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    5 4 the graphs of quadratic equations 5 4 the graphs of quadratic equations Presentation Transcript

    • Graphs of Quadratic Equations
      Frank Ma © 2011
    • Graphs of Quadratic Equations
      We start with an example of a graph gives the general shape of the graphs of 2nd (quadratic) degree equations.
    • Graphs of Quadratic Equations
      We start with an example of a graph gives the general shape of the graphs of 2nd (quadratic) degree equations.
      Example A. Graph y = –x2
    • Graphs of Quadratic Equations
      We start with an example of a graph gives the general shape of the graphs of 2nd (quadratic) degree equations.
      Example A. Graph y = –x2
      Make a table
    • Graphs of Quadratic Equations
      We start with an example of a graph gives the general shape of the graphs of 2nd (quadratic) degree equations.
      Example A. Graph y = –x2
      Make a table
      x
      -4
      -3
      -2
      -1
      0
      1
      2
      3
      4
      y
    • Graphs of Quadratic Equations
      We start with an example of a graph gives the general shape of the graphs of 2nd (quadratic) degree equations.
      Example A. Graph y = –x2
      Make a table
      x
      -4
      -3
      -2
      -1
      0
      1
      2
      3
      4
      y
      -16
      -9
      -4
      -1
      0
      -1
      -4
      -9
      -16
    • Graphs of Quadratic Equations
      We start with an example of a graph gives the general shape of the graphs of 2nd (quadratic) degree equations.
      Example A. Graph y = –x2
      Make a table
      x
      -4
      -3
      -2
      -1
      0
      1
      2
      3
      4
      y
      -16
      -9
      -4
      -1
      0
      -1
      -4
      -9
      -16
    • 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 Equations
      The graphs of 2nd (quadratic) equations are called parabolas. Parabolas describe the paths of thrown objects (or the upside-down paths).
      Properties of Parabolas:
      • Parabolas are symmetric with respect to a center line
    • 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).
      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.
    • 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).
      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.
    • 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).
      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.
      -b
      Vertex Formula: The vertex of y = ax2 + bx + c is at x = .
      2a
    • 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.
    • 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 = x2 – 4x – 12
    • 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 = x2 – 4x – 12
      Vertex: set x = = 2
      –(–4)
      2(1)
    • 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 = x2 – 4x – 12
      Vertex: set x = = 2
      –(–4)
      2(1)
      Make a table centered at x = 2.
      x y
      0
      1
      2
      3
      4
    • 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 = x2 – 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
    • 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 = x2 – 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
    • 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 = x2 – 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
      Note the y values are symmetric around the vertex just as the points.
    • 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 = x2 – 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
      Note the y values are symmetric around the vertex just as the points. (If they are not, check your calculation.)
    • 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 = x2 – 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
      Note the y values are symmetric around the vertex just as the points. (If they are not, check your calculation.)
      (2, -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 = x2 – 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
      Note the y values are symmetric around the vertex just as the points. (If they are not, check your calculation.)
      (3, -15)
      (2, -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 = x2 – 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
      Note the y values are symmetric around the vertex just as the points. (If they are not, check your calculation.)
      (4, -12)
      (3, -15)
      (2, -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 = x2 – 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
      Note the y values are symmetric around the vertex just as the points. (If they are not, check your calculation.)
      (4, -12)
      (3, -15)
      (1, -15)
      (2, -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 = x2 – 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
      Note the y values are symmetric around the vertex just as the points. (If they are not, check your calculation.)
      (4, -12)
      (0, -12)
      (3, -15)
      (1, -15)
      (2, -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 = x2 – 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
      Note the y values are symmetric around the vertex just as the points. (If they are not, check your calculation.)
      (4, -12)
      (0, -12)
      (3, -15)
      (1, -15)
      (2, -16)
    • Graphs of Quadratic Equations
      When graphing parabolas, we must also give the x-intercepts and the y-intercept.
    • Graphs of Quadratic Equations
      When graphing parabolas, we must also give the x-intercepts and the y-intercept.
