Linear Relationships Nonlinear Relationships Linear · PDF fileEC1008 Handout 1: FUNCTIONS AND...

EC1008 Handout 1: FUNCTIONS AND GRAPHS

• Linear Relationships • Nonlinear Relationships

Linear Equations

Y = a + bX a, b positive parameters

Key feature: slope is not a function of position on the line (i.e., not a function of the value of X).

Y

X

a b

Nonlinear Relationships

Key difference between linear and nonlinear relationships: slope becomes a function of X. Four important nonlinear relationships: • Quadratic functions • Power functions • Logarithmic functions • Exponential functions

Quadratic Function

2cXbXaY +++= a, b, c positive parameters

Y

Power Functions

Y = a + bXc, a, b, c parameters. Suppose a, b, c positive, c > 1.

Exponential Function

Y = exp(X) = eX, e ≈ 2.7183

Y

X

a

X

eX

X

Logarithmic Function

Y = ln(X)

Summary

Generic representation of a function: Y = f(X) Five important examples: linear functions: bXaY += quadratic functions 2cXbXaY +++=

power functions: CbXaY += exponential function: ))(exp( XfY = logarithmic function: ))(ln( XfY = Logarithmic transformations: ,aXYGiven b= ).Xln(b)aln()Yln( •+= ),Xexp(aYGiven •= .X)aln()Yln( +=

ln(X)

X 1

A horizontal line has zero slope

0

10

20

30

0 5 10x

yy = 18

slope = 0

as x increases, y does not change

Positive slope, zero intercept

0

250

500

0 25 50x

y

y = 9xas x increases,

y increases

slope = 9

line passes through the origin

Negative slope, positive intercept

0

10

20

30

40

50

60

0 5 10 15x

y

y = 50 - 4x

larger x values go with smaller y values

slope = - 4

Positive slope, negative intercept

-30-20-10

010203040

x

y

10 20

y = -25 + 3x

line cuts y axis below the origin

slope = 3

as x increases, y increases

A vertical line has infinite slope

0

10

20

30

40

0 5 10 15 20x

y x = 15y increases but x does not change

slope = ∞

1Copyright©2001 Teresa Bradley and John Wiley & Sons Ltd

Table 2.3 Demand scheduleQuantity (x) Price (y)0 10040 8080 60120 40160 20200 0Intercept (vertical) = 100Slope = - 0.5: Horizontal intercept = 200

Demand Function P =100 - 0.5Q :

Join the points and label the graph

P =100 - 0.5Q

Quantity Q

PriceP

: Demand

P = 100 - 0.5Q

D

0

20

40

60

80

100

120

0 40 80 120

160

200

240

Figure 2.17


The General Linear Demand function such as

0

20

40

60

80

100

120

0

40 80 120

160

200

240

Figure 2.19 Demand function, P = 100 - 0.5Q

a = 100

P = 100 - 0.5Q

D

Q

P

Slope = - b = - 0.5

ab= 200

P = a - bQ P =100 - 0.5Q


Supply Function P = 10 + 0.5QCalculate and plot the supply schedule P = 10 + 0.5Q

0

10

20

30

40

50

60

70

-20 0 20 40 60 80 100

Table 2.4 Supply scheduleQ: Quantity P: P = 10 +0.5Q0 10: P =10 +0.5(0)20 20: P = 10+0.5(20)40 3060 4080 50100 60Intercept (vertical) = 10Slope = 0.5Horizontal intercept = - 20

P = 10 + 0.5Q

NOTE: the supply function may be plotted by simply joining the intercepts

Figure 2.22

c = 10

P = 10 + 0.5Q

Slope = 0.5

Q

P

S


The equation of a line

Putting it another way:the equation of a line may be described as the formula that allows you to calculate the y co-ordinate for any point on the line, when given the value of the x co-ordinate.

