Chapter 5

Functions and their Graphs

Function Notation

f(t) = h

Independent Variable Dependent Variable

Example h = f(t) = 1454 –16t2

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

Ch 5.1 (pg 251) Definition and Notation

Example –To rent a plane flying lessons cost $ 800 plus $30 per hourSuppose 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 independent variable, and C is the dependent variable, because its values are determined by the value of t

This type of relationship is called a function

A function is a relationship between two variables for which a unique value of the dependent variable can be determined from a value of the independent variable

t c

0 800

4 920

10 1100

(t, c)

(0, 800)

(4, 920)

(10, 1100)

Table Ordered Pair

Using Graphing Calculator Pg 258

Enter Y1= 5 – x3 Press 2nd and table Enter graph

Ex5.1, pg 264-265No. 40

g(t) = 5t – 3

a) g(1) = 5(1) – 3 = 2

b) g(-4) = 5(-4) – 3 = -20 – 3= -23

c) g(14.1) = 5( 14.1) – 3= 70.5 – 3= 67.5

d) g = 5 – 3 = - 3 =

No. 51. The velocity of a car that brakes suddenly can be determined from the length of its skid marks, d, by v(d) = , where d is in feet and v is in miles per hour. Complete the table of values.

Solution. V(20) Similarly put all values of d and find v

4

3

4

34

15

4

3

d12

d 20 50 80 100

v 15.5 24.5 31.0 34.6

5.15

240

)20(12

Ch 5.2 Graphs of Functions (Pg 266)Reading Function Values from a Graph

12 13 14 15 16 19 20 21 22 23 October 1987

Dow

Jon

es I

nd

ust

rial

Ave

rage

D

epen

den

t V

aria

ble

2500

2400

2300

2200

2100

2000

1900

1800

P (15, 2412)

Q (20, 1726)

Time Independent Variable

f(15) = 2412f(20) = 1726

Vertical Line Test ( pg 269)

A graph represents a function if and only if every vertical line intersects the graph in at most one point

Function Not a functionGo through all example 4 ( pg 270)

Some basic Graphs

• b = 3 a if b 3 = a

Absolute Value

-10 - 9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

Six Units Six Units

So absolute value of a number x as followsx = x if x > 0 - x if x< 0

Graphs of Eight Basic Functions

x

3 xy = x2 g(x) = x3 f(x) = f(x) =

f(x) = 1/ x g(x) = 1/x3 f(x) = x

g(x) = -x g(x) = x

x

No 15( pg 285)

- 1 0 1

f(x) = x3

Guide pointGuide point

5.4 Domain and Range

Enter y Enter window Press graph

Domain

Ran

ge

STEP FUNCTION

1 2 3 4 5 6 7

5

4

3

2

1

Range

Domain

5.5 Variation

Direct Variation

Two variables are directly proportional if the ratios of their corresponding values are always equal

Gallons of gasoline

Total Price

Price /gallons

4 $4.60 4.60/4 = 1.15

6 $6.90 6.90/6= 1.15

8 $9.20 9.20/8 = 1.15

12 $13.80 13.80/12 = 1.15

15 $17.25 17.25/15 = 1.15

The ratio = total price /number of gallons

5 10 15

20

10

Other Type of Direct Variation

• General equation, y = f(x) = kxn

y= kx3 K > 0

y = kx2 K> 0

= kxK> 0

Inverse Variation

y = n where k is positive constant and n> 0

y is inversely proportional to xn

x

k

No 4, Ex 5.5 ( pg 309)The force of gravity( F ) on a 1-kg mass is inversely proportional to the

square of the object’s distance (D) from the center of the earth F

F= k/d2 ( k = constant of proportionality)

a) Fd2 = k

= 9.8(1)2

K = 9.8

b) F= 9.8/d2 substitute k

d

1

2

1 2 4

9.8 2.45 0.6125

Distance Earth Radii

Force (Newtons)

1 2 3 4 5 Distance d

8

6

4

2

Force

Graph

Pg 311, No 11The weight of an object on the moon varies directly with its

weight on earth

a) m w where m = weight of object, on

moon and w= wt . Of object on earth

m = kw

m = 24.75 pounds, w = 150 pounds

K = 24.75/150 = 0.165

m = 0.165w , substitute k

b) m = 0.165( 120) = 19.8 pounds

c) w= m/k = 30/0.165 = 303.03 pound

w 100 150 200 400

m 16.5 24.75 33 66

Wt. on moon (m)

Wt. on earth (W)

d)

Functions as Mathematical Models(Shape of the graph)

Time Elapsed

Time ElapsedTime Elapsed

Dis

tanc

e fr

om H

ome

Dis

tanc

e fr

om H

ome

Dis

tanc

e fr

om H

ome

walk

wait

bus

Example 5 , Pg 322

Gas Station Mall Highway 17

15 miles

Miles in highway

0 5 10 15 20 25 30

Miles from Mall

15 10 5 0 5 10 15

10 20 30 Miles in Highway

15

10

5

Mil

es f

rom

Mal

l

f(x)= - x + 15 x - 15

When 0 < x < 15

When x > 15

0 10 20 30

10 < x < 20

x – 15 < 5

15

10

5

0 10 20 30

15

10

5

Miles on highway

Mile

s fr

om M

all

x – 15 > 10

The solution is x < 5 or x > 25

x

y

x

y

Pg 323