      The y-intercept is (0, c) obtained by setting x = 0.
    • Graphs of Quadratic Equations
      When graphing parabolas, we must also give the x-intercepts and the y-intercept.
      The y-intercept is (0, c) obtained by setting x = 0.
      The x-intercept is obtained by setting y = 0 and solve the equation 0 = ax2 + bx + c
    • Graphs of Quadratic Equations
      When graphing parabolas, we must also give the x-intercepts and the y-intercept.
      The y-intercept is (0, c) obtained by setting x = 0.
      The x-intercept is obtained by setting y = 0 and solve the equation 0 = ax2 + bx + c which may or may not have real number solutions.
    • Graphs of Quadratic Equations
      When graphing parabolas, we must also give the x-intercepts and the y-intercept.
      The y-intercept is (0, c) obtained by setting x = 0.
      The x-intercept is obtained by setting y = 0 and solve the equation 0 = ax2 + bx + c which may or may not have real number solutions. Hence there might not be any x-intercept.
    • Graphs of Quadratic Equations
      When graphing parabolas, we must also give the x-intercepts and the y-intercept.
      The y-intercept is (0, c) obtained by setting x = 0.
      The x-intercept is obtained by setting y = 0 and solve the equation 0 = ax2 + bx + c which may or may not have real number solutions. Hence there might not be any x-intercept.
      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-intercept.
      The y-intercept is (0, c) obtained by setting x = 0.
      The x-intercept is obtained by setting y = 0 and solve the equation 0 = ax2 + bx + c which may or may not have real number solutions. Hence there might not be any x-intercept.
      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-intercept.
      The y-intercept is (0, c) obtained by setting x = 0.
      The x-intercept is obtained by setting y = 0 and solve the equation 0 = ax2 + bx + c which may or may not have real number solutions. Hence there might not be any x-intercept.
      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-intercept.
      The y-intercept is (0, c) obtained by setting x = 0.
      The x-intercept is obtained by setting y = 0 and solve the equation 0 = ax2 + bx + c which may or may not have real number solutions. Hence there might not be any x-intercept.
      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
      (2nd way) To graph a parabola y = ax2 + bx + c.
    • Graphs of Quadratic Equations
      (2nd way) To graph a parabola y = ax2 + bx + c.
      Set x = in the equation to find the vertex.
      -b
      2a
    • Graphs of Quadratic Equations
      (2nd way) To graph a parabola y = ax2 + bx + c.
      Set x = in the equation to find the vertex.
      2. Find another point, use the y-intercept (0, c) if feasible.
      -b
      2a
    • Graphs of Quadratic Equations
      (2nd way) To graph a parabola y = ax2 + bx + c.
      Set x = in the equation to find the vertex.
      2. Find another point, use the y-intercept (0, c) if feasible.
      3. Locate its reflection across the center line. These three points form the tip of the parabola.
      -b
      2a
    • Graphs of Quadratic Equations
      (2nd way) To graph a parabola y = ax2 + bx + c.
      Set x = in the equation to find the vertex.
      2. Find another point, use the y-intercept (0, c) if feasible.
      3. Locate its reflection across the center line. These three points form the tip of the parabola. Trace the parabola.
      -b
      2a
    • Graphs of Quadratic Equations
      (2nd way) To graph a parabola y = ax2 + bx + c.
      Set x = in the equation to find the vertex.
      2. Find another point, use the y-intercept (0, c) if feasible.
      3. Locate 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
      (2nd way) To graph a parabola y = ax2 + bx + c.
      Set x = in the equation to find the vertex.
      2. Find another point, use the y-intercept (0, c) if feasible.
      3. Locate 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
      Example C. Graph y = –x2 + 2x + 15
    • Graphs of Quadratic Equations
      (2nd way) To graph a parabola y = ax2 + bx + c.
      Set x = in the equation to find the vertex.
