slope and b is the y intercept, and the form is called the slope-intercept form. The slope m is also the ratio of the change in the output vs. the change in the input. 2 3 2 3 From the examples above, a. 2x – 3y = 12 y = x – 4, so the slope = b. -3y = 12 y = 0x – 4, so the slope = 0 c. 2x = 12, the slope is undefined since we can't solve for y.
- 33. First Degree Functions Given Ax + By = C with B = 0, treating x as the input and y as the output, we may solve for y and put the equation in a function form: y = f(x) = mx + b, m is called the slope and b is the y intercept, and the form is called the slope-intercept form. The slope m is also the ratio of the change in the output vs. the change in the input. If two points on the line are given, the slope may be calculated via the following formula. 2 3 2 3 From the examples above, a. 2x – 3y = 12 y = x – 4, so the slope = b. -3y = 12 y = 0x – 4, so the slope = 0 c. 2x = 12, the slope is undefined since we can't solve for y.
- 34. Slope Formula: Let (x1, y1) and (x2, y2) be two points on a line, then the slope Δy Δxm = First Degree Functions
- 35. Slope Formula: Let (x1, y1) and (x2, y2) be two points on a line, then the slope Δy Δxm = First Degree Functions (The Greek letter Δ means "the difference".)
- 36. Slope Formula: Let (x1, y1) and (x2, y2) be two points on a line, then the slope Δy Δx y2 – y1 x2 – x1 m = = First Degree Functions (The Greek letter Δ means "the difference".)
- 37. Slope Formula: Let (x1, y1) and (x2, y2) be two points on a line, then the slope Δy Δx y2 – y1 x2 – x1 m = rise run = = First Degree Functions (The Greek letter Δ means "the difference".)
- 38. Slope Formula: Let (x1, y1) and (x2, y2) be two points on a line, then the slope Δy Δx y2 – y1 x2 – x1 m = rise run = = (x1, y1) (x2, y2) First Degree Functions (The Greek letter Δ means "the difference".)
- 39. Slope Formula: Let (x1, y1) and (x2, y2) be two points on a line, then the slope Δy Δx y2 – y1 x2 – x1 m = rise run = = (x1, y1) (x2, y2) Δy=y2–y1=rise First Degree Functions (The Greek letter Δ means "the difference".)
- 40. Slope Formula: Let (x1, y1) and (x2, y2) be two points on a line, then the slope Δy Δx y2 – y1 x2 – x1 m = rise run = = (x1, y1) (x2, y2) Δy=y2–y1=rise Δx=x2–x1=run First Degree Functions (The Greek letter Δ means "the difference".)
- 41. Slope Formula: Let (x1, y1) and (x2, y2) be two points on a line, then the slope Δy Δx y2 – y1 x2 – x1 m = rise run = = (x1, y1) (x2, y2) Δy=y2–y1=rise Δx=x2–x1=run The Point Slope Formula: Let y = f(x) be a first degree equation with slope m, and (x1, y1) is a point on the line, then y = f(x) = m(x – x1) + y1 First Degree Functions (The Greek letter Δ means "the difference".)
- 42. Slope Formula: Let (x1, y1) and (x2, y2) be two points on a line, then the slope Δy Δx y2 – y1 x2 – x1 m = rise run = = (x1, y1) (x2, y2) Δy=y2–y1=rise Δx=x2–x1=run The Point Slope Formula: Let y = f(x) be a first degree equation with slope m, and (x1, y1) is a point on the line, then y = f(x) = m(x – x1) + y1 First degree functions are also called linear functions because their graphs are straight lines. First Degree Functions (The Greek letter Δ means "the difference".)
- 43. Slope Formula: Let (x1, y1) and (x2, y2) be two points on a line, then the slope Δy Δx y2 – y1 x2 – x1 m = rise run = = (x1, y1) (x2, y2) Δy=y2–y1=rise Δx=x2–x1=run The Point Slope Formula: Let y = f(x) be a first degree equation with slope m, and (x1, y1) is a point on the line, then y = f(x) = m(x – x1) + y1 First degree functions are also called linear functions because their graphs are straight lines. We will use linear functions to approximate other functions just like we use line segments to approximate a curve. First Degree Functions (The Greek letter Δ means "the difference".)