Example: y = x is a line which has a slope = 1, intercept = 0

Example:y = x + 2 is the line which has a slope = 1 , intercept = 2

The equation of a line may be written in terms of the two characteristics, m (slope) and c (intercept) . y = mx + c


How to plot any Linear Budget Constraint

Rearrange the equation in the form y = mx + c (see above)Plot y on the vertical axis, against x on the horizontal axisCalculate and plot the vertical and horizontal interceptsJoin the points and label the graph

xP yP M y MP

PP

xX YY

X

Y+ = → = −

⎛

⎝⎜

⎞

⎠⎟

MP X

MPYQuantity of good Y , y

Quantity of good X, x0

10

20

30

40

0 30 60 90


Plot the Budget Constraint: 2x + 6y = 180where PX =£2: PY = £6: M = 180: x(2) + y(6) = 180

Plot the horizontal intercept: x = 90Plot the vertical intercept: y = 30Join these points

0

10

20

30

40

0 30 60 90

Quantity of good Y , y

Quantity of good X, x

y = 30 - 0.33 x

MP X

MPY

Slope = −PP

X

Y

Figure 2.39


1) If Px = 3, PxX+PyY=M: 3x+ 6y=180 : y=30-0.5x2) Now adjust the graph of the Budget Constraint(y=30 - 0.5x) when the price of good X decreases

Intercept is the same: Slope has changed from -0.5 to -0.25

Figure 2.40 ∆PX and its effect on the Budget constraint

0

10

20

30

0 20 40 60 80 100 120

y = 30 - 0.5x Original Constraint

x

y

y = 30 - 0.25x Constraint after P changed


Adjust the graph of the Budget Constraint: y = 30 - 0.5xwhen the per unit price of good Y decreases from 6 to 3

The adjusted constraint pivots upwards from the unchanged horizontal intercept (see Figure 2.41)Comment: When Y decreases in price, more units of Y are affordable

Figure 2.41 ∆Py and its effect on the budget constraint

0

10

20

30

40

50

60

70

0 20 40 60

y = 30 - 0.5 x

y = 60 - x

x

y


Summary: Change in the graph of the Budget Constraint:y = 30 - 0.5x when the budget limit increases

Slope is the same: intercept has changed from 30 to 40When the budget limit increases, the constraint moves upwards, parallel to the original constraintComment: When the budget limit increases, more units of both X and Y are affordable

Figure 2.42 ∆Y and its effect on the Budget constraint

0

10

20

30

40

50

0 20 40 60 80x

y

Budget = 240

Budget = 180


Quadratic Function

The basic quadratic is y = x2

The quadratic, y = ax2 : (a > 0) is wider than y = x2 when a < 1:is narrower than y = x2 when a > 1:

A quadratic equation takes the form ax2 + bx + c = 0

You can solve it graphically or by using the formula

aacbb

242 −±−

=x


Worked Example 4.4(a)y = 0.6 x2 is wider than y = x2Worked Example 4.4(a)y = 0.6 x2 is wider than y = x2

5

15

25

35

-4 -3 -2 -1 0 1 2 3 4 x

y

Figure 4.5

y = x2

y = 2x2

y = 0.6x2

x -3 -2 -1 0 1 2 3

y =0.6x 2 5.4 2.4 0.6 0 0.6 2.4 5.4


Worked Example 4.3 The graph of y = - x2 is the reflection of the graph of y = x2 through the x-axis

x -4 -3 -2 -1 0 1 2 3 4y = x2 16 9 4 1 0 1 4 9 16y = -x2 -16 -9 -4 -1 0 -1 -4 -9 -16

y = - x 2-20-15-10-5

5101520

-4 -3 -2 1 2 3 4x

y y = x2

Figure 4.4


Properties of quadratic functions, illustrated graphically

If c<0 intercept is below x-axis, if c>0 aboveThe quadratic is a minimum type if a > 0, a maximum type if a < 0The graph is symmetrical about the vertical line drawn through the maximum or minimum pointThe roots are at the points of intersection with the x-axisThe roots are equidistant, (one greater, one smaller), from the x-coordinate of the maximum or minimum point

