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Chapter 1
Functions and
Linear Models
Sections 1.3 and 1.4
Linear Function
A linear function can be expressed in the form
f ( x)  mx  b
Function notation
y  mx  b
Equation notation
where m and b are fixed numbers.
Graph of a Linear Function
The graph of a linear function is a straight line.
This means that we need only two points to
completely determine its graph.
y  f ( x)  mx  b
m is called the slope of the line and
b is the y-intercept of the line.
Example: Sketch the graph of f (x) = 3x – 1
y-axis
Arbitrary point
(1,2)
(0,-1)
y-intercept
x-axis
Role of m and b in f (x) = mx + b
The Role of m (slope)
f changes m units for each one-unit change in x.
The Role of b (y-intercept)
When x = 0, f (0) = b
Role of m and b in f (x) = mx + b
To see how f changes, consider a unit change in x.
f ( x)  mx  b
f ( x  1)  m( x  1)  b
Then, the change in f is given by
f ( x  1)  f ( x)  m( x  1)  b ( mx  b)
f ( x  1)  f ( x)  mx  m  b  mx  b
f ( x  1)  f ( x)  m
Role of m and b in f (x) = mx + b
Role of m and b in f (x) = mx + b
Role of m and b in f (x) = mx + b
The graph of a Linear Function:
Slope and y-Intercept
Example: Sketch the graph of f (x) = 3x – 1
y-axis
(1,2)
Slope = 3/1
x-axis
y-intercept
Graphing a Line Using Intercepts
Example: Sketch 3x + 2y = 6
y-axis
x-intercept
(y = 0)
x-axis
y-intercept
(x = 0)
Delta Notation
If a quantity q changes from q1 to q2 , the change
in q is denoted by q and it is computed as
 q  q2  q1
Example: If x is changed from 2 to 5, we write
x  x2  x1  5  2  3
Delta Notation
Example: the slope of a non-vertical line that
passes through the points (x1 , y1) and (x2 , y2) is
given by:
y y2  y1
m 
x x2  x1
Example: Find the slope of the line that passes through
the points (4,0) and (6, -3)
y 3  0 3
3
m 
 
x 6  4 2
2
Delta Notation
Zero Slope and Undefined Slope
Example: Find the slope of the line that passes
through the points (4,5) and (2, 5).
y 5  5 0
m 
 0
x 2  4 2
This is a horizontal
line
Example: Find the slope of the line that passes
through the points (4,1) and (4, 3).
y 3  1 2
m 

x 4  4 0
Undefined
This is a vertical
line
Examples
Estimate the slope of all line segments in the figure
Point-Slope Form of the Line
An equation of a line that passes through the point
(x1 , y1) with slope m is given by:
y  y1  m  x  x1 
Example: Find an equation of the line that passes
through (3,1) and has slope m = 4
y  y1  m  x  x1 
y  1  4  x  3
y  4 x  11
Horizontal Lines
Can be expressed in the form y = b
y=2
Vertical Lines
Can be expressed in the form x = a
x=3
Linear Models:
Applications of linear
Functions
First,
General Definitions
Cost Function
 A cost function specifies the cost C as a
function of the number of items x produced.
Thus, C(x) is the cost of x items.
 The cost functions is made up of two parts:
C(x)= “variable costs” + “fixed costs”
Cost Function
 If the graph of a cost function is a straight
line, then we have a Linear Cost Function.
 If the graph is not a straight line, then we
have a Nonlinear Cost Function.
Linear Cost Function
Dollars
Dollars
Cost
Units
Cost
Units
Non-Linear Cost Function
Dollars
Dollars
Cost
Units
Cost
Units
Revenue Function
 The revenue function specifies the total payment
received R from selling x items. Thus, R(x) is the
revenue from selling x items.
 A revenue function may be Linear or Nonlinear
depending on the expression that defines it.
Linear Revenue Function
Dollars
Revenue
Units
Nonlinear Revenue Functions
Dollars
Dollars
Revenue
Units
Revenue
Units
Profit Function
 The profit function specifies the net proceeds P.
P represents what remains of the revenue when
costs are subtracted. Thus, P(x) is the profit from
selling x items.
Profit = Revenue – Cost
 A profit function may be linear or nonlinear
depending on the expression that defines it.
Linear Profit Function
Dollars
Profit
Units
Nonlinear Profit Functions
Dollars
Dollars
Profit
Units
Profit
Units
The Linear Models are
Cost Function:
C ( x)  mx  b
** m is the marginal cost (cost per item), b is fixed cost.
Revenue Function:
R ( x)  mx
** m is the marginal revenue.
Profit Function: P ( x)  R( x)  C ( x)
where x = number of items (produced and sold)
Break-Even Analysis
The break-even point is the level of production that
results in no profit and no loss.
To find the break-even point we set the profit function
equal to zero and solve for x.
P( x)  0
R( x)  C ( x)  0
Break-Even Analysis
The break-even point is the level of production that
results in no profit and no loss.
Profit = 0
Dollars
means
Revenue = Cost
Revenue
profit
loss
Break-even
Revenue
Cost
Units
Break-even point
Example: A shirt producer has a fixed monthly cost of
$3600. If each shirt has a cost of $3 and sells for $12
find:
a.
The cost function
C (x) = 3x + 3600 where x is the number of
shirts produced.
b.
The revenue function
R (x) = 12x where x is the number of shirts
sold.
c.
The profit from 900 shirts
P (x) = R(x) – C(x)
P (x) = 12x – (3x + 3600) = 9x – 3600
P(900) = 9(900) – 3600 = $4500
Example: A shirt producer has a fixed monthly cost of
$3600. If each shirt has a cost of $3 and sells for $12
find the break-even point.
The break even point is the solution of the equation
C (x) = R (x)
12 x  3x  3600
x  400
and R(400)  4800
Therefore, at 400 units the break-even revenue is $4800
7000
C, R, P in $
C(x)
6000
5000
4000
3000
2000
1000
x
-200
-1000
-2000
-3000
-4000
200
400
600
800
1000 1200 1400
x = number of shirts sold
7000
C, R, P in $
R(x)
C(x)
6000
5000
4000
3000
2000
1000
x
-200
-1000
-2000
-3000
-4000
200
400
600
800
1000 1200 1400
x = number of shirts sold
7000
C, R, P in $
R(x)
C(x)
6000
5000
P(x)
4000
3000
2000
1000
x
-200
-1000
-2000
-3000
-4000
200
400
600
800
1000 1200 1400
x = number of shirts sold
Demand Function
 A demand function or demand equation expresses
the number q of items demanded as a function of
the unit price p (the price per item).
 Thus, q(p) is the number of items demanded when
the price of each item is p.
 As in the previous cases we have linear and
nonlinear demand functions.
Linear Demand Function
q = items
demanded
Price p
Nonlinear Demand Functions
q = items
demanded
q = items
demanded
Price p
Price p
Supply Function
 A supply function or supply equation expresses
the number q of items, a supplier is willing to
make available, as a function of the unit price p
(the price per item).
 Thus, q(p) is the number of items supplied when
the price of each item is p.
 As in the previous cases we have linear and
nonlinear supply functions.
Linear Supply Function
q = items
supplied
Price p
Nonlinear Supply Functions
q = items
supplied
q = items
supplied
Price p
Price p
Market Equilibrium
Market Equilibrium occurs when the quantity produced
is equal to the quantity demanded.
supply
curve
q
shortage
surplus
demand
curve
p
Equilibrium Point
Market Equilibrium
Market Equilibrium occurs when the quantity produced
is equal to the quantity demanded.
q
supply curve
surplus
shortage
Equilibrium
demand
demand curve
p
Equilibrium price
Market Equilibrium
 To find the Equilibrium price set the demand
equation equal to the supply equation and solve
for the price p.
 To find the Equilibrium demand evaluate the
demand (or supply) function at the equilibrium
price found in the previous step.
Example of Linear Demand
The quantity demanded of a particular computer game
is 5000 games when the unit price is $6. At $10 per
unit the quantity demanded drops to 3400 games.
Find a linear demand equation relating the price p, and
the quantity demanded, q (in units of 100).
( p1 , q1 )  (6,50) and ( p2 , q2 )  (10, 34)
q q2  q1 34  50 16
m



