Unit 4 Lecture 32
Quadratics Applications &
Their Graphs
Lecture 32
1
Objectives
• Formulate a quadratic equation from a
problem situation
• Solve a quadratic equation by using the
quadratic formula
• Graph the quadratic equation
• Interpret the graph in terms of the problem
situation
Lecture 32
2
Earl Black makes tea bags. The cost of making x
million teas bags per month is C = x2 –38 x+400.
The revenue from selling x million tea bags per
month is R = 78x - x2
Find the equation for Profit
Profit = Revenue - Cost
Revenue: R = 78x - x2
Cost: C = x2 –38 x + 400
Lecture 32
3
The equation for profit is,
Profit = Revenue - Cost
Revenue:
R = 78x - x2
Cost:
C = x2 –38 x + 400
P = ( 78x – x2) – (x2 –38 x + 400)
Lecture 32
4
The equation for profit is,
Profit = Revenue - Cost
Revenue:
R = 78x - x2
Cost:
C = x2 –38 x + 400
P = ( 78x – x2) – (x2 –38 x + 400)
P = 78x – x2 – x2 +38 x - 400
Lecture 32
5
The equation for profit is,
Profit = Revenue - Cost
Revenue:
R = 78x - x2
Cost:
C = x2 –38 x + 400
P = ( 78x – x2) – (x2 –38 x + 400)
P = 78x – x2 – x2 + 38 x - 400
P = – 2x2
Lecture 32
6
The equation for profit is,
Profit = Revenue - Cost
Revenue:
R = 78x - x2
Cost:
C = x2 –38 x + 400
P = ( 78x – x2) – (x2 –38 x + 400)
P = 78x – x2 – x2 + 38 x - 400
P = – 2x2 + 116 x
Lecture 32
7
The equation for profit is,
Profit = Revenue - Cost
Revenue:
R = 78x - x2
Cost:
C = x2 –38 x + 400
P = ( 78x – x2) – (x2 –38 x + 400)
P = 78x – x2 – x2 + 38 x - 400
P = – 2x2 + 116 x - 400
Lecture 32
8
Graph the Profit Equation
P = – 2x2 + 116 x - 400
Lecture 32
9
Graph the Profit Equation
P = – 2x2 + 116 x - 400
Vertex:
b
x
2a
Lecture 32
10
Graph the Profit Equation
P = – 2x2 + 116 x - 400
Vertex:
b
x
2a
116 116
x

 29
2(2) 4
Lecture 32
11
Graph the Profit Equation
P = – 2x2 + 116 x - 400
Vertex:
b
x
2a
116 116
x

 29
2(2) 4
P = – 2(29)2 + 116 (29) – 400=1282
Lecture 32
12
Graph the Profit Equation
P = – 2x2 + 116 x - 400
Vertex:
b
x
2a
116 116
x

 29
2(2) 4
P = – 2(29)2 + 116 (29) – 400=1282
Vertex: (29,1282)
x=29, P = $1282
Lecture 32
13
Graph the Profit Equation
P = – 2x2 + 116 x - 400
X-Intercepts: when P = 0, what is x?
Lecture 32
14
Graph the Profit Equation
P = – 2x2 + 116 x - 400
X-Intercepts: when P = 0, what is x?
P = – 2x2 + 116 x - 400
0 = – 2x2 + 116 x - 400
Lecture 32
15
Graph the Profit Equation
P = – 2x2 + 116 x - 400
X-Intercepts: when P = 0, what is x?
P = – 2x2 + 116 x - 400
0 = – 2x2 + 116 x - 400
b  b  4ac
x
2a
2
Lecture 32
16
Graph the Profit Equation
P = – 2x2 + 116 x - 400
X-Intercepts: when P = 0, what is x?
P = – 2x2 + 116 x - 400
0 = – 2x2 + 116 x - 400
b  b  4ac
x
2a
2
116  116  4( 2)( 400)
x
2( 2)
2
Lecture 32
17
Graph the Profit Equation
P = – 2x2 + 116 x - 400
X-Intercepts: when P = 0, what is x?
2
b  b  4ac
x
2a
2
116  116  4( 2)( 400)
x
2( 2)
116  13456  3200
x
4
Lecture 32
18
Graph the Profit Equation
P = – 2x2 + 116 x - 400
X-Intercepts: when P = 0, what is x?
2
b  b  4ac
x
2a
2
116  116  4( 2)( 400)
x
2( 2)
116  13456  3200
x
4
116  10256
x
4
Lecture 32
19
Graph the Profit Equation
P = – 2x2 + 116 x - 400
X-Intercepts: when P = 0, what is x?
2
116  116  4( 2)( 400)
x
2( 2)
116  13456  3200
x
4
116  10256
x
4
116  101.3
x
4
Lecture 32
20
Use the Quadratic Formula to solve:
0 = -2x2 + 116x - 400
116  101.3
x
4
116  101.3 14.7

