Ch 1.1

Functions and Graphs

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 toAnalyze dataIdentify trendsPredict 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

General Form For a Linear Equation

The graph of any equation

Ax + By = C

Where A and B are not both equal to zero, is a straight lines

Linear Equations

All the models for examples have equations with a similar formY = ( starting value ) + ( rate of change ) . XGraph will be straight linesSo we called these are Linear EquationsFor Example

C = 6 + 5tThis can be written equivalently as - 5t + C = 6 (subtract 5t from both sides )So It is a linear equation

Note : General Form for a Linear EquationThe graph of any equationAx + By = CWhere A and B are not both equal to zero, is a straight line

Intercepts of a Graph

Intercepts of a graph

The points where a graph crosses the axes are called the intercepts ofthe graph

1. To find the x-intercepts, set y = 0 and solve for x2. To find the y-intercepts, set x = 0 and solve for y

To Graph a Line Using the Intercept Method:

3. Find the intercepts of the linea. To find the x-intercept, set y = 0 and solve for xb. To find the y-intercept, set x = 0 and solve for y.

2. Plot the intercepts.3. Choose a value for x and find a third point on the line.4. Draw a line through the points

Example 17 , Pg = 16

(0, -4)

(9, 0)

Find the intercepts of the graph and graph the equation by the intercept method

y-intercept

x-intercept

x _ y = 1

9 4

Solution, Set 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-intercept is the point (9, 0)

Graphing an Equation

To graph an equation: ( in Graphing Calculator)

1. Press Y = and enter the equation you wish to graph

2. Press WINDOW and select a suitable graphing window

3. Press GRAPH4. Press 2nd and Table

Using Graphing Calculator to solve the equation, Equation 572 – 23x = 181

Press Window and enter Press Y1 and Y2 and enter Press 2nd and Table

Press Graph and Trace

Y1 Y2

Ex 1.1 , 39 ( Pg 18)a) Solve the equation for y in terms of xb) Graph the equation on your calculator in the specified windowc) Make a pencil and paper sketch of the graphLabel the scales on your axes, and the coordinates of the intercepts

3x - 4y = 1200Xmin = - 1000 Ymin = - 1000Xmax = 1000 Y max = 1000XSc1 = 1 YSc 1 = 1Solution –3x – 4y = 1200Find y in the given equation-4y = - 3x + 1200 ( Isolate y)y = -3/-4 x + 1200/-4y = ¾ x – 300Y1 = ¾ x - 300 Hit Y, Enter

Y1 = ¾ x - 300 Hit GraphHit Window , Enter the values

Graphing CalculatorPress Y = key, EnterY1 = 1454 –16X2

Press 2nd WINDOW to access the Tbl Set start from 0

And the increment of one unit in the x values ,

Press 2nd and graph for table Press graph Press WINDOW

Ex4( pg 15)Leon’s camper has a 20-gallon gas tank and he gets 12 miles to the

gallon. (that is , he uses gallon per mile)Complete the table of values for the amount of gas, g, left in Leon’s tankafter driving m miles

Solution

a) Write an equation that expresses the amount of gas, g, in Leon’s fuel tank in terms of the number of miles, m, he has driven.

b) Graph the equationc) How much gas will Leon use between 8am, when his odometer reads 96 miles, and 9 a.m, when the

odometer reads 144 miles ? Illustrate the graphd) If Leon has less than 5 gallons of gas left, how many miles has he driven ? Illustrate on the graph.

m 0 48 96 144 192

g

12

1

P.T.O

Exercise 1.1 ( Example 4, pg – 15)

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

g

mTable

0

4

8

1

2

16

2

0

24

200

180

175

150

125

100

75 50

25

Function Notation

f(x) = y

Input Variable Output Variable

Example y = f(x) = 1454 –16x2

When x= 1, y= f(1)= 1438, We read as “f of 1 equals 1438”When x = 2, y = f(2) =1390, We read as “ f of 2 equals 1390 ”

Ch 1.2 Function Notation

Independent Variable Dependent VariableThis function t = Input variable and h is the output variableExample h = f(t) = 1454 –16t2