      2. Find another point, use the y-intercept (0, c) if feasible.
      3. Locate 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
      Example C. Graph y = –x2 + 2x + 15
      The vertex is at x = 1, y = 16
    • Graphs of Quadratic Equations
      (2nd way) To graph a parabola y = ax2 + bx + c.
      Set x = in the equation to find the vertex.
      2. Find another point, use the y-intercept (0, c) if feasible.
      3. Locate 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
      Example C. Graph y = –x2 + 2x + 15
      The vertex is at x = 1, y = 16
      y-intercept is at (0, 15)
    • Graphs of Quadratic Equations
      (2nd way) To graph a parabola y = ax2 + bx + c.
      Set x = in the equation to find the vertex.
      2. Find another point, use the y-intercept (0, c) if feasible.
      3. Locate 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
      Example C. Graph y = –x2 + 2x + 15
      The vertex is at x = 1, y = 16
      y-intercept is at (0, 15)
      Plot its reflection (2, 15)
    • Graphs of Quadratic Equations
      (2nd way) To graph a parabola y = ax2 + bx + c.
      Set x = in the equation to find the vertex.
      2. Find another point, use the y-intercept (0, c) if feasible.
      3. Locate 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
      Example C. Graph y = –x2 + 2x + 15
      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:
      –x2 + 2x + 15 = 0
    • Graphs of Quadratic Equations
      (2nd way) To graph a parabola y = ax2 + bx + c.
      Set x = in the equation to find the vertex.
      2. Find another point, use the y-intercept (0, c) if feasible.
      3. Locate 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
      Example C. Graph y = –x2 + 2x + 15
      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:
      –x2 + 2x + 15 = 0
      x2 – 2x – 15 = 0
    • Graphs of Quadratic Equations
      (2nd way) To graph a parabola y = ax2 + bx + c.
      Set x = in the equation to find the vertex.
      2. Find another point, use the y-intercept (0, c) if feasible.
      3. Locate 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
      Example C. Graph y = –x2 + 2x + 15
      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:
      –x2 + 2x + 15 = 0
      x2 – 2x – 15 = 0
      (x + 3)(x – 5) = 0
      x = –3, x = 5
    • Graphs of Quadratic Equations
      (2nd way) To graph a parabola y = ax2 + bx + c.
      Set x = in the equation to find the vertex.
      2. Find another point, use the y-intercept (0, c) if feasible.
      3. Locate 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
      Example C. Graph y = –x2 + 2x + 15
      (1, 16)
      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:
      –x2 + 2x + 15 = 0
      x2 – 2x – 15 = 0
      (x + 3)(x – 5) = 0
      x = –3, x = 5
    • Graphs of Quadratic Equations
      (2nd way) To graph a parabola y = ax2 + bx + c.
      Set x = in the equation to find the vertex.
      2. Find another point, use the y-intercept (0, c) if feasible.
      3. Locate 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
      Example C. Graph y = –x2 + 2x + 15
      (1, 16)
      (0, 15)
      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:
      –x2 + 2x + 15 = 0
      x2 – 2x – 15 = 0
      (x + 3)(x – 5) = 0
      x = –3, x = 5
    • Graphs of Quadratic Equations
      (2nd way) To graph a parabola y = ax2 + bx + c.
      Set x = in the equation to find the vertex.
      2. Find another point, use the y-intercept (0, c) if feasible.
      3. Locate 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
      Example C. Graph y = –x2 + 2x + 15
      (1, 16)
      (0, 15)
      (2, 15)
      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:
      –x2 + 2x + 15 = 0
      x2 – 2x – 15 = 0
      (x + 3)(x – 5) = 0
      x = –3, x = 5
    • Graphs of Quadratic Equations
      (2nd way) To graph a parabola y = ax2 + bx + c.
      Set x = in the equation to find the vertex.