 3.7
116  101.3 
x
4
4

4

Lecture 32
21
Use the Quadratic Formula to solve:
0 = -2x2 + 116x - 400
116  101.3
x
4
116  101.3 14.7

 3.7
116  101.3 

4

4
x

4
 116  101.3  217.3  54.3
4
4
Lecture 32
22
Use the Quadratic Formula to solve:
0 = -2x2 + 116x - 400
116  101.3
x
4
116  101.3 14.7

 3.7
116  101.3 

4

4
x

4
 116  101.3  217.3  54.3
4
4
Two solutions to make P = 0:
x = 3.7 and x = 54.3 items
Lecture 32
23
Graph the Profit Equation
P = – 2x2 + 116 x - 400
P-Intercept: when x = 0, what is P?
P = – 2x2 + 116 x - 400
Lecture 32
24
Graph the Profit Equation
P = – 2x2 + 116 x - 400
P-Intercept: when x = 0, what is P?
P = – 2x2 + 116 x - 400
P = – 2(0)2 + 116 (0) – 400 = - 400
Lecture 32
25
Graph the Profit Equation
P = – 2x2 + 116 x - 400
P-Intercept: when x = 0, what is P?
P = – 2x2 + 116 x - 400
P = – 2(0)2 + 116 (0) – 400 = - 400
P = - 400
Lecture 32
26
Graph the Profit Equation
P = – 2x2 + 116 x - 400
1. Because a = -2, parabola opens down
2. Vertex: ( 29,1282)
3. X-Intercepts: (3.7, 0) and (54.3,0)
4. P-Intercept: (0,-400)
Lecture 32
27
Graph:
P = – 2x2 + 116 x - 400
(29,1282)
1200
800
400
-400
-800
10
20
30
40
Lecture 32
50
60
28
Graph:
P = – 2x2 + 116 x - 400
(29,1282)
1200
800
400
-400
-800
(54.3,0)
(3.7,0)
10
20
30
40
Lecture 32
50
60
29
Graph:
P = – 2x2 + 116 x - 400
(29,1282)
1200
800
400
(54.3,0)
(3.7,0)
-400
10 20
-800 (0,-400)
30
40
Lecture 32
50
60
30
Graph:
P = – 2x2 + 116 x - 400
(29,1282)
1200
800
400
(54.3,0)
(3.7,0)
-400
10 20
-800 (0,-400)
30
40
Lecture 32
50
60
31
How many items must be made and sold
to generate $500 profit.
P = – 2x2 + 116 x - 400
Lecture 32
32
How many items must be made and sold
to generate $500 profit.
P = – 2x2 + 116 x - 400
500 = -2x2 + 116x - 400
Lecture 32
33
How many items must be made and sold
to generate $500 profit.
P = – 2x2 + 116 x - 400
500 = -2x2 + 116x - 400 Subtract 500 from
both sides
Lecture 32
34
How many items must be made and sold
to generate $500 profit.
P = – 2x2 + 116 x - 400
500 = -2x2 + 116x - 400 Subtract 500 from
both sides
0 = -2x2 + 116x - 900
Lecture 32
35
Use the Quadratic Formula to solve:
0 = -2x2 + 116x - 900
0=
ax2 +
b  b  4ac
x
2a
2
bx + c
Lecture 32
36
Use the Quadratic Formula to solve:
0 = -2x2 + 116x - 900
0=
ax2 +
a = -2
b  b  4ac
x
2a
2
bx + c
b = 116
c = -900
Lecture 32
37
Use the Quadratic Formula to solve:
0 = -2x2 + 116x - 900
0=
ax2 +
a = -2
b  b  4ac
x
2a
2
bx + c
b = 116
c = -900
116  116  4( 2)( 900)
x
2( 2)
2
Lecture 32
38
Use the Quadratic Formula to solve:
0 = -2x2 + 116x - 900
0=
ax2 +
a = -2
b  b  4ac
x
2a
2
bx + c
b = 116
c = -900
116  116  4( 2)( 900)
x
2( 2)
2
116  13456  7200
x
4
Lecture 32
39
Use the Quadratic Formula to solve:
0 = -2x2 + 116x - 900
b  b  4ac
0=
bx + c x 
2a
2
116  116  4( 2)( 900)
x
2( 2)
116  13456  7200
x
4
116  6256
x
4
2
ax2 +
Lecture 32
40
Use the Quadratic Formula to solve:
0 = -2x2 + 116x - 900
b  b  4ac
0=
bx + c x 
2a
2
116  116  4( 2)( 900)
x
2( 2)
116  13456  7200
x
4
116  79.1
116  6256
x
x
4
4
2
ax2 +
Lecture 32
41
Use the Quadratic Formula to solve:
0 = -2x2 + 116x - 900
116  79.1
x
4
116  79.1 36.9

