The document discusses two forms of quadratic equations: standard form (y = ax^2 + bx + c) and vertex form (y = a(x - h)^2 + k). It shows that vertex form can be rewritten as standard form by expanding the expression, with a = a, b = -2ah, and c = ah^2 + k. This allows the vertex (h, k) to be determined directly from the standard form equation by solving for h and k in terms of a, b, and c. An example demonstrates rewriting a vertex form equation into standard form and finding the vertex (-1, -4).

1. SECTION 6-4
Graphing y = ax2 + bx + c

2. TWO FORMS OF A
QUADRATIC
Standard form: Vertex Form:

3. TWO FORMS OF A
QUADRATIC
Standard form: Vertex Form:
2
y = ax + bx + c

4. TWO FORMS OF A
QUADRATIC
Standard form: Vertex Form:
2 2
y = ax + bx + c y = a(x − h) + k

5. TWO FORMS OF A
QUADRATIC
Standard form: Vertex Form:
2 2
y = ax + bx + c y = a(x − h) + k
How do these equations relate to each other? How does it
apply to the real world?

6. TWO FORMS OF A
QUADRATIC
Standard form: Vertex Form:
2 2
y = ax + bx + c y = a(x − h) + k
How do these equations relate to each other? How does it
apply to the real world?
If we have a standard form equation, how do we know
where to start graphing it?

7. TWO FORMS OF A
QUADRATIC
Standard form: Vertex Form:
2 2
y = ax + bx + c y = a(x − h) + k
How do these equations relate to each other? How does it
apply to the real world?
If we have a standard form equation, how do we know
where to start graphing it?
VERTEX!!!

8. EXAMPLE 1
Rewrite in standard form.
2
y = 3(x + 1) − 4

9. EXAMPLE 1
Rewrite in standard form.
2
y = 3(x + 1) − 4
2
y = 3(x + 2x + 1) − 4

10. EXAMPLE 1
Rewrite in standard form.
2
y = 3(x + 1) − 4
2
y = 3(x + 2x + 1) − 4
2
y = 3x + 6x + 3 − 4

11. EXAMPLE 1
Rewrite in standard form.
2
y = 3(x + 1) − 4
2
y = 3(x + 2x + 1) − 4
2
y = 3x + 6x + 3 − 4
2
y = 3x + 6x − 1

12. EXAMPLE 1
Rewrite in standard form.
2
y = 3(x + 1) − 4
2
y = 3(x + 2x + 1) − 4
2
y = 3x + 6x + 3 − 4
2
y = 3x + 6x − 1

13. EXAMPLE 1
Rewrite in standard form.
2
y = 3(x + 1) − 4
Vertex form:
2
y = 3(x + 2x + 1) − 4
2
y = 3x + 6x + 3 − 4
2
y = 3x + 6x − 1

14. EXAMPLE 1
Rewrite in standard form.
2
y = 3(x + 1) − 4
Vertex form:
2
y = 3(x + 2x + 1) − 4
2
y = 3x + 6x + 3 − 4
2
y = 3x + 6x − 1
Standard form:

15. EXAMPLE 1
Rewrite in standard form.
2
y = 3(x + 1) − 4
Vertex form:
2
y = 3(x + 2x + 1) − 4
2
y = 3x + 6x + 3 − 4
2
y = 3x + 6x − 1
Standard form:
Both are translations of y = 3x2

16. FURTHER EXAMINATION
Rewrite the generic vertex form equation into standard
form and compare.
2
y = a(x − h) + k

17. FURTHER EXAMINATION
Rewrite the generic vertex form equation into standard
form and compare.
2
y = a(x − h) + k
2 2
y = a(x − 2hx + h ) + k

18. FURTHER EXAMINATION
Rewrite the generic vertex form equation into standard
form and compare.
2
y = a(x − h) + k
2 2
y = a(x − 2hx + h ) + k
2 2
y = ax − 2ahx + ah + k

19. FURTHER EXAMINATION
Rewrite the generic vertex form equation into standard
form and compare.
2
y = a(x − h) + k
2 2
y = a(x − 2hx + h ) + k
2 2
y = ax − 2ahx + ah + k
2
y = ax + bx + c

20. FURTHER EXAMINATION
Rewrite the generic vertex form equation into standard
form and compare.
2
y = a(x − h) + k
2 2
y = a(x − 2hx + h ) + k
2 2
y = ax − 2ahx + ah + k
2
y = ax + bx + c

21. FURTHER EXAMINATION
Rewrite the generic vertex form equation into standard
form and compare.
2
y = a(x − h) + k
2 2
y = a(x − 2hx + h ) + k
2 2
y = ax − 2ahx + ah + k
2
y = ax + bx + c

