Download Week 5

Linear and Quadratic Functions
On completion of this module you should be able to:
 define the terms function, domain, range, gradient,
independent/dependent variable
 use function notation
 recognise the relationship between functions and
equations
 graph linear and quadratic functions
 calculate the function given initial values (gradient, 1 or 2
coordinates)
 solve problems using functions
 model elementary supply and demand curves using
functions and solve associated problems
1
Functions
A function describes the relationship that exists
between two sets of numbers.
Put another way, a function is a rule applied to one
set of numbers to produce a second set of
numbers.
2
Example: Converting Fahrenheit to Celsius
5
C   F  32 
9
This rule operates on values of F to produce
values of C.
The values of F are called input values and the
set of possible input values is called the domain.
The values of C are called output values and the
set of output values produced by the domain is
called the range.
3
Function Notation
Consider the function
x
f ( x) 
x 3
The x are the input values and f(x), read f of x, are
the output values.
The domain is the set of positive real numbers
including 0 and excepting 3. (Why?) The output
values produced by the domain is the range.
Sometimes the symbol y is used instead of f(x).
4
Function and Equations
An equation is produced when a function takes on
a specific output value.
eg f(x) = 3x + 6 is a function.
When f(x) = 0, then the equation becomes
0 = 3x + 6
which can be easily solved (to give x=-2)
5
This is shown graphically as follows:
f (x )
(0,6)
(2,0)
x
f ( x)  0
6
Graphing Functions
Input and output values form coordinate pairs:
x, f(x) or (x,y).
x values measure the distance from the origin in the
horizontal direction and f(x) values the distance
from the origin in the vertical direction.
To plot a straight line (linear function), 2 sets of
coordinates (3 sets is better) must be calculated. For
other functions, a selection of x values should be
made and coordinates calculated.
7
Example: Linear Function
Graph f(x) = 2x - 4
x
3
0
3
f ( x)
2(3)  4  10
2(0)  4  4
2(3)  4  2
(3,10)
(0,4)
(3, 2)
8
f (x )
2
3
3
x
4
f ( x)  2 x  4
9
Example: Quadratic Function
Graph the function:
f ( x)  2 x 2  5 x  2
At the y-intercept, x = 0, so
f ( x)  2  0   5  0   2  2
2
and the coordinate is (0,2).
10
At the x-intercept, f(x) = 0, so
0  2 x 2  5x  2
5  (5) 2  4  2  2 
x
2 2
53

4
 2 or 0.5
and the coordinates are (2,0) and (0.5,0).
11
b
Vertex: x 
2a
  5 

 1.25
2(2)
When x  1.25,
2
f ( x)  2(1.25)  5(1.25)  2  1.125
The coordinates of the vertex are: (1.25, -1.125).
12
f (x )
2 (0,2)
f ( x)  2 x 2  5 x  2
(0.5,0)
1
2
(2,0)
x
-1
(1.25, -1.125)
13
Linear Functions
All linear functions (or equations) have the
following features:
 a slope or gradient (m)
 a y-intercept (b)
If (x1, y1) and (x2, y2) are two points on the line then
the gradient is given by:
y2  y1
m
x2  x1
14
Gradient is a measure of the steepness of the
line.


If m>0, then the line rises from left to right.

If m<0, the line falls from left to right.
 A horizontal
line has a gradient of 0; a vertical
line has an undefined gradient.
The y-intercept is calculated by substituting
x = 0 into the equation for the line.

15
All straight line functions can be expressed in
the form
y = mx + b
Note: The standard form equation for linear
functions is Ax + By + C = 0.
Equations in this form are not as useful as when
expressed as y = mx + b.
Equations can be derived in the following way,
depending on what information is given.
16
Deriving Straight Line Functions
1. Given (x1, y1) and (x2, y2)
y  y1 y 2  y1

x  x1 x2  x1
2. Given m and (x1, y1)
y  y1  m( x  x1 )
3. Given m and b
y  mx  b
17
Problem: Depreciation
A tractor costs $60 000 to purchase and has a
useful life of 10 years.
It then has a scrap value of $15 000.
Find the equation for the book value of the
tractor and its value after 6 years.
18
V
60 000
?
15 000
6
10
t
19
Value (V) depends on time (t).
t is called the independent variable and
V the dependent variable.
The independent variable is always plotted on the
horizontal axis and the dependent variable on the
vertical axis.
20
When t  0, V  60000
 (0, 60 000)
( x1 , y1 )
When t  10, V  15000
 (10, 15 000)
( x2 , y2 )
Given two points, the equation becomes:
y  60000 15000  60000

