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# 1.2 the graphs of quadratic equations

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• 2. Graphs of Quadratic Equations We start with an example of a graph gives the general shape of the graphs of 2 nd (quadratic) degree equations.
• 3. Graphs of Quadratic Equations We start with an example of a graph gives the general shape of the graphs of 2 nd (quadratic) degree equations. Example A. Graph y = –x 2
• 4. Graphs of Quadratic Equations We start with an example of a graph gives the general shape of the graphs of 2 nd (quadratic) degree equations. Make a table Example A. Graph y = –x 2
• 5. x -4 -3 -2 -1 0 1 2 3 4 Graphs of Quadratic Equations We start with an example of a graph gives the general shape of the graphs of 2 nd (quadratic) degree equations. Make a table Example A. Graph y = –x 2 y
• 6. x -4 -3 -2 -1 0 1 2 3 4 Graphs of Quadratic Equations We start with an example of a graph gives the general shape of the graphs of 2 nd (quadratic) degree equations. Make a table Example A. Graph y = –x 2 y -16 -9 -4 -1 0 -1 -4 -9 -16
• 7. Graphs of Quadratic Equations We start with an example of a graph gives the general shape of the graphs of 2 nd (quadratic) degree equations. Make a table Example A. Graph y = –x 2 x -4 -3 -2 -1 0 1 2 3 4 y -16 -9 -4 -1 0 -1 -4 -9 -16
• 8. The graphs of 2 nd (quadratic) equations are called parabolas . Parabolas describe the paths of thrown objects (or the upside-down paths). Graphs of Quadratic Equations
• 9. The graphs of 2 nd (quadratic) equations are called parabolas . Parabolas describe the paths of thrown objects (or the upside-down paths). Graphs of Quadratic Equations
• Properties of Parabolas:
• Parabolas are symmetric with respect to a center line
• 10. The graphs of 2 nd (quadratic) equations are called parabolas . Parabolas describe the paths of thrown objects (or the upside-down paths). Graphs of Quadratic Equations
• Properties of Parabolas:
• Parabolas are symmetric with respect to a center line
• The highest or lowest point of the parabola sits on the
• center line.
• 11. The graphs of 2 nd (quadratic) equations are called parabolas . Parabolas describe the paths of thrown objects (or the upside-down paths). Graphs of Quadratic Equations
• Properties of Parabolas:
• Parabolas are symmetric with respect to a center line
• The highest or lowest point of the parabola sits on the
• center line. This point is called the vertex .
• 12. The graphs of 2 nd (quadratic) equations are called parabolas . Parabolas describe the paths of thrown objects (or the upside-down paths). Graphs of Quadratic Equations
• Properties of Parabolas:
• Parabolas are symmetric with respect to a center line
• The highest or lowest point of the parabola sits on the
• center line. This point is called the vertex .
Vertex Formula : The vertex of y = ax 2 + bx + c is at x = . -b 2a
• 13. Graphs of Quadratic Equations One way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically.