When t= 1, i.e after 1 second the book’s height, h= f(1)= 1454 – 16 (1)2

=1438 feet, We read as “f of 1 equals 1438”When t = 2, h = f(2) = 1454 – 16(2)2 = 1390 feet, We read as” f of 2 equals 1390 “

As of 2006, the Sears Tower in Chicago is the nation’s tallest building, at 1454 feet. If an algebra book is dropped from the top of the Sears Tower, its height above the ground after t seconds is given by the equation h = 1454 –16t2

f(t) = h

Definition of function ( Pg 19)

A function is a relationship between two variables for which a unique value of the output variable can be determined from a value of the input variable.

Function Notation f(x) = y

Input variable Output Variable

Ch 1.2 (pg 19) Definition and Function

Example –To rent a plane flying lessons cost $ 800 plus $30 per hour

Suppose C = 30 t + 800 (t > 0)

When t = 0, C = 30(0) + 800= 800When t = 4, C = 30(4) + 800 = 920When t = 10, C = 30(10) + 800 = 1100

The variable t in Equation is called the input or independentvariable, and C is the output or dependent variable, because its values are determined bythe value of t.

This type of relationship is called a function

t c

0 800

4 920

10 1100

(t, c)

(0, 800)

(4, 920)

(10, 1100)

Table Ordered Pair

InputOutput

1.2 Functions defined by Tables (24 Pg 31)Check the following tables whether the second variable as a

functionof the first variable? Explain why or why not ?

Inflation rate ( I) Unemployment rate ( U)

1972 5.6% 5.1%

1973 6.2% 4.5%

1974 10.1% 4.9%

1975 9.2% 7.4%

1976 5.8% 6.7%

1977 5.6% 6.8%

1978 6.7% 7.4%

Not a function: Some values of table I have more than one value of table U

1972 and 1977 same inflationRate 5.6 %But two differentUnemployment rates 5.1% and 6.8%

No26Cost of merchandise (M) Shipping charge (C)

$0.01 – 10.00 $2.50

10.01 – 20.00 3.75

20.01 – 35.00 4.85

30.01 – 50.00 5.95

50.01 – 75.00 6.95

75.01 – 100.00 7.95

Over 100.00 8.95

It is a function: Each value of M has a unique value of C

Graph of the function

No35

12

5

-1 0 1 2 3 4 5 6 7 8

151413121110987654321

C

t measured in years

a) When did 2000 students consider themselves computer literate ? Ans- In 1991

b) How long did it take that number to double?Ans Value of C doubled from 2 to 4 in one year

c) How long did it take for the number to double again?Ans- From 4 to 8 in one year

d) How many students became computer literarate

Ans- t starts from January 1990In the beginning January 1992, t = 2 and C = 4, so 4000 studentswere computer literate. In the beginning of June 1993, t = 3 = 3.4 and C = 11. So 11,000 students were computer literate.

Thus 11,000 – 4000 = 7000 students became computer literate between January 1992 and June 1993

No of students in thousands

1990

Evaluate each function for the given values (pg 34 )

41 f(x) = 6 -2xa) f(3) = 6 – 2(3)=0b) f(-2)= 6 – 2(-2)= 6 + 4= 10c) f(12.7)= 6 -2(12.7) = 6 – 25.4 = -19.4d) f( ) = 6 – 2( ) = 6 - = 4

48D( r) = 5 – re) d(4) = 5 -4 = 1f) d( - 3) = 5 – (-3) = = 2.828g) d(-9)= = = 3.742h) d(4.6)= = = 0.632

3

2

3

2

3

23

4

8

5 ( 9) 14

5 4.6 .4

1.2 pg 37

61. g(x) = 8a) g(2) = 8b) g(8) = 8c) g(a+1) = 8d) g(-x) = 8

64. Q(t) = 5t3

e) Q(2t) = 5(2t)3= 40 t3

f) 2Q(t) = 2.5t3 = 10t3

g) Q(t2) = 5 (t2)3= 5t6

h) [Q(t)]2 = (5t3)2 = 25t6

Evaluate the function and simplify