- 44. Linear Equations and Lines Example A. A river floods regularly, and on a rock by the river there is a mark indicating the highest point the water level ever recorded. At 12 pm July 11, the water level is 28 inches below this mark. At 8 am July 12 the water is 18 inches below this mark.
- 45. Linear Equations and Lines Example A. A river floods regularly, and on a rock by the river there is a mark indicating the highest point the water level ever recorded. At 12 pm July 11, the water level is 28 inches below this mark. At 8 am July 12 the water is 18 inches below this mark. Let x = time, y = distance between the water level and the mark. Find the linear function y = f(x) = mx + b of the distance y in terms of time x.
- 46. Linear Equations and Lines Example A. A river floods regularly, and on a rock by the river there is a mark indicating the highest point the water level ever recorded. At 12 pm July 11, the water level is 28 inches below this mark. At 8 am July 12 the water is 18 inches below this mark. Let x = time, y = distance between the water level and the mark. Find the linear function y = f(x) = mx + b of the distance y in terms of time x. Since how the time was measured is not specified, we may select the stating time to the time of the first observation.
- 47. Linear Equations and Lines Example A. A river floods regularly, and on a rock by the river there is a mark indicating the highest point the water level ever recorded. At 12 pm July 11, the water level is 28 inches below this mark. At 8 am July 12 the water is 18 inches below this mark. Let x = time, y = distance between the water level and the mark. Find the linear function y = f(x) = mx + b of the distance y in terms of time x. Since how the time was measured is not specified, we may select the stating time to the time of the first observation. By setting x = 0 (hr) at 12 pm July 11, then x = 20 at 8 am of July 12.
- 48. Equations of Lines In particular, we are given that at x = 0 →y = 28, and at x = 20 → y = 18
- 49. Equations of Lines In particular, we are given that at x = 0 →y = 28, and at x = 20 → y = 18 and that we want the equation y = m(x – x1) + y1 of the line that contains the points (0, 28) and (20, 18).
- 50. Δy Δx 28 – 18 0 – 20 Equations of Lines In particular, we are given that at x = 0 →y = 28, and at x = 20 → y = 18 and that we want the equation y = m(x – x1) + y1 of the line that contains the points (0, 28) and (20, 18). =The slope m = = –1/2
- 51. Using the point (0, 28) and the point–slope formula, y = – ½ (x – 0) + 28 or that Δy Δx 28 – 18 0 – 20 Equations of Lines In particular, we are given that at x = 0 →y = 28, and at x = 20 → y = 18 and that we want the equation y = m(x – x1) + y1 of the line that contains the points (0, 28) and (20, 18). =The slope m = = –1/2 – xy = + 28 2
- 52. Using the point (0, 28) and the point–slope formula, y = – ½ (x – 0) + 28 or that Δy Δx 28 – 18 0 – 20 Equations of Lines In particular, we are given that at x = 0 →y = 28, and at x = 20 → y = 18 and that we want the equation y = m(x – x1) + y1 of the line that contains the points (0, 28) and (20, 18). =The slope m = = –1/2 – xy = + 28 2 The linear equation that we found is also called the trend line.
