Topic 1:
Elementary Linear Functions
Jacques Text Book:
section 1.1 and 1.2
(Basic Revision)
section 1.3
Linear Functions
The function f is a rule that assigns
an incoming number x, a uniquely
defined outgoing number y.
y = f(x)
The Variable x takes on different
values…...
The function maps out how different
values of x affect the outgoing
number y.
A Constant remains fixed when we
study a relationship between the
incoming and outgoing variables
Simplest Linear Relationship takes
the form:
y = a+bx  independent
dependent 
variable
variable

intercept
This represents a straight line on a
graph i.e. a linear function has a
constant slope
*
b = slope of the line = the change in
the dependent variable y, given a
change in the independent variable
x.
Example:
y = a + bx
y : is the final grade,
x : is number of hours studied,
a%: guaranteed
y=5+0x
output= constant
y
a
5
5
5
5
5
5
5
5
5
5
5
5
slope
b
0
0
0
0
0
0
input
x
0
1
2
3
4
5
Linear Functions
Dependent Y Variable
60
50
40
30
20
10
0
0
1
2
Independent X Variable
3
4
y=5+15x
output= constant
y
a
5
5
20
5
35
5
50
5
65
5
slope
b
15
15
15
15
15
input
X
0
1
2
3
4
Linear Functions
65
Dependent Y Variable
55
45
35
25
15
5
-5 0
1
2
Independent X Variable
3
4
If x = 4, what grade will you get?
Y = 5 + (4 * 15) = 65
Demand Function: D=a-bP
D= 10 -2P
D
a
10
10
8
10
6
10
4
10
-b
-2
-2
-2
-2
P
0
1
2
3
Demand Function
12
Q Demand
10
8
6
4
2
0
0
1
2
3
4
Price
If p =5, how much will be demanded?
D = 10 - (2 * 5) = 0
5
Inverse Functions:
If y = f(x)
then x = g(y)
f and g are inverse functions
 Let y = 5 + 15x
If a student gets y= 80% for their final
grade, how many hours per week did they
study?
Express x as a function of y:
The Inverse Function is:
x = (y-5)/15
solving, x = (80-5 / 15)
x = 5 hours per week
If D = a – bP then the inverse demand
curve is given by P = (a/b) – (1/b)D
So the inverse demand curve of the
function D= 10 -2P is P= 5 - 0.5D
Example
The demand for a pint of Guinness
in the Student bar on a Friday
evening is a linear function of price.
When the price per pint is £2, the
demand is 6 pints. When the price is
£3, the demand is only 4 pints. Find
the function D = a + bP?
6 = a + 2b
4 = a + 3b
=> a = 6-2b
=> a = 4-3b
6-2b = 4-3b
Solving we find that b = -2
If b = -2, then a = 6-(-4) = 10
The function is D = 10 – 2P
What does this tell us?
Note, the inverse Function is
P = 5- 0.5D
(a simpler version of topic 1,
question 3- note: D= a +bP + Y)
A Tax Example….
1. Let £4000 be set as the target income.
All income above the target is taxed at
40%.
2. Income taxed at 40%, but for every £1
below £4000, the worker gets a negative
income tax (subsidy) of 40%.
Write out the linear function between
take-home pay and earnings for 1 & 2?
THP = E – 0.4 (E – 4000)
THP = E + 0.4 (4000 - E)
In both cases, THP = 1600 +0.6E
If the hourly wage rate is equal to £3 per
hour, rewrite take home pay in terms of
number of hours worked?
THP = 1600 +0.6E
Total Earnings E =
(no. hours worked X hourly wage)
THP = 1600 + 0.6(3H) = 1600 + 1.8H
Now add a (tax free) family allowance of
£100 per child to the function
THP = 1600 + 0.6E + 100Z
Now assume that all earners are given a
£100 supplement that is not taxable,
THP = 1600 + 0.6E + 100Z + 100
= 1700 + 0.6E + 100Z
First Question for Tutorial:
Topic 1, Q2:
Note: to find ‘breakeven income’,
set E = THP
Topic : Quadratic Functions
Lecture Notes:
section 1.2
Jacques Text Book:
section 2.1
Quadratic Functions
Represent Non-Linear Relationships
y = ax2+bx+c where a0
Intercept
a, b and c are constants
So the graph is U-Shaped if a>0,
And ‘Hill-Shaped’ if a<0
And a Linear Function if a=0
e.g. y = -x2+4x+5
y = - 1. x2 + 4
. x+ 5
y
a
x2
b
X
C
-7
-1
4
4
-2
5
5
-1
0
4
0
5
9
5
-7
-1
-1
-1
4
16
36
4
4
4
2
4
6
5
5
5
Since a<0 => ‘Hill Shaped Graph’
Quadratic Functions
y=0, then x= +5
OR x = -1
10
8
Intercept = 5
6
4
2
0
-2
-2 0
2
-4
-6
-8
Independent X Variable
4
6
Special Case: a=1, b=0 and c=0
So y = ax2+bx+c
y = x2
=>
y=
a
x2
b
x
c
16
1
16
0
-4
0
4
1
4
0
-2
0
0
1
0
0
0
0
4
16
36
1
1
1
4
16
36
0
0
0
2
4
6
0
0
0
Quadratic Functions
40
35
30
25
Y = X2
20
15
10
5
0
-4
-2
-5 0
2
4
6
-10
Intercept = 0
Independent X Variable
Min. Point: (0,0)
If the curve cuts the x-axis in 2
places: there are always TWO
values of x that yield the same
value of y when y=0
If it cuts x-axis only once: when
y=0 there is a unique value of x
If it never cuts the x-axis: when
y=0 there is no solution for x
How to find Value(s), if any, of x
when y=0?
Solve General Quadratic Equations:
x
b
b
2
 4ac

