Download Calculus Section 1.2 powerpoint lesson

Chapter 1
Linear Equations
and Graphs
Section 2
Graphs and Lines
Part I
Learning Objectives for Section 1.2
Graphs and Lines
 The student will be able to identify and work with the
Cartesian coordinate system.
 The student will be able to draw graphs for equations of the
form Ax + By = C.
 The student will be able to calculate the slope of a line.
 The student will be able to graph special forms of equations
of lines.
 The student will be able to solve applications of linear
equations.
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The Cartesian Coordinate System
 The Cartesian coordinate system was named after René Descartes. It
consists of two real number lines, the horizontal axis (x-axis) and the
vertical axis (y-axis) which meet in a right angle at a point called the
origin. The two number lines divide the plane into four areas called
quadrants.
 The quadrants are numbered using Roman numerals as shown on the
next slide. Each point in the plane corresponds to one and only one
ordered pair of numbers (x,y). Two ordered pairs are shown.
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The Cartesian Coordinate System
(continued)
II
I
(3,1)
x
III
(–1,–1)
Two points, (–1,–1)
and (3,1), are plotted.
Four quadrants are as
labeled.
IV
y
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Linear Equations in Two Variables
 A linear equation in two variables is an equation that can
be written in the standard form Ax + By = C, where A, B,
and C are constants (A and B not both 0), and x and y are
variables.
 A solution of an equation in two variables is an ordered
pair of real numbers that satisfy the equation. For example,
(4,3) is a solution of 3x - 2y = 6.
 The solution set of an equation in two variables is the set
of all solutions of the equation.
 The graph of an equation is the graph of its solution set.
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Linear Equations in Two Variables
(continued)
 If A is not equal to zero and B is not equal to zero, then
Ax + By = C can be rewritten as
𝐴
𝐶
𝑦 = − x+
→ 𝑦 = mx + b
𝐵
𝐵
 This is known as slope-intercept form.
 If A = 0 and B is not equal to zero,
then the graph is a horizontal line
C
y
B
 If A is not equal to zero and B = 0,
then the graph is a vertical line
C
x
A
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Using Intercepts to Graph a Line
Graph 2x – 6y = 12.
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Using Intercepts to Graph a Line
Graph 2x – 6y = 12.
x
y
0
–2
y-intercept
6
0
x-intercept
3
–1 check point
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Special Cases
 The graph of x = k is the graph of a vertical line k units from
the y-axis.
 The graph of y = k is the graph of the horizontal line k units
from the x-axis.
 Examples:
1. Graph x = –7 (this cannot be graphed on some calculators)
2. Graph y = 4
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Solutions
x = –7
y=4
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Slope of a Line
 Slope of a line:
y2  y1 rise
m

x2  x1 run
 x1 , y1 
rise
 x2 , y2 
run
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Slope-Intercept Form
The equation
y = mx+b
is called the slope-intercept form of an equation of a line.
The letter m represents the slope and b represents the
y-intercept.
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Find the Slope and Intercept
from the Equation of a Line
Example: Find the slope and y intercept of the line
whose equation is 5x – 2y = 10.
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Find the Slope and Intercept
from the Equation of a Line
Example: Find the slope and y intercept of the line
whose equation is 5x – 2y = 10.
Solution: Solve the equation
for y in terms of x. Identify the
coefficient of x as the slope and
the y intercept as the constant
term.
5𝑥 − 2𝑦 = 10
−2𝑦 = −5𝑥 + 10
5
𝑦 = 𝑥−5
2
Therefore: the slope is 5/2 and
the y intercept is –5.
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Point-Slope Form
The point-slope form of the equation of a line is
y  y1  m( x  x1 )
where m is the slope and (x1, y1) is a given point.
It is derived from the definition of the slope of a line:
y2  y1
m
x2  x1
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Cross-multiply and
substitute the more
general x for x2
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Example
Find the equation of the line through the points (–5, 7) and (4, 16).
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Example
Find the equation of the line through the points (–5, 7) and (4, 16).
Solution:
16  7 9
m
 1
4  (5) 9
Now use the point-slope form with m = 1 and (x1, x2) = (4, 16).
(We could just as well have used (–5, 7)).
𝑦 − 16 = 1(𝑥 − 4)
𝑦 = 𝑥 − 4 + 16
𝑦 = 𝑥 + 12
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Homework
#1-2A: Pg 23
(3-51 mult of 3)
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Chapter 1
Linear Equations
and Graphs
Section 2
Graphs and Lines
Part II
Using a Graphing Calculator
Graph 2x – 6y = 12 on a graphing calculator and find the intercept.
