1. Quadratic functions or second degree functions are
functions of the form y = f(x) = ax2 + bx + c, a = 0.
Second Degree Functions

2. Quadratic functions or second degree functions are
functions of the form y = f(x) = ax2 + bx + c, a = 0.
Second Degree Functions
Below is the graph of y = f(x) = –x2.

3. x y
–4
–3
–2
–1
0
1
2
3
4
Quadratic functions or second degree functions are
functions of the form y = f(x) = ax2 + bx + c, a = 0.
Second Degree Functions
Below is the graph of y = f(x) = –x2.

4. Quadratic functions or second degree functions are
functions of the form y = f(x) = ax2 + bx + c, a = 0.
Second Degree Functions
Below is the graph of y = f(x) = –x2.
x y
–4
–3
–2
–1
0 0
1 –1
2 –4
3 –9
4 –16

5. Quadratic functions or second degree functions are
functions of the form y = f(x) = ax2 + bx + c, a = 0.
Second Degree Functions
Below is the graph of y = f(x) = –x2.
x y
–4 –16
–3 –9
–2 –4
–1 –1
0 0
1 –1
2 –4
3 –9
4 –16

6. Quadratic functions or second degree functions are
functions of the form y = f(x) = ax2 + bx + c, a = 0.
Second Degree Functions
Below is the graph of y = f(x) = –x2 .
x y
–4 –16
–3 –9
–2 –4
–1 –1
0 0
1 –1
2 –4
3 –9
4 –16

7. The graphs of 2nd (quadratic)
equations are called parabolas.
Second Degree Functions

8. The graphs of 2nd (quadratic)
equations are called parabolas.
Parabolas describe the paths of
thrown objects (or the upside–
down paths).
Second Degree Functions

9. The graphs of 2nd (quadratic)
equations are called parabolas.
Parabolas describe the paths of
thrown objects (or the upside–
down paths).
Second Degree Functions

10. Second Degree Functions
Properties of Parabolas:

11. Second Degree Functions
Properties of Parabolas:
I. A parabola is symmetric to a center line.

12. Second Degree Functions
Properties of Parabolas:
I. A parabola is symmetric to a center line.
II. The highest or lowest point of the parabola sits on
the center line.

13. Second Degree Functions
Properties of Parabolas:
I. A parabola is symmetric to a center line.
II. The highest or lowest point of the parabola sits on
the center line. This point is called the vertex.

14. Second Degree Functions
Properties of Parabolas:
I. A parabola is symmetric to a center line.
II. The highest or lowest point of the parabola sits on
the center line. This point is called the vertex.
III. If the positions of the vertex and another point on
the parabola are known, the parabola is determined.

15. Second Degree Functions
The vertex position is given by the following formula.
Properties of Parabolas:
I. A parabola is symmetric to a center line.
II. The highest or lowest point of the parabola sits on
the center line. This point is called the vertex.
III. If the positions of the vertex and another point on
the parabola are known, the parabola is determined.

16. Vertex Formula: The vertex of the
parabola y = ax2 + bx + c is at x = –b
2a
Second Degree Functions

17. Vertex Formula: The vertex of the
parabola y = ax2 + bx + c is at x = –b
2a
Second Degree Functions
Moreover, if a > 0, the parabola opens up,
if a < 0 the parabola opens down.

18. Vertex Formula: The vertex of the
parabola y = ax2 + bx + c is at x = –b
2a
To graph a parabola y = ax2 + bx + c:
Second Degree Functions
Moreover, if a > 0, the parabola opens up,
if a < 0 the parabola opens down.

19. Vertex Formula: The vertex of the
parabola y = ax2 + bx + c is at x = –b
2a
To graph a parabola y = ax2 + bx + c:
1. Set x = in the equation to find the vertex.–b
2a
Second Degree Functions
Moreover, if a > 0, the parabola opens up,
if a < 0 the parabola opens down.

20. Vertex Formula: The vertex of the
parabola y = ax2 + bx + c is at x = –b
2a
To graph a parabola y = ax2 + bx + c:
1. Set x = in the equation to find the vertex.
2. Find another point.
–b
2a
Second Degree Functions
Moreover, if a > 0, the parabola opens up,
if a < 0 the parabola opens down.

21. Vertex Formula: The vertex of the
parabola y = ax2 + bx + c is at x = –b
2a
To graph a parabola y = ax2 + bx + c:
1. Set x = in the equation to find the vertex.
2. Find another point. (Use x=0 to get the y intercept
if its not the vertex.)
–b
2a
Second Degree Functions
Moreover, if a > 0, the parabola opens up,
if a < 0 the parabola opens down.