22. FURTHER EXAMINATION
Rewrite the generic vertex form equation into standard
form and compare.
2
y = a(x − h) + k
2 2
y = a(x − 2hx + h ) + k
2 2
y = ax − 2ahx + ah + k
2
y = ax + bx + c
a=a

23. FURTHER EXAMINATION
Rewrite the generic vertex form equation into standard
form and compare.
2
y = a(x − h) + k
2 2
y = a(x − 2hx + h ) + k
2 2
y = ax − 2ahx + ah + k
2
y = ax + bx + c
a=a

24. FURTHER EXAMINATION
Rewrite the generic vertex form equation into standard
form and compare.
2
y = a(x − h) + k
2 2
y = a(x − 2hx + h ) + k
2 2
y = ax − 2ahx + ah + k
2
y = ax + bx + c
a=a

25. FURTHER EXAMINATION
Rewrite the generic vertex form equation into standard
form and compare.
2
y = a(x − h) + k
2 2
y = a(x − 2hx + h ) + k
2 2
y = ax − 2ahx + ah + k
2
y = ax + bx + c
a=a b = -2ah

26. FURTHER EXAMINATION
Rewrite the generic vertex form equation into standard
form and compare.
2
y = a(x − h) + k
2 2
y = a(x − 2hx + h ) + k
2 2
y = ax − 2ahx + ah + k
2
y = ax + bx + c
a=a b = -2ah

27. FURTHER EXAMINATION
Rewrite the generic vertex form equation into standard
form and compare.
2
y = a(x − h) + k
2 2
y = a(x − 2hx + h ) + k
2 2
y = ax − 2ahx + ah + k
2
y = ax + bx + c
a=a b = -2ah

28. FURTHER EXAMINATION
Rewrite the generic vertex form equation into standard
form and compare.
2
y = a(x − h) + k
2 2
y = a(x − 2hx + h ) + k
2 2
y = ax − 2ahx + ah + k
2
y = ax + bx + c
a=a b = -2ah c= ah2 + k

29. Can we now find the vertex from our standard form
equation?

30. Can we now find the vertex from our standard form
equation?
YES!!!!

31. Can we now find the vertex from our standard form
equation?
YES!!!!
Find h and k in the standard form equation found in
example 1 and state the vertex.

32. Can we now find the vertex from our standard form
equation?
YES!!!!
Find h and k in the standard form equation found in
example 1 and state the vertex.
2
y = 3x + 6x − 1

33. Can we now find the vertex from our standard form
equation?
YES!!!!
Find h and k in the standard form equation found in
example 1 and state the vertex.
2
y = 3x + 6x − 1
a=3

34. Can we now find the vertex from our standard form
equation?
YES!!!!
Find h and k in the standard form equation found in
example 1 and state the vertex.
2
y = 3x + 6x − 1
a=3 b = -2ah

35. Can we now find the vertex from our standard form
equation?
YES!!!!
Find h and k in the standard form equation found in
example 1 and state the vertex.
2
y = 3x + 6x − 1
a=3 b = -2ah
6 = -2(3)h

36. Can we now find the vertex from our standard form
equation?
YES!!!!
Find h and k in the standard form equation found in
example 1 and state the vertex.
2
y = 3x + 6x − 1
a=3 b = -2ah
6 = -2(3)h
6 = -6h

37. Can we now find the vertex from our standard form
equation?
YES!!!!
Find h and k in the standard form equation found in
example 1 and state the vertex.
2
y = 3x + 6x − 1
a=3 b = -2ah
6 = -2(3)h
6 = -6h
h = -1

38. Can we now find the vertex from our standard form
equation?
YES!!!!
Find h and k in the standard form equation found in
example 1 and state the vertex.
2
y = 3x + 6x − 1
a=3 b = -2ah c = ah2 + k
6 = -2(3)h
6 = -6h
h = -1

39. Can we now find the vertex from our standard form
equation?
YES!!!!
Find h and k in the standard form equation found in
example 1 and state the vertex.
2
y = 3x + 6x − 1
a=3 b = -2ah c = ah2 + k
6 = -2(3)h -1 = 3(-1)2 + k
6 = -6h
h = -1

40. Can we now find the vertex from our standard form
equation?
YES!!!!
Find h and k in the standard form equation found in
example 1 and state the vertex.
2
y = 3x + 6x − 1
a=3 b = -2ah c = ah2 + k
6 = -2(3)h -1 = 3(-1)2 + k
6 = -6h -1 = 3 + k
h = -1

41. Can we now find the vertex from our standard form
equation?
YES!!!!
Find h and k in the standard form equation found in
example 1 and state the vertex.
2
y = 3x + 6x − 1
a=3 b = -2ah c = ah2 + k
6 = -2(3)h -1 = 3(-1)2 + k
6 = -6h -1 = 3 + k
h = -1 k = -4