 9.2
116  79.1
x
4
4

4

Lecture 32
42
Use the Quadratic Formula to solve:
0 = -2x2 + 116x - 900
116  79.1
x
4
116  79.1 36.9

 9.2
116  79.1

4

4
x

4
 116  79.1  195.1  48.8
4
4
Lecture 32
43
Use the Quadratic Formula to solve:
0 = -2x2 + 116x - 900
116  79.1
x
4
116  79.1 36.9

 9.2
116  79.1

4

4
x

4
 116  79.1  195.1  48.8
4
4
Lecture 32
44
Use the Quadratic Formula to solve:
0 = -2x2 + 116x - 900
116  79.1
x
4
116  79.1 36.9

 9.2
116  79.1

4

4
x

4
 116  79.1  195.1  48.8
4
4
Two solutions to make P = $500:
x = 9.2 and x = 48.8 items
Lecture 32
45
Graph:
P = – 2x2 + 116 x - 400
1200
800
400
-400
-800
10
20
30
40
Lecture 32
50
60
46
Graph:
P = – 2x2 + 116 x - 400
1200
800
400
-400
-800
10
20
30
40
Lecture 32
50
60
47
Graph:
1200
800
400
-400
-800
P = – 2x2 + 116 x - 400
(48.8,500)
(9.2,500)
10
20
30
40
Lecture 32
50
60
48
An angry student stands on the top of a 250 foot cliff
and throws his book upward with a velocity of 46 feet
per second. The height of the book from the ground is
given by the equation: h = -16t2 + 46t + 250
h is in feet and t is in seconds. Graph the equation.
Lecture 32
49
Graph:
h = -16t2 + 46t + 250
Vertex:
b
x
2a
Lecture 32
50
Graph:
h = -16t2 + 46t + 250
Vertex:
b
x
2a
46
46
t