      2. Find another point, use the y-intercept (0, c) if feasible.
      3. Locate 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
      Example C. Graph y = –x2 + 2x + 15
      (1, 16)
      (0, 15)
      (2, 15)
      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:
      –x2 + 2x + 15 = 0
      x2 – 2x – 15 = 0
      (x + 3)(x – 5) = 0
      x = –3, x = 5
    • Graphs of Quadratic Equations
      (2nd way) To graph a parabola y = ax2 + bx + c.
      Set x = in the equation to find the vertex.
      2. Find another point, use the y-intercept (0, c) if feasible.
      3. Locate 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
      Example C. Graph y = –x2 + 2x + 15
      (1, 16)
      (0, 15)
      (2, 15)
      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:
      –x2 + 2x + 15 = 0
      x2 – 2x – 15 = 0
      (x + 3)(x – 5) = 0
      x = –3, x = 5
      (-3, 0)
      (5, 0)
    • Graphs of Quadratic Equations
      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.
      if a < 0, then the parabola opens downward.
    • Graphs of Quadratic Equations
      Exercise A. 1. Practice drawing the following parabolas with
      paper and pencil. Visualize them as the paths of thrown
      objects. Make sure pay attention to the symmetry.
    • Graphs of Quadratic Equations
      Exercise B. Graph the parabolas by taking a table around the
      vertex that reflect the symmetry. Find the x and y intercepts.
      2. y = –x2
      3. y = x2
      4. y = x2 – 4
      5. y = –x2 + 4
      6. y = x2 + 4
      7. y = –x2 – 4
      8. y = x2 – 2x – 3
      9. y = –x2 + 2x + 3
      10. y = x2 + 2x – 3
      11. y = –x2 – 2x + 3
      12. y = x2 – 2x – 8
      13. y = –x2 + 2x + 8
      14. y = x2 + 2x – 8
      15. y = –x2 – 2x + 8
      16. a. y = x2 – 4x – 5
      b. y = –x2 + 4x + 5
      17. a. y = x2 + 4x – 5
      b. y = –x2 – 4x + 5
      19. y = x2 + 4x – 21
      20. y = x2 – 4x – 45
      21. y = x2 – 6x – 27
      22. y = – x2 – 6x + 27
    • Graphs of Quadratic Equations
      Exercise C. Graph the following parabolas by plotting the
      vertex point, the y–intercept and its reflection.
      Find the x intercepts.
      24. y = –x2 + 2x + 3
      23. y = x2 – 2x – 3
      26. y = –x2 – 2x + 3
      25. y = x2 + 2x – 3
      28. y = –x2 + 2x + 8
      27. y = x2 – 2x – 8
      30. y = –x2 – 2x + 8
      29. y = x2 + 2x – 8
      32. y = x2 – 4x – 45
      31. y = x2 + 4x – 21
      34. y = – x2 – 6x + 27
      33. y = x2 – 6x – 27
      Exercise D. Graph the following parabolas by 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 = x2 – 2x + 8
      36. y = –x2 + 2x – 5
      37. y = x2 + 2x + 3
      38. y = –x2 – 3x – 4
      39. y = 2x2 + 3x + 4
      40. y = x2 – 4x + 32
    • Graphs of Quadratic Equations
      3. y = x2
      5. y = –x2 + 4
      7. y = –x2 – 4
      (0, 4)
      (0, –4)
      (2, 0)
      (–2, 0)
      (0,0)
      15. y = –x2 – 2x + 8
      17. a. y = x2 + 4x – 5
      b. y = –x2 – 4x + 5
      (–2 , 9)
      (–1, 9)
      (1, 0)
      (–5, 0)
      (0, 5)
      (1, 0)
      (–5, 0)
      (2, 0)
      (–4 , 0)
      (0, –5)
      (–2 , –9)
      19. y = x2 + 4x – 21
      21. y = x2 – 6x – 27
      23. y = x2 – 2x – 3
      (3, 0)
      (–1, 0)
      (3, 0)
      (–7, 0)
      (9, 0)
      (–3, 0)
      (0, –3)
      (0, –21)
      (0 , –27)
      (0, –21)
      (–2 , –25)
      (3 , –36)
      (–1 , –4)