Lecture 5 - Graph Transformationsand Maths for Economics

C2 Foundation Mathematics (Standard Track)

Dr Linda Stringer Dr Simon Craikl.stringer@uea.ac.uk s.craik@uea.ac.uk

INTO City/UEA London

Lecture 5 skills

I Sketch the line with equations y = xI Sketch the curves with equations y = x2, y = x3 and y = 1

x

I Sketch a given curve after transformationI Find the new equation of a given curve after transformation

I Differentiate a total revenue function and hence findmaximum revenue

Sketch a curve after transformation

Question: The dashed curve below is shifted up by 2. Sketchthe new curve on the same axes. Label the stationary pointsand the y-intercept.

1 2 3 4

2

4

6

•(0,1)

•(1,5)

•(3,1) x

y

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

Sketch a curve after transformation

Question: The dashed line below is reflected in the y-axisSketch the new line on the same axes. Label the intercepts.

−3 −2 −1 1 2 3

2

4

•(0,2)

•(1,0) x

y

Sketch a curve after transformation

Answer:

−3 −2 −1 1 2 3

2

4

•(0,2)

•(−1,0) x

y

The graph of y = x2

−4 −2 2 4

−4

−2

2

4

x

y

Let f (x) = x2

Then y = f (x) is thesame as y = x2

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 curvey = f (x) in the x-axis,then we get the curvey = −f (x).

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 curvey = f (x) verticallyupwards by 1,then we get the curvey = f (x) + 1.

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 curvey = f (x) verticallydownwards by 4,then we get the curvey = f (x)− 4.

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 curvey = f (x) left by 2,then we get the curvey = f (x + 2).

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 curvey = f (x) right by 2,then we get the curvey = f (x − 3).

Graph transformations

Let f (x) be a function and let c be a constant. Consider thegraph y = f (x).

y = Transformation−f (x) Reflection in the x-axisf (−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 cf (x + c) Shift left by cf (x − c) Shift right by c

cf (x) Vertical stretch by a factor of cf (cx) Horizontal stretch by a factor of 1

c

Graph transformations

Let f (x) be a function and let c be a constant.I The graph of y = −f (x) is the graph of y = f (x) reflected in

the x-axis.I The graph of y = f (−x) is the graph of y = f (x) reflected in

the y-axis.I The graph of y = f (x) + c is the graph of y = f (x) shifted

up by c.I The graph of y = f (x + c) is the graph of y = f (x) shifted

to the left by c.I 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).I The graph of y = f (cx) is the graph of y = f (x) stretched

horizontally by a factor of 1c (the y-axis remains fixed).

The graph of y = x3

−2 −1 1 2

−5

5

x

y

y = x3

The graph of y = −x3

The graph of y = −x3 is the graph of y = x3 reflected in thex-axis OR in the y-axis

−2 −1 1 2

−5

5

x

y

Let f (x) = x3.Then −f (x) = −x3, and alsof (−x) = (−x)3 = −x ×−x ×−x = −x3.So the graph of y = −f (x) (reflection in x-axis) is the same asthe graph y = f (−x) (reflection in y-axis).

Transformations of the graph y = x3 − xI The dashed line shows y = 5(x3 − x). This is the graph of

y = x3 − x stretched vertically by factor 5.I The dotted line shows y = (3x)3 − (3x). This is the graph

of y = x3 − x stretched horizontally by factor 13 . (Note that

stretching by a factor less than 1 is the same ascompressing).

−1 1

−4

−2

2

4

x

y

The graph of y = 1x

−10 −5 5 10

−4

−2

2

4

x

y

y = 1x

The graph of y = 1x + 3

This is the graph of y = 1x shifted upwards by 3.

−10 −5 5 10

−4

−2

2

4

6

8

x

y

y = 1x + 3

The graph of y = 1x+3

This is the graph of y = 1x shifted left by 3.

−10 −5 5 10

−4

−2

2

4

x

y

y = 1x+3

ExamplesI Question: A curve has equation y = x4 + 9x2. The curve is

shifted vertically upward by 3. What is the equation of thenew curve?

I 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.

I Question: A curve has equation y = x2 +2x +5. The curveis shifted left by 3. What is the equation of the new curve?

I 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 equationy = (x + 3)2 + 2(x + 3) + 5.Now simplify the right sidey = (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.