- 53. Using the point (0, 28) and the point–slope formula, y = – ½ (x – 0) + 28 or that Δy Δx 28 – 18 0 – 20 Equations of Lines In particular, we are given that at x = 0 →y = 28, and at x = 20 → y = 18 and that we want the equation y = m(x – x1) + y1 of the line that contains the points (0, 28) and (20, 18). =The slope m = = –1/2 – xy = + 28 2 The linear equation that we found is also called the trend line. So if at 4 pm July 12, i.e. when x = 28, we measured that y = 12”
- 54. Using the point (0, 28) and the point–slope formula, y = – ½ (x – 0) + 28 or that Δy Δx 28 – 18 0 – 20 Equations of Lines In particular, we are given that at x = 0 →y = 28, and at x = 20 → y = 18 and that we want the equation y = m(x – x1) + y1 of the line that contains the points (0, 28) and (20, 18). =The slope m = = –1/2 – xy = + 28 2 The linear equation that we found is also called the trend line. So if at 4 pm July 12, i.e. when x = 28, we measured that y = 12” but based on the formula prediction that y should be – 28/2 + 28 = 14”,
- 55. Using the point (0, 28) and the point–slope formula, y = – ½ (x – 0) + 28 or that Δy Δx 28 – 18 0 – 20 Equations of Lines In particular, we are given that at x = 0 →y = 28, and at x = 20 → y = 18 and that we want the equation y = m(x – x1) + y1 of the line that contains the points (0, 28) and (20, 18). =The slope m = = –1/2 – xy = + 28 2 The linear equation that we found is also called the trend line. So if at 4 pm July 12, i.e. when x = 28, we measured that y = 12” but based on the formula prediction that y should be – 28/2 + 28 = 14”, we may conclude that the flood is intensifying.
- 56. More Facts on Slopes: First Degree Functions
- 57. More Facts on Slopes: • Parallel lines have the same slope. First Degree Functions
- 58. More Facts on Slopes: • Parallel lines have the same slope. • Slopes of perpendicular lines are the negative reciprocal of each other. First Degree Functions
- 59. More Facts on Slopes: • Parallel lines have the same slope. • Slopes of perpendicular lines are the negative reciprocal of each other. Example B. Find the equation of the line L that passes through (2, –4) and is perpendicular to 4x – 3y = 5. First Degree Functions
- 60. More Facts on Slopes: • Parallel lines have the same slope. • Slopes of perpendicular lines are the negative reciprocal of each other. Example B. Find the equation of the line L that passes through (2, –4) and is perpendicular to 4x – 3y = 5. Solve for y to find the slope of 4x – 3y = 5 First Degree Functions
- 61. More Facts on Slopes: • Parallel lines have the same slope. • Slopes of perpendicular lines are the negative reciprocal of each other. Example B. Find the equation of the line L that passes through (2, –4) and is perpendicular to 4x – 3y = 5. Solve for y to find the slope of 4x – 3y = 5 4x – 5 = 3y First Degree Functions
- 62. More Facts on Slopes: • Parallel lines have the same slope. • Slopes of perpendicular lines are the negative reciprocal of each other. Example B. Find the equation of the line L that passes through (2, –4) and is perpendicular to 4x – 3y = 5. Solve for y to find the slope of 4x – 3y = 5 4x – 5 = 3y 4x/3 – 5/3 = y First Degree Functions
- 63. More Facts on Slopes: • Parallel lines have the same slope. • Slopes of perpendicular lines are the negative reciprocal of each other. Example B. Find the equation of the line L that passes through (2, –4) and is perpendicular to 4x – 3y = 5. Solve for y to find the slope of 4x – 3y = 5 4x – 5 = 3y 4x/3 – 5/3 = y Hence the slope of 4x – 2y = 5 is 4/3. First Degree Functions
- 64. More Facts on Slopes: • Parallel lines have the same slope. • Slopes of perpendicular lines are the negative reciprocal of each other. Example B. Find the equation of the line L that passes through (2, –4) and is perpendicular to 4x – 3y = 5. Solve for y to find the slope of 4x – 3y = 5 4x – 5 = 3y 4x/3 – 5/3 = y Hence the slope of 4x – 2y = 5 is 4/3. Therefore L has slope –3/4. First Degree Functions
- 65. More Facts on Slopes: • Parallel lines have the same slope. • Slopes of perpendicular lines are the negative reciprocal of each other. Example B. Find the equation of the line L that passes through (2, –4) and is perpendicular to 4x – 3y = 5. Solve for y to find the slope of 4x – 3y = 5 4x – 5 = 3y 4x/3 – 5/3 = y Hence the slope of 4x – 2y = 5 is 4/3. Therefore L has slope –3/4. So the equation of L is First Degree Functions y = (–3/4)(x – 2) + (–4) or y = –3x/4 – 5/2.
