1. Graphs of Quadratic Equations

2. Graphs of Quadratic Equations
The graphs of the equations of the form
y = ax2 + bx + c and x = ay2 + bx + c
are parabolas.

3. Graphs of Quadratic Equations
The graphs of the equations of the form
y = ax2 + bx + c and x = ay2 + bx + c
are parabolas.
Each parabola has a vertex and the center line that contains
the vertex.

4. Graphs of Quadratic Equations
The graphs of the equations of the form
y = ax2 + bx + c and x = ay2 + bx + c
are parabolas.
Each parabola has a vertex and the center line that contains
the vertex.

5. Graphs of Quadratic Equations
The graphs of the equations of the form
y = ax2 + bx + c and x = ay2 + bx + c
are parabolas.
Each parabola has a vertex and the center line that contains
the vertex. Suppose we know another point on the parabola,

6. Graphs of Quadratic Equations
The graphs of the equations of the form
y = ax2 + bx + c and x = ay2 + bx + c
are parabolas.
Each parabola has a vertex and the center line that contains
the vertex. Suppose we know another point on the parabola,

7. Graphs of Quadratic Equations
The graphs of the equations of the form
y = ax2 + bx + c and x = ay2 + bx + c
are parabolas.
Each parabola has a vertex and the center line that contains
the vertex. Suppose we know another point on the parabola,
then the reflection of the point across the center line is also
on the parabola.

8. Graphs of Quadratic Equations
The graphs of the equations of the form
y = ax2 + bx + c and x = ay2 + bx + c
are parabolas.
Each parabola has a vertex and the center line that contains
the vertex. Suppose we know another point on the parabola,
then the reflection of the point across the center line is also
on the parabola.

9. Graphs of Quadratic Equations
The graphs of the equations of the form
y = ax2 + bx + c and x = ay2 + bx + c
are parabolas.
Each parabola has a vertex and the center line that contains
the vertex. Suppose we know another point on the parabola,
then the reflection of the point across the center line is also
on the parabola. There is exactly one parabola that goes
through these three points.

10. Graphs of Quadratic Equations
The graphs of the equations of the form
y = ax2 + bx + c and x = ay2 + bx + c
are parabolas.
Each parabola has a vertex and the center line that contains
the vertex. Suppose we know another point on the parabola,
then the reflection of the point across the center line is also
on the parabola. There is exactly one parabola that goes
through these three points.

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

12. Graphs of Quadratic Equations
When graphing parabolas, we must also give the x-intercepts
and the y-intercept.
The y-intercept is obtained by setting x = 0 and solve for y.

13. Graphs of Quadratic Equations
When graphing parabolas, we must also give the x-intercepts
and the y-intercept.
The y-intercept is obtained by setting x = 0 and solve for y.
The x-intercept is obtained by setting y = 0 and solve for x.

14. Graphs of Quadratic Equations
When graphing parabolas, we must also give the x-intercepts
and the y-intercept.
The y-intercept is obtained by setting x = 0 and solve for y.
The x-intercept is obtained by setting y = 0 and solve for x.
The graphs of y = ax2 + bx = c are up-down parabolas.

15. Graphs of Quadratic Equations
When graphing parabolas, we must also give the x-intercepts
and the y-intercept.
The y-intercept is obtained by setting x = 0 and solve for y.
The x-intercept is obtained by setting y = 0 and solve for x.
The graphs of y = ax2 + bx = c are up-down parabolas.
If a > 0, the parabola opens up.
a>0

16. Graphs of Quadratic Equations
When graphing parabolas, we must also give the x-intercepts
and the y-intercept.
The y-intercept is obtained by setting x = 0 and solve for y.
The x-intercept is obtained by setting y = 0 and solve for x.
The graphs of y = ax2 + bx = c are up-down parabolas.
If a > 0, the parabola opens up.
If a < 0, the parabola opens down.
a>0 a<0
y = x2 y = –x2

