1. Lecture 5 - Graph Transformations
and Maths for Economics
C2 Foundation Mathematics (Standard Track)
Dr Linda Stringer Dr Simon Craik
l.stringer@uea.ac.uk s.craik@uea.ac.uk
INTO City/UEA London

2. Lecture 5 skills
Sketch the line with equations y = x
Sketch the curves with equations y = x2, y = x3 and y = 1
x
Sketch a given curve after transformation
Find the new equation of a given curve after transformation
Differentiate a total revenue function and hence find
maximum revenue

3. Sketch a curve after transformation
Question: The dashed curve below is shifted up by 2. Sketch
the new curve on the same axes. Label the stationary points
and the y-intercept.
1 2 3 4
2
4
6
•(0, 1)
•
(1, 5)
•
(3, 1) x
y

4. Sketch a curve after transformation
Answer:
1 2 3 4
2
4
6
•(0, 1)
•
(1, 5)
•
(3, 1)
•(0, 3)
•
(1, 7)
•
(3, 3)
x
y

5. Sketch a curve after transformation
Question: The dashed line below is reflected in the y-axis
Sketch the new line on the same axes. Label the intercepts.
−3 −2 −1 1 2 3
2
4
•(0, 2)
•
(1, 0) x
y

6. Sketch a curve after transformation
Answer:
−3 −2 −1 1 2 3
2
4
•(0, 2)
•
(−1, 0) x
y

7. The graph of y = x2
−4 −2 2 4
−4
−2
2
4
x
y
Let f(x) = x2
Then y = f(x) is the
same as y = x2

8. The graph of y = −x2
If we reflect the curve y = x2 in the x-axis,
then we get the curve y = −x2.
−4 −2 2 4
−4
−2
2
4
x
y Let f(x) = x2,
(so −f(x) = −x2.)
If we reflect the curve
y = f(x) in the x-axis,
then we get the curve
y = −f(x).

9. The graph of y = x2
+ 1
If we shift the curve y = x2 vertically upwards by 1,
then we get the curve y = x2 + 1.
−4 −2 2 4
−4
−2
2
4
x
y Let f(x) = x2,
(so f(x) + 1 = x2 + 1.)
If we shift the curve
y = f(x) vertically
upwards by 1,
then we get the curve
y = f(x) + 1.

10. The graph of y = x2
− 4
If we shift the curve y = x2 vertically downwards by 4,
then we get the curve y = x2 − 4.
−4 −2 2 4
−4
−2
2
4
x
y Let f(x) = x2,
(so f(x) − 4 = x2 − 4.)
If we shift the curve
y = f(x) vertically
downwards by 4,
then we get the curve
y = f(x) − 4.

11. The graph of y = (x + 2)2
If we shift the curve y = x2 left by 2,
then we get the curve y = (x + 2)2.
−4 −2 2 4
−4
−2
2
4
x
y Let f(x) = x2,
so f(x + 2) = (x + 2)2.
If we shift the curve
y = f(x) left by 2,
then we get the curve
y = f(x + 2).

12. The graph of y = (x − 3)2
If we shift the curve y = x2 right by 3,
then we get the curve y = (x − 3)2.
−4 −2 2 4
−4
−2
2
4
x
y Let f(x) = x2,
so f(x − 3) = (x − 3)2.
If we shift the curve
y = f(x) right by 2,
then we get the curve
y = f(x − 3).

13. Graph transformations
Let f(x) be a function and let c be a constant. Consider the
graph y = f(x).
y = Transformation
−f(x) Reflection in the x-axis
f(−x) Reflection in the y-axis
f(x) + c Shift up by c (translation of +c in the y-direction)
f(x) − c Shift down by c
f(x + c) Shift left by c
f(x − c) Shift right by c
cf(x) Vertical stretch by a factor of c
f(cx) Horizontal stretch by a factor of 1
c

14. Graph transformations
Let f(x) be a function and let c be a constant.
The graph of y = −f(x) is the graph of y = f(x) reflected in
the x-axis.
The graph of y = f(−x) is the graph of y = f(x) reflected in
the y-axis.
The graph of y = f(x) + c is the graph of y = f(x) shifted
up by c.
The graph of y = f(x + c) is the graph of y = f(x) shifted
to the left by c.
The graph of y = cf(x) is the graph of y = f(x) stretched
vertically by a factor of c (the x-axis remains fixed).
The graph of y = f(cx) is the graph of y = f(x) stretched
horizontally by a factor of 1
c (the y-axis remains fixed).

15. The graph of y = x3
−2 −1 1 2
−5
5
x
y
y = x3

16. The graph of y = −x3
The graph of y = −x3 is the graph of y = x3 reflected in the
x-axis OR in the y-axis
−2 −1 1 2
−5
5
x
y
Let f(x) = x3.
Then −f(x) = −x3, and also
f(−x) = (−x)3 = −x × −x × −x = −x3.
So the graph of y = −f(x) (reflection in x-axis) is the same as
the graph y = f(−x) (reflection in y-axis).