- 66. Quadratic functions or second degree functions are functions of the form y = f(x) = ax2 + bx + c, a = 0. Second Degree Functions
- 67. Quadratic functions or second degree functions are functions of the form y = f(x) = ax2 + bx + c, a = 0. Second Degree Functions Below is the graph of y = f(x) = -x2.
- 68. x y -4 -3 -2 -1 0 1 2 3 4 Quadratic functions or second degree functions are functions of the form y = f(x) = ax2 + bx + c, a = 0. Second Degree Functions Below is the graph of y = f(x) = -x2.
- 69. x y -4 -3 -2 -1 0 0 1 -1 2 -4 3 -9 4 -16 Quadratic functions or second degree functions are functions of the form y = f(x) = ax2 + bx + c, a = 0. Second Degree Functions Below is the graph of y = f(x) = -x2.
- 70. x y -4 -16 -3 -9 -2 -4 -1 -1 0 0 1 -1 2 -4 3 -9 4 -16 Quadratic functions or second degree functions are functions of the form y = f(x) = ax2 + bx + c, a = 0. Second Degree Functions Below is the graph of y = f(x) = -x2.
- 71. x y -4 -16 -3 -9 -2 -4 -1 -1 0 0 1 -1 2 -4 3 -9 4 -16 Quadratic functions or second degree functions are functions of the form y = f(x) = ax2 + bx + c, a = 0. Second Degree Functions Below is the graph of y = f(x) = -x2.
- 72. The graphs of 2nd (quadratic) equations are called parabolas. Second Degree Functions
- 73. The graphs of 2nd (quadratic) equations are called parabolas. Parabolas describe the paths of thrown objects (or the upside- down paths). Second Degree Functions
- 74. The graphs of 2nd (quadratic) equations are called parabolas. Parabolas describe the paths of thrown objects (or the upside- down paths). Second Degree Functions
- 75. Second Degree Functions Properties of Parabolas:
- 76. Second Degree Functions Properties of Parabolas: I. A parabola is symmetric to a center line.
- 77. Second Degree Functions Properties of Parabolas: I. A parabola is symmetric to a center line. II. The highest or lowest point of the parabola sits on the center line.
- 78. Second Degree Functions Properties of Parabolas: I. A parabola is symmetric to a center line. II. The highest or lowest point of the parabola sits on the center line. This point is called the vertex.
- 79. Second Degree Functions Properties of Parabolas: I. A parabola is symmetric to a center line. II. The highest or lowest point of the parabola sits on the center line. This point is called the vertex. III. If the positions of the vertex and another point on the parabola are known, the parabola is determined.
- 80. Second Degree Functions The vertex position is given by the following formula. Properties of Parabolas: I. A parabola is symmetric to a center line. II. The highest or lowest point of the parabola sits on the center line. This point is called the vertex. III. If the positions of the vertex and another point on the parabola are known, the parabola is determined.
- 81. Vertex Formula: The vertex of the parabola y = ax2 + bx + c is at x = -b 2a Second Degree Functions
- 82. Vertex Formula: The vertex of the parabola y = ax2 + bx + c is at x = -b 2a Second Degree Functions Moreover, if a > 0, the parabola opens up, if a < 0 the parabola opens down.
- 83. Vertex Formula: The vertex of the parabola y = ax2 + bx + c is at x = -b 2a To graph a parabola y = ax2 + bx + c: Second Degree Functions Moreover, if a > 0, the parabola opens up, if a < 0 the parabola opens down.
- 84. Vertex Formula: The vertex of the parabola y = ax2 + bx + c is at x = -b 2a To graph a parabola y = ax2 + bx + c: 1. Set x = in the equation to find the vertex.-b 2a Second Degree Functions Moreover, if a > 0, the parabola opens up, if a < 0 the parabola opens down.