 1.4
2(16) 32
Lecture 32
51
Graph:
h = -16t2 + 46t + 250
Vertex:
b
x
2a
46
46
t

 1.4
2(16) 32
h = – 16(1.4)2 + 46 (1.4) + 250 = 283
Lecture 32
52
Graph:
h = -16t2 + 46t + 250
Vertex:
b
x
2a
46
46
t

 1.4
2(16) 32
h = – 16(1.4)2 + 46 (1.4) + 250 = 283
Vertex: (1.4, 283)
x=1.4 sec, h = 283 feet
Lecture 32
53
Graph:
h = -16t2 + 46t + 250
t-Intercepts: when h = 0, what is t?
Lecture 32
54
Graph:
h = -16t2 + 46t + 250
t-Intercepts: when h = 0, what is t?
h = – 16t2 + 46 t + 250
Lecture 32
55
Graph:
h = -16t2 + 46t + 250
t-Intercepts: when h = 0, what is t?
h = – 16t2 + 46 t + 250
0 = – 16t2 + 46 t + 250
Lecture 32
56
Graph:
h = -16t2 + 46t + 250
t-Intercepts: when h = 0, what is t?
h = – 16t2 + 46 t + 250
0 = – 16t2 + 46 t + 250
b  b  4ac
x
2a
2
Lecture 32
57
Graph:
h = -16t2 + 46t + 250
t-Intercepts: when h = 0, what is t?
h = – 16t2 + 46 t + 250
0 = – 16t2 + 46 t + 250
b  b  4ac
x
2a
2
46  46  4( 16)(250)
t
2( 16)
2
Lecture 32
58
Graph:
h = -16t2 + 46t + 250
t-Intercepts: when P = 0, what is t?
2
46  46  4( 16)(250)
t
2( 16)
Lecture 32
59
Graph:
h = -16t2 + 46t + 250
t-Intercepts: when P = 0, what is t?
2
46  46  4( 16)(250)
t
2( 16)
46  2116  16000
t
32
Lecture 32
60
Graph:
h = -16t2 + 46t + 250
t-Intercepts: when P = 0, what is t?
2
46  46  4( 16)(250)
t
2( 16)
46  2116  16000
t
32
46  18116
t
32
Lecture 32
61
Graph:
h = -16t2 + 46t + 250
t-Intercepts: when P = 0, what is t?
2
46  46  4( 16)(250)
t
2( 16)
46  2116  16000
t
32
46  18116
t
32
46  134.6
t
32
Lecture 32
62
Graph:
h = -16t2 + 46t + 250
t-Intercepts: when P = 0, what is t?
2
46  46  4( 16)(250)
t
2( 16)
46  2116  16000
t
32
46  18116
t
32
46  134.6
t
32
Lecture 32
63
Use the Quadratic Formula to solve:
0 = -16t2 + 46t + 250
46  134.6
t
32
46  134.6 88.6

   2.8
46  134.6 
t
32
32

32

Lecture 32
64
Use the Quadratic Formula to solve:
0 = -16t2 + 46t + 250
46  134.6
t
32
46  134.6 88.6

 2.8
46  134.6 
32
32
t

32
 46  134.6 180.6

 5.6
32
32
Lecture 32
65
Use the Quadratic Formula to solve:
0 = -16t2 + 46t + 250
46  134.6
t
32
46  134.6 88.6

