1. Topic 1
Topic 1 : Elementary functions
Reading: Jacques
Section 1.1 - Graphs of linear equations
Section 2.1 – Quadratic functions
Section 2.2 – Revenue, Cost and Profit
1

2. 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 f 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
2

3. Simplest Linear Relationship:
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 = change in the dependent
variable y, given a change in the independent
variable x.
• Slope of a line = ∆y / ∆x
= (y2-y1) / (x2-x1)
3

4. Example: Student grades
• Example: • Consider the function:
• y = a + bx • y = 5+ 0x
• y : is the final grade, • What does this tell
• x : is number of hours us?
studied, • Assume different
• a%: guaranteed values of x ………
4

5. Example Continued: What grade if
you study 0 hours? 5 hours?
• y=5+0x
Linear Functions
Output = constant slope Input
y a b X 60
50
5 5 0 0
Dependent Y Variable
40
5 5 0 1
30
5 5 0 2
20
5 5 0 3 10
5 5 0 4 0
0 1 2 3 4
5 5 0 5 Independent X Variable
5

6. Example Continued…. output= constant slope input
y a b X
5 5 15 0
• y=5+15x 20 5 15 1
35 5 15 2
50 5 15 3
65 5 15 4
Linear Functions
65
55
If x = 4, what grade
Dependent Y Variable
45
35
will you get?
25
Y = 5 + (4 * 15) = 65
15
5
-5 0 1 2 3 4
Independent X Variable
6

7. Demand functions: The relationship
between price and quantity
Demand Function: D=a-bP
D= 10 -2P
D a -b P
10 10 -2 0
8 10 -2 1
6 10 -2 2
4 10 -2 3
Demand Function
12
10
If p =5, how much will be
8
demanded?
Q Demand
6 D = 10 - (2 * 5) = 0
4
2
0
0 1 2 3 4 5
7
Price

8. Inverse Functions:
• Definition
• If y = f(x)
• then x = g(y)
• f and g are inverse functions
• Example
• Let y = 5 + 15x
• If y is 80, how many hours per week did they
study?
8

9. Example continued…..
• If y is 80, how many hours per week did
they study?
• Express x as a function of y: 15x = y – 5....
• So the Inverse Function is: x = (y-5)/15
• Solving for value of y = 80
x = (80-5 / 15)
x = 5 hours per week
9

10. An inverse demand function
• If D = a – bP then the inverse demand
curve is given by P = (a/b) – (1/b)D
• E.g. to find the inverse demand curve
of the function D= 10 -2P ……
First, re-write P as a function of D
2P = 10 – D
Then, simplify
So P= 5 – 0.5D is the inverse function
10

11. More Variables:
• Student grades again: • Example:
y = a + bx + cz • If y = 5+ 15x + 3z, and a
• y : is the final grade, student studies 4 hours per
week and completes 5
• x : is number of hours questions per week, what
studied, is the final grade?
• z: number of • Answer:
questions completed • y = 5 + 15x + 3z
• a%: guaranteed • y = 5 + (15*4) + (3*5)
• y = 5+60+15 = 80

12. Another example: Guinness
Demand.
• The demand for a pint of • 6 = a + 2b
Guinness in the Student => a = 6-2b
bar on a Friday evening
• 4 = a + 3b
is a linear function of
=> a = 4-3b
price. When the price
• 6-2b = 4-3b
per pint is €2, the
• Solving we find that b = -2
demand ‘is €6 pints.
When the price is €3, • If b = -2, then a = 6-(-4) = 10
the demand is only 4 • The function is D = 10 – 2P
pints. Find the function • What does this tell us??
D = a + bP • Note, the inverse Function is
• P = 5- 0.5D
12

13. A Tax Example….
• Answer:
• let €4000 be set as the • THP = E – 0.4 (E – 4000)
target income. All income if E>4000
above the target is taxed
at 40%. For every €1 • THP = E + 0.4 (4000-E)
below the target, the if E<4000
worker gets a negative • In both cases,
income tax (subsidy) of THP = 1600 +0.6E So
40%.
i) If E = 4000 =>
• Write out the linear
function between take- THP = 1600+2400=4000
home pay and earnings. ii) If E = 5000 =>
THP = 1600+3000=4600
iii) If E = 3000 =>
THP = 1600+1800=3400
13

14. Tax example continued….
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?
• 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
• THP = 1600 + 0.6E + 100Z (where z is number of children)
Now assume that all earners are given a €100
supplement that is not taxable,
• THP = 1600 + 0.6E + 100Z + 100
= 1700 + 0.6E + 100Z 14

15. Topic 1 continued:
Non- linear Equations
Jacques Text Book:
Sections 2.1 and 2.2
15

16. Quadratic Functions
• Represent Non-Linear Relationships
y = ax2+bx+c where a≠0, c=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
16