17. Transformations of the graph y = x3
− x
The dashed line shows y = 5(x3 − x). This is the graph of
y = x3 − x stretched vertically by factor 5.
The dotted line shows y = (3x)3 − (3x). This is the graph
of y = x3 − x stretched horizontally by factor 1
3 . (Note that
stretching by a factor less than 1 is the same as
compressing).
−1 1
−4
−2
2
4
x
y

18. The graph of y = 1
x
−10 −5 5 10
−4
−2
2
4
x
y
y = 1
x

19. The graph of y = 1
x + 3
This is the graph of y = 1
x shifted upwards by 3.
−10 −5 5 10
−4
−2
2
4
6
8
x
y
y = 1
x + 3

20. The graph of y = 1
x+3
This is the graph of y = 1
x shifted left by 3.
−10 −5 5 10
−4
−2
2
4
x
y
y = 1
x+3

21. Examples
Question: A curve has equation y = x4 + 9x2. The curve is
shifted vertically upward by 3. What is the equation of the
new curve?
Answer: If we shift a curve with equation y = f(x) up by 3,
then we get a curve with equation y = f(x) + 3.
Let f(x) = x4 + 9x2.
Then the new curve has equation y = x4 + 9x2 + 3.
Question: A curve has equation y = x2 + 2x + 5. The curve
is shifted left by 3. What is the equation of the new curve?
Answer: If we shift a curve with equation y = f(x) left by 3,
then we get a curve with equation y = f(x + 3).
Let f(x) = x2 + 2x + 5.
Then the new curve has equation
y = (x + 3)2 + 2(x + 3) + 5.
Now simplify the right side
y = (x + 3)(x + 3) + 2x + 6 + 5 =
x2 + 6x + 9 + 2x + 6 + 5 = x2 + 8x + 20.
The new curve has equation y = x2 + 8x + 20.

22. Sketching graphs to solve simultaneous equations
Question: Graphically solve the simultaneous equations
−x + y = −1
x + 2y = 4
Answer: Rearrange the equations and sketch the two lines
y = x − 1
y = −1
2 x + 2
−1 1 2 3
−1
1
2
3
•
(2, 1)
The lines cross at the point (2, 1) and so the solutions to these
simultaneous equations are x = 2, y = 1

23. Maths for economics
Linear equations and straight lines in economics
Graph transformations in economics
Calculus in economics - differentiating to find maximum
total revenue
Economics students only - marginal revenue
(differentiation and integration)
Economics students only - an example (average costs,
marginal revenue, elasticity)

24. The demand function and the inverse demand function
The demand function expresses the quantity of a product
demanded, in terms of the market price per unit.
Q = −2P + 20 (demand function)
As the market price increases, the quantity demanded by
consumers decreases.
In economics, P is always on the vertical axis, and in
maths we plot the subject of the equation on the vertical
axes, so rearrange the demand function to make P the
subject of the equation.
Q = −2P + 20
Q + 2P = 20
2P = −Q + 20
P = −0.5Q + 10 (inverse demand function)
Now price is expressed in terms of quantity.

25. The demand curve
Q = −2P + 20 (demand function)
P = −0.5Q + 10 (inverse demand function)
5 10 15 20
5
10
Demand curve
0
Q, quantity
P, price per item (£)
The gradient of a demand curve is usually negative - the line
slopes down towards the right.
(The exception is when we consider the demand curve for a
small firm in a perfectly competitive market - in that case the
demand curve is horizontal, so the gradient is zero).

26. Zero demand and free items
Find the P-intercept and the Q-intercept.
(At what price do items cease to be bought? What quantity
is demanded when the item is free?)
P = −0.5Q + 10
0
Q
P
P-intercept: where the line crosses the P-axis
Q = 0, so P = (−0.5 × 0) + 10 = 10
(Items cease to be bought when the price is £10)
Q-intercept: where the line crosses the Q-axis
P = 0, so −0.5Q + 10 = 0, and Q = 20
(20 units are demanded when the item is free)

27. The supply function and the inverse supply function
The supply function expresses the quantity that companies
supply in terms of the market price per unit.
Q = 10P − 4 (supply function)
The quantity produced by companies increases as the
market price increases.
Rearrange the supply function to make P the subject of the
equation.
Q = 10P − 4
Q − 10P = −4
−10P = −Q − 4
P = 0.1Q + 0.4 (inverse supply function)
Now price is expressed in terms of quantity.

28. The supply curve
Q = 10P − 4 (supply function)
P = 0.1Q + 0.4 (inverse supply function)
5 10 15 20
5
10
Supply curve
0
Q
P
The gradient of the supply curve is positive - the line slopes up
towards the right.
(What does the P-intercept of the supply curve tell us?)