Sketching graphs to solve simultaneous equationsQuestion: Graphically solve the simultaneous equations

−x + y = −1x + 2y = 4

Answer: Rearrange the equations and sketch the two lines

y = x − 1y = −1

2x + 2

−1 1 2 3−1

1

2

3

•(2,1)

The lines cross at the point (2,1) and so the solutions to thesesimultaneous equations are x = 2, y = 1

Maths for economics

I Linear equations and straight lines in economicsI Graph transformations in economicsI Calculus in economics - differentiating to find maximum

total revenueI Economics students only - marginal revenue

(differentiation and integration)I Economics students only - an example (average costs,

marginal revenue, elasticity)

The demand function and the inverse demand function

I The demand function expresses the quantity of a productdemanded, in terms of the market price per unit.

I Q = −2P + 20 (demand function)I As the market price increases, the quantity demanded by

consumers decreases.I In economics, P is always on the vertical axis, and in

maths we plot the subject of the equation on the verticalaxes, so rearrange the demand function to make P thesubject of the equation.

I Q = −2P + 20Q + 2P = 202P = −Q + 20P = −0.5Q + 10 (inverse demand function)Now price is expressed in terms of quantity.

The demand curveQ = −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 lineslopes down towards the right.(The exception is when we consider the demand curve for asmall firm in a perfectly competitive market - in that case thedemand curve is horizontal, so the gradient is zero).

Zero demand and free itemsI Find the P-intercept and the Q-intercept.

(At what price do items cease to be bought? What quantityis demanded when the item is free?)

P = −0.5Q + 10

0Q

P

I P-intercept: where the line crosses the P-axisQ = 0, so P = (−0.5 × 0) + 10 = 10(Items cease to be bought when the price is £10)

I Q-intercept: where the line crosses the Q-axisP = 0, so −0.5Q + 10 = 0, and Q = 20(20 units are demanded when the item is free)

The supply function and the inverse supply function

I The supply function expresses the quantity that companiessupply in terms of the market price per unit.

I Q = 10P − 4 (supply function)I The quantity produced by companies increases as the

market price increases.I Rearrange the supply function to make P the subject of the

equation.I Q = 10P − 4

Q − 10P = −4−10P = −Q − 4P = 0.1Q + 0.4 (inverse supply function)Now price is expressed in terms of quantity.

The supply curve

Q = 10P − 4 (supply function)P = 0.1Q + 0.4 (inverse supply function)

5 10 15 20

5

10

Supply curve

0Q

P

The gradient of the supply curve is positive - the line slopes uptowards the right.(What does the P-intercept of the supply curve tell us?)

Market equilibrium

I Solve for P and QQ = 10P − 4 (supply function)Q = −2P + 20 (demand function)(Find equilibrium price and quantity)

I Equate the right hand side of both equations

10P − 4 = −2P + 2012P = 24P = 2

Substitute P = 2 into either equation to find Q

Q = 10 × 2 − 4 = 16

(Equilibrium price is £2, the corresponding quantity is 16).

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).

Fixed tax changes the supply functions and supplycurve

I P = 0.1Q + 0.4 (S, inverse supply function)I 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)

I P = (0.1Q + 0.4) + 1.2P = 0.1Q + 1.6 (S∗, new inverse supply function)

5 10 15 20

0.4

1.6S

S∗ (with tax)

Q

P

I This is the same as replacing P by P − 1.2 in the supplyfunctions.

Fixed tax changes the supply functions and supplycurve

I P = 0.1Q + 0.4 (S, inverse supply function)I 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

I Previously, if the market price was £1.40, companies wouldsupply 10 items. Now the government takes £1.20 peritem, so for companies to supply 10 items, the market pricemust be £2.60.

Fixed tax changes the supply functions and supplycurve

I P = 0.1Q + 0.4 (S, inverse supply function)I 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

I Previously, for a market price of £2.00, companies wouldsupply 16 items. Now, for a market price of £2.00,companies will only supply 4 items.

Equilibrium after tax

I 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

SS∗ (with tax)•

D

Q

P

I Solve D and S∗ (find the new equilibrium after tax)I −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

Shifts of the supply and demand curve

I Fixed tax shifts the supply curve upI Government subsidies shift the supply curve downI Advertising shifts the demand curve rightI Decreased costs of substitute goods shifts the demand

curve leftI What else shifts the curves?

10 20 30

−5

5

10

S

D

Q

P

Proportionate taxI P = 0.1Q + 0.4 (S, inverse supply function)I 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

3S

S∗ (with tax)

Q

P

I P = 1.25(0.1Q + 0.4)P = 0.125Q + 0.5 (S∗, new inverse supply function)

I This is the same as replacing P by 0.8P in the supplyfunctions.

Use calculus to find the maximum total revenueI TR = −2Q2 + 100Q.I What is the maximum value of TR (total revenue), and

what is the corresponding value of Q?

20 40

500

1,000

1,500

0Q

TR (£)

I 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.