2a
e.g. y = -x2+4x+5
hence, a = -1; b=4; c=5
x
x
4
16  4(1 5) 
2
 4  16  20

2
46

2
Hence, x = +5 or x = -1 when y=0
Function cuts x-axis at +5 and –1
Example 1:
y = x2-4x+4
hence, a = 1; b= - 4; c=4
If y = 0
x
x
4
16  4(1 4) 
2
4 0

2
x = 2 when y = 0
Function only cuts x-axis at one
point, where x=2
Example 2
y = 3x2-5x+6
hence, a = 3; b= - 5; c=6
If y = 0
x
x
5
25  4(3  6) 
6
4  25  72 4   47


6
6
when y = 0 there is no solution –
The quadratic function does not
intersect the x-axis
Understanding Quadratic Functions
 intercept where x=0 is c
 a>0 then graph is U-shaped
 a<0 then graph is inverse-U
 a = 0 then graph is linear
 b2 – 4ac > 0 : cuts x-axis twice
 b2 – 4ac = 0 : cuts x-axis once
 b2 – 4ac < 0 : no solution
Essential equations for Economic
Examples:
Total Costs = TC = FC + VC
Total Revenue = TR = P * Q
 = Profit = TR – TC
Break even:  = 0, or TR = TC
Profit Maximisation: MR = MC
An Applied Problem
A firm has MC = 3Q2-32Q+96
And MR = 236 – 16Q
What is the profit Maximising Output?
Maximise profit where MR = MC
3Q2-32Q+96 = 236 – 16Q
3Q2-32Q+16Q +96 – 236 = 0
3Q2-16Q –140 = 0
Solve the quadratic using the formula
where a = 3; b = -16 and c = -140
Solution:
Q = +10 or Q = -4.67
Profit maximising output is +10
350
300
MC
MR
250
MR and MC
200
150
100
50
0
-5
-4
-3
-2
-1
0
1
2
3
4
Q
5
6
7
8
9
10
11
12
Another Example….
If fixed costs are 10 and variable costs per
unit are 2, then given the inverse demand
function P = 14-2Q:
1.Obtain an expression for the profit
function in terms of Q
2.Determine the values of Q for which the
firm breaks even.
3.Sketch the graph of the profit function
against Q
Solution:
1. Profit Function
 = (14 - 2Q)Q – (2Q + 10)
 = -2Q2 + 12Q – 10
2. Breakeven
Apply formula to solve quadratic where
=0
-2Q2 + 12Q – 10 = 0
Solution: at Q = 1 or Q = 5 the firm
breaks even
3. Graphing Profit Function
STEP 1: coefficient on the squared term
determines the shape of the curve
STEP 2: constant term determines where
the graph crosses the vertical axis
STEP 3: Solution where  = 0 is where
the graph crosses the horizontal axis
20
Profit
10
0
-2
-1
0
1
2
3
Profit
-10
-20
-30
-40
-50
Q
4
5
6
7
8