42. Can we now find the vertex from our standard form
equation?
YES!!!!
Find h and k in the standard form equation found in
example 1 and state the vertex.
2
y = 3x + 6x − 1
a=3 b = -2ah c = ah2 + k
6 = -2(3)h -1 = 3(-1)2 + k (h, k) =
6 = -6h -1 = 3 + k
h = -1 k = -4

43. Can we now find the vertex from our standard form
equation?
YES!!!!
Find h and k in the standard form equation found in
example 1 and state the vertex.
2
y = 3x + 6x − 1
a=3 b = -2ah c = ah2 + k
6 = -2(3)h -1 = 3(-1)2 + k (h, k) = (-1, -4)
6 = -6h -1 = 3 + k
h = -1 k = -4

44. Can we now find the vertex from our standard form
equation?
YES!!!!
Find h and k in the standard form equation found in
example 1 and state the vertex.
2
y = 3x + 6x − 1
a=3 b = -2ah c = ah2 + k
6 = -2(3)h -1 = 3(-1)2 + k (h, k) = (-1, -4)
6 = -6h -1 = 3 + k
h = -1 k = -4
2
y = 3(x + 1) − 4

45. EXAMPLE 2
Find the vertex of the parabola for
2
y = −2x − 12x − 22

46. EXAMPLE 2
Find the vertex of the parabola for
2
y = −2x − 12x − 22
a = -2

47. EXAMPLE 2
Find the vertex of the parabola for
2
y = −2x − 12x − 22
a = -2 b = -2ah

48. EXAMPLE 2
Find the vertex of the parabola for
2
y = −2x − 12x − 22
a = -2 b = -2ah
-12 = -2(-2)h

49. EXAMPLE 2
Find the vertex of the parabola for
2
y = −2x − 12x − 22
a = -2 b = -2ah
-12 = -2(-2)h
-12 = 4h

50. EXAMPLE 2
Find the vertex of the parabola for
2
y = −2x − 12x − 22
a = -2 b = -2ah
-12 = -2(-2)h
-12 = 4h
h = -3

51. EXAMPLE 2
Find the vertex of the parabola for
2
y = −2x − 12x − 22
a = -2 b = -2ah c = ah2 + k
-12 = -2(-2)h
-12 = 4h
h = -3

52. EXAMPLE 2
Find the vertex of the parabola for
2
y = −2x − 12x − 22
a = -2 b = -2ah c = ah2 + k
-12 = -2(-2)h -22 = (-2)(-3)2 + k
-12 = 4h
h = -3

53. EXAMPLE 2
Find the vertex of the parabola for
2
y = −2x − 12x − 22
a = -2 b = -2ah c = ah2 + k
-12 = -2(-2)h -22 = (-2)(-3)2 + k
-12 = 4h -22 = (-2)(9) + k
h = -3

54. EXAMPLE 2
Find the vertex of the parabola for
2
y = −2x − 12x − 22
a = -2 b = -2ah c = ah2 + k
-12 = -2(-2)h -22 = (-2)(-3)2 + k
-12 = 4h -22 = (-2)(9) + k
h = -3 -22 = -18 + k

55. EXAMPLE 2
Find the vertex of the parabola for
2
y = −2x − 12x − 22
a = -2 b = -2ah c = ah2 + k
-12 = -2(-2)h -22 = (-2)(-3)2 + k
-12 = 4h -22 = (-2)(9) + k
h = -3 -22 = -18 + k
k = -4

56. EXAMPLE 2
Find the vertex of the parabola for
2
y = −2x − 12x − 22
a = -2 b = -2ah c = ah2 + k
-12 = -2(-2)h -22 = (-2)(-3)2 + k
-12 = 4h -22 = (-2)(9) + k
h = -3 -22 = -18 + k
k = -4
Vertex:

57. EXAMPLE 2
Find the vertex of the parabola for
2
y = −2x − 12x − 22
a = -2 b = -2ah c = ah2 + k
-12 = -2(-2)h -22 = (-2)(-3)2 + k
-12 = 4h -22 = (-2)(9) + k
h = -3 -22 = -18 + k
k = -4
Vertex: (-3, -4)

58. NEWTON’S FORMULA
2
h=− gt + v0t + h0
1
2

59. NEWTON’S FORMULA
2
h=− gt + v0t + h0
1
2
h = Height

60. NEWTON’S FORMULA
2
h=− gt + v0t + h0
1
2
h = Height g = Acceleration due to gravity

61. NEWTON’S FORMULA
2
h=− gt + v0t + h0
1
2
h = Height g = Acceleration due to gravity
32 ft/sec2 or 9.8 m/sec2

62. NEWTON’S FORMULA
2
h=− gt + v0t + h0
1
2
h = Height g = Acceleration due to gravity
32 ft/sec2 or 9.8 m/sec2
t = Time

