the graphs of quadratic equations

1. Graphs of Quadratic Equations

2. The graphs of 1st degree equations Ax + By = C
are straight lines hence they are called linear equations.
Graphs of Quadratic Equations

3. The graphs of 1st degree equations Ax + By = C
are straight lines hence they are called linear equations.
If we have a linear function in the form of y = f(x) = mx + b,
then we may eyeball the graph, starting from the y-intercept b,
Graphs of Quadratic Equations

4. For example, for y = -2x + 3, we start at the point (0, 3).
the graph of y = -2x + 3
The graphs of 1st degree equations Ax + By = C
are straight lines hence they are called linear equations.
If we have a linear function in the form of y = f(x) = mx + b,
then we may eyeball the graph, starting from the y-intercept b,
Graphs of Quadratic Equations

5. For example, for y = -2x + 3, we start at the point (0, 3).
The graphs of 1st degree equations Ax + By = C
are straight lines hence they are called linear equations.
If we have a linear function in the form of y = f(x) = mx + b,
then we may eyeball the graph, starting from the y-intercept b,
eyeball the tilt using the slope m.
Graphs of Quadratic Equations
the graph of y = -2x + 3

6. For example, for y = -2x + 3, we start at the point (0, 3).
The slope m = -2/1 means we may set ∆y = -2 with ∆x = 1,
The graphs of 1st degree equations Ax + By = C
are straight lines hence they are called linear equations.
If we have a linear function in the form of y = f(x) = mx + b,
then we may eyeball the graph, starting from the y-intercept b,
eyeball the tilt using the slope m.
Graphs of Quadratic Equations
the graph of y = -2x + 3

7. For example, for y = -2x + 3, we start at the point (0, 3).
The slope m = -2/1 means we may set ∆y = -2 with ∆x = 1,
i.e. another point is located 2 down and 1 to the right from (0, 3)
The graphs of 1st degree equations Ax + By = C
are straight lines hence they are called linear equations.
If we have a linear function in the form of y = f(x) = mx + b,
then we may eyeball the graph, starting from the y-intercept b,
eyeball the tilt using the slope m.
Graphs of Quadratic Equations
the graph of y = -2x + 3

8. (0, 3)
∆y = -2
the graph of y = -2x + 3
The graphs of 1st degree equations Ax + By = C
are straight lines hence they are called linear equations.
If we have a linear function in the form of y = f(x) = mx + b,
then we may eyeball the graph, starting from the y-intercept b,
eyeball the tilt using the slope m.
Graphs of Quadratic Equations
For example, for y = -2x + 3, we start at the point (0, 3).
The slope m = -2/1 means we may set ∆y = -2 with ∆x = 1,
i.e. another point is located 2 down and 1 to the right from (0, 3)

9. (0, 3)
(1, 1)∆y = -2
∆x = 1
the graph of y = -2x + 3
The graphs of 1st degree equations Ax + By = C
are straight lines hence they are called linear equations.
If we have a linear function in the form of y = f(x) = mx + b,
then we may eyeball the graph, starting from the y-intercept b,
eyeball the tilt using the slope m.
Graphs of Quadratic Equations
For example, for y = -2x + 3, we start at the point (0, 3).
The slope m = -2/1 means we may set ∆y = -2 with ∆x = 1,
i.e. another point is located 2 down and 1 to the right from (0, 3)

10. (0, 3)
(1, 1)∆y = -2
∆x = 1
the graph of y = -2x + 3
The graphs of 1st degree equations Ax + By = C
are straight lines hence they are called linear equations.
If we have a linear function in the form of y = f(x) = mx + b,
then we may eyeball the graph, starting from the y-intercept b,
eyeball the tilt using the slope m.
Graphs of Quadratic Equations
For example, for y = -2x + 3, we start at the point (0, 3).
The slope m = -2/1 means we may set ∆y = -2 with ∆x = 1,
i.e. another point is located 2 down and 1 to the right from (0, 3)
and the line contains both points.

11. For example, for y = -2x + 3, we start at the point (0, 3).
The slope m = -2/1 means we may set ∆y = -2 with ∆x = 1,
i.e. another point is located 2 down and 1 to the right from (0, 3)
and the line contains both points.
The graphs of 2nd degree equations
y = ax2 + bx + c
are not straight.
They are curves.
We will develop a method
for eyeballing the graphs of
y = ax2 + bx + c in this section.
(0, 3)
(1, 1)∆y = -2
∆x = 1
the graph of y = -2x + 3
The graphs of 1st degree equations Ax + By = C
are straight lines hence they are called linear equations.
If we have a linear function in the form of y = f(x) = mx + b,
then we may eyeball the graph, starting from the y-intercept b,
eyeball the tilt using the slope m.
Graphs of Quadratic Equations

12. Graphs of Quadratic Equations
We start with an example of a graph gives the general
shape of the graphs of 2nd (quadratic) degree equations.

13. 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

14. Graphs of Quadratic Equations
We start with an example of a graph gives the general
shape of the graphs of 2nd (quadratic) degree equations.
Make a table
Example A. Graph y = –x2

15. 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 2nd (quadratic) degree equations.
Make a table
Example A. Graph y = –x2
y

16. 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 2nd (quadratic) degree equations.
Make a table
Example A. Graph y = –x2
y
-16
-9
-4
-1
0
-1
-4
-9
-16

17. 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 2nd (quadratic) degree equations.
Make a table
Example A. Graph y = –x2
y
-16
-9
-4
-1
0
-1
-4
-9
-16

18. Graphs of Quadratic Equations
We start with an example of a graph gives the general
shape of the graphs of 2nd (quadratic) degree equations.
Make a table
Example A. Graph y = –x2
x
-4
-3
-2
-1
0
1
2
3
4
y
-16
-9
-4
-1
0
-1
-4
-9
-16

19. The graphs of 2nd (quadratic) equations are called parabolas.
Graphs of Quadratic Equations

20. 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

21. 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
Properties of Parabolas:

22. 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
Properties of Parabolas:
• Parabolas are symmetric with respect to a center line

23. 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
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, or the tip.