22. Vertex Formula: The vertex of the
parabola y = ax2 + bx + c is at x = –b
2a
To graph a parabola y = ax2 + bx + c:
1. Set x = in the equation to find the vertex.
2. Find another point. (Use x=0 to get the y intercept
if its not the vertex.)
3. Locate the reflection across the center line, these
three points form the tip of the parabola.
–b
2a
Second Degree Functions
Moreover, if a > 0, the parabola opens up,
if a < 0 the parabola opens down.

23. Vertex Formula: The vertex of the
parabola y = ax2 + bx + c is at x = –b
2a
To graph a parabola y = ax2 + bx + c:
1. Set x = in the equation to find the vertex.
2. Find another point. (Use x=0 to get the y intercept
if its not the vertex.)
3. Locate the reflection across the center line, these
three points form the tip of the parabola. Trace the
parabola.
–b
2a
Second Degree Functions
Moreover, if a > 0, the parabola opens up,
if a < 0 the parabola opens down.

24. Vertex Formula: The vertex of the
parabola y = ax2 + bx + c is at x = –b
2a
To graph a parabola y = ax2 + bx + c:
1. Set x = in the equation to find the vertex.
2. Find another point. (Use x=0 to get the y intercept
if its not the vertex.)
3. Locate the reflection across the center line, these
three points form the tip of the parabola. Trace the
parabola.
4. Set y=0 to find the x intercept.
–b
2a
Second Degree Functions
Moreover, if a > 0, the parabola opens up,
if a < 0 the parabola opens down.

25. Vertex Formula: The vertex of the
parabola y = ax2 + bx + c is at x = –b
2a
To graph a parabola y = ax2 + bx + c:
1. Set x = in the equation to find the vertex.
2. Find another point. (Use x=0 to get the y intercept
if its not the vertex.)
3. Locate the reflection across the center line, these
three points form the tip of the parabola. Trace the
parabola.
4. Set y=0 to find the x intercept. If no real solution
exists, there is no x intercept.
–b
2a
Second Degree Functions
Moreover, if a > 0, the parabola opens up,
if a < 0 the parabola opens down.

26. Example C. Graph y = x2 – 4x – 12
Second Degree Functions

27. Example C. Graph y = x2 – 4x – 12
Vertex: set x =
–(–4)
2(1)
Second Degree Functions
To graph a parabola y = ax2 + bx + c:
1. Set x = in the equation to find the vertex.–b
2a

28. Example C. Graph y = x2 – 4x – 12
Vertex: set x = = 2
–(–4)
2(1)
Second Degree Functions
To graph a parabola y = ax2 + bx + c:
1. Set x = in the equation to find the vertex.–b
2a

29. Second Degree Functions
Example C. Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y = 22 – 4*2 – 12 = –16,
–(–4)
2(1)
To graph a parabola y = ax2 + bx + c:
1. Set x = in the equation to find the vertex.–b
2a

30. Second Degree Functions
Example C. Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y = 22 – 4*2 – 12 = –16,
–(–4)
2(1)
so v = (2, –16).
To graph a parabola y = ax2 + bx + c:
1. Set x = in the equation to find the vertex.–b
2a

31. Second Degree Functions
Example C. Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y = 22 – 4*2 – 12 = –16,
–(–4)
2(1)
so v = (2, –16).
To graph a parabola y = ax2 + bx + c:
1. Set x = in the equation to find the vertex.–b
2a
(2, –16)

32. Second Degree Functions
Example C. Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y = 22 – 4*2 – 12 = –16,
–(–4)
2(1)
so v = (2, –16).
To graph a parabola y = ax2 + bx + c:
1. Set x = in the equation to find the vertex.
2. Find another point.
(Use x=0 or the y intercept if possible.)
–b
2a
(2, –16)

33. Another point:
Set x = 0
Second Degree Functions
Example C. Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y = 22 – 4*2 – 12 = –16,
–(–4)
2(1)
so v = (2, –16).
To graph a parabola y = ax2 + bx + c:
1. Set x = in the equation to find the vertex.
2. Find another point.
(Use x=0 or the y intercept if possible.)
–b
2a
(2, –16)

