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22 the graphs of quadratic equations
 

22 the graphs of quadratic equations

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    22 the graphs of quadratic equations 22 the graphs of quadratic equations Presentation Transcript

    • Graphs of Quadratic EquationsFrank Ma © 2011
    • Graphs of Quadratic EquationsWe start with an example of a graph gives the generalshape of the graphs of 2nd (quadratic) degree equations.
    • Graphs of Quadratic EquationsWe start with an example of a graph gives the generalshape of the graphs of 2nd (quadratic) degree equations.Example A. Graph y = –x2
    • Graphs of Quadratic EquationsWe start with an example of a graph gives the generalshape of the graphs of 2nd (quadratic) degree equations.Make a tableExample A. Graph y = –x2
    • x-4-3-2-101234Graphs of Quadratic EquationsWe start with an example of a graph gives the generalshape of the graphs of 2nd (quadratic) degree equations.Make a tableExample A. Graph y = –x2y
    • x-4-3-2-101234Graphs of Quadratic EquationsWe start with an example of a graph gives the generalshape of the graphs of 2nd (quadratic) degree equations.Make a tableExample A. Graph y = –x2y-16-9-4-10-1-4-9-16
    • Graphs of Quadratic EquationsWe start with an example of a graph gives the generalshape of the graphs of 2nd (quadratic) degree equations.Make a tableExample A. Graph y = –x2x-4-3-2-101234y-16-9-4-10-1-4-9-16
    • The graphs of 2nd (quadratic) equations are called parabolas.Parabolas describe the paths of thrown objects (or the upside-down paths).Graphs of Quadratic Equations
    • The graphs of 2nd (quadratic) equations are called parabolas.Parabolas describe the paths of thrown objects (or the upside-down paths).Graphs of Quadratic EquationsProperties of Parabolas:• Parabolas are symmetric with respect to a center line
    • The graphs of 2nd (quadratic) equations are called parabolas.Parabolas describe the paths of thrown objects (or the upside-down paths).Graphs of Quadratic EquationsProperties of Parabolas:• Parabolas are symmetric with respect to a center line• The highest or lowest point of the parabola sits on thecenter line.
    • The graphs of 2nd (quadratic) equations are called parabolas.Parabolas describe the paths of thrown objects (or the upside-down paths).Graphs of Quadratic EquationsProperties of Parabolas:• Parabolas are symmetric with respect to a center line• The highest or lowest point of the parabola sits on thecenter line. This point is called the vertex.
    • The graphs of 2nd (quadratic) equations are called parabolas.Parabolas describe the paths of thrown objects (or the upside-down paths).Graphs of Quadratic EquationsProperties of Parabolas:• Parabolas are symmetric with respect to a center line• The highest or lowest point of the parabola sits on thecenter line. This point is called the vertex.Vertex Formula: The vertex of y = ax2 + bx + c is at x = .-b2a
    • Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around thevertex so the points will be plotted symmetrically.
    • Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around thevertex so the points will be plotted symmetrically.Example B. Graph y = x2 – 4x – 12
    • Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around thevertex so the points will be plotted symmetrically.Example B. Graph y = x2 – 4x – 12Vertex: set x = = 2–(–4)2(1)
    • Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around thevertex so the points will be plotted symmetrically.Example B. Graph y = x2 – 4x – 12Vertex: set x = = 2–(–4)2(1)Make a table centered at x = 2.x y01234
    • Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around thevertex so the points will be plotted symmetrically.Example B. Graph y = x2 – 4x – 12Vertex: set x = = 2–(–4)2(1)Make a table centered at x = 2.x y01234–12–15–16–15–12
    • Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around thevertex so the points will be plotted symmetrically.Example B. Graph y = x2 – 4x – 12Vertex: set x = = 2–(–4)2(1)Make a table centered at x = 2.x y01234–12–15–16–15–12
    • Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around thevertex so the points will be plotted symmetrically.Example B. Graph y = x2 – 4x – 12Vertex: set x = = 2–(–4)2(1)Make a table centered at x = 2.x y01234Note the y values are symmetricaround the vertex just as thepoints.–12–15–16–15–12
    • Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around thevertex so the points will be plotted symmetrically.Example B. Graph y = x2 – 4x – 12Vertex: set x = = 2–(–4)2(1)Make a table centered at x = 2.x y01234Note the y values are symmetricaround the vertex just as thepoints. (If they are not, checkyour calculation.)–12–15–16–15–12
    • Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around thevertex so the points will be plotted symmetrically.Example B. Graph y = x2 – 4x – 12Vertex: set x = = 2–(–4)2(1)Make a table centered at x = 2.x y01234Note the y values are symmetricaround the vertex just as thepoints. (If they are not, checkyour calculation.)(2, -16)–12–15–16–15–12
    • Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around thevertex so the points will be plotted symmetrically.Example B. Graph y = x2 – 4x – 12Vertex: set x = = 2–(–4)2(1)Make a table centered at x = 2.x y01234Note the y values are symmetricaround the vertex just as thepoints. (If they are not, checkyour calculation.)(2, -16)(3, -15)–12–15–16–15–12
    • Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around thevertex so the points will be plotted symmetrically.Example B. Graph y = x2 – 4x – 12Vertex: set x = = 2–(–4)2(1)Make a table centered at x = 2.x y01234Note the y values are symmetricaround the vertex just as thepoints. (If they are not, checkyour calculation.)(2, -16)(4, -12)(3, -15)–12–15–16–15–12
    • Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around thevertex so the points will be plotted symmetrically.Example B. Graph y = x2 – 4x – 12Vertex: set x = = 2–(–4)2(1)Make a table centered at x = 2.x y01234Note the y values are symmetricaround the vertex just as thepoints. (If they are not, checkyour calculation.)(2, -16)(4, -12)(3, -15)(1, -15)–12–15–16–15–12
    • Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around thevertex so the points will be plotted symmetrically.Example B. Graph y = x2 – 4x – 12Vertex: set x = = 2–(–4)2(1)Make a table centered at x = 2.x y01234Note the y values are symmetricaround the vertex just as thepoints. (If they are not, checkyour calculation.)(2, -16)(0, -12) (4, -12)(3, -15)(1, -15)–12–15–16–15–12
    • Graphs of Quadratic EquationsExample B. Graph y = x2 – 4x – 12Vertex: set x = = 2–(–4)2(1)(2, -16)(0, -12) (4, -12)Make a table centered at x = 2.Note the y values are symmetricaround the vertex just as thepoints. (If they are not, checkyour calculation.)(3, -15)(1, -15)One way to graph a parabola is to make a table around thevertex so the points will be plotted symmetrically.x y01234–12–15–16–15–12
    • Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-interceptsand the y-intercept.
    • Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-interceptsand the y-intercept.The y-intercept is (o, c) obtained by setting x = 0.
    • Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-interceptsand the y-intercept.The y-intercept is (o, c) obtained by setting x = 0.The x-intercept is obtained by setting y = 0 and solve theequation 0 = ax2 + bx + c
    • Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-interceptsand the y-intercept.The y-intercept is (o, c) obtained by setting x = 0.The x-intercept is obtained by setting y = 0 and solve theequation 0 = ax2 + bx + c which may or may not have realnumber solutions.
    • Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-interceptsand the y-intercept.The y-intercept is (o, c) obtained by setting x = 0.The x-intercept is obtained by setting y = 0 and solve theequation 0 = ax2 + bx + c which may or may not have realnumber solutions. Hence there might not be any x-intercept.
    • The center line is determined by the vertex.Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-interceptsand the y-intercept.The y-intercept is (o, c) obtained by setting x = 0.The x-intercept is obtained by setting y = 0 and solve theequation 0 = ax2 + bx + c which may or may not have realnumber solutions. Hence there might not be any x-intercept.
    • The center line is determined by the vertex. Suppose weknow another point on the parabola,Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-interceptsand the y-intercept.The y-intercept is (o, c) obtained by setting x = 0.The x-intercept is obtained by setting y = 0 and solve theequation 0 = ax2 + bx + c which may or may not have realnumber solutions. Hence there might not be any x-intercept.
    • The center line is determined by the vertex. Suppose weknow another point on the parabola, the reflection of the pointacross the center is also on the parabola.Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-interceptsand the y-intercept.The y-intercept is (o, c) obtained by setting x = 0.The x-intercept is obtained by setting y = 0 and solve theequation 0 = ax2 + bx + c which may or may not have realnumber solutions. Hence there might not be any x-intercept.
