Introductory Mathematical Analysis - Chapter 3 : Lines , Parabolas and Systems

1. INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL
ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences
©2007 Pearson Education Asia
Chapter 3Chapter 3
Lines, Parabolas and SystemsLines, Parabolas and Systems

2. ©2007 Pearson Education Asia
INTRODUCTORY MATHEMATICAL
ANALYSIS
0. Review of Algebra
1. Applications and More Algebra
2. Functions and Graphs
3. Lines, Parabolas, and Systems
4. Exponential and Logarithmic Functions
5. Mathematics of Finance
6. Matrix Algebra
7. Linear Programming
8. Introduction to Probability and Statistics

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9. Additional Topics in Probability
10. Limits and Continuity
11. Differentiation
12. Additional Differentiation Topics
13. Curve Sketching
14. Integration
15. Methods and Applications of Integration
16. Continuous Random Variables
17. Multivariable Calculus
INTRODUCTORY MATHEMATICAL
ANALYSIS

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• To develop the notion of slope and different forms of
equations of lines.
• To develop the notion of demand and supply curves and
to introduce linear functions.
• To sketch parabolas arising from quadratic functions.
• To solve systems of linear equations in both two and
three variables by using the technique of elimination by
addition or by substitution.
• To use substitution to solve nonlinear systems.
• To solve systems describing equilibrium and break-even
points.
Chapter 3: Lines, Parabolas and Systems
Chapter ObjectivesChapter Objectives

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Lines
Applications and Linear Functions
Quadratic Functions
Systems of Linear Equations
Nonlinear Systems
Applications of Systems of Equations
3.1)
3.2)
3.3)
3.4)
3.5)
Chapter 3: Lines, Parabolas and Systems
Chapter OutlineChapter Outline
3.6)

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Slope of a Line
• The slope of the line is for two different points
(x1, y1) and (x2, y2) is
Chapter 3: Lines, Parabolas and Systems
3.1 Line3.1 Line






=
−
−
=
changehorizontal
changevertical
12
12
xx
yy
m

7. ©2007 Pearson Education Asia
The line in the figure shows the relationship
between the price p of a widget (in dollars) and the
quantity q of widgets (in thousands) that consumers
will buy at that price. Find and interpret the slope.
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 1 – Price-Quantity Relationship

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Solution:
The slope is
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 1 – Price-Quantity Relationship
2
1
28
41
12
12
−=
−
−
=
−
−
=
qq
pp
m
Equation of line
• A point-slope form of an equation of the line
through (x1, y1) with slope m is
( )1212
12
12
xxmyy
m
xx
yy
−=−
=
−
−

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Find an equation of the line passing through (−3, 8)
and (4, −2).
Solution:
The line has slope
Using a point-slope form with (−3, 8) gives
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 3 – Determining a Line from Two Points
( ) 7
10
34
82
−=
−−
−−
=m
( )[ ]
026710
3010567
3
7
10
8
=−+
−−=−
−−−=−
yx
xy
xy

10. ©2007 Pearson Education Asia
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 5 – Find the Slope and y-intercept of a Line
• The slope-intercept form of an equation of the line
with slope m and y-intercept b is . cmxy +=
Find the slope and y-intercept of the line with
equation y = 5(3-2x).
Solution:
Rewrite the equation as
The slope is −10 and the y-intercept is 15.
( )
1510
1015
235
+−=
−=
−=
xy
xy
xy

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a.Find a general linear form of the line whose
slope-intercept form is
Solution:
By clearing the fractions, we have
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 7 – Converting Forms of Equations of Lines
4
3
2
+−= xy
01232
04
3
2
=−+
=−+
yx
yx

12. ©2007 Pearson Education Asia
b. Find the slope-intercept form of the line having a
general linear form
Solution:
We solve the given equation for y,
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 7 – Converting Forms of Equations of Lines
0243 =−+ yx
2
1
4
3
234
0243
+−=
+−=
=−+
xy
xy
yx

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Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Parallel and Perpendicular Lines
• Parallel Lines are two lines that have the same
slope.
• Perpendicular Lines are two lines with slopes
m1 and m2 perpendicular to each other only if
2
1
1
m
m −=

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Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 9 – Parallel and Perpendicular Lines
The figure shows two lines passing through (3, −2).
One is parallel to the line y = 3x + 1, and the other is
perpendicular to it. Find the equations of these lines.