• 14. Graphs of Quadratic Equations One way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically. Example B. Graph y = x 2 – 4x – 12
• 15. Graphs of Quadratic Equations One way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically. Example B. Graph y = x 2 – 4x – 12 Vertex: set x = = 2 – ( – 4) 2(1)
• 16. Graphs of Quadratic Equations One way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically. Example B. Graph y = x 2 – 4x – 12 Vertex: set x = = 2 – ( – 4) 2(1) Make a table centered at x = 2 . x y 0 1 2 3 4
• 17. Graphs of Quadratic Equations One way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically. Example B. Graph y = x 2 – 4x – 12 Vertex: set x = = 2 – ( – 4) 2(1) Make a table centered at x = 2 . x y 0 1 2 3 4 – 12 – 15 – 16 – 15 – 12
• 18. Graphs of Quadratic Equations One way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically. Example B. Graph y = x 2 – 4x – 12 Vertex: set x = = 2 – ( – 4) 2(1) Make a table centered at x = 2 . x y 0 1 2 3 4 – 12 – 15 – 16 – 15 – 12
• 19. Graphs of Quadratic Equations One way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically. Example B. Graph y = x 2 – 4x – 12 Vertex: set x = = 2 – ( – 4) 2(1) Make a table centered at x = 2 . x y 0 1 2 3 4 Note the y values are symmetric around the vertex just as the points. – 12 – 15 – 16 – 15 – 12
• 20. Graphs of Quadratic Equations One way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically. Example B. Graph y = x 2 – 4x – 12 Vertex: set x = = 2 – ( – 4) 2(1) Make a table centered at x = 2 . x y 0 1 2 3 4 Note the y values are symmetric around the vertex just as the points. (If they are not, check your calculation.) – 12 – 15 – 16 – 15 – 12
• 21. Graphs of Quadratic Equations One way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically. Example B. Graph y = x 2 – 4x – 12 Vertex: set x = = 2 – ( – 4) 2(1) Make a table centered at x = 2 . x y 0 1 2 3 4 Note the y values are symmetric around the vertex just as the points. (If they are not, check your calculation.) (2, -16) – 12 – 15 – 16 – 15 – 12
• 22. Graphs of Quadratic Equations One way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically. Example B. Graph y = x 2 – 4x – 12 Vertex: set x = = 2 – ( – 4) 2(1) Make a table centered at x = 2 . x y 0 1 2 3 4 Note the y values are symmetric around the vertex just as the points. (If they are not, check your calculation.) (2, -16) (3, -15) – 12 – 15 – 16 – 15 – 12
• 23. Graphs of Quadratic Equations One way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically. Example B. Graph y = x 2 – 4x – 12 Vertex: set x = = 2 – ( – 4) 2(1) Make a table centered at x = 2 . x y 0 1 2 3 4 Note the y values are symmetric around the vertex just as the points. (If they are not, check your calculation.) (2, -16) (4, -12) (3, -15) – 12 – 15 – 16 – 15 – 12
• 24. Graphs of Quadratic Equations One way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically. Example B. Graph y = x 2 – 4x – 12 Vertex: set x = = 2 – ( – 4) 2(1) Make a table centered at x = 2 . x y 0 1 2 3 4 Note the y values are symmetric around the vertex just as the points. (If they are not, check your calculation.) (2, -16) (4, -12) (3, -15) (1, -15) – 12 – 15 – 16 – 15 – 12
• 25. Graphs of Quadratic Equations One way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically. Example B. Graph y = x 2 – 4x – 12 Vertex: set x = = 2 – ( – 4) 2(1) Make a table centered at x = 2 . x y 0 1 2 3 4 Note the y values are symmetric around the vertex just as the points. (If they are not, check your calculation.) (2, -16) (0, -12) (4, -12) (3, -15) (1, -15) – 12 – 15 – 16 – 15 – 12
• 26. Graphs of Quadratic Equations Example B. Graph y = x 2 – 4x – 12 Vertex: set x = = 2 – ( – 4) 2(1) (2, -16) (0, -12) (4, -12) Make a table centered at x = 2. Note the y values are symmetric around the vertex just as the points. (If they are not, check your calculation.) (3, -15) (1, -15) One way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically. x y 0 1 2 3 4 – 12 – 15 – 16 – 15 – 12
• 27. Graphs of Quadratic Equations When graphing parabolas, we must also give the x-intercepts and the y-intercepts.
• 28. Graphs of Quadratic Equations When graphing parabolas, we must also give the x-intercepts and the y-intercepts. The y-intercept is (o, c) obtained by setting x = 0.
• 29. Graphs of Quadratic Equations When graphing parabolas, we must also give the x-intercepts and the y-intercepts. The y-intercept is (o, c) obtained by setting x = 0. The x-intercept is obtained by setting y = 0 and solving the equation 0 = ax 2 + bx + c
• 30. Graphs of Quadratic Equations When graphing parabolas, we must also give the x-intercepts and the y-intercepts. The y-intercept is (o, c) obtained by setting x = 0. The x-intercept is obtained by setting y = 0 and solving the equation 0 = ax 2 + bx + c which may or may not have real number solutions.