 2.8
46  134.6 
32
32
t

32
 46  134.6 180.6

 5.6
32
32
Two solutions to make h = 0:
t = -2.8 and t = 5.6 seconds
t = -2.8 does not make sense
Lecture 32
66
Graph:
h = -16t2 + 46t + 250
h-Intercept: when t = 0, what is h?
Lecture 32
67
Graph:
h = -16t2 + 46t + 250
h-Intercept: when t = 0, what is h?
h = – 16t2 + 46 t + 250
Lecture 32
68
Graph:
h = -16t2 + 46t + 250
h-Intercept: when t = 0, what is h?
h = – 16t2 + 46 t + 250
h = – 16(0)2 + 46 (0) + 250
Lecture 32
69
Graph:
h = -16t2 + 46t + 250
h-Intercept: when t = 0, what is h?
h = – 16t2 + 46 t + 250
h = – 16(0)2 + 46 (0) + 250
h = 250 feet
Lecture 32
70
Graph:
h = -16t2 + 46t + 250
1. Because a = -16, parabola opens down
2. Vertex: ( 1.4, 283)
3. t-Intercept: (5.6, 0)
4. h-Intercept: (0, 250)
Lecture 32
71
h = -16t2 + 46t + 250
Graph:
Height
(feet)
(1.4,283)
300
250
200
150
100
50
1
2
3
4
Lecture 32
5
6
Time (sec)
72
h = -16t2 + 46t + 250
Graph:
Height
(feet)
(1.4,283)
300
250
200
150
100
50
(5.6,0)
1
2
3
4
Lecture 32
5
6
Time (sec)
73
h = -16t2 + 46t + 250
Graph:
Height
(feet)
(1.4,283)
300
250 (0,250)
200
150
100
50
(5.6,0)
1
2
3
4
Lecture 32
5
6
Time (sec)
74
h = -16t2 + 46t + 250
Graph:
Height
(feet)
(1.4,283)
300
250 (0,250)
200
150
100
50
(5.6,0)
1
2
3
4
Lecture 32
5
6
Time (sec)
75
Ms. Piggie wants to enclose two adjacent chicken
coops of equal size against the hen house wall. She
has 66 feet of chicken-wire fencing and would like to
make the chicken coup as large as possible. Find the
formula for the area of the chicken coops. Wall
W
W
W
L
Lecture 32
76
Ms. Piggie wants to enclose two adjacent chicken
coops of equal size against the hen house wall. She
has 66 feet of chicken-wire fencing and would like to
make the chicken coup as large as possible. Find the
formula for the area of the chicken coops. Wall
Width Length
Area
W
W
W
5
L
10
15
X
Lecture 32
77
Ms. Piggie wants to enclose two adjacent chicken
coops of equal size against the hen house wall. She
has 66 feet of chicken-wire fencing and would like to
make the chicken coup as large as possible. Find the
formula for the area of the chicken coops. Wall
Width Length
Area
W
5
66-3(5)=56
W
W
5(51)=251
L
10
15
X
Lecture 32
78
Ms. Piggie wants to enclose two adjacent chicken
coops of equal size against the hen house wall. She
has 66 feet of chicken-wire fencing and would like to
make the chicken coup as large as possible. Find the
formula for the area of the chicken coops. Wall
Width Length
Area
W
5
66-3(5)=56
10
66-3(10)=36 10(36)=360
W
W
5(51)=251
L
15
X
Lecture 32
79
Ms. Piggie wants to enclose two adjacent chicken
coops of equal size against the hen house wall. She
has 66 feet of chicken-wire fencing and would like to
make the chicken coup as large as possible. Find the
formula for the area of the chicken coops. Wall
Width Length
Area
W
5
66-3(5)=56
10
66-3(10)=36 10(36)=360
15
66-3(15)=21 15(21)=315
W
W
5(51)=251
L
X
Lecture 32
80
Ms. Piggie wants to enclose two adjacent chicken
coops of equal size against the hen house wall. She
has 66 feet of chicken-wire fencing and would like to
make the chicken coup as large as possible. Find the
formula for the area of the chicken coops. Wall
Width Length
Area
W
5
66-3(5)=56
10
66-3(10)=36 10(36)=360
15
66-3(15)=21 15(21)=315
X
66-3(X)
W
W
5(51)=251
L
X(66-3X)=
66X-3X2
Lecture 32
81
Graph:
A = - 3X2 + 66X
Vertex:
b
x
2a
Lecture 32
82
Graph:
A = - 3X2 + 66X
Vertex:
b
x
2a
66 66
x