63. NEWTON’S FORMULA
2
h=− gt + v0t + h0
1
2
h = Height g = Acceleration due to gravity
32 ft/sec2 or 9.8 m/sec2
t = Time v0 = Initial velocity

64. NEWTON’S FORMULA
2
h=− gt + v0t + h0
1
2
h = Height g = Acceleration due to gravity
32 ft/sec2 or 9.8 m/sec2
t = Time v0 = Initial velocity h0 = Initial height

65. NEWTON’S FORMULA
2
h=− gt + v0t + h0
1
2
h = Height g = Acceleration due to gravity
32 ft/sec2 or 9.8 m/sec2
t = Time v0 = Initial velocity h0 = Initial height
This formula gives a highly accurate approximation of
any object in motion (thrown or free fall)

66. EXAMPLE 3
Matt Mitarnowski is on a bridge 22 feet above thewater.
Suppose he drops a ball over the 3 foot-high railing. Write an
equation, graph it, then estimate how long it will take to hit
the water.

67. EXAMPLE 3
Matt Mitarnowski is on a bridge 22 feet above thewater.
Suppose he drops a ball over the 3 foot-high railing. Write an
equation, graph it, then estimate how long it will take to hit
the water.
2
h=− gt + v0t + h0
1
2

68. EXAMPLE 3
Matt Mitarnowski is on a bridge 22 feet above thewater.
Suppose he drops a ball over the 3 foot-high railing. Write an
equation, graph it, then estimate how long it will take to hit
the water.
2
h=− gt + v0t + h0
1
2
0 = − 12

69. EXAMPLE 3
Matt Mitarnowski is on a bridge 22 feet above thewater.
Suppose he drops a ball over the 3 foot-high railing. Write an
equation, graph it, then estimate how long it will take to hit
the water.
2
h=− gt + v0t + h0
1
2
0 = − 1 2 (32)

70. EXAMPLE 3
Matt Mitarnowski is on a bridge 22 feet above thewater.
Suppose he drops a ball over the 3 foot-high railing. Write an
equation, graph it, then estimate how long it will take to hit
the water.
2
h=− gt + v0t + h0
1
2
2
0 = − 2 (32) t +
1

71. EXAMPLE 3
Matt Mitarnowski is on a bridge 22 feet above thewater.
Suppose he drops a ball over the 3 foot-high railing. Write an
equation, graph it, then estimate how long it will take to hit
the water.
2
h=− gt + v0t + h0
1
2
2
0 = − 2 (32) t + 0t +
1

72. EXAMPLE 3
Matt Mitarnowski is on a bridge 22 feet above thewater.
Suppose he drops a ball over the 3 foot-high railing. Write an
equation, graph it, then estimate how long it will take to hit
the water.
2
h=− gt + v0t + h0
1
2
2
0 = − 2 (32) t + 0t + 25
1

73. EXAMPLE 3
Matt Mitarnowski is on a bridge 22 feet above thewater.
Suppose he drops a ball over the 3 foot-high railing. Write an
equation, graph it, then estimate how long it will take to hit
the water.
2
h=− gt + v0t + h0
1
2
2
0 = − 2 (32) t + 0t + 25
1
2
0 = −16t + 25

74. EXAMPLE 3
Matt Mitarnowski is on a bridge 22 feet above thewater.
Suppose he drops a ball over the 3 foot-high railing. Write an
equation, graph it, then estimate how long it will take to hit
the water.
2
h=− gt + v0t + h0
1
2
2
0 = − 2 (32) t + 0t + 25
1
2
0 = −16t + 25

75. EXAMPLE 3
Matt Mitarnowski is on a bridge 22 feet above thewater.
Suppose he drops a ball over the 3 foot-high railing. Write an
equation, graph it, then estimate how long it will take to hit
the water.
2
h=− gt + v0t + h0
1
2
2
0 = − 2 (32) t + 0t + 25
1
2
0 = −16t + 25

76. 2
0 = −16t + 25

77. 2
0 = −16t + 25
2
16t = 25

78. 2
0 = −16t + 25
2
16t = 25
25
2
t=
16

79. 2
0 = −16t + 25
2
16t = 25
25
2
t=
16
25
2
t =±
16

80. 2
0 = −16t + 25
2
16t = 25
25
2
t=
16
25
2
t =±
16
5
t=±
4

81. 2
0 = −16t + 25
2
16t = 25
25
2
t=
16
25
2
t =±
16
5
t=±
4
It will take about 1 and 1/4 seconds for the
ball to hit the water.

82. HOMEWORK

83. HOMEWORK
p. 367 #1 - 27
“A ship is safe in harbor, but that’s not what ships are for.”
- Unknown