34. Another point:
Set x = 0 then y = –12
or (0, –12)
Second Degree Functions
Example C. Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y = 22 – 4*2 – 12 = –16,
–(–4)
2(1)
so v = (2, –16).
To graph a parabola y = ax2 + bx + c:
1. Set x = in the equation to find the vertex.
2. Find another point.
(Use x=0 or the y intercept if possible.)
–b
2a
(2, –16)

35. Another point:
Set x = 0 then y = –12
or (0, –12)
Second Degree Functions
(2, –16)
(0, –12)
Example C. Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y = 22 – 4*2 – 12 = –16,
–(–4)
2(1)
so v = (2, –16).
To graph a parabola y = ax2 + bx + c:
1. Set x = in the equation to find the vertex.
2. Find another point.
(Use x=0 or the y intercept if possible.)
–b
2a

36. Another point:
Set x = 0 then y = –12
or (0, –12)
Second Degree Functions
(2, –16)
(0, –12)
Example C. Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y = 22 – 4*2 – 12 = –16,
–(–4)
2(1)
so v = (2, –16).
To graph a parabola y = ax2 + bx + c:
1. Set x = in the equation to find the vertex.
2. Find another point.
(Use x=0 or the y intercept if possible.)
3. Locate the reflection across the center line,
these three points form the tip of the parabola.
–b
2a

37. (2, –16)
(0, –12) (4, –12)
Second Degree Functions
Another point:
Set x = 0 then y = –12
or (0, –12)
It's reflection across the
mid–line is (4, –12)
Example C. Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y = 22 – 4*2 – 12 = –16,
–(–4)
2(1)
so v = (2, –16).
To graph a parabola y = ax2 + bx + c:
1. Set x = in the equation to find the vertex.
2. Find another point.
(Use x=0 or the y intercept if possible.)
3. Locate the reflection across the center line,
these three points form the tip of the parabola.
–b
2a

38. (2, –16)
(0, –12) (4, –12)
Second Degree Functions
Another point:
Set x = 0 then y = –12
or (0, –12)
It's reflection across the
mid–line is (4, –12)
Example C. Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y = 22 – 4*2 – 12 = –16,
–(–4)
2(1)
so v = (2, –16).
To graph a parabola y = ax2 + bx + c:
1. Set x = in the equation to find the vertex.
2. Find another point.
(Use x=0 or the y intercept if possible.)
3. Locate the reflection across the center line,
these three points form the tip of the parabola.
Trace the parabola.
–b
2a

39. (2, –16)
(0, –12) (4, –12)
Second Degree Functions
Another point:
Set x = 0 then y = –12
or (0, –12)
It's reflection across the
mid–line is (4, –12)
Set y = 0 and get x–int:
x2 – 4x – 12 = 0
Example C. Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y = 22 – 4*2 – 12 = –16,
–(–4)
2(1)
so v = (2, –16).

40. (2, –16)
(0, –12) (4, –12)
Second Degree Functions
Another point:
Set x = 0 then y = –12
or (0, –12)
It's reflection across the
mid–line is (4, –12)
Set y = 0 and get x–int:
x2 – 4x – 12 = 0
(x + 2)(x – 6) = 0
x = –2, x = 6
Example C. Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y = 22 – 4*2 – 12 = –16,
–(–4)
2(1)
so v = (2, –16).

41. Example C. Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y = 22 – 4*2 – 12 = –16,
–(–4)
2(1)
so v = (2, –16).
(2, –16)
(0, –12) (4, –12)
Second Degree Functions
Following is an example of maximization via the vertex.
Another point:
Set x = 0 then y = –12
or (0, –12)
It's reflection across the
mid–line is (4, –12)
Set y = 0 and get x–int:
x2 – 4x – 12 = 0
(x + 2)(x – 6) = 0
x = –2, x = 6

42. Example D. The "Crazy Chicken" can sell 120 whole
roasted chicken for $8 in one day. For every $0.50
increases in the price, 4 less chickens are sold. Find
the price that will maximize the revenue (the sale).
Second Degree Functions

43. Example D. The "Crazy Chicken" can sell 120 whole
roasted chicken for $8 in one day. For every $0.50
increases in the price, 4 less chickens are sold. Find
the price that will maximize the revenue (the sale).
Second Degree Functions
Set x to be the number of $0.50 increments above the
base price of $8.00.

44. Example D. The "Crazy Chicken" can sell 120 whole
roasted chicken for $8 in one day. For every $0.50
increases in the price, 4 less chickens are sold. Find
the price that will maximize the revenue (the sale).
Second Degree Functions
Set x to be the number of $0.50 increments above the
base price of $8.00.
So the price is (8 + 0.5x) or (8 + x/2).