    • The center line is determined by the vertex. Suppose weknow another point on the parabola, the reflection of the pointacross the center is also on the parabola. There is exactlyone parabola that goes through these three points.Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-interceptsand the y-intercept.The y-intercept is (o, c) obtained by setting x = 0.The x-intercept is obtained by setting y = 0 and solve theequation 0 = ax2 + bx + c which may or may not have realnumber solutions. Hence there might not be any x-intercept.
    • (2nd way) To graph a parabola y = ax2 + bx + c.Graphs of Quadratic Equations
    • (2nd way) To graph a parabola y = ax2 + bx + c.1. Set x = in the equation to find the vertex.-b2aGraphs of Quadratic Equations
    • (2nd way) To graph a parabola y = ax2 + bx + c.1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if possible.-b2aGraphs of Quadratic Equations
    • (2nd way) To graph a parabola y = ax2 + bx + c.1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if possible.3. Locate the its reflection across the center line, these threepoints form the tip of the parabola.-b2aGraphs of Quadratic Equations
    • (2nd way) To graph a parabola y = ax2 + bx + c.1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if possible.3. Locate the its reflection across the center line, these threepoints form the tip of the parabola. Trace the parabola.-b2aGraphs of Quadratic Equations
    • (2nd way) To graph a parabola y = ax2 + bx + c.1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if possible.3. Locate the its reflection across the center line, these threepoints form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.-b2aGraphs of Quadratic Equations
    • (2nd way) To graph a parabola y = ax2 + bx + c.1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if possible.3. Locate the its reflection across the center line, these threepoints form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.-b2aGraphs of Quadratic EquationsExample C. Graph y = –x2 + 2x + 15
    • (2nd way) To graph a parabola y = ax2 + bx + c.1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if possible.3. Locate the its reflection across the center line, these threepoints form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.-b2aGraphs of Quadratic EquationsThe vertex is at x = 1, y = 16Example C. Graph y = –x2 + 2x + 15
    • (2nd way) To graph a parabola y = ax2 + bx + c.1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if possible.3. Locate the its reflection across the center line, these threepoints form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.-b2aGraphs of Quadratic EquationsThe vertex is at x = 1, y = 16y-intercept is at (0, 15)Example C. Graph y = –x2 + 2x + 15
    • (2nd way) To graph a parabola y = ax2 + bx + c.1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if possible.3. Locate the its reflection across the center line, these threepoints form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.-b2aGraphs of Quadratic EquationsThe vertex is at x = 1, y = 16y-intercept is at (0, 15)Plot its reflection (2, 15)Example C. Graph y = –x2 + 2x + 15
    • (2nd way) To graph a parabola y = ax2 + bx + c.1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if possible.3. Locate the its reflection across the center line, these threepoints form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.-b2aGraphs of Quadratic EquationsThe vertex is at x = 1, y = 16y-intercept is at (0, 15)Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0Example C. Graph y = –x2 + 2x + 15
    • (2nd way) To graph a parabola y = ax2 + bx + c.1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if possible.3. Locate the its reflection across the center line, these threepoints form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.-b2aGraphs of Quadratic EquationsThe vertex is at x = 1, y = 16y-intercept is at (0, 15)Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0x2 – 2x – 15 = 0Example C. Graph y = –x2 + 2x + 15
    • (2nd way) To graph a parabola y = ax2 + bx + c.1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if possible.3. Locate the its reflection across the center line, these threepoints form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.-b2aGraphs of Quadratic EquationsThe vertex is at x = 1, y = 16y-intercept is at (0, 15)Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0x2 – 2x – 15 = 0(x + 3)(x – 5) = 0x = –3, x = 5Example C. Graph y = –x2 + 2x + 15