17. Solving Quadratic Equations:
1) Graphical Approach: To find Value(s), if any,
of x when y=0, plot the function and see where it
cuts the x-axis
• 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
17

18. e.g. y = -x2+4x+5
2
y a x b X C
-7 -1 4 4 -2 5
5 -1 0 4 0 5
9 -1 4 4 2 5
5 -1 16 4 4 5
-7 -1 36 4 6 5
Since a<0 => ‘Hill Shaped Graph’
18

19. The graph
Quadratic Functions
10
8 y=0, then x= +5
6 OR x = -1
4
Y = X2
2
0
-2 -2 0 2 4 6
-4
-6
-8
Independent X Variable
19

20. Special Case: a=1, b=0 and c=0
So y = ax2+bx+c => y = x2
2 Quadratic Functions
y= a x b x c
40
16 1 16 0 -4 0 35
4 1 4 0 -2 0 30
25
0 1 0 0 0 0
20
2
4 1 4 0 2 0 Y=X
15
16 1 16 0 4 0 10
36 1 36 0 6 0 5
0
-4 -2 -5 0 2 4 6
-10 Min. Point: (0,0)
Intercept = 0
Independent X Variable
20

21. Practice examples
• Plot the graphs for the following functions
and note (i) the intercept value (ii) the
value(s), if any, where the quadratic
function cuts the x-axis
• y = x2-4x+4
• y = 3x2-5x+6
21

22. Solving Quadratic Equations:
• 2) Algebraic Approach: find the value(s),
if any, of x when y=0 by applying a simple
formula…
x=
−b ± (b 2
− 4ac )
2a
22

23. Example
• e.g. y = -x2+4x+5
• hence, a = -1; b=4; c=5
−4 ± (16 − 4(−1×5) )
x=
−2
−4 ± (16 + 20) −4 ±6
x = =
−2 −2
• Hence, x = +5 or x = -1 when y=0
• Function cuts x-axis at +5 and –1
23

24. Example 2
• y = x2-4x+4 18
• hence, a = 1; b= - 4; 16 y
c=4 14
• If y = 0 12
10
Y
8
+4± (16 − 4(1× 4) ) 6
x= 4
2 2
4± 0
=
0
x -2 -1 0 1 2 3 4 5
2 X
Function only cuts x-axis at one point,
x = 2 when y = 0 where x=2
24

25. Example 3
• y = 3x2-5x+6 120
y
• hence, a = 3; 100
b= - 5; c=6 80
• If y = 0 Y 60
+ 5 ± ( 25 − 4(3 × 6) )
x= 40
6 20
4 ± − 47
= 0
6 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5
X
when y = 0 there is no solution
The quadratic function does not intersect
the x-axis
25

26. 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
26

27. Essential equations for
Economic Examples:
• Total Costs = TC = FC + VC
• Total Revenue = TR = P * Q
∀ π = Profit = TR – TC
• Break even: π = 0, or TR = TC
• Marginal Revenue = MR = change in total
revenue from a unit increase in output Q
• Marginal Cost = MC = change in total cost
from a unit increase in output Q
• Profit Maximisation: MR = MC
27

28. An Applied Problem
• A firm has MC = 3Q2- 32Q+96
• And MR = 236 – 16Q
• What is the profit Maximising Output?
Solution
• Maximise profit where MR = MC
3Q2 – 32Q + 96 = 236 – 16Q
3Q2 – 32Q+16Q +96 – 236 = 0
3Q2 – 16Q –140 = 0 − b ± ( b 2 − 4ac )
Q=
• Solve the quadratic using the formula 2a
where a = 3; b = -16 and c = -140
• Solution:
Q = +10 or Q = -4.67
• Profit maximising output is +10 (negative Q inadmissable)
28

29. 350
Graphically
300
MC
MR
250
200
MR and MC
150
100
50
0
-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12
Q
29

30. 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
30

31. Solution:
1. Profit = TR – TC = P.Q – (FC + VC)
π = (14 - 2Q)Q – (2Q + 10)
π = -2Q2 + 12Q – 10
2. Breakeven: where Profit = 0
Apply formula to solve quadratic where π = 0
so solve -2Q2 + 12Q – 10 = 0 with Q = −b± (b
− 4ac ) 2
2a
• Solution: at Q = 1 or Q = 5 the firm breaks
even
31

32. 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 π = 0 is where
the graph crosses the horizontal axis
32

33. 20
Profit
10
0
-2 -1 0 1 2 3 4 5 6 7 8
-10
Profit
-20
-30
-40
-50
Q 33

34. Questions Covered on Topic 1:
Elementary Functions
• Linear Functions and Tax……
• Finding linear Demand functions
• Plotting various types of functions
• Solving Quadratic Equations
• Solving Simultaneous Linear (more in next
lecture)
• Solving quadratic functions
34