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Chapter 4
Applications of Quadratic Models
To graph the quadratic equation
• y = ax2+ bx +c
 Use vertex formula
xv = -b/2a
 Find the y-coordinate of the vertex by substituting x, into
the equation of parabola
 Locate x-intercepts by setting y= 0
 Locate y-intercept by evaluating y for x = 0
 Locate axis of symmetry
 Vertex form for a Quadratic Formula
vertex of the graph is
xv , yv
y = a(x – xv ) 2 + yv
where the
The x- coordinate of the vertex of the graph of y = ax2 +
bx
y = 2x2+ 8x + 6
xv = - 8/2(2) Substitute – 2 for x
=-2
yv = 2(-2) + 8( -2) + 6
• xv = -b/2a
6
= 8 – 16 + 6 = -2
So the vertex is the point (-2, -2)
The x-intercepts of the graph by
setting y equal to zero
-3
0 = 2x2+ 8x + 6 = 2(x + 1)(x + 3)
x + 1 = 0 or x + 3 = 0
x = -1, x = -3
-1
The x-intercepts are the points (-1, 0)
and (-3, 0)
And y-intercept = 6
(-2, -2)
-5
Problem 39, Pg 211
a
b). The price of a room is 20 + 2x, the number of1500
rooms rented is
60 – 3x
The total revenue earned at that price is
(20 + 2x) (60 – 3x)
c). Enter Y1 = 20 + 2x
Y2 = 60 – 3x
Y3 = (20 + 2x)(60 – 3x) in your calculator 0
Tbl start = 0
Tb1 = 1
The values in the calculator’s table should match with table
d). If x = 20, the total revenue is 0
e). Graph
f). The owner must charge atleast $24 but no more than $36 per
room to make a revenue atleast $1296 per night
g). The maximum revenue from night is $1350, which is obtained
by charging $30 per room and renting 45 rooms at this price
20
No of price
increases
•
•
•
•
•
•
•
•
•
•
•
•
•
0
1
2
3
4
5
6
7
8
10
12
16
20
x be the no of price increases
Price of room = 20 + 2x
No. of rooms rented = 60 – 3x
Total revenue = (20 + 2x)(60 – 3x)
Price of room No. of rooms rented
Total revenue
20
22
24
26
28
30
32
34
36
40
44
52
60
60
57
54
Lowest 51
48
45
42
39
36
Highest 30
24
12
0
1200
1254
1296
1326
1344
1350
1344
1326
1296
1200
1056
624
0
Max. Revenue
4.2 Using Calculator for Quadratic Regression
Pg 216, Ex 4
STAT Enter datas
Store in Y1by pressing
STAT right 5 VARS right 1, 1 Enter
Press Y = and select Plot 1 then press ZOOM 9
Graph
• 4.2 Ex 13, Page 221
Tower
500
Tower
Cable
20
0
2000
4000
The vertex is (2000, 20) and another point on the cable is (0, 500).
2
Using vertex form, y = a(x – 2000) + 20
Use point (0, 500)
500 = a(0 – 2000) + 20,
500 = 4,000,000a + 20
2
480 = 4,000,000a a = 0.000012 The shape of the cable is given by the equation y = 0.00012(x – 2000) + 20
4.3 Maximum and Minimum Values
Example 1
a) Revenue = (price of one item) (number of items sold)
R = x(600 – 15x)
R = 600x – 15x 2
b) Graph is a parabola
6000
c) x
v = - b/2a = -600/2(-15) = 20
Rv= 600(20) – 15(20) 2 = 6000
5000
R = 600x – 15x 2
20
40
Late Nite Blues should charge $20 for a pair of jeans in order to maximize revenue at
$6000 a week
Solving Systems with the Graphing Calculator
Example 2 Page 225
Enter Y1=
Pg 226
Enter Y1, Y2
Press 2nd , table
Enter Window
Press window
Press graph
press graph
Press 2nd and calc
4.4 Quadratic Inequalities
Example 1 , Pg 235h
h = 256 t – 16t2
h = 256 t – 16t2
256 t – 16 t2 > 800
1000
800
500
5
The solution set
4.25 < t < 11.75
10
Compound inequality
15
t
Interval Notation
An interval is a set that consists of all the real numbers
between two numbers a and b.
If the set includes both of the end points a and b ,
so that a< x < b, then the set is called a closed
interval
Denoted by [a, b]
If the set does not include its endpoints, so that a<x<b,
then it is called an open interval, and is denoted by
(a, b)
-2
-1
0
x < -2
(-  , -2 ) U ( 2,  )
1
2
x>2
3
Examples
• Half open or half closed interval
• 3<x<6
0
1
2
3
4
5
6
-5
-4
-3
-2
3
4
5
6
7
x>9
[ -9, )
-9
x < 1 or x > 4
-1
( - , 1] U ( 4,  )
-8
0
-7
1
-6
2
-1
0
4.4 Ex 20, Pg 241
*
Y1 =16 – 6x - x^2
Y2 = 21
X min = -9.4 Xmax = 9.4
Ymin = -25 Ymax = 25
-8
a.
b.
c.
d.
-1
0
2
y > 0 for –8 < x < 2. Using interval notation [-8, 2]
y< 0 for x < - 8 or x > 2. Using interval notation ( - , -8) U ( - 1, )
y< 21 for x < -5 or x > -1. Using interval notation ( , -5) U (- 1,  )
y > 21 for –5 < x < -2. Using interval notation (-5, -2)