- 85. Vertex Formula: The vertex of the parabola y = ax2 + bx + c is at x = -b 2a To graph a parabola y = ax2 + bx + c: 1. Set x = in the equation to find the vertex. 2. Find another point. -b 2a Second Degree Functions Moreover, if a > 0, the parabola opens up, if a < 0 the parabola opens down.
- 86. Vertex Formula: The vertex of the parabola y = ax2 + bx + c is at x = -b 2a To graph a parabola y = ax2 + bx + c: 1. Set x = in the equation to find the vertex. 2. Find another point. (Use x=0 to get the y intercept if its not the vertex.) -b 2a Second Degree Functions Moreover, if a > 0, the parabola opens up, if a < 0 the parabola opens down.
- 87. Vertex Formula: The vertex of the parabola y = ax2 + bx + c is at x = -b 2a To graph a parabola y = ax2 + bx + c: 1. Set x = in the equation to find the vertex. 2. Find another point. (Use x=0 to get the y intercept if its not the vertex.) 3. Locate the reflection across the center line, these three points form the tip of the parabola. -b 2a Second Degree Functions Moreover, if a > 0, the parabola opens up, if a < 0 the parabola opens down.
- 88. Vertex Formula: The vertex of the parabola y = ax2 + bx + c is at x = -b 2a To graph a parabola y = ax2 + bx + c: 1. Set x = in the equation to find the vertex. 2. Find another point. (Use x=0 to get the y intercept if its not the vertex.) 3. Locate the reflection across the center line, these three points form the tip of the parabola. Trace the parabola. -b 2a Second Degree Functions Moreover, if a > 0, the parabola opens up, if a < 0 the parabola opens down.
- 89. Vertex Formula: The vertex of the parabola y = ax2 + bx + c is at x = -b 2a To graph a parabola y = ax2 + bx + c: 1. Set x = in the equation to find the vertex. 2. Find another point. (Use x=0 to get the y intercept if its not the vertex.) 3. Locate the reflection across the center line, these three points form the tip of the parabola. Trace the parabola. 4. Set y=0 to find the x intercept. -b 2a Second Degree Functions Moreover, if a > 0, the parabola opens up, if a < 0 the parabola opens down.
- 90. Vertex Formula: The vertex of the parabola y = ax2 + bx + c is at x = -b 2a To graph a parabola y = ax2 + bx + c: 1. Set x = in the equation to find the vertex. 2. Find another point. (Use x=0 to get the y intercept if its not the vertex.) 3. Locate the reflection across the center line, these three points form the tip of the parabola. Trace the parabola. 4. Set y=0 to find the x intercept. If no real solution exists, there is no x intercept. -b 2a Second Degree Functions Moreover, if a > 0, the parabola opens up, if a < 0 the parabola opens down.