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Using a Graphing Calculator
Graph 2x – 6y = 12 on a graphing calculator and find the intercept.
Solution: First, we solve the equation for y.
2x – 6y = 12
Subtract 2x from each side.
–6y = –2x + 12
Divide both sides by –6
y = (1/3)x – 2
Now we enter the right side of this equation in Y=, then Zoom
Standard for the window dimensions, and graph the line.
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Using a Graphing Calculator
(continued)
• To find the x-intercept
• Go to CALC
• Select 2: zero
• Left bound? Position the cursor somewhere left
of the x-intercept and hit ENTER
• Right bound? Position the cursor somewhere
right of the x-intercept and hit ENTER
• Guess? Hit ENTER
• X-intercept value appears at bottom of screen
• To “plug in” a value for x
• Go to TRACE
• Enter x value (try 0), then ENTER
• Y-coordinate appears at bottom of screen
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Depreciation
 What is depreciation? Why do products depreciate?
 Give an example of a product that depreciates quickly.
 Give an example of a product that depreciates slowly.
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Cost Analysis
Office equipment was purchased brand new for $20,000
and will depreciate to a value of $2,000 after 10 years.
1. If its value depreciates linearly, find an equation that
represents the depreciated value in dollars after t
years.
2. What is the value of the equipment after 11 years?
3. When will the office equipment be worth half as much
as it was brand new?
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Cost Analysis (solution)
1. If its value depreciates linearly, find an equation that represents
the depreciated value in dollars after t years.
Answer: When t = 0, V = 20,000 and when t = 10, V = 2,000.
Thus, we have two ordered pairs (0, $20000) and (10, $2000).
Now, find the slope using the two points:
2000−20000
10−0
𝑚=
= −1800
The y intercept is already known (when t = 0, V = 20000)
b = 20000
Therefore, our equation is: V = –1,800t + 20,000.
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Cost Analysis (solution)
2. What is the value of the office equipment after 11 years?
• Set-up: t = 11 , V = ?
V = -1800(11) + 20000
• Answer: V= $200
3. When will the office equipment be worth half as much as
it was brand new?
• Set-up: V = 10000, t = ?
10000 = -1800t + 20000
• Answer: t  5.6 years
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Supply and Demand
 In a free competitive market, the price of a product is
determined by the relationship between supply and
demand.
• If there is a surplus, the price tends to come down.
• If there is a shortage, the price tends to go up.
• The price tends to stabilize at the point of intersection
of the demand and supply equations.
 This point of intersection is called the equilibrium point.
 The corresponding price is called the equilibrium price.
 The common value of supply and demand is called the
equilibrium quantity.
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What is a bushel?
A bushel is a term used in
agriculture.
8 quarts = 1 peck
4 pecks = 1 bushel
? Quarts = 1 bushel
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Example
At a price of $1.94 per bushel, the supply of corn is 9.8 billion
bushels and the demand is 9.3 billion bushels.
At a price of $1.82 per bushel, the supply of corn is 9.4 billion
bushels and the demand is 9.5 billion bushels.
1. Find a price-supply equation of the form p = mx + b
2. Find a price-demand equation of the form p = mx + b
3. Find the equilibrium point algebraically.
4. Graph the equations on your graphing calculator and
verify your answer in step 3.
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Example (solution)
1. Find a price-supply equation of the form p = mx + b
Supply data: 9.8 billion bushels  $1.94
9.4 billion bushels  $1.82
Ordered pairs: (9.8, $1.94) (9.4, $1.82)
1.82 − 1.94
𝑚=
= 0.3
9.4 − 9.8
Use the point-slope form to find the equation of the line:
𝑦 − 𝑦1 = 𝑚 𝑥 − 𝑥1
𝑦 − 1.94 = 0.3(𝑥 − 9.8)
𝑦 = 0.3𝑥 − 1
So, price-supply equation is: p = 0.3x – 1
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Example (solution)
2. Find a price-demand equation of the form p = mx + b
Demand data: 9.3 billion bushels  $1.94
9.5 billion bushels  $1.82
Ordered pairs: (9.3, $1.94) (9.5, $1.82)
1.82 − 1.94
𝑚=
= −0.6
9.5 − 9.3
Use the point-slope form to find the equation of the line:
𝑦 − 𝑦1 = 𝑚 𝑥 − 𝑥1
𝑦 − 1.94 = −0.6(𝑥 − 9.3)
𝑦 = −0