• 31. Graphs of Quadratic Equations When graphing parabolas, we must also give the x-intercepts and the y-intercepts. The y-intercept is (o, c) obtained by setting x = 0. The x-intercept is obtained by setting y = 0 and solving the equation 0 = ax 2 + bx + c which may or may not have real number solutions. Hence there might not be any x-intercept.
• 32. The center line is determined by the vertex. Graphs of Quadratic Equations When graphing parabolas, we must also give the x-intercepts and the y-intercepts. The y-intercept is (o, c) obtained by setting x = 0. The x-intercept is obtained by setting y = 0 and solving the equation 0 = ax 2 + bx + c which may or may not have real number solutions. Hence there might not be any x-intercept.
• 33. The center line is determined by the vertex. Suppose we know another point on the parabola, Graphs of Quadratic Equations When graphing parabolas, we must also give the x-intercepts and the y-intercepts. The y-intercept is (o, c) obtained by setting x = 0. The x-intercept is obtained by setting y = 0 and solving the equation 0 = ax 2 + bx + c which may or may not have real number solutions. Hence there might not be any x-intercept.
• 34. The center line is determined by the vertex. Suppose we know another point on the parabola, the reflection of the point across the center is also on the parabola. Graphs of Quadratic Equations When graphing parabolas, we must also give the x-intercepts and the y-intercepts. The y-intercept is (o, c) obtained by setting x = 0. The x-intercept is obtained by setting y = 0 and solving the equation 0 = ax 2 + bx + c which may or may not have real number solutions. Hence there might not be any x-intercept.
• 35. The center line is determined by the vertex. Suppose we know another point on the parabola, the reflection of the point across the center is also on the parabola. There is exactly one parabola that goes through these three points. Graphs of Quadratic Equations When graphing parabolas, we must also give the x-intercepts and the y-intercepts. The y-intercept is (o, c) obtained by setting x = 0. The x-intercept is obtained by setting y = 0 and solving the equation 0 = ax 2 + bx + c which may or may not have real number solutions. Hence there might not be any x-intercept.
• 36. (2 nd way) To graph a parabola y = ax 2 + bx + c. Graphs of Quadratic Equations
• 37.
• (2 nd way) To graph a parabola y = ax 2 + bx + c.
• Set x = in the equation to find the vertex.
-b 2a Graphs of Quadratic Equations
• 38.
• (2 nd way) To graph a parabola y = ax 2 + bx + c.
• Set x = in the equation to find the vertex.
• 2. Find another point, use the y-intercept (0, c) if possible.
-b 2a Graphs of Quadratic Equations
• 39.
• (2 nd way) To graph a parabola y = ax 2 + bx + c.
• 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 three points form the tip of the parabola.
-b 2a Graphs of Quadratic Equations
• 40.
• (2 nd way) To graph a parabola y = ax 2 + bx + c.
• 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 three points form the tip of the parabola. Trace the parabola.
-b 2a Graphs of Quadratic Equations
• 41.
• (2 nd way) To graph a parabola y = ax 2 + bx + c.
• 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 three points form the tip of the parabola. Trace the parabola.
• 4. Set y = 0 and solve to find the x intercept.
-b 2a Graphs of Quadratic Equations
• 42.
• (2 nd way) To graph a parabola y = ax 2 + bx + c.
• 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 three points form the tip of the parabola. Trace the parabola.
• 4. Set y = 0 and solve to find the x intercept.
-b 2a Graphs of Quadratic Equations Example C. Graph y = –x 2 + 2x + 15
• 43.
• (2 nd way) To graph a parabola y = ax 2 + bx + c.
• 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 three points form the tip of the parabola. Trace the parabola.
• 4. Set y = 0 and solve to find the x intercept.
-b 2a Graphs of Quadratic Equations The vertex is at x = 1, y = 16 Example C. Graph y = –x 2 + 2x + 15
• 44.
• (2 nd way) To graph a parabola y = ax 2 + bx + c.
• 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 three points form the tip of the parabola. Trace the parabola.
• 4. Set y = 0 and solve to find the x intercept.