15. ©2007 Pearson Education Asia
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 9 – Parallel and Perpendicular Lines
Solution:
The line through (3, −2) that is parallel to y = 3x + 1
also has slope 3.
For the line perpendicular to y = 3x + 1,
( ) ( )
113
932
332
−=
−=+
−=−−
xy
xy
xy
( ) ( )
1
3
1
1
3
1
2
3
3
1
2
−−=
+−=+
−−=−−
xy
xy
xy

16. ©2007 Pearson Education Asia
Chapter 3: Lines, Parabolas and Systems
3.2 Applications and Linear Functions3.2 Applications and Linear Functions
Example 1 – Production Levels
Suppose that a manufacturer uses 100 lb of material
to produce products A and B, which require 4 lb and
2 lb of material per unit, respectively.
Solution:
If x and y denote the number of units produced of A
and B, respectively,
Solving for y gives
0,where10024 ≥=+ yxyx
502 +−= xy

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Chapter 3: Lines, Parabolas and Systems
3.2 Applications and Linear Functions
Demand and Supply Curves
• Demand and supply curves have the following
trends:

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Chapter 3: Lines, Parabolas and Systems
3.2 Applications and Linear Functions
Example 3 – Graphing Linear Functions
Linear Functions
• A function f is a linear function which can be
written as ( ) 0where ≠+= abaxxf
Graph and .
Solution:
( ) 12 −= xxf ( )
3
215 t
tg
−
=

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Chapter 3: Lines, Parabolas and Systems
3.2 Applications and Linear Functions
Example 5 – Determining a Linear Function
If y = f(x) is a linear function such that f(−2) = 6 and
f(1) = −3, find f(x).
Solution:
The slope is .
Using a point-slope form:
( )
3
21
63
12
12
−=
−−
−−
=
−
−
=
xx
yy
m
( )
( )[ ]
( ) xxf
xy
xy
xxmyy
3
3
236
11
−=
−=
−−−=−
−=−

20. ©2007 Pearson Education Asia
Chapter 3: Lines, Parabolas and Systems
3.3 Quadratic Functions3.3 Quadratic Functions
Example 1 – Graphing a Quadratic Function
Graph the quadratic function .
Solution: The vertex is .
• Quadratic function is written as
where a, b and c are constants and
( ) 22
++= bxaxxf
0≠a
( ) 1242
+−−= xxxf
( )
2
12
4
2
−=
−
−
−=−
a
b
( )( )
2and6
260
1240 2
−=
−+=
+−−=
x
xx
xx
The points are

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Chapter 3: Lines, Parabolas and Systems
3.3 Quadratic Functions
Example 3 – Graphing a Quadratic Function
Graph the quadratic function .
Solution:
( ) 762
+−= xxxg
( )
3
12
6
2
=−=−
a
b
23 ±=x

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Chapter 3: Lines, Parabolas and Systems
3.3 Quadratic Functions
Example 5 – Finding and Graphing an Inverse
From determine the inverse
function for a = 2, b = 2, and c = 3.
Solution:
( ) cbxaxxfy ++== 2

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Chapter 3: Lines, Parabolas and Systems
3.4 Systems of Linear Equations3.4 Systems of Linear Equations
Two-Variable Systems
• There are three different linear systems:
• Two methods to solve simultaneous equations:
a) elimination by addition
b) elimination by substitution
Linear system
(one solution)
Linear system
(no solution)
Linear system
(many solutions)

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Chapter 3: Lines, Parabolas and Systems
3.4 Systems of Linear Equations
Example 1 – Elimination-by-Addition Method
Use elimination by addition to solve the system.
Solution: Make the y-component the same.
Adding the two equations, we get . Use to
find
Thus,



=+
=−
323
1343
xy
yx



=+
=−
12128
39129
yx
yx
3=x
( )
1
391239
−=
=−
y
y



−=
=
1
3
y
x
3=x

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Chapter 3: Lines, Parabolas and Systems
3.4 Systems of Linear Equations
Example 3 – A Linear System with Infinitely Many Solutions
Solve
Solution: Make the x-component the same.
Adding the two equations, we get .
The complete solution is