    • (2nd way) To graph a parabola y = ax2 + bx + c.1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if possible.3. Locate the its reflection across the center line, these threepoints form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.-b2aGraphs of Quadratic EquationsThe vertex is at x = 1, y = 16y-intercept is at (0, 15)Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0x2 – 2x – 15 = 0(x + 3)(x – 5) = 0x = –3, x = 5Example C. Graph y = –x2 + 2x + 15 (1, 16)
    • (2nd way) To graph a parabola y = ax2 + bx + c.1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if possible.3. Locate the its reflection across the center line, these threepoints form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.-b2aGraphs of Quadratic EquationsThe vertex is at x = 1, y = 16y-intercept is at (0, 15)Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0x2 – 2x – 15 = 0(x + 3)(x – 5) = 0x = –3, x = 5Example C. Graph y = –x2 + 2x + 15 (1, 16)(0, 15)
    • (2nd way) To graph a parabola y = ax2 + bx + c.1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if possible.3. Locate the its reflection across the center line, these threepoints form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.-b2aGraphs of Quadratic EquationsThe vertex is at x = 1, y = 16y-intercept is at (0, 15)Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0x2 – 2x – 15 = 0(x + 3)(x – 5) = 0x = –3, x = 5Example C. Graph y = –x2 + 2x + 15 (1, 16)(0, 15) (2, 15)
    • (2nd way) To graph a parabola y = ax2 + bx + c.1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if possible.3. Locate the its reflection across the center line, these threepoints form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.-b2aGraphs of Quadratic EquationsThe vertex is at x = 1, y = 16y-intercept is at (0, 15)Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0x2 – 2x – 15 = 0(x + 3)(x – 5) = 0x = –3, x = 5Example C. Graph y = –x2 + 2x + 15 (1, 16)(0, 15) (2, 15)
    • (2nd way) To graph a parabola y = ax2 + bx + c.1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if possible.3. Locate the its reflection across the center line, these threepoints form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.-b2aGraphs of Quadratic EquationsThe vertex is at x = 1, y = 16y-intercept is at (0, 15)Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0x2 – 2x – 15 = 0(x + 3)(x – 5) = 0x = –3, x = 5Example C. Graph y = –x2 + 2x + 15 (1, 16)(0, 15) (2, 15)(-3, 0) (5, 0)
    • Finally, we make the observation that given y = ax2 + …,if a > 0, then the parabola opens upward.Graphs of Quadratic Equations
    • Finally, we make the observation that given y = ax2 + …,if a > 0, then the parabola opens upward.Graphs of Quadratic Equationsif a < 0, then the parabola opens downward.
    • Exercise A. Practice drawing the following parabolas withpaper and pencil. Visualize them as the paths of thrownobjects and make sure pay attention to the symmetry.Graphs of Quadratic Equations
    • Exercise B. Graph the following parabolas by making a tablearound the x–vertex value –b/(2a) that reflect the symmetry ofthe parabolas. Find the x and y intercepts.Graphs of Quadratic Equations4. y = x2 – 4 5. y = –x2 + 42. y = x2 3. y = –x26. y = x2 + 4 7. y = –x2 – 48. y = x2 – 2x – 3 9. y = –x2 + 2x + 310. y = x2 + 2x – 3 11. y = –x2 – 2x + 312. y = x2 – 2x – 8 13. y = –x2 + 2x + 814. y = x2 + 2x – 8 15. y = –x2 – 2x + 816. a. y = x2 – 4x – 5 b. y = –x2 + 4x + 517. a. y = x2 + 4x – 5 b. y = –x2 – 4x + 519. y = x2 + 4x – 21 20. y = x2 – 4x – 4521. y = x2 – 6x – 27 22. y = – x2 – 6x + 27
    • Exercise C. Graph the following parabolas by finding theplotting the vertex point, the y–intercept and its reflection.Find the x intercepts.Graphs of Quadratic Equations23. y = x2 – 2x – 3 24. y = –x2 + 2x + 325. y = x2 + 2x – 3 26. y = –x2 – 2x + 327. y = x2 – 2x – 8 28. y = –x2 + 2x + 829. y = x2 + 2x – 8 30. y = –x2 – 2x + 831. y = x2 + 4x – 21 32. y = x2 – 4x – 4535. y = x2 – 6x – 27 34. y = – x2 – 6x + 27Exercise C. Graph the following parabolas by finding theplotting the vertex point, the y–intercept and its reflection.Verify that there is no x intercepts (i.e. they have complex roots).35. y = x2 – 2x + 8 36. y = –x2 + 2x – 537. y = x2 + 2x + 3 38. y = –x2 – 3x – 439. y = 2x2 + 3x + 4 40. y = x2 – 4x + 32