Use algebra to find maximum total revenue

ECONOMICS STUDENTS ONLYMarginal revenue

Marginal revenue is the increase in total revenue for each extraitem sold.There are two methods of finding marginal revenue.

Marginal revenue betweentwo quantities (no calculus)= increase in total revenue

increase in quantity = ∆TR∆Q

•

•

Q

TR

The gradient of the line joiningtwo points on the TR curve.

Marginal revenue for aparticular quantity (calculus)= d(TR)

dQ

•

Q

TR

The gradient of the TR curve ata point on the curve.

ECONOMICS STUDENTS ONLYMarginal revenue (no calculus)

Marginal revenue (no calculus)= increase in total revenueincrease in quantity = ∆TR

∆QThe 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

I Calculate marginal revenue between Q = 15 and Q = 25.When Q = 15, TR = (−2 × 152) + (100 × 15) = 1050When Q = 25, TR = (−2 × 252) + (100 × 25) = 1250

I Marginal revenue is 1250−105025−15 = 20

I Between Q = 15 and Q = 25, each extra item sold earnsthe firm £20 in revenue.

ECONOMICS STUDENTS ONLYMarginal 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

I Calculate marginal revenue for Q = 15d(TR)

dQ = −4Q + 100 (the gradient function)I For Q = 15, the gradient is d(TR)

dQ = (−4 × 15) + 100 = 40I For Q = 15, each extra item sold earns the firm £40 in

revenue.I (For Q = 25, we have maximum total revenue and

marginal revenue is zero)

ECONOMICS STUDENTS ONLYIntegration in economics

I Marginal revenue (using calculus), MR = d(TR)dQ

I Differentiate total revenue to find marginal revenueI Integrate marginal revenue to find total revenue

Differentiate

ytotal revenue, TR

marginal revenue, MR

x Integrate

I MR = d(TR)dQ

I TR =∫(MR)dQ

ECONOMICS STUDENTS ONLYIntegration in economics

I Question: MR = 18Q2 + 5. Integrate MR with respect to Qto find TR.

I Solution:TR =

∫(MR)dQ

TR =∫(18Q2 + 5)dQ

= 18 × Q3

3 + 5Q + CTR = 6Q3 + 5Q + C

When Q = 0, we have TR = 0 (since there is no revenuewhen no products are sold).Substitute these values to find C

0 = 6 × 03 + 5 × 0 + C

so C = 0 andTR = 6Q3 + 5Q

ECONOMICS STUDENTS ONLYEconomics example - maximum profit

Profit =total revenue (TR) - total cost (TC)

TR = −2Q2 + 100QTC = 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

0Q

TC and TR (£)

ECONOMICS STUDENTS ONLYEconomics example - maximum profit

10 20 30 40 50

500

1,000

0Q

TC and TR (£)

10 20 30 40 50

−100

100

0Q

MC and MR (£)

I TR = −2Q2 + 100QMR = d(TR)

dQ = −4Q + 100I TC = 20Q + 330

MC = d(TC)dQ = 20

I Maximum profit occurs when TR − TCis a maximum,that is when d(TR−TC)

dQ = 0d(TR)

dQ − d(TC)dQ = 0

MR − MC = 0MR = MC.So maximum profit occurs whenmarginal cost = marginal revenue.

I −4Q + 100= 20Q = 20.

ECONOMICS STUDENTS ONLYEconomics example - elasticity of demand

10 20 30 40 50

TR

10 20 30 40 50

MR

AR

10 20 30 40 50

Ped

−1

I Inverse demand functionP = −2Q + 100

I Total revenue= quantity × price per itemTR = −2Q2 + 100Q

I MR = d(TR)dQ = −4Q + 100

AR = TRQ = −2Q + 100 (average revenue)

I Point price elasticity of demand = 1dP/dQ

× PQ

Ped = 1−2 × −2Q+100

QPed = −50Q−1 + 1

I When Q < 25, we have Ped < −1so demand is elastic.

I When Q > 25, we have −1 < Ped < 0so demand is inelastic.

ECONOMICS STUDENTS ONLYEconomics example - summary

10 20 30 40 50TR

TC

0

10 20 30 40 50MR

ARAC

0

I TR = −2Q2 + 100QMR = d(TR)

dQ = −4Q + 100AR = TR

Q = −2Q + 100I TC = 20Q + 330

MC = d(TC)dQ = 20

AC = TCQ = 20 + 330

Q

I At Q = 25 maximum total revenueoccurs, marginal revenue=0 anddemand changes from elastic toinelastic.

I At Q = 20 maximum profit occurs,marginal cost = marginal revenue.

I At Q = 5 and 35 there is no profit,average cost = average revenue.