Transcript Chapter 3 Lines, Parabolas, and Systems

INTRODUCTORY MATHEMATICAL ANALYSIS
For Business, Economics, and the Life and Social Sciences
Chapter 3
Lines, Parabolas, and Systems
2011 Pearson Education, Inc.
Chapter 3: Lines, Parabolas and Systems
Chapter Objectives
• 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.
2011 Pearson Education, Inc.
Chapter 3: Lines, Parabolas and Systems
Chapter Outline
3.1) Lines
3.2) Applications and Linear Functions
3.3) Quadratic Functions
3.4) Systems of Linear Equations
3.5) Nonlinear Systems
3.6) Applications of Systems of Equations
2011 Pearson Education, Inc.
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Slope of a Line
• The slope of the line is for two different points
(x1, y1) and (x2, y2) is
y 2  y1 
verticalchange 
 

m
x2  x1  horizontal change 
2011 Pearson Education, Inc.
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 1 – Price-Quantity Relationship
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.
2011 Pearson Education, Inc.
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 1 – Price-Quantity Relationship
Solution:
The slope is
p2  p1 1  4
1
m


q2  q1 8  2
2
Equations of lines
• A point-slope form of an equation of the line
through (x1, y1) with slope m is
y 2  y1
m
x2  x1
y 2  y1  mx2  x1 
2011 Pearson Education, Inc.
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 3 – Determining a Line from Two Points
Find an equation of the line passing through (−3, 8)
and (4, −2).
Solution:
 2 8
10

The line has slope m 
4   3
7
Using a point-slope form with (−3, 8) gives
10
y  8   x   3
7
7 y  56  10x  30
10x  7 y  26  0
2011 Pearson Education, Inc.
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
• The slope-intercept form of an equation of the
line with slope m and y-intercept b is y  mx  c.
Example 5 – Find the Slope and y-intercept of a Line
Find the slope and y-intercept of the line with
equation y = 5(3  2x).
Solution:
Rewrite the equation as y  53  2x 
y  15  10 x
y  10 x  15
The slope is −10 and the y-intercept is 15.
2011 Pearson Education, Inc.
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 7 – Converting Forms of Equations of Lines
a. Find a general linear form of the line whose
slope-intercept form is
2
y  x4
3
Solution:
By clearing the fractions, we have
2
xy 40
3
2 x  3 y  12  0
2011 Pearson Education, Inc.
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 7 – Converting Forms of Equations of Lines
b. Find the slope-intercept form of the line having a
general linear form
3 x  4y  2  0
Solution:
We solve the given equation for y,
3 x  4y  2  0
4 y  3 x  2
y 
2011 Pearson Education, Inc.
3
1
x
4
2
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
1
m1  
m2
2011 Pearson Education, Inc.
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.
2011 Pearson Education, Inc.
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.
y   2  3x  3
y  2  3x  9
y  3 x  11
For the line perpendicular to y = 3x + 1,
y   2  
2011 Pearson Education, Inc.
1
x  3 
3
1
y  2   x 1
3
1
y   x 1
3
Chapter 3: Lines, Parabolas and Systems
3.2 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,
4x  2y  100
w herex, y  0
Solving for y gives
y  2x  50
2011 Pearson Education, Inc.
Chapter 3: Lines, Parabolas and Systems
3.2 Applications and Linear Functions
Demand and Supply Curves
• Demand and supply curves have the following
trends:
2011 Pearson Education, Inc.
Chapter 3: Lines, Parabolas and Systems
3.2 Applications and Linear Functions
Linear Functions
• A function f is a linear function which can be
written as f x   ax  b w herea  0
Example 3 – Graphing Linear Functions
15  2t
Graph f x   2x  1 and g t  
.
3
Solution:
2011 Pearson Education, Inc.
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:
y 2  y1  3  6
The slope is m 

 3 .
x2  x1
1   2
Using a point-slope form:
y  y 1  mx  x1 
y  6  3x   2
y  3 x
f x   3 x
2011 Pearson Education, Inc.
Chapter 3: Lines, Parabolas and Systems
3.3 Quadratic Functions
• Quadratic function is written as f x   ax 2  bx  2
where a, b and c are constants and a  0
Example 1 – Graphing a Quadratic Function
2


f
x


x
 4x  12 .
Graph the quadratic function
b
4

 2 .
Solution: The vertex is 
2a
2 1
The points are
0   x 2  4 x  12
0  x  6x  2
x  6 and 2
2011 Pearson Education, Inc.
Chapter 3: Lines, Parabolas and Systems
3.3 Quadratic Functions
Example 3 – Graphing a Quadratic Function
Graph the quadratic function g x   x 2  6x  7.
Solution:
b
6