- 91. Example C. Graph y = x2 – 4x – 12 Second Degree Functions
- 92. Example C. Graph y = x2 – 4x – 12 Vertex: set x = -(-4) 2(1) Second Degree Functions
- 93. Example C. Graph y = x2 – 4x – 12 Vertex: set x = = 2 -(-4) 2(1) Second Degree Functions
- 94. Example C. Graph y = x2 – 4x – 12 Vertex: set x = = 2 then y=22 – 4*2 – 12 -(-4) 2(1) Second Degree Functions
- 95. Example C. Graph y = x2 – 4x – 12 Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16, -(-4) 2(1) so v=(2, -16). Second Degree Functions
- 96. Example C. Graph y = x2 – 4x – 12 Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16, -(-4) 2(1) so v=(2, -16). Another point: Set x = 0 Second Degree Functions
- 97. Example C. Graph y = x2 – 4x – 12 Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16, -(-4) 2(1) so v=(2, -16). Another point: Set x = 0 then y = -12 or (0, -12) Second Degree Functions
- 98. Example C. Graph y = x2 – 4x – 12 Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16, -(-4) 2(1) so v=(2, -16). Another point: Set x = 0 then y = -12 or (0, -12) Second Degree Functions (2, -16)
- 99. Example C. Graph y = x2 – 4x – 12 Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16, -(-4) 2(1) so v=(2, -16). Another point: Set x = 0 then y = -12 or (0, -12) Second Degree Functions (2, -16) (0, -12)
- 100. Example C. Graph y = x2 – 4x – 12 Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16, -(-4) 2(1) so v=(2, -16). (2, -16) Second Degree Functions Another point: Set x = 0 then y = -12 or (0, -12) It's reflection across the mid-line is (4, -12) (0, -12)
- 101. Example C. Graph y = x2 – 4x – 12 Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16, -(-4) 2(1) so v=(2, -16). (2, -16) (0, -12) (4, -12) Second Degree Functions Another point: Set x = 0 then y = -12 or (0, -12) It's reflection across the mid-line is (4, -12)
- 102. Example C. Graph y = x2 – 4x – 12 Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16, -(-4) 2(1) so v=(2, -16). (2, -16) (0, -12) (4, -12) Second Degree Functions Another point: Set x = 0 then y = -12 or (0, -12) It's reflection across the mid-line is (4, -12)
- 103. Example C. Graph y = x2 – 4x – 12 Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16, -(-4) 2(1) so v=(2, -16). (2, -16) (0, -12) (4, -12) Second Degree Functions Another point: Set x = 0 then y = -12 or (0, -12) It's reflection across the mid-line is (4, -12) Set y = 0 and get x-int: x2 – 4x – 12 = 0
- 104. Example C. Graph y = x2 – 4x – 12 Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16, -(-4) 2(1) so v=(2, -16). (2, -16) (0, -12) (4, -12) Second Degree Functions Another point: Set x = 0 then y = -12 or (0, -12) It's reflection across the mid-line is (4, -12) Set y = 0 and get x-int: x2 – 4x – 12 = 0 (x + 2)(x – 6) = 0 x = -2, x = 6
- 105. Example C. Graph y = x2 – 4x – 12 Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16, -(-4) 2(1) so v=(2, -16). (2, -16) (0, -12) (4, -12) Second Degree Functions Following is an example of maximization via the vertex. Another point: Set x = 0 then y = -12 or (0, -12) It's reflection across the mid-line is (4, -12) Set y = 0 and get x-int: x2 – 4x – 12 = 0 (x + 2)(x – 6) = 0 x = -2, x = 6
- 106. Example D. The "Crazy Chicken" can sell 120 whole roasted chicken for $8 in one day. For every $0.50 increases in the price, 4 less chicken are sold. Find the price that will maximize the revenue (the sale). Second Degree Functions
- 107. Example D. The "Crazy Chicken" can sell 120 whole roasted chicken for $8 in one day. For every $0.50 increases in the price, 4 less chicken are sold. Find the price that will maximize the revenue (the sale). Second Degree Functions Set x to be the number of $0.50 increments above the base price of $8.00.
- 108. Example D. The "Crazy Chicken" can sell 120 whole roasted chicken for $8 in one day. For every $0.50 increases in the price, 4 less chicken are sold. Find the price that will maximize the revenue (the sale). Second Degree Functions Set x to be the number of $0.50 increments above the base price of $8.00. So the price is (8 + 0.5x) or (8 + x/2).
- 109. Example D. The "Crazy Chicken" can sell 120 whole roasted chicken for $8 in one day. For every $0.50 increases in the price, 4 less chicken are sold. Find the price that will maximize the revenue (the sale). Second Degree Functions Set x to be the number of $0.50 increments above the base price of $8.00. So the price is (8 + 0.5x) or (8 + x/2). The number of chicken sold at this price is (120 – 4x).