x0
10  0
21
y  60000
 4500
x
y  60000  4500 x
y  4500 x  60000
or more correctly
V  4500t  60000
When t  6, V  4500  6  60000  33000
The book value of the tractor after 6 years is $33000.
22
Example
Suppose a manufacturer of shoes will place on the
market 50 (thousand pairs) when the price is $35
(per pair) and 35 (thousand pairs) when the price is
$30 (per pair).
Find the supply equation, assuming that price p
and quantity q are linearly related.
(50, 35)  ( x1 , y )
1
(35, 30)  ( x2 , y2 )
23
(50, 35)  ( x1 , y1 )
(35, 30)  ( x2 , y2 )
y  y1 y2  y1

x  x1 x2  x1
y  35 30  35

x  50 35  50
y  35 1

x  50 3
1
y  35  ( x  50)
3
24
y  35  0.33x  16.67
y  0.33x  16.67  35
y  0.33x  18.33
The supply equation is
p  0.33q  18.33
25
Example
For sheep maintained at high environmental
temperatures, respiratory rate r (per minute)
increases as wool length l (in centimetres) decreases.
Suppose sheep with a wool length of 2cm have an
(average) respiratory rate of 160, and those with a
wool length of 4cm have a respiratory rate of 125.
Assume that r and l are linearly related.
(a) Find an equation that gives r in terms of l.
(b) Find the respiratory rate of sheep with a wool
length of 1cm.
26
(a) Find r in terms of l
 l is independent
r is dependent
Coordinates will be of the form: (l, r).
(2,160)  ( x1 , y1 )
(4,125)  ( x2 , y2 )
y 2  y1 125  160
m

 17.5
x2  x1
42
27
(2, 160), (4, 125)
m  17.5
y  y1  m( x  x1 )
y  160  17.5( x  2)
y  17.5 x  35  160
y  17.5 x  195
r  17.5l  195
28
(b) When l=1
r  17.5(1)  195
 177.5
When wool length is 1cm, average respiratory
rate will be 177.5 per minute.
29
Quadratic Functions
All quadratic functions can be written in the
form
f ( x)  ax 2  bx  c
where a, b and c are constants and a  0.
30
Elementary Supply and Demand
In general, the higher the price, the smaller the
demand is for some item and as the price falls
demand will increase.
p
Demand curve
q
31
Concerning supply, the higher the price, the larger
the quantity of some item producers are willing to
supply and as the price falls, supply decreases.
p
Supply curve
q
32
Note that these descriptions of supply and
demand imply that they are dependent on
price (that is, price is the independent
variable) but it is a business standard to plot
supply and demand on the horizontal axis
and price on the vertical axis.
33
Example: Equilibrium price
The supply of radios is given as a function of
price by
S  p   2 p 2  8 p  12,
2 p5
and demand by
D  p   p 2  18 p  68,
0 p5
Find the equilibrium price.
34
Graphically,
70
equilibrium
price
0
0
1
2
3
4
5
p
Note the restricted domains.
35
Algebraically,
D(p) = S(p)
p  18 p  68  2 p  8 p  12
2
2
p  2 p  18 p  8 p  68  12  0
2
2
 p  10 p  56  0
a  1, b  10, c  56
2
10  (10) 2  4(1)(56)
p
2  1
p  14 or 4
36
-14 is not in the domain of the functions and so
is rejected.
The equilibrium price is $4.
37
Example: Maximising profit
If an apple grower harvests the crop now, she will
pick on average 50kg per tree and will receive
$0.89 per kg.
For each week she waits, the yield per tree
increases by 5kg while the price decreases by
$0.03 per kg.
How many weeks should she wait to maximise
sales revenue?
38
Weight and Price can both be expressed as
functions of time (t).
W(t) = 50 + 5t
P(t) = 0.89 - 0.03t
Revenue  Weight  Price
 W (t ) P(t )
 (50  5t )(0.89  0.03t )
 44.5  1.5t  4.45t  0.15t 2
 0.15t 2  2.95t  44.5
39
a  0.15  0 
Maximum occurs at
b
 2.95
t

 9.83
2a 2(0.15)
She should wait 9.83 weeks
(  10 weeks) for maximum revenue.
(R = $59 per tree)
40