Modeling
The ability to model problems or phenomena by algebraic expressions and equations is the ultimate goal of any algebra course
The examples are designed with interactive investigation and to show how mathematical techniques are used for learning
get opportunity to explore open-minded modeling problems.
understand new situation
Functions
Functions are useful not only in Calculus but in nearby every field students may pursue. We employ celebrated “ Rule of Four” all problems should
be considered using Algebraic Method Verbal Algebraic Expression Numerical Graphical
Students learn to write algebraic expression from verbal description, to recognize trends in a table of data, extract & interpret information from the graph of a function
Graphs
No tool for conveying information about a
system is more powerful than a graph
Large number of examples are explained
plotting by hand and using graphing
calculator(TI 83/84)
Ch 1 - Linear Model
Mathematical techniques are used to• Analyze data• Identify trends• Predict the effects of change
This quantitative methods are the concepts of skills of algebra
You will use skills you learned in elementary algebra to solve problems and to study a variety of phenomena
the description of relationships between variables by using equations, graphs, and table of values. This process is called Mathematical Modeling
Some examples of Linear Models ( Example 1 pg – 3)
0 1 2 3 4 5 Length of Rentals
20
15
10
5
Length of Cost of
Rental rental (dollars)
(hours)
0 5
1 8
2 11
3 14
(t, C)
(0, 5)
(1, 8)
(2, 11)
(3, 14)
C = 5 + 3(0)C = 5 + 3(1)C = 5 + 3(2)C = 5 + 3(3)
Equation
C = ($ 5 isurance fee) + $ 3 x (number of hours)C = 5 + 3. t
Cost ofRentals
Exercise 1.1
( Example 2, pg – 10)
0 4 8 12 16 20 24
200
180
175
150
125
100
75
50
25
The Equation g = 20 – 1 m 12
m 0 48 96 144 192
g 20 16 12 8 4 Let g = 5, 5 = 20 – 1 m 12 60 = 240 – m (Multiply by 12 both sides ) - - 180 = - m (multiply by – 1 both sides)m = 180
Leon has traveled more than 180 miles if he has less than 5 gallons of gas left
4 gallons
m
gTable
Intercepts
Consider the graph of the equation3y – 4x = 12
The y-coordinate of the x-intercept is zero, so we set y = 0
in the equation to get 3(0) – 4x = 12
x = - 3
The x-intercept is the point (-3, 0).
Also, the x-coordinate of the y-intercept is zero, so we set
x = 0 in the equation to get3y – 4(0) = 12
y = 4
The y-intercept is (0, 4)( -3, 0)
(0, 4)
Example 9 , Pg = 10
x - y = 1 9 4Set x = 0, 0 y 9 – 4 = 1 - y 4 = 1, y = - 4The y-intercept is the point (0, - 4)
Set y = 0, x - 0 = 1 9 4
x = 1, x = 99The x-int is the point (9, 0)
(0, -4)
(9, 0)
Find the intercepts and graph the equation
1.2 Using Graphing CalculatorSolving y in terms of x
Press Y1=2x - 5 Press 2nd and Table Press Zoom and 6
Pg 12, 2x – y = 5
Press Y1 = -1.5x + 0.6 Press window and Enter
Pg 14, Example 2, 3x + 2y = 16Press Zoom and 6
Finding Coordinates with a Graphing Calculator (Pg 15)
Press Y1 enter Press 2nd and Table Press Graph and then press, Trace and enter
“Bug” begins flashing on the display. The coordinates of the bug appear at the bottom of the display.Use the left and right arrows to move the bug along the graph
Graph the equation Y = -2.6x – 5.4
X min = -5, X max = 4.4, Y min = - 20 , Y max = 15
Using Graphing Calculator to solve the equation, (Pg 19)
Equation 572 – 23x = 181
Press Window and enter Press Y1 and Y2 and enter Press 2nd and Table
Press Graph and Trace
Y1 Y2
Graphical Solution of Inequalities (Pg – 17)
Consider the inequality 285 – 15x > 150
x 0 2 4 6 8 10 12
285 – 15x 285 255 225 195 165 135 105
10 25
300
200150100
- 100
y = 285 – 15x
1.3 Measuring Steepness ( pg 23)
Which path is more strenuous ?
5 ft
2 ft
Steepness measures how sharply the altitude increases..To compare the steepness of two inclined paths, we compute the ratio of changein horizontal distance for each path
1.3 Slope (Ex- 3, Pg 25)
Definition of Slope: The slope of a line is the ratio
Change in y- Coordinate
Change in x- coordinate
A B
0 2 3 4
54321
Slope = Change in y-coordinate = 5 - 4 = 1 Change in x- coordinate 4 – 2 2
Significance of the slope (Ex 5, Pg 28)
1 2 3 4 5 t
250
200
150
100
50
t = 2
D = 100
H (4, 200)
G (2, 100)
D Change in distance = 100miles = 50 miles per hour
T Change in time 2 hoursSlope m =
No of hours
Dis
tan
ce i
n m
iles
tra
vele
d
The slope represents the trucker’s average speed or velocity
1.4 Equations of Lines ( Pg 40)
y = 2x
y = ½ x
y = -2x
These lines have the same y- intercept but different slopes
y = 2x + 3
y = 2x + 1
y = 2x - 2
These lines have the same slope but different y - intercepts
1.4 Slope – Intercept Formy = mx + b ( m is slope of a line, and b is the y- intercept)
3x + 4y = 6 Subtract 3x from both sides
4y = - 3x + 6 Divide both sides by 4
y = - 3x + 3
4 2
m= -3 and b = 3
4 2
General
Slope- Intercept Method of Graphing b Start here
New point
y = mx + b
x
y
A Formula for Slope
Slope Formula m = y2 – y1 = 9-(-6) -15 = - 3
x2 – x1 7 – 2 5
m = Y x
P1 (2, 9)
P2 (7, -6)
10
5
-5
10
The slope of the line passing through the points P1(x1, y1) and P2 ( x2, y2) is given by
1.5 Linear Regression (Lines of Best Fit)
The datas in the scatterplot are roughly linear, we can estimate the location of imaginary”lines best fit” that passes as close as possible to the data points
We can make the predictions about the data. The process of predicting a value
of y based on a straightline that fits the data is called a linear regression, and the line itself called the regression line.