17. Graphs of Quadratic Equations
When graphing parabolas, we must also give the x-intercepts
and the y-intercept.
The y-intercept is obtained by setting x = 0 and solve for y.
The x-intercept is obtained by setting y = 0 and solve for x.
The graphs of y = ax2 + bx = c are up-down parabolas.
If a > 0, the parabola opens up.
If a < 0, the parabola opens down.
a>0 a<0
Vertex Formula (up-down parabolas) The x-coordinate of
the vertex of the parabola y = ax2 + bx + c is at x = -b .
2a

18. Graphs of Quadratic Equations
Following are the steps to graph the parabola y = ax2 + bx + c.

19. Graphs of Quadratic Equations
Following are the steps to graph the parabola y = ax2 + bx + c.
3. Set x = -b in the equation to find the vertex.
2a

20. Graphs of Quadratic Equations
Following are the steps to graph the parabola y = ax2 + bx + c.
3. Set x = -b in the equation to find the vertex.
2a
2. Find another point, use the y-intercept (0, c) if possible.

21. Graphs of Quadratic Equations
Following are the steps to graph the parabola y = ax2 + bx + c.
3. Set x = -b in the equation to find the vertex.
2a
2. Find another point, use the y-intercept (0, c) if possible.
3. Locate its reflection across the center line. These three
points form the tip of the parabola. Trace the parabola.

22. Graphs of Quadratic Equations
Following are the steps to graph the parabola y = ax2 + bx + c.
3. Set x = -b in the equation to find the vertex.
2a
2. Find another point, use the y-intercept (0, c) if possible.
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.

23. Graphs of Quadratic Equations
Following are the steps to graph the parabola y = ax2 + bx + c.
3. Set x = -b in the equation to find the vertex.
2a
2. Find another point, use the y-intercept (0, c) if possible.
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.
Example A. Graph y = –x2 + 2x + 15

24. Graphs of Quadratic Equations
Following are the steps to graph the parabola y = ax2 + bx + c.
3. Set x = -b in the equation to find the vertex.
2a
2. Find another point, use the y-intercept (0, c) if possible.
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.
Example A. Graph y = –x2 + 2x + 15
–(2)
The vertex is at x =2(–1) = 1

25. Graphs of Quadratic Equations
Following are the steps to graph the parabola y = ax2 + bx + c.
3. Set x = -b in the equation to find the vertex.
2a
2. Find another point, use the y-intercept (0, c) if possible.
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.
Example A. Graph y = –x2 + 2x + 15
–(2)
The vertex is at x =2(–1) = 1y = 16.

26. Graphs of Quadratic Equations
Following are the steps to graph the parabola y = ax2 + bx + c.
3. Set x = -b in the equation to find the vertex.
2a
2. Find another point, use the y-intercept (0, c) if possible.
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.
Example A. Graph y = –x2 + 2x + 15 (1, 16)
–(2)
The vertex is at x =2(–1) = 1y = 16.

27. Graphs of Quadratic Equations
Following are the steps to graph the parabola y = ax2 + bx + c.
3. Set x = -b in the equation to find the vertex.
2a
2. Find another point, use the y-intercept (0, c) if possible.
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.
Example A. Graph y = –x2 + 2x + 15 (1, 16)
–(2)
The vertex is at x =2(–1) =1y = 16.
The y-intercept is at (0, 15)

28. Graphs of Quadratic Equations
Following are the steps to graph the parabola y = ax2 + bx + c.
3. Set x = -b in the equation to find the vertex.
2a
2. Find another point, use the y-intercept (0, c) if possible.
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.
Example A. Graph y = –x2 + 2x + 15 (1, 16)
–(2)
The vertex is at x =2(–1) =1y = 16. (0, 15)
The y-intercept is at (0, 15)

29. Graphs of Quadratic Equations
Following are the steps to graph the parabola y = ax2 + bx + c.
3. Set x = -b in the equation to find the vertex.
2a
2. Find another point, use the y-intercept (0, c) if possible.
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.
Example A. Graph y = –x2 + 2x + 15 (1, 16)
–(2)
The vertex is at x =2(–1) = 1y = 16. (0, 15)
The y-intercept is at (0, 15)
Plot its reflection (2, 15).