29. Market equilibrium
Solve for P and Q
Q = 10P − 4 (supply function)
Q = −2P + 20 (demand function)
(Find equilibrium price and quantity)
Equate the right hand side of both equations
10P − 4 = −2P + 20
12P = 24
P = 2
Substitute P = 2 into either equation to find Q
Q = 10 × 2 − 4 = 16
(Equilibrium price is £2, the corresponding quantity is 16).

30. Market equilibrium
P = 0.1Q + 0.4 (inverse supply function)
P = −0.5Q + 10 (inverse demand function)
5 10 15 20
5
10
Supply curve
Demand curve
•
(16, 2)
Q
P
The lines cross at (16, 2).
(Equilibrium price is £2, the corresponding quantity is 16).

31. Fixed tax changes the supply functions and supply
curve
P = 0.1Q + 0.4 (S, inverse supply function)
Shift line S upwards by 1.2 to a new line, S∗.
Find the equation of line S∗.
(A government tax of £1.20 per item is introduced)
P = (0.1Q + 0.4) + 1.2
P = 0.1Q + 1.6 (S∗, new inverse supply function)
5 10 15 20
0.4
1.6
S
S∗ (with tax)
Q
P
This is the same as replacing P by P − 1.2 in the supply
functions.

32. Fixed tax changes the supply functions and supply
curve
P = 0.1Q + 0.4 (S, inverse supply function)
P = 0.1Q + 1.6 (S∗, new inverse supply function)
(A government tax of £1.20 per item is introduced)
5 10 15 20
1.4
2.6 S
S∗ (with tax)
Q
P
Previously, if the market price was £1.40, companies would
supply 10 items. Now the government takes £1.20 per
item, so for companies to supply 10 items, the market price
must be £2.60.

33. Fixed tax changes the supply functions and supply
curve
P = 0.1Q + 0.4 (S, inverse supply function)
P = 0.1Q + 1.6 (S∗, new inverse supply function)
(A government tax of £1.20 per item is introduced)
5 10 15 20
2
S
S∗ (with tax)
Q
P
Previously, for a market price of £2.00, companies would
supply 16 items. Now, for a market price of £2.00,
companies will only supply 4 items.

34. Equilibrium after tax
P = −0.5Q + 10 (D, inverse demand function)
P = 0.1Q + 0.4 (S, inverse supply function)
P = 0.1Q + 1.6 (S∗, new inverse supply function)
5 10 15 20
5
10
S
S∗ (with tax)•
D
Q
P
Solve D and S∗ (find the new equilibrium after tax)
−0.5Q + 10 = 0.1Q + 1.6, so 8.4 = 0.6Q
Q = 8.4 ÷ 0.6 = 14, and P = 0.1 × 14 + 1.6 = 3

35. Shifts of the supply and demand curve
Fixed tax shifts the supply curve up
Government subsidies shift the supply curve down
Advertising shifts the demand curve right
Decreased costs of substitute goods shifts the demand
curve left
What else shifts the curves?
10 20 30
−5
5
10
S
D
Q
P

36. Proportionate tax
P = 0.1Q + 0.4 (S, inverse supply function)
Stretch line S in the vertical direction by a factor of 1.25 to
a new line, S∗. Find the equation of line S∗.
(A government tax of 20% is introduced)
10 20
1
2
3
S
S∗ (with tax)
Q
P
P = 1.25(0.1Q + 0.4)
P = 0.125Q + 0.5 (S∗, new inverse supply function)
This is the same as replacing P by 0.8P in the supply
functions.

37. Use calculus to find the maximum total revenue
TR = −2Q2 + 100Q.
What is the maximum value of TR (total revenue), and
what is the corresponding value of Q?
20 40
500
1,000
1,500
0
Q
TR (£)
The maximum value of TR occurs where the gradient = 0.
d(TR)
dQ = −4Q + 100 (the gradient function)
The maximum value of TR occurs where d(TR)
dQ = 0.
If −4Q + 100 = 0, then Q = 25.
The maximum value of TR is
(−2 × 252) + (100 × 25) =£1250,
and the corresponding value of Q is 25 units.

38. Use algebra to find maximum total revenue

39. ECONOMICS STUDENTS ONLY
Marginal revenue
Marginal revenue is the increase in total revenue for each extra
item sold.
There are two methods of finding marginal revenue.
Marginal revenue between
two quantities (no calculus)
= increase in total revenue
increase in quantity = ∆TR
∆Q
•
•
Q
TR
The gradient of the line joining
two points on the TR curve.
Marginal revenue for a
particular quantity (calculus)
= d(TR)
dQ
•
Q
TR
The gradient of the TR curve at
a point on the curve.