The equation of the regression line is usually used(instead of graph) to predict values
Example of Linear Regression ( pg 55)
1000 2000Boiling Point 0 C
200
100
Hea
t of
Vap
oriz
atio
n (k
J)
(900, 100)
(1560, 170)
a) Slope = m= 170 – 100 = 0.106 1560 – 900The equation of regression line isy- y1 = m(x – x1)y – 100 = 0.106(x – 900) , y = 0.106x + 4.6
b) Regression equation for potassium bromide , x = 1435y = 0.106(1435) + 4.6
a) Estimate a line of best fit and find the equation of the regression lineb) Use the regression line to predict the heat of vaporization of potassium bromide, whose boiling temperature is 14350C
Choose two points in the regression line
112
96
80
64
48
20 40 60 80 100 Age (months)
The graph is not linear because her rate of growth is not constant; her growth slows down as she approaches her adult height. The short time of interval the graph is close to a line, and that line can be used to approximate the coordinates of points on the curve.
Hei
ght (
cm)
Interpolation- The process of estimating between known data points
Extrapolation- Making predictions beyond the range of known data
Ex 1.5 No 13, Pg 65
Loss in Mass (mg)
Vol. Of GasCubic cm
20 40 60 80 100
100
80
60
40
20
a) Student E made a mistake in the experiment since for a loss in mass of 88mg according to a line passing through the other points, volume should have been about 68cm3 instead of 76 cm3
b) The line of best fit passes through points (0,0) and (80, 60). The slope is m = 60/80 = 0.75 the equation of line is y = 0.75xc) Let y = 1000d) 1000= 0.75x , x = 1333 1/3 The mass of 1000cm3 of the gas is about 1333mge) The density of unknown gas is 1333mg = 1333mg = 1333mg = 1333 mg/liter 1000cm3 1000militers 1 literSince oxygen has a density of 1330mg/liter, oxygen is the most likely gas
Hydrogen 8mg/literNitrogen 1160 mg/literOxygen 1330 mg/literCarbon dioxide 1830 mg/liter
Using Graphing Calculator for Linear RegressionPg 59
Step 1 Press Stat , choose 1 press Enter
Step 3 Stat, right arrow go to 4And enter
Step 4Press Vars 5 ,Right,Right , Enter
Step 5 Press 2nd, Stat Plot and enter
Step 2 Enter Y = 1.95x – 7.86
Step 6 the graph ( Pg 60)
1.6 Additional Properties of Lines ( pg 68)
Horizontal Line
y = 4 Horizontal line4
X axis
Y axis
- 5 5
( - 1, 4) (2, 4)
y = k( const) Horizontal LineSlope = 0
Vertical Line
X axis
Y axis
- 5 5
5
-5
x = 3
(3, 1)
(3, 3)
x = k (const) Vertical line Slope is undefined
Perpendicular Lines
Two lines with slopes m1 and m2 are perpendicular if and only if m2 = - 1/m 1
5-5
-5
5
y = 2/3 x -2y = -3/2 x + 3 3
- 2
X axis
Y axis
m 1 = 2/3m2 = -3/2= - 1/ m1
m 1m2
Parallel Lines
-5 5
- 5
5 y = 2/3 x + 2
y = 2/3 x -2
Two lines with slopes m1 and m2 are parallel if and only if m1 = m 2
X axis
Y axis
Show that the triangle with vertices A(0, 8), B(6, 2) and C(-4, 4) is a right triangle
(-4, 4) (6, 2)
(0, 8)A
B
C
Slope of AB= m1 = 2 – 8 = - 6 = - 1 6 - 0 6Slope of AC =m2 = 4 – 8 = - 4 = 1 - 1 = -1 = 1 = m2
- 4 – 0 - 4 m1 -1
AB is Perpendicular to AC
X axis
Y axis
Example 33 Pg 77
• A) Sketch the triangle with vertices A(-6, -3) , B ( - 6, 3), and C(4, 5).
• B) Find the slope of the side AC.• C) Find the slope of the side of the altitude from point B to
side AC• D) Find an equation for the line that includes the altitude
from point B to side AC
A ( - 6, - 3)
C ( 4, 5)B ( - 6, 3 )
6
-6
-6 6B ) Slope of side AC = - 3 – 5 = - 8 = 4 - 6 – 4 - 10 5C) Slope of the altitude from Pt. B to side AC is perpendicular to AC and therefore – 5 4D) y – 3 = -5/4 ( x + 6) , y = 3/2 x + 15/2