30. Graphs of Quadratic Equations
Following are the steps to graph the parabola y = ax2 + bx + c.
3. Set x = -b in the equation to find the vertex.
2a
2. Find another point, use the y-intercept (0, c) if possible.
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.
Example A. Graph y = –x2 + 2x + 15 (1, 16)
–(2)
The vertex is at x =2(–1) = 1y = 16. (0, 15) (2, 15)
The y-intercept is at (0, 15)
Plot its reflection (2, 15).

31. Graphs of Quadratic Equations
Following are the steps to graph the parabola y = ax2 + bx + c.
3. Set x = -b in the equation to find the vertex.
2a
2. Find another point, use the y-intercept (0, c) if possible.
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.
Example A. Graph y = –x2 + 2x + 15 (1, 16)
–(2)
The vertex is at x =2(–1) = 1y = 16. (0, 15) (2, 15)
The y-intercept is at (0, 15)
Plot its reflection (2, 15)
Draw,

32. Graphs of Quadratic Equations
Following are the steps to graph the parabola y = ax2 + bx + c.
3. Set x = -b in the equation to find the vertex.
2a
2. Find another point, use the y-intercept (0, c) if possible.
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.
Example A. Graph y = –x2 + 2x + 15 (1, 16)
–(2)
The vertex is at x =2(–1) = 1y = 16. (0, 15) (2, 15)
The y-intercept is at (0, 15)
Plot its reflection (2, 15)
Draw, set y = 0 to get x-int:

33. Graphs of Quadratic Equations
Following are the steps to graph the parabola y = ax2 + bx + c.
3. Set x = -b in the equation to find the vertex.
2a
2. Find another point, use the y-intercept (0, c) if possible.
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.
Example A. Graph y = –x2 + 2x + 15 (1, 16)
–(2)
The vertex is at x =2(–1) = 1y = 16. (0, 15) (2, 15)
The y-intercept is at (0, 15)
Plot its reflection (2, 15)
Draw, set y = 0 to get x-int:
–x2 + 2x + 15 = 0

34. Graphs of Quadratic Equations
Following are the steps to graph the parabola y = ax2 + bx + c.
3. Set x = -b in the equation to find the vertex.
2a
2. Find another point, use the y-intercept (0, c) if possible.
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.
Example A. Graph y = –x2 + 2x + 15 (1, 16)
–(2)
The vertex is at x =2(–1) = 1y = 16. (0, 15) (2, 15)
The 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

35. Graphs of Quadratic Equations
Following are the steps to graph the parabola y = ax2 + bx + c.
3. Set x = -b in the equation to find the vertex.
2a
2. Find another point, use the y-intercept (0, c) if possible.
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.
Example A. Graph y = –x2 + 2x + 15 (1, 16)
–(2)
The vertex is at x =2(–1) = 1y = 16. (0, 15) (2, 15)
The 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

36. Graphs of Quadratic Equations
Following are the steps to graph the parabola y = ax2 + bx + c.
3. Set x = -b in the equation to find the vertex.
2a
2. Find another point, use the y-intercept (0, c) if possible.
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.
Example A. Graph y = –x2 + 2x + 15 (1, 16)
–(2)
The vertex is at x =2(–1) = 1y = 16. (0, 15) (2, 15)
The y-intercept is at (0, 15)
Plot its reflection (2, 15)
Draw, set y = 0 to get x-int:
(-3, 0) (5, 0)
–x2 + 2x + 15 = 0
x2 – 2x – 15 = 0
(x + 3)(x – 5) = 0
x = –3, x = 5

37. Graphs of Quadratic Equations
The graphs of the equations
x = ay2 + by + c
are parabolas that open sideway.

38. Graphs of Quadratic Equations
The graphs of the equations
x = ay2 + by + c
are parabolas that open sideway.
If a>0, the parabola opens
to the right.
x = y2

39. Graphs of Quadratic Equations
The graphs of the equations
x = ay2 + by + c
are parabolas that open sideway.
If a>0, the parabola opens If a<0, the parabola open
to the right. to the left.
x = y2 x = –y2

40. Graphs of Quadratic Equations
The graphs of the equations
x = ay2 + by + c
are parabolas that open sideway.
If a>0, the parabola opens If a<0, the parabola open
to the right. to the left.
x = y2 x = –y2
Each sideway parabola is symmetric to a horizontal center
line.