-b 2a Graphs of Quadratic Equations The vertex is at x = 1, y = 16 y-intercept is at (0, 15 ) Example C. Graph y = –x 2 + 2x + 15
• 45.
• (2 nd way) To graph a parabola y = ax 2 + bx + c.
• 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 three points form the tip of the parabola. Trace the parabola.
• 4. Set y = 0 and solve to find the x intercept.
-b 2a Graphs of Quadratic Equations The vertex is at x = 1, y = 16 y-intercept is at (0, 15 ) Plot its reflection (2, 15 ) Example C. Graph y = –x 2 + 2x + 15
• 46.
• (2 nd way) To graph a parabola y = ax 2 + bx + c.
• 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 three points form the tip of the parabola. Trace the parabola.
• 4. Set y = 0 and solve to find the x intercept.
-b 2a Graphs of Quadratic Equations The vertex is at x = 1, y = 16 y-intercept is at (0, 15 ) Plot its reflection (2, 15 ) Draw, set y = 0 to get x-int: – x 2 + 2x + 15 = 0 Example C. Graph y = –x 2 + 2x + 15
• 47.
• (2 nd way) To graph a parabola y = ax 2 + bx + c.
• 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 three points form the tip of the parabola. Trace the parabola.
• 4. Set y = 0 and solve to find the x intercept.
-b 2a Graphs of Quadratic Equations The vertex is at x = 1, y = 16 y-intercept is at (0, 15 ) Plot its reflection (2, 15 ) Draw, set y = 0 to get x-int: – x 2 + 2x + 15 = 0 x 2 – 2x – 15 = 0 Example C. Graph y = –x 2 + 2x + 15
• 48.
• (2 nd way) To graph a parabola y = ax 2 + bx + c.
• 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 three points form the tip of the parabola. Trace the parabola.
• 4. Set y = 0 and solve to find the x intercept.
-b 2a Graphs of Quadratic Equations The vertex is at x = 1, y = 16 y-intercept is at (0, 15 ) Plot its reflection (2, 15 ) Draw, set y = 0 to get x-int: – x 2 + 2x + 15 = 0 x 2 – 2x – 15 = 0 (x + 3)(x – 5) = 0 x = – 3, x = 5 Example C. Graph y = –x 2 + 2x + 15
• 49.
• (2 nd way) To graph a parabola y = ax 2 + bx + c.
• 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 three points form the tip of the parabola. Trace the parabola.
• 4. Set y = 0 and solve to find the x intercept.
-b 2a Graphs of Quadratic Equations The vertex is at x = 1, y = 16 y-intercept is at (0, 15 ) Plot its reflection (2, 15 ) Draw, set y = 0 to get x-int: – x 2 + 2x + 15 = 0 x 2 – 2x – 15 = 0 (x + 3)(x – 5) = 0 x = – 3, x = 5 Example C. Graph y = –x 2 + 2x + 15 (1, 16)
• 50.
• (2 nd way) To graph a parabola y = ax 2 + bx + c.
• 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 three points form the tip of the parabola. Trace the parabola.
• 4. Set y = 0 and solve to find the x intercept.
-b 2a Graphs of Quadratic Equations The vertex is at x = 1, y = 16 y-intercept is at (0, 15 ) Plot its reflection (2, 15 ) Draw, set y = 0 to get x-int: – x 2 + 2x + 15 = 0 x 2 – 2x – 15 = 0 (x + 3)(x – 5) = 0 x = – 3, x = 5 Example C. Graph y = –x 2 + 2x + 15 (1, 16) (0, 15)
• 51.
• (2 nd way) To graph a parabola y = ax 2 + bx + c.
• 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 three points form the tip of the parabola. Trace the parabola.
• 4. Set y = 0 and solve to find the x intercept.
-b 2a Graphs of Quadratic Equations The vertex is at x = 1, y = 16 y-intercept is at (0, 15 ) Plot its reflection (2, 15 ) Draw, set y = 0 to get x-int: – x 2 + 2x + 15 = 0 x 2 – 2x – 15 = 0 (x + 3)(x – 5) = 0 x = – 3, x = 5 Example C. Graph y = –x 2 + 2x + 15 (1, 16) (0, 15) (2, 15)
• 52.