These point are illustrated in Worked Example 4.5, for the quadratic: y = 2x2 - 7x -9


Measure the distance (along the horizontal axis) between the roots and the vertical line thro’ the minimum point: y = 2x2 - 7x + 9

2.752.75root = 4.5root = -1

(1.75, - 17)Minimum point

Figure 4.6 Graph for Worked Example 4.5

x

-17

-9

9

18

-4 -3 -2 -1 0 1 3 4 5 6

y = 2 x 2 - 7 x - 9

y


Find the roots of 2x2 - 7x - 9 = 0using the ‘-b’ formula

2x2 - 7x - 9 = 0

Thus, confirming the answer obtained graphically

x = − − ± − − −( ) ( ) ( )( )( )

7 7 4 2 92 2

2

=± +7 49 72

4

4=

±=

±7 121 7 114

4 4x = +

= =7 11 18 4.5 x =

−=−

= −7 11

44

41or


Total Revenue TR = 50Q - 2Q2:: Summary

From the graph, the maximum TR = 312.5, when Q = 12.5From the graph, TR = 0 at Q = 0 and Q = 25 (roots)

Roots

Figure 4.8 Total revenue function TR = 50Q - 2Q2

0

120

240

360

0 5 10 15 20 25

TR= 312.5

Q =12.5

30

Maximum

Q


Rule 1

To multiply numbers with

the same base, add the

indices.

a m × a n = a m + n 5 2 × 5 3 = 5 2 + 3 = 5 5 ...by rule

(5 × 5) × (5 × 5 × 5) = 5 5 ...directly

Rule 2

To divide numbers with the

same base, subtract the

indices.

aa a

m

nm n= − 3

33 3

3 3 3 33 3

3 31

3

4

24 2 2

2

= =

× × ××

= × =

− . . .

. . .

by rule

directly

Rule 3

To raise an exponential to

a power, multiply the

indices.

( ) ka am m k= × ( ) by rule

(2 2 2)(2 2 2) = 2 directly

2

6

2 2 23 3 2 6= =

× × × ×

× . . .

. . .

Rules for indices


Converting from Index to Log

for example:

(b) 8 23= , then

log 2 (8) = 3

(c)

0 125 8 1. = = −− then log (0.125) 18

Number basepower=

Number basepower= the base of the index becomesthe base of the log, the power

then log (Number) powerbase =drops down.

So, logs are powers!


Rules for LogsOperation Notation Example

Rule 1Add

log b M + log b N ⇔ log b MN ln(4) + ln(29) = ln(4 × 29)

1.3863 + 3.3673 = ln(116)

4.7536 = 4.7536

Rule 2Subtract

log b M - log b N ⇔ log bMN

⎛⎝⎜

⎞⎠⎟ log(90) - log(26) = log 90

26⎛⎝⎜

⎞⎠⎟

1.9542 - 1.4150 = log(3.4615)

0.5392 = 0.53926

Rule 3Log of an

exponentiallog b(M z) ⇔ z log b (M )

log(5 3) = 3 log(5)

log(125) = 3(0.69897)

2.09691 = 2.09691

Rule 4Change of base log b

x

x

NNb

( )log ( )log ( )

⇔

the new base, x , is usually

10 or e since both are readily

available on the calculator.

log ( ) log( )log( )

.

..2 16 16

21 20410 3010

4 0= = =

or change to base e

log ( ) ln( )ln( )

.

..2 16 16

22 77260 6931

4 0= = =


Antilogs: Going from log form to index form

Number basepower=

Number basepower= the base of the power becomes

then log (Number) powerbase = the base of the log, the power

drops downthen, taking antilogs is described as reversing the the process of taking logs,

logbase (Number) power= ...the base of the log becomes

(Number) basepower= the base of the power, the power goes up.