41. Graphs of Quadratic Equations
The graphs of the equations
x = ay2 + by + c
are parabolas that open sideway.
If a>0, the parabola opens If a<0, the parabola open
to the right. to the left.
x = y2 x = –y2
Each sideway parabola is symmetric to a horizontal center
line. The vertex of the parabola is on this line.

42. Graphs of Quadratic Equations
The graphs of the equations
x = ay2 + by + c
are parabolas that open sideway.
If a>0, the parabola opens If a<0, the parabola open
to the right. to the left.
x = y2 x = –y2
Each sideway parabola is symmetric to a horizontal center
line. The vertex of the parabola is on this line. If we know the
location of the vertex and another point on the parabola, the
parabola is completely determined.

43. Graphs of Quadratic Equations
The graphs of the equations
x = ay2 + by + c
are parabolas that open sideway.
If a>0, the parabola opens If a<0, the parabola open
to the right. to the left.
x = y2 x = –y2
Each sideway parabola is symmetric to a horizontal center
line. The vertex of the parabola is on this line. If we know the
location of the vertex and another point on the parabola, the
parabola is completely determined. The vertex formula is the
same as before except it's for the y coordinate.

44. Graphs of Quadratic Equations
Vertex Formula (sideway parabolas)
The y coordinate of the vertex of the parabola
x = ay2 + by + c
is at y = –b .
2a

45. Graphs of Quadratic Equations
Vertex Formula (sideway parabolas)
The y coordinate of the vertex of the parabola
x = ay2 + by + c
is at y = –b .
2a
Following are the steps to graph the parabola x = ay2 + by +
c.

46. Graphs of Quadratic Equations
Vertex Formula (sideway parabolas)
The y coordinate of the vertex of the parabola
x = ay2 + by + c
is at y = –b .
2a
Following are the steps to graph the parabola x = ay2 + by +
c. –b
2a
• Set y = in the equation to find the x coordinate of the
vertex.

47. Graphs of Quadratic Equations
Vertex Formula (sideway parabolas)
The y coordinate of the vertex of the parabola
x = ay2 + by + c
is at y = –b .
2a
Following are the steps to graph the parabola x = ay2 + by +
c. –b
2a
• Set y = in the equation to find the x coordinate of the
vertex.
2. Find another point; use the x intercept (c, 0) if it's not the
vertex.

48. Graphs of Quadratic Equations
Vertex Formula (sideway parabolas)
The y coordinate of the vertex of the parabola
x = ay2 + by + c
is at y = –b .
2a
Following are the steps to graph the parabola x = ay2 + by +
c. –b
2a
• Set y = in the equation to find the x coordinate of the
vertex.
2. Find another point; use the x intercept (c, 0) if it's not the
vertex.
3. Locate the reflection of the point across the horizontal
center line. These three points form the tip of the
parabola. Trace the parabola.

49. Graphs of Quadratic Equations
Vertex Formula (sideway parabolas)
The y coordinate of the vertex of the parabola
x = ay2 + by + c
is at y = –b .
2a
Following are the steps to graph the parabola x = ay2 + by +
c. –b
2a
• Set y = in the equation to find the x coordinate of the
vertex.
2. Find another point; use the x intercept (c, 0) if it's not the
vertex.
3. Locate the reflection of the point across the horizontal
center line. These three points form the tip of the
parabola. Trace the parabola.
4. Set x = 0 to find the y intercept.

50. Graphs of Quadratic Equations
Example B. Graph x = –y2 + 2y + 15

51. Graphs of Quadratic Equations
Example B. Graph x = –y2 + 2y + 15
–(2)
Vertex: set y = 2(–1) =1

52. Graphs of Quadratic Equations
Example B. Graph x = –y2 + 2y + 15
–(2)
Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16

53. Graphs of Quadratic Equations
Example B. Graph x = –y2 + 2y + 15
–(2)
Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16
so v = (16, 1).

54. Graphs of Quadratic Equations
Example B. Graph x = –y2 + 2y + 15
–(2)
Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16
so v = (16, 1).
Another point:
Set y = 0 then x = 15
or (15, 0).