• (2 nd way) To graph a parabola y = ax 2 + bx + c.
• 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 three points form the tip of the parabola. Trace the parabola.
• 4. Set y = 0 and solve to find the x intercept.
-b 2a Graphs of Quadratic Equations The vertex is at x = 1, y = 16 y-intercept is at (0, 15 ) Plot its reflection (2, 15 ) Draw, set y = 0 to get x-int: – x 2 + 2x + 15 = 0 x 2 – 2x – 15 = 0 (x + 3)(x – 5) = 0 x = – 3, x = 5 Example C. Graph y = –x 2 + 2x + 15 (1, 16) (0, 15) (2, 15)
• 53.
• (2 nd way) To graph a parabola y = ax 2 + bx + c.
• 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 three points form the tip of the parabola. Trace the parabola.
• 4. Set y = 0 and solve to find the x intercept.
-b 2a Graphs of Quadratic Equations The vertex is at x = 1, y = 16 y-intercept is at (0, 15 ) Plot its reflection (2, 15 ) Draw, set y = 0 to get x-int: – x 2 + 2x + 15 = 0 x 2 – 2x – 15 = 0 (x + 3)(x – 5) = 0 x = – 3, x = 5 Example C. Graph y = –x 2 + 2x + 15 (1, 16) (0, 15) (2, 15) (-3, 0) (5, 0)
• 54. Finally, we make the observation that given y = ax 2 + …, if a > 0, then the parabola opens upward. Graphs of Quadratic Equations
• 55. Finally, we make the observation that given y = ax 2 + …, if a > 0, then the parabola opens upward. Graphs of Quadratic Equations if a < 0, then the parabola opens downward.
• 56. Exercise A. Practice drawing the following parabolas with paper and pencil. Visualize them as the paths of thrown objects and make sure pay attention to the symmetry. Graphs of Quadratic Equations
• 57. Exercise B. Graph the following parabolas by making a table around the x–vertex value –b/(2a) that reflect the symmetry of the parabolas. Find the x and y intercepts. Graphs of Quadratic Equations 4. y = x 2 – 4 5. y = –x 2 + 4 2. y = x 2 3. y = –x 2 6. y = x 2 + 4 7. y = –x 2 – 4 8. y = x 2 – 2x – 3 9. y = –x 2 + 2x + 3 10. y = x 2 + 2x – 3 11. y = –x 2 – 2x + 3 12. y = x 2 – 2x – 8 13. y = –x 2 + 2x + 8 14. y = x 2 + 2x – 8 15. y = –x 2 – 2x + 8 16. a. y = x 2 – 4x – 5 b. y = –x 2 + 4x + 5 17. a. y = x 2 + 4x – 5 b. y = –x 2 – 4x + 5 19. y = x 2 + 4x – 21 20. y = x 2 – 4x – 45 21. y = x 2 – 6x – 27 22. y = – x 2 – 6x + 27
• 58. Exercise C. Graph the following parabolas by finding the plotting the vertex point, the y–intercept and its reflection. Find the x intercepts. Graphs of Quadratic Equations 23. y = x 2 – 2x – 3 24. y = –x 2 + 2x + 3 25. y = x 2 + 2x – 3 26. y = –x 2 – 2x + 3 27. y = x 2 – 2x – 8 28. y = –x 2 + 2x + 8 29. y = x 2 + 2x – 8 30. y = –x 2 – 2x + 8 31. y = x 2 + 4x – 21 32. y = x 2 – 4x – 45 35. y = x 2 – 6x – 27 34. y = – x 2 – 6x + 27 Exercise C. Graph the following parabolas by finding the plotting the vertex point, the y–intercept and its reflection. Verify that there is no x intercepts (i.e. they have complex roots). 35. y = x 2 – 2x + 8 36. y = –x 2 + 2x – 5 37. y = x 2 + 2x + 3 38. y = –x 2 – 3x – 4 39. y = 2x 2 + 3x + 4 40. y = x 2 – 4x + 32