45. Example D. The "Crazy Chicken" can sell 120 whole
roasted chicken for $8 in one day. For every $0.50
increases in the price, 4 less chickens are sold. Find
the price that will maximize the revenue (the sale).
Second Degree Functions
Set x to be the number of $0.50 increments above the
base price of $8.00.
So the price is (8 + 0.5x) or (8 + x/2).
The number of chicken sold at this price is (120 – 4x).

46. Example D. The "Crazy Chicken" can sell 120 whole
roasted chicken for $8 in one day. For every $0.50
increases in the price, 4 less chickens are sold. Find
the price that will maximize the revenue (the sale).
Second Degree Functions
Set x to be the number of $0.50 increments above the
base price of $8.00.
So the price is (8 + 0.5x) or (8 + x/2).
The number of chicken sold at this price is (120 – 4x).
Hence the revenue, depending on the number of
times of $0.50 price hikes x, is
R(x) = (8 + x/2)(120 – 4x)

47. Example D. The "Crazy Chicken" can sell 120 whole
roasted chicken for $8 in one day. For every $0.50
increases in the price, 4 less chickens are sold. Find
the price that will maximize the revenue (the sale).
Second Degree Functions
Set x to be the number of $0.50 increments above the
base price of $8.00.
So the price is (8 + 0.5x) or (8 + x/2).
The number of chicken sold at this price is (120 – 4x).
Hence the revenue, depending on the number of
times of $0.50 price hikes x, is
R(x) = (8 + x/2)(120 – 4x) = 960 + 28x – 2x2.

48. Example D. The "Crazy Chicken" can sell 120 whole
roasted chicken for $8 in one day. For every $0.50
increases in the price, 4 less chickens are sold. Find
the price that will maximize the revenue (the sale).
Second Degree Functions
Set x to be the number of $0.50 increments above the
base price of $8.00.
So the price is (8 + 0.5x) or (8 + x/2).
The number of chicken sold at this price is (120 – 4x).
Hence the revenue, depending on the number of
times of $0.50 price hikes x, is
R(x) = (8 + x/2)(120 – 4x) = 960 + 28x – 2x2.
This a 2nd degree equation whose graph is a
parabola that opens downward.

49. Example D. The "Crazy Chicken" can sell 120 whole
roasted chicken for $8 in one day. For every $0.50
increases in the price, 4 less chickens are sold. Find
the price that will maximize the revenue (the sale).
Second Degree Functions
Set x to be the number of $0.50 increments above the
base price of $8.00.
So the price is (8 + 0.5x) or (8 + x/2).
The number of chicken sold at this price is (120 – 4x).
Hence the revenue, depending on the number of
times of $0.50 price hikes x, is
R(x) = (8 + x/2)(120 – 4x) = 960 + 28x – 2x2.
This a 2nd degree equation whose graph is a
parabola that opens downward. The vertex is the
highest point on the graph where R is the largest.

50. Second Degree Functions
x
R
R = 960 + 28x – 2x2.

51. Second Degree Functions
x
RThe vertex of this
parabola is at
x = = 7.–(28)
2(–2)
R = 960 + 28x – 2x2.

52. Second Degree Functions
x
RThe vertex of this
parabola is at
x = = 7.
So R(7) = $1058 gives
the maximum revenue.
–(28)
2(–2)
R = 960 + 28x – 2x2.
v = (7, 1058)

53. Second Degree Functions
x
RThe vertex of this
parabola is at
x = = 7.
So R(7) = $1058 gives
the maximum revenue.
–(28)
2(–2)
R = 960 + 28x – 2x2.
v = (7, 1058)
More precisely, raise
the price x = 7 times to
8 + 7(0.50) = 11.50 per
chicken,chicken,

54. Second Degree Functions
x
RThe vertex of this
parabola is at
x = = 7.
So R(7) = $1058 gives
the maximum revenue.
–(28)
2(–2)
R = 960 + 28x – 2x2.
v = (7, 1058)
More precisely, raise
the price x = 7 times to
8 + 7(0.50) = 11.50 per
chicken,chicken, and we can sell 120 – 4(7) = 92 chickens
per day with the revenue of 11.5(92) = 1058.