3
2a
21
x 3 2
2011 Pearson Education, Inc.
Chapter 3: Lines, Parabolas and Systems
3.3 Quadratic Functions
Example 5 – Finding and Graphing an Inverse
2


y

f
x

ax
 bx  c determine the inverse
From
function for a = 2, b = 2, and c = 3.
Solution:
2011 Pearson Education, Inc.
Chapter 3: Lines, Parabolas and Systems
3.4 Systems of Linear Equations
Two-Variable Systems
•
There are three different linear systems:
Linear system
(one solution)
•
Linear system
(no solution)
Linear system
(many solutions)
Two methods to solve simultaneous equations:
a) elimination by addition
b) elimination by substitution
2011 Pearson Education, Inc.
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.
 3 x  4y  13

 3y  2x  3
Solution: Make the y-component the same.
9 x  12y  39

8 x  12y  12
Adding the two equations, we get x  3. Use
find
93  12y  39
Thus,
x 3

y  1
2011 Pearson Education, Inc.
y  1
x 3
to
Chapter 3: Lines, Parabolas and Systems
3.4 Systems of Linear Equations
Example 3 – A Linear System with Infinitely Many Solutions
x  5y  2
1x  5 y 1

2
2
Solve 

Solution: Make the x-component the same.
 x  5y  2

 x  5y  2
Adding the two equations, we get 0  0 .
The complete solution is
x  2  5r
y r
2011 Pearson Education, Inc.
Chapter 3: Lines, Parabolas and Systems
3.4 Systems of Linear Equations
Example 5 – Solving a Three-Variable Linear System
Solve
 2x  y  z  3

 x  2 y  2z  1
 x  y  3z  6

Solution: By substitution, we get
 3 y  7 z  15

 y  z  5
 x  y  3z  6

Since y = -5 + z, we can find z = 3 and y = -2. Thus,
 z3

 y  2
 x 1

2011 Pearson Education, Inc.
Chapter 3: Lines, Parabolas and Systems
3.4 Systems of Linear Equations
Example 7 – Two-Parameter Family of Solutions
Solve the system
 x  2y  z  4

2x  4y  2z  8
Solution:
Multiply the 2nd equation by 1/2 and add to the 1st
equation,
 x  2y  z  4


00
Setting y = r and z = s, the solutions are
x  4  2r  s
y r
zs
2011 Pearson Education, Inc.
Chapter 3: Lines, Parabolas and Systems
3.5 Nonlinear Systems
•
A system of equations with at least one nonlinear
equation is called a nonlinear system.
Example 1 – Solving a Nonlinear System
Solve
x 2  2x  y  7  0

 3x  y  1  0
(1)
(2)
Solution: Substitute Eq (2) into (1),
x 2  2 x  3 x  1  7  0
x2  x  6  0
x  3x  2  0
x  3 or x  2
2011 Pearson Education, Inc.
y  8 or y  7
Chapter 3: Lines, Parabolas and Systems
3.6 Applications of Systems of Equations
Equilibrium
•
The point of equilibrium is where demand and
supply curves intersect.
2011 Pearson Education, Inc.
Chapter 3: Lines, Parabolas and Systems
3.6 Applications of Systems of Equations
Example 1 – Tax Effect on Equilibrium
p
8
q  50
100
Let
be the supply equation for a
manufacturer’s product, and suppose the demand
equation is p   7 q  65.
100
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.
2011 Pearson Education, Inc.
Chapter 3: Lines, Parabolas and Systems
3.6 Applications of Systems of Equations
Example 1 – Tax Effect on Equilibrium
Solution:
a. By substitution,
7
8

q  65 
q  50
100
100
and
q  100
After new tax,
8
100  51.50  58.70
p
100
8
100  51.50   7 q  65
100
100
q  90
2011 Pearson Education, Inc.
and
8
100  50  58
p
100
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
yTR  pq  58100  5800
After tax,
yTR  pq  58.7090  5283
2011 Pearson Education, Inc.
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
yTC  yVC  y FC
•
The break-even point is where TR = TC.
2011 Pearson Education, Inc.
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.
2011 Pearson Education, Inc.
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 yTR  8q
yTC  yVC  y FC 
22
q  5000
9
At break-even point,
yTR  yTC
22
q  5000
9
q  900
8q 
and
b.
yTR  8900  7200
yTR  yTC
 22

 81800   1800  5000  5000
9

The profit is $5000.
2011 Pearson Education, Inc.