55. Graphs of Quadratic Equations
Example B. Graph x = –y2 + 2y + 15
–(2)
Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16
so v = (16, 1).
Another point:
Set y = 0 then x = 15
or (15, 0). (16, 1)

56. Graphs of Quadratic Equations
Example B. Graph x = –y2 + 2y + 15
–(2)
Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16
so v = (16, 1).
Another point:
Set y = 0 then x = 15
or (15, 0). (16, 1)
(15, 0)

57. Graphs of Quadratic Equations
Example B. Graph x = –y2 + 2y + 15
–(2)
Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16
so v = (16, 1).
Another point:
Set y = 0 then x = 15
or (15, 0). (16, 1)
Plot its reflection.
(15, 0)

58. Graphs of Quadratic Equations
Example B. Graph x = –y2 + 2y + 15
–(2)
Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16
so v = (16, 1).
Another point:
Set y = 0 then x = 15
(15, 2)
or (15, 0). (16, 1)
Plot its reflection.
It's (15, 2). (15, 0)

59. Graphs of Quadratic Equations
Example B. Graph x = –y2 + 2y + 15
–(2)
Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16
so v = (16, 1).
Another point:
Set y = 0 then x = 15
(15, 2)
or (15, 0). (16, 1)
Plot its reflection.
It's (15, 2). (15, 0)
Draw.

60. Graphs of Quadratic Equations
Example B. Graph x = –y2 + 2y + 15
–(2)
Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16
so v = (16, 1).
Another point:
Set y = 0 then x = 15
(15, 2)
or (15, 0). (16, 1)
Plot its reflection.
It's (15, 2) (15, 0)
Draw.

61. Graphs of Quadratic Equations
Example B. Graph x = –y2 + 2y + 15
–(2)
Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16
so v = (16, 1).
Another point:
Set y = 0 then x = 15
(15, 2)
or (15, 0). (16, 1)
Plot its reflection.
It's (15, 2) (15, 0)
Draw. Get y-int:
–y2 + 2y + 15 = 0

62. Graphs of Quadratic Equations
Example B. Graph x = –y2 + 2y + 15
–(2)
Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16
so v = (16, 1).
Another point:
Set y = 0 then x = 15
(15, 2)
or (15, 0). (16, 1)
Plot its reflection.
It's (15, 2) (15, 0)
Draw. Get y-int:
–y2 + 2y + 15 = 0
y2 – 2y – 15 = 0
(y – 5) (y + 3) = 0
y = 5, -3

63. Graphs of Quadratic Equations
Example B. Graph x = –y2 + 2y + 15
–(2)
Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16
so v = (16, 1).
Another point:
Set y = 0 then x = 15
(15, 2)
or (15, 0). (16, 1)
Plot its reflection.
It's (15, 2) (15, 0)
Draw. Get y-int: (0, -3)
–y2 + 2y + 15 = 0
y2 – 2y – 15 = 0
(y – 5) (y + 3) = 0
y = 5, -3

64. Graphs of Quadratic Equations
When graphing parabolas, we must also give the x-
interceptand 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 the
equation 0 = ax2 + bx + c which may or may not have real
number solutions. Hence there might not be any x-intercept.
Following are the steps to graph a parabola y = ax2 + bx + c.
3. Set x = -b in the equation to find the vertex.
2a
2. Find another point, use the y-intercept (0, c) if possible.
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.
The graph of y = ax2 + bx = c are up-down parabolas.
If a > 0, the parabola opens up.
If a < 0, the parabola opens down.

65. Graphs of Quadratic Equations
Exercise A. Graph the following parabolas by finding and
plotting the vertex point and intercepts.
1. y = x2 – 2x – 3 and x = y2 – 2y – 3
2. y = –x2 + 2x + 3 and x = –y2 + 2y + 3
3. y = x2 + 2x – 3 and x = y2 + 2y – 3
4. y = –x2 – 2x + 3 and y = –x2 – 2x + 3
5. x = y2 – 2y – 8 6. x = –y2 + 2y – 5
7. x = y2 + 4y + 3 8. x = –y2 – 3y – 4
9. x = y + 2y – 8
2 10. x = –y2 – 2y + 8
11. x = – y2 – 6y + 27 12. x = y2 – 6y + 60