55. Second Degree Functions
x
RThe vertex of this
parabola is at
x = = 7.
So R(7) = $1058 gives
the maximum revenue.
–(28)
2(–2)
R = 960 + 28x – 2x2.
v = (7, 1058)
More precisely, raise
the price x = 7 times to
8 + 7(0.50) = 11.50 per
chicken,chicken, and we can sell 120 – 4(7) = 92 chickens
per day with the revenue of 11.5(92) = 1058.
Any inverse square law in science is 2nd degree.
Laws related to distance or area in mathematics are
2nd degree. That’s why 2nd equations are important.

56. 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.

57. Exercise A. 1. Practice drawing the parabolas y = ax2 + bx + c.
Graphs of Quadratic Equations
a > 0
a < 0

58. Graphs of Quadratic Equations
A. Graph the parabolas by plotting the vertex and using
the symmetry. Find the x and y intercepts, if any.
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

59. Graphs of Quadratic Equations
Ex B. Graph the following parabolas by plotting the vertex
point, the y–intercept and its reflection.
Verify that there are no x intercepts
(i.e. they have complex roots).
1. y = x2 – 2x + 8 2. y = –x2 + 2x – 5
3. y = x2 + 2x + 3 4. y = –x2 – 3x – 4
5. y = 2x2 + 3x + 4 6. y = x2 – 4x + 32

60. Graphs of Quadratic Equations
Ex C. Carry out the following steps to find the answer for 1 to 3.
a. Let x be the number of times we would raise the price at
each time from the base price of
so the new price would be .
b. Each price hike drop out so the number of customers is .
c. The revenue R(x) is the product of the formulas in a and b which is
2nd degree formula. Find it’s vertex which gives the maximum revenue.
1. Mom’s Kitchen can sell 60 Mom’s Famous apple pies for $12 each
everyday. For every fifty-cent increase in the price, the number of pie sold
drops by 1 pie per day. Use this information to maximize the revenue.
What should you charge?
2. Popi’s Dinner can sell 72 Popi’s Special meal at $24 per person.
For each dollar rise in the price, 2 less Popi’s Special meals would be sold.
Use this information to maximize the revenue.
What should you charge?
3. Jr. Eatery can sell 480 fried chickens at $8 each everyday.
For every 25 cents increase in the price, four less chickens are sold.
Use this information to maximize the revenue.
What should you charge?

61. Graphs of Quadratic Equations
Carry out the following steps to find the answer for 4 to 6.
a. Let (x, y) be any point on the line, solving y in terms of x,
the coordinates may be expressed as .
b. Using the distance formula, the distance between P and a
point on the line is.
c. The minimum distance happens at the vertex of the 2nd
degree equation in the square-root (why?).
5. The distance from the origin to the line y = 2x + 1. Draw.
4. The distance from the point (2, 3) to the line
y = –2x + 3. Draw.
6. The distance from the point (–2, 1) to the line
y = –2x + 2. Draw.

62. (Answers to odd problems) Exercise A.
5. y = –x2 + 4
(0,0)
(0, 4)
7. y = –x2 – 4
(0, –4)
3. y = -x2
11. y = –x2 – 2x + 3 13. y = x2 + 2x – 89. y = x2+2x+3
(-1, 2)
(0, 3)
(-1, 4)
(0, 3)
(-3, 0) (1, 0)
(-1, -9)
(0, -8)
(2, 0)(-4, 0)
(2, 0)(–4 , 0)
(–1, 9)
15. y = –x2 – 2x + 8 17. a. y = x2 + 4x – 5 b. y = –x2 – 4x + 5
(1, 0)(–5, 0)
(–2 , –9)
(0, –5) (1, 0)(–5, 0)
(–2 , 9)
(0, 5)
Graphs of Quadratic Equations

63. 19. y = x2 + 4x – 21 21. y = x2 – 6x – 27
(3, 0)(–7, 0)
(–2 , –25)
(0, –21)
(9, 0)(–3, 0)
(0 , –27) (0, –21)
(3 , –36)
Exercise B.
1. y = x2 – 2x + 8 3. y = x2 + 2x + 3 5. y = 2x2 + 3x + 4
(1, 7)
(0, 8)
(-1, 7)
(0, 8)
(-3/4, 23/8)
(0, 4)
Exercise C.
1. The maximum
revenue is
R(18)=$882. The
price should be
$21 per pie.
3. The maximum
revenue is is
R(44)=5776. The
price should be
$19 per chicken.
5. The minimum distance
happens at (-2/3, 1/3)
(2, 8) (-2, 8) (-1.5, 4)
(-2/3, 1/3)