40. ECONOMICS STUDENTS ONLY
Marginal revenue (no calculus)
Marginal revenue (no calculus)= increase in total revenue
increase in quantity = ∆TR
∆Q
The gradient of the line joining two points on the TR curve.
10 20 30 40
1,000
1,200
1,400
TR = −2Q2 + 100Q
•
•
Q
TR
Calculate marginal revenue between Q = 15 and Q = 25.
When Q = 15, TR = (−2 × 152) + (100 × 15) = 1050
When Q = 25, TR = (−2 × 252) + (100 × 25) = 1250
Marginal revenue is 1250−1050
25−15 = 20
Between Q = 15 and Q = 25, each extra item sold earns

41. ECONOMICS STUDENTS ONLY
Marginal revenue (using calculus)
Marginal revenue (using calculus) = d(TR)
dQ
The gradient of the TR curve at a point on the curve.
10 20 30 40
1,000
1,200
1,400
TR = −2Q2 + 100Q
•
Q
TR
Calculate marginal revenue for Q = 15
d(TR)
dQ = −4Q + 100 (the gradient function)
For Q = 15, the gradient is d(TR)
dQ = (−4 × 15) + 100 = 40
For Q = 15, each extra item sold earns the firm £40 in
revenue.
(For Q = 25, we have maximum total revenue and
marginal revenue is zero)

42. ECONOMICS STUDENTS ONLY
Integration in economics
Marginal revenue (using calculus), MR = d(TR)
dQ
Differentiate total revenue to find marginal revenue
Integrate marginal revenue to find total revenue
Differentiate





total revenue, TR
marginal revenue, MR





Integrate
MR = d(TR)
dQ
TR = (MR) dQ

43. ECONOMICS STUDENTS ONLY
Integration in economics
Question: MR = 18Q2 + 5. Integrate MR with respect to Q
to find TR.
Solution:
TR = (MR) dQ
TR = (18Q2 + 5) dQ
= 18 × Q3
3 + 5Q + C
TR = 6Q3 + 5Q + C
When Q = 0, we have TR = 0 (since there is no revenue
when no products are sold).
Substitute these values to find C
0 = 6 × 03
+ 5 × 0 + C
so C = 0 and
TR = 6Q3
+ 5Q

44. ECONOMICS STUDENTS ONLY
Economics example - maximum profit
Profit =total revenue (TR) - total cost (TC)
TR = −2Q2 + 100Q
TC = 20Q + 330
For which value of Q is
1. there profit? (TR > TC)
2. there a loss? (TR < TC)
3. total revenue a maximum?
4. profit a maximum? 10 20 30 40 50
500
1,000
0
Q
TC and TR (£)

45. ECONOMICS STUDENTS ONLY
Economics example - maximum profit
10 20 30 40 50
500
1,000
0
Q
TC and TR (£)
10 20 30 40 50
−100
100
0
Q
MC and MR (£)
TR = −2Q2 + 100Q
MR = d(TR)
dQ = −4Q + 100
TC = 20Q + 330
MC = d(TC)
dQ = 20
Maximum profit occurs when TR − TC
is a maximum,
that is when d(TR−TC)
dQ = 0
d(TR)
dQ − d(TC)
dQ = 0
MR − MC = 0
MR = MC.
So maximum profit occurs when
marginal cost = marginal revenue.
−4Q + 100= 20
Q = 20.

46. ECONOMICS STUDENTS ONLY
Economics example - elasticity of demand
10 20 30 40 50
TR
10 20 30 40 50
MR
AR
10 20 30 40 50
Ped
−1
Inverse demand function
P = −2Q + 100
Total revenue= quantity × price per item
TR = −2Q2 + 100Q
MR = d(TR)
dQ = −4Q + 100
AR = TR
Q = −2Q + 100 (average revenue)
Point price elasticity of demand = 1
dP/dQ
× P
Q
Ped = 1
−2 × −2Q+100
Q
Ped = −50Q−1 + 1
When Q < 25, we have Ped < −1
so demand is elastic.
When Q > 25, we have −1 < Ped < 0
so demand is inelastic.

47. ECONOMICS STUDENTS ONLY
Economics example - summary
10 20 30 40 50
TR
TC
0
10 20 30 40 50
MR
AR
AC
0
TR = −2Q2 + 100Q
MR = d(TR)
dQ = −4Q + 100
AR = TR
Q = −2Q + 100
TC = 20Q + 330
MC = d(TC)
dQ = 20
AC = TC
Q = 20 + 330
Q
At Q = 25 maximum total revenue
occurs, marginal revenue=0 and
demand changes from elastic to
inelastic.
At Q = 20 maximum profit occurs,
marginal cost = marginal revenue.
At Q = 5 and 35 there is no profit,
average cost = average revenue.