 4
p p2  p1 10  6
4
q  50  4( p  6)
q  4 p  74
Example: The maker of a plastic container has determined
that the demand for its product is 400 units if the unit price
is $3 and 900 units if the unit price is $2.50.
The manufacturer will not supply any containers for less
than $1 but for each $0.30 increase in unit price above the
$1, the manufacturer will market an additional 200 units.
Assume that the supply and demand functions are linear.
Let p be the price in dollars, q be in units of 100 and find:
a. The demand function
b. The supply function
c. The equilibrium price and equilibrium demand
a. The demand function
m
 p1, q1   3, 4 and  p2 , q2    2.5,9 ;
q  4  10  p  3
q
94
p

2.5  3
 10
q  10 p  34
b. The supply function
 p1, q1   1,0 and  p2 , q2   1.3, 2 ;
q 2  0 20
m


p 1.3  1 3
20
20
q
p
3
3
c. The equilibrium price and equilibrium demand
Demand  Supply
10 p  34  (1/ 3)(20 p  20)
30 p  102  20 p  20
p  2.44
q  10(2.44)  34  9.6
The equilibrium demand is 960 units at a price of
$2.44 per unit.
Linear Change over Time
A quantity q, as a linear function of time t:
q(t )  mt  b
Rate of
change of q
Quantity at
time t = 0
If q represents the position of a moving object,
then the rate of change is velocity.
Linear Regression
We have seen how to find a linear model given
two data points. We find the equation of the line
passing through them.
However, we usually have more than two data
points, and they will rarely all lie on a single
straight line, but may often come close to doing
so.
The problem is to find the line coming closest to
passing through all of the points.
Linear Regression
We use the method of least squares to determine a
straight line that best fits a set of data points when
the points are scattered about a straight line.
least squares
line
The Method of Least Squares
Given the following n data points:
( x1 , y1 ), ( x2 , y2 ),..., ( xn , yn )
The least-squares (regression) line for the data is given
by y = mx + b, where m and b satisfy:
m
n   xy     x   y 
n
and
 x    x
2
y  m  x

b
n
2
Example: Find the equation of least-squares for
the data (1,2), (2,3), (3,7).
The scatter plot of the points is
Solution: We complete the following table
Sum:
m
b
x
y
xy
x2
1
2
2
1
2
3
6
4
3
7
21
9
6
12
29
14
3  29    6 12 
3 14    6 
12  2.5  6 
3
2
 1
 2.5
y  2.5 x  1
Example: Find the equation of least-squares for
the data (1,2), (2,3), (3,7).
The scatter plot of the points and the least squares line
is
y  2.5 x  1
Coefficient of Correlation
A measurement of the closeness of fit of the least
squares line. Denoted r, it is between –1 and 1, the
better the fit, the closer it is to 1 or –1.
n   xy     x   y 
r
n
 x    x
2
2
 n
 y    y 
2
2
Example: Find the correlation coefficient for
the least-squares line from the last example.
Points:
n   xy     x   y 
r
n

(1 , 2), (2 , 3), (3 , 7)
 x    x
2
2
 n
 y    y 
2
3  29    6 12 
3 14    6   3  62   12 
= 0.9449
2
2
2