24. 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
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, or the tip.
the vertex
the vertex

25. 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
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.
The vertex is the starting point for graphing a parabola,
i.e. the graph of y = ax2 + bx + c, a 2nd degree function.
the vertex
the vertex

26. 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
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 = ax2 + bx + c is at x = .
–b
2a
The vertex is the starting point for graphing a parabola,
i.e. the graph of y = ax2 + bx + c, a 2nd degree function.
the vertex
the vertex

27. 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.

28. 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

29. 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)

30. 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

31. 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

32. 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

33. 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
Note the y values are symmetric
around the vertex just as the
points.
–12
–15
–16
–15
–12

34. 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
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

35. 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
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

36. 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
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

37. 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
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

38. 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
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

39. 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
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

40. Graphs of Quadratic Equations
Example B. Graph y = x2 – 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

41. Graphs of Quadratic Equations
When graphing parabolas, we must also give the x-intercepts
and the y-intercept.

42. 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.

43. 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

44. 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.

45. 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.

46. 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.

47. 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.

48. 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.

49. 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-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.

50. (2nd way) To graph a parabola y = ax2 + bx + c.
Graphs of Quadratic Equations

51. (2nd way) To graph a parabola y = ax2 + bx + c.
1. Set x = in the equation to find the vertex.
-b
2a
Graphs of Quadratic Equations

52. (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 feasible.
-b
2a
Graphs of Quadratic Equations

53. (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 feasible.
3. Locate its reflection across the center line. These three
points form the tip of the parabola.
-b
2a
Graphs of Quadratic Equations

54. (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 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

55. (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 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

56. (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 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
Example C. Graph y = –x2 + 2x + 15

57. (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 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
The vertex is at x = 1, y = 16
Example C. Graph y = –x2 + 2x + 15

58. (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 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
The vertex is at x = 1, y = 16
y-intercept is at (0, 15)
Example C. Graph y = –x2 + 2x + 15

59. (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 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
The vertex is at x = 1, y = 16
y-intercept is at (0, 15)
Plot its reflection (2, 15)
Example C. Graph y = –x2 + 2x + 15

60. (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 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
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
Example C. Graph y = –x2 + 2x + 15

61. (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 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
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
Example C. Graph y = –x2 + 2x + 15

62. (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 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
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
Example C. Graph y = –x2 + 2x + 15

63. (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 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
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
Example C. Graph y = –x2 + 2x + 15 (1, 16)

64. (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 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
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
Example C. Graph y = –x2 + 2x + 15 (1, 16)
(0, 15)

65. (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 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
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
Example C. Graph y = –x2 + 2x + 15 (1, 16)
(0, 15) (2, 15)

66. (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 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
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
Example C. Graph y = –x2 + 2x + 15 (1, 16)
(0, 15) (2, 15)

67. (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 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
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
Example C. Graph y = –x2 + 2x + 15 (1, 16)
(0, 15) (2, 15)
(-3, 0) (5, 0)

68. Finally, we make the observation that given y = ax2 + …,
if a > 0, then the parabola opens upward.
Graphs of Quadratic Equations

69. Finally, we make the observation that given y = ax2 + …,
if a > 0, then the parabola opens upward.
Graphs of Quadratic Equations
if a < 0, then the parabola opens downward.

70. 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

71. Exercise B. Graph the parabolas by taking a table around the
vertex that reflect the symmetry. Find the x and y intercepts.
Graphs of Quadratic Equations
4. y = x2 – 4 5. y = –x2 + 4
2. y = –x2 3. y = x2
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

72. Exercise C. Graph the following parabolas by plotting the
vertex point, the y–intercept and its reflection.
Find the x intercepts.
Graphs of Quadratic Equations
23. y = x2 – 2x – 3 24. y = –x2 + 2x + 3
25. y = x2 + 2x – 3 26. y = –x2 – 2x + 3
27. y = x2 – 2x – 8 28. y = –x2 + 2x + 8
29. y = x2 + 2x – 8 30. y = –x2 – 2x + 8
31. y = x2 + 4x – 21 32. y = x2 – 4x – 45
33. y = x2 – 6x – 27 34. 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

73. Graphs of Quadratic Equations
Answers to odd problems.
5. y = –x2 + 4
(0,0)
(0, 4)
(2, 0)
7. y = –x2 – 4
(0, –4)
(2, 0)(–4 , 0)
(–1, 9)
(–2, 0)
15. y = –x2 – 2x + 8 17. a. y = x2 + 4x – 5 b. y = –x2 – 4x + 5
19. y = x2 + 4x – 21 21. y = x2 – 6x – 27
(1, 0)(–5, 0)
(–2 , –9)
(0, –5)
(3, 0)(–7, 0)
(–2 , –25)
(0, –21)
(1, 0)(–5, 0)
(–2 , 9)
(0, 5)
(9, 0)(–3, 0)
(0 , –27) (0, –21)
(3 , –36)
23. y = x2 – 2x – 3
(3, 0)(–1, 0)
(–1 , –4)
(0, –3)
3. y = x2