Figure 4.15: As the size of the base increases, the graph becomes steeper For example, base a increases from a = 2 to a = e (2.718..) to a = 3.5,

0

1

2

3

4

5

6

-3 -2 -1 0 1 2 3

y = ( 2 ) x

y = ( 3 . 5 ) x y = e x

x

y

Figure 4.15 Graphs for Tables 4.10 and 4.11


The graphs of ex and e-x: summary

From the Table: when x is replaced by -x, points for e-x are reflected to ex and vice-versa

y = ex

The sketch below confirms that the graphs are mirror images through the y-axis

x -2.00 -1.50 -1.00 -0.50 0.00 0.50 1.00 1.50 2 2.5

0.14 0.22 0.37 0.61 1.00 1.65 2.72 4.48 7.39 12.18

7.39 4.48 2.72 1.65 1.00 0.61 0.37 0.22 0.14 0.08y = e-x

-2.0

0

-1.5

0

-1.0

0

-0.5

0

0.00

0.50

1.00

1.50

y = ex

y = e-x

2.00

4.00

6.00

8.00

10.00

12.00

14.00

2

2.5x

(growth curve)(decay curve)


Worked Example 4.15(b) Algebraic solution

The population in any year is calculated from the equation

where t = 0 at the beginning of 1980

To calculate the population in 1990, (a) determine the value of t: t = 1990 - 1980 = 10(b) substitute t = 10 into the equation for population

= 1016.4 (million)Similarly, in 2000 (a) t = 20 (b) Population = 1372 (million)

P e t= 753 0 03.

P e= 753 0 03 10. ( )


Worked Example 4.15(c)

Start with the population of 1750, on the vertical axisDraw a horizontal line across to the graphFrom this point on the graph, drop a vertical line down to the horizontal axisRead off the year. The population is 1750 in the year 2008 approximately

250

2250

1980 1985 1990 1995 2000 2005 2010

1750

753

population =

Year

FUNCTIONS AND GRAPHS: SUMMARY

Let pxy =

If 1−=p 1−= xy exponential function

If 0=p 1=y linear function (b=0)

If 1=p xy = linear function (a=0)

If 2=p 2xy = quadratic function

• Functions of One Variable bxay +=

2cxbxay ++= xey =

• Functions of Two (or more) Variables

czbxay ++=

Functions of More Than 1 Variable

• Multivariate function: the dependent variable, y, is a function of more than one independent variable

• If y = f(x,z) y is a function of the two variables x and z

• We substitute values for x and z to find the value of the function

• If we hold one variable constant and investigate the effect on y of changing the other, this is a form of comparative staticsanalysis

Total and Average Revenue

• TR = P.Q• AR = TR/Q = P

• A downward sloping linear demand curve implies a total revenue curve which has an inverted U shape

• Symmetric: the shape of one half of the curve is the mirror image of the other half

Total and Average Cost

• Total Cost is denoted TC • Fixed Cost: FC is the constant term in TC• Variable Cost: VC = TC – FC• Average Cost per unit output:

AC = TC/Q• Average Variable Cost:

AVC = VC/Q• Average Fixed Cost:

AFC = FC/Q

Profit

• Profit = π = TR – TC• If TC = 120 + 45Q – Q2 + 0.4Q3

• and TR = 240Q – 20Q2

• π = TR – TC substitute using brackets • π = 240Q – 20Q2 – (120 + 45Q – Q2 + 0.4Q3)

Taking the minus sign through the brackets and applying it to each term in turn gives

• π = 240Q – 20Q2 – 120 – 45Q + Q2 – 0.4Q3

and collecting like terms we find

• π = – 120 + 195Q – 19Q2 – 0.4Q3

Linear Relationships Nonlinear Relationships Linear · PDF fileEC1008 Handout 1: FUNCTIONS AND...

Documents

Transcript of Linear Relationships Nonlinear Relationships Linear · PDF fileEC1008 Handout 1: FUNCTIONS AND...