 11
2(3) 6
Lecture 32
83
Graph:
A = - 3X2 + 66X
Vertex:
b
x
2a
66 66
x

 11
2(3) 6
A = – 3(11)2 + 66 (11) = 363
Lecture 32
84
Graph:
A = - 3X2 + 66X
Vertex:
b
x
2a
66 66
x

 11
2(3) 6
A = – 3(11)2 + 66 (11) = 363
Vertex: (11, 363)
x=11 feet, A = 363 sq. feet
Lecture 32
85
Graph:
A = - 3X2 + 66X
Vertex:
b
x
2a
66 66
x

 11
2(3) 6
A = – 3(11)2 + 66 (11) = 363
Vertex: (11, 363)
x=11 feet, A = 363 sq. feet
Lecture 32
86
Graph:
A = - 3X2 + 66X
x-Intercepts: when A = 0, what is x?
Lecture 32
87
Graph:
A = - 3X2 + 66X
x-Intercepts: when A = 0, what is x?
A = - 3X2 + 66X
Lecture 32
88
Graph:
A = - 3X2 + 66X
x-Intercepts: when A = 0, what is x?
A = - 3X2 + 66X
0 = - 3X2 + 66X
Lecture 32
89
Graph:
A = - 3X2 + 66X
x-Intercepts: when A = 0, what is x?
A = - 3X2 + 66X
0 = - 3X2 + 66X
0 = - 3X(X - 22)
Lecture 32
90
Graph:
A = - 3X2 + 66X
x-Intercepts: when A = 0, what is x?
A = - 3X2 + 66X
0 = - 3X2 + 66X
0 = - 3X(X - 22)
0 = - 3X
or
Lecture 32
91
Graph:
A = - 3X2 + 66X
x-Intercepts: when A = 0, what is x?
A = - 3X2 + 66X
0 = - 3X2 + 66X
0 = - 3X(X - 22)
0 = - 3X
or
0 = (X - 22)
Lecture 32
92
Graph:
A = - 3X2 + 66X
x-Intercepts: when A = 0, what is x?
A = - 3X2 + 66X
0 = - 3X2 + 66X
0 = - 3X(X - 22)
0 = - 3X
0=X
or
or
0 = (X - 22)
X = 22
Lecture 32
93
Graph:
A = - 3X2 + 66X
A-Intercept: when x = 0, what is A?
Lecture 32
94
Graph:
A = - 3X2 + 66X
A-Intercept: when x = 0, what is A?
A = - 3X2 + 66X
Lecture 32
95
Graph:
A = - 3X2 + 66X
A-Intercept: when x = 0, what is A?
A = - 3X2 + 66X
A = – 3(0)2 + 66 (0)
Lecture 32
96
Graph:
A = - 3X2 + 66X
A-Intercept: when x = 0, what is A?
A = - 3X2 + 66X
A = – 3(0)2 + 66 (0)
A = 0 sq. feet
Lecture 32
97
Graph:
A = - 3X2 + 66X
1. Because a = -3, parabola opens down
2. Vertex: ( 11, 363)
3. x-Intercepts: (0, 0) and (22, 0)
4. A-Intercept: (0, 0)
Lecture 32
98
A = - 3X2 + 66X
Graph:
Area
(sq. feet)
(11,363)
350
300
250
200
150
100
50
3
6
9
12
15
Lecture 32
18
21
Width (feet)
99
Graph:
A = - 3X2 + 66X
Area
(sq. feet)
(11,363)
350
300
250
200
150
100
50
(0,0)
3
6
9
12
15
Lecture 32
(22,0)
18 21
Width (feet)
100
Graph:
A = - 3X2 + 66X
Area
(sq. feet)
(11,363)
350
300
250
200
150
100
50
(0,0)
3
6
9
12
15
Lecture 32
(22,0)
18 21
Width (feet)
101
Graph:
A = - 3X2 + 66X
Area
(sq. feet)
(11,363)
350
300
250
200
150
100
50
(0,0)
3
6
9
12
15
Lecture 32
(22,0)
18 21
Width (feet)
102
Find the dimensions of the coop if the area can
only be 360 sq feet.
When A = 360, what is x?
Lecture 32