- 110. Example D. The "Crazy Chicken" can sell 120 whole roasted chicken for $8 in one day. For every $0.50 increases in the price, 4 less chicken are sold. Find the price that will maximize the revenue (the sale). Second Degree Functions Set x to be the number of $0.50 increments above the base price of $8.00. So the price is (8 + 0.5x) or (8 + x/2). The number of chicken sold at this price is (120 – 4x). Hence the revenue, depending on the number of times of $0.50 price hikes x, is R(x) = (8 + x/2)(120 – 4x)
- 111. Example D. The "Crazy Chicken" can sell 120 whole roasted chicken for $8 in one day. For every $0.50 increases in the price, 4 less chicken are sold. Find the price that will maximize the revenue (the sale). Second Degree Functions Set x to be the number of $0.50 increments above the base price of $8.00. So the price is (8 + 0.5x) or (8 + x/2). The number of chicken sold at this price is (120 – 4x). Hence the revenue, depending on the number of times of $0.50 price hikes x, is R(x) = (8 + x/2)(120 – 4x) = 960 + 28x – 2x2.
- 112. Example D. The "Crazy Chicken" can sell 120 whole roasted chicken for $8 in one day. For every $0.50 increases in the price, 4 less chicken are sold. Find the price that will maximize the revenue (the sale). Second Degree Functions Set x to be the number of $0.50 increments above the base price of $8.00. So the price is (8 + 0.5x) or (8 + x/2). The number of chicken sold at this price is (120 – 4x). Hence the revenue, depending on the number of times of $0.50 price hikes x, is R(x) = (8 + x/2)(120 – 4x) = 960 + 28x – 2x2. This a 2nd degree equation whose graph is a parabola that opens downward.
- 113. Example D. The "Crazy Chicken" can sell 120 whole roasted chicken for $8 in one day. For every $0.50 increases in the price, 4 less chicken are sold. Find the price that will maximize the revenue (the sale). Second Degree Functions Set x to be the number of $0.50 increments above the base price of $8.00. So the price is (8 + 0.5x) or (8 + x/2). The number of chicken sold at this price is (120 – 4x). Hence the revenue, depending on the number of times of $0.50 price hikes x, is R(x) = (8 + x/2)(120 – 4x) = 960 + 28x – 2x2. This a 2nd degree equation whose graph is a parabola that opens downward. The vertex is the highest point on the graph where R is the largest.
- 114. Second Degree Functions x R R = 960 + 28x – 2x2.
- 115. Second Degree Functions x RThe vertex of this parabola is at x = = 7.-(28) 2(-2) R = 960 + 28x – 2x2.
- 116. Second Degree Functions x RThe vertex of this parabola is at x = = 7. So R(7) = $1058 gives the maximum revenue. -(28) 2(-2) R = 960 + 28x – 2x2. v = (7, 1058)
- 117. Second Degree Functions x RThe vertex of this parabola is at x = = 7. So R(7) = $1058 gives the maximum revenue. -(28) 2(-2) R = 960 + 28x – 2x2. v = (7, 1058) More precisely, raise the price x = 7 times to 8 + 7(0.50) = 11.50 per chicken,chicken,
- 118. Second Degree Functions x RThe vertex of this parabola is at x = = 7. So R(7) = $1058 gives the maximum revenue. -(28) 2(-2) R = 960 + 28x – 2x2. v = (7, 1058) More precisely, raise the price x = 7 times to 8 + 7(0.50) = 11.50 per chicken,chicken, and we can sell 120 – 4(7) = 92 chickens per day with the revenue of 11.5(92) = 1058.
- 119. Second Degree Functions x RThe vertex of this parabola is at x = = 7. So R(7) = $1058 gives the maximum revenue. -(28) 2(-2) R = 960 + 28x – 2x2. v = (7, 1058) More precisely, raise the price x = 7 times to 8 + 7(0.50) = 11.50 per chicken,chicken, and we can sell 120 – 4(7) = 92 chickens per day with the revenue of 11.5(92) = 1058. Any inverse square law in science is 2nd degree. Laws related to distance or area in mathematics are 2nd degree. That’s why 2nd equations are important.

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