=+
=+
1
2
5
2
1
25
yx
yx



−=−+−
=+
25
25
yx
yx
00 =
ry
rx
=
−= 52

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Chapter 3: Lines, Parabolas and Systems
3.4 Systems of Linear Equations
Example 5 – Solving a Three-Variable Linear System
Solve
Solution: By substitution, we get
Since y = -5 + z, we can find z = 3 and y = -2. Thus,





−=−−
=++−
=++
63
122
32
zyx
zyx
zyx





−+=
−=−
=+
63
5
1573
zyx
zy
zy





=
−=
=
1
2
3
x
y
z

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Chapter 3: Lines, Parabolas and Systems
3.4 Systems of Linear Equations
Example 7 – Two-Parameter Family of Solutions
Solve the system
Solution:
Multiply the 2nd
equation by 1/2 and add to the 1st
equation,
Setting y = r and z = s, the solutions are



=++
=++
8242
42
zyx
zyx



=
=++
00
42 zyx
sz
ry
srx
=
=
−−= 24

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Chapter 3: Lines, Parabolas and Systems
3.5 Nonlinear Systems3.5 Nonlinear Systems
Example 1 – Solving a Nonlinear System
• A system of equations with at least one nonlinear
equation is called a nonlinear system.
Solve (1)
(2)
Solution: Substitute Eq (2) into (1),



=+−
=−+−
013
0722
yx
yxx
( )
( )( )
7or8
2or3
023
06
07132
2
2
=−=
=−=
=−+
=−+
=−++−
yy
xx
xx
xx
xxx

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Chapter 3: Lines, Parabolas and Systems
3.6 Applications of Systems of Equations3.6 Applications of Systems of Equations
Equilibrium
• The point of equilibrium is where demand and
supply curves intersect.

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Chapter 3: Lines, Parabolas and Systems
3.6 Applications of Systems of Equations
Example 1 – Tax Effect on Equilibrium
Let be the supply equation for a
manufacturer’s product, and suppose the demand
equation is .
a. If a tax of $1.50 per unit is to be imposed on the
manufacturer, how will the original equilibrium price
be affected if the demand remains the same?
b. Determine the total revenue obtained by the
manufacturer at the equilibrium point both before and
after the tax.
50
100
8
+= qp
65
100
7
+−= qp

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Chapter 3: Lines, Parabolas and Systems
3.6 Applications of Systems of Equations
Example 1 – Tax Effect on Equilibrium
Solution:
a. By substitution,
and
After new tax,
and
100
50
100
8
65
100
7
=
+=+−
q
qq ( ) 5850100
100
8
=+=p
( ) 70.5850.51100
100
8
=+=p
( )
90
65
100
7
50.51100
100
8
=
+−=+
q
q

32. ©2007 Pearson Education Asia
Chapter 3: Lines, Parabolas and Systems
3.6 Applications of Systems of Equations
Example 1 – Tax Effect on Equilibrium
Solution:
b. Total revenue given by
After tax,
( )( ) 580010058 === pqyTR
( )( ) 52839070.58 === pqyTR

33. ©2007 Pearson Education Asia
Chapter 3: Lines, Parabolas and Systems
3.6 Applications of Systems of Equations
Break-Even Points
• Profit (or loss) = total revenue(TR) – total cost(TC)
• Total cost = variable cost + fixed cost
• The break-even point is where TR = TC.
FCVCTC yyy +=

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Chapter 3: Lines, Parabolas and Systems
3.6 Applications of Systems of Equations
Example 3 – Break-Even Point, Profit, and Loss
A manufacturer sells a product at $8 per unit, selling
all that is produced. Fixed cost is $5000 and variable
cost per unit is 22/9 (dollars).
a. Find the total output and revenue at the break-even
point.
b. Find the profit when 1800 units are produced.
c. Find the loss when 450 units are produced.
d. Find the output required to obtain a profit of
$10,000.

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Chapter 3: Lines, Parabolas and Systems
3.6 Applications of Systems of Equations
Example 3 – Break-Even Point, Profit, and Loss
Solution:
a. We have
At break-even point,
and
b.
The profit is $5000.
5000
9
22
8
+=+=
=
qyyy
qy
FCVCTC
TR
900
5000
9
22
8
=
+=
=
q
qq
yy TCTR
( ) 72009008 ==TRy
( ) ( ) 500050001800
9
22
18008 =





+−=− TCTR yy