103
Find the dimensions of the coop if the area can
only be 360 sq feet.
When A = 360, what is x?
A = - 3X2 + 66X
Lecture 32
104
Find the dimensions of the coop if the area can
only be 360 sq feet.
When A = 360, what is x?
A = - 3X2 + 66X
360 = - 3X2 + 66X
Lecture 32
105
Find the dimensions of the coop if the area can
only be 360 sq feet.
When A = 360, what is x?
A = - 3X2 + 66X
360 = - 3X2 + 66X
Subtract 360 from
both sides
Lecture 32
106
Find the dimensions of the coop if the area can
only be 360 sq feet.
When A = 360, what is x?
A = - 3X2 + 66X
360 = - 3X2 + 66X
0 = - 3X2 + 66X - 360
Lecture 32
Subtract 360 from
both sides
107
Find the dimensions of the coop if the area can
only be 360 sq feet.
When A = 360, what is x?
A = - 3X2 + 66X
360 = - 3X2 + 66X
0 = - 3X2 + 66X - 360
Subtract 360 from
both sides
0 = - 3(X2 – 22X + 120)
Lecture 32
108
Find the dimensions of the coop if the area can
only be 360 sq feet.
When A = 360, what is x?
A = - 3X2 + 66X
360 = - 3X2 + 66X
0 = - 3X2 + 66X - 360
Subtract 360 from
both sides
0 = - 3(X2 – 22X + 120)
0 = - 3(X – 10)(X – 12)
Lecture 32
109
Find the dimensions of the coop if the area can
only be 360 sq feet.
When A = 360, what is x?
A = - 3X2 + 66X
360 = - 3X2 + 66X
0 = - 3X2 + 66X - 360
Subtract 360 from
both sides
0 = - 3(X2 – 22X + 120)
0 = - 3(X – 10)(X – 12)
0 = (X - 10) or
Lecture 32
110
Find the dimensions of the coop if the area can
only be 360 sq feet.
When A = 360, what is x?
A = - 3X2 + 66X
360 = - 3X2 + 66X
0 = - 3X2 + 66X - 360
Subtract 360 from
both sides
0 = - 3(X2 – 22X + 120)
0 = - 3(X – 10)(X – 12)
0 = (X - 10) or
0 = (X - 12)
Lecture 32
111
Find the dimensions of the coop if the area can
only be 360 sq feet.
When A = 360, what is x?
A = - 3X2 + 66X
360 = - 3X2 + 66X
0 = - 3X2 + 66X - 360
Subtract 360 from
both sides
0 = - 3(X2 – 22X + 120)
0 = - 3(X – 10)(X – 12)
0 = (X - 10) or
0 = (X - 12)
X =10
or
X = 12
Lecture 32
112
Graph:
A = - 3X2 + 66X
Area
(sq. feet)
(11,363)
350
300
250
200
150
100
50
(0,0)
3
6
9
12
15
Lecture 32
(22,0)
18 21
Width (feet)
113
A = - 3X2 + 66X
Graph:
Area
(sq. feet)
350
(11,363)
(10,360)
(12,360)
300
250
200
150
100
50
(0,0)
3
6
9
12
15
Lecture 32
(22,0)
18 21
Width (feet)
114
A = - 3X2 + 66X
Graph:
Area
(sq. feet)
350
(11,363)
(10,360)
(12,360)
300
250
200
150
100
50
(0,0)
3
6
9
12
15
Lecture 32
(22,0)
18 21
Width (feet)
115
Lecture 32
116
Lecture 32
117
Lecture 32
118
Lecture 32
119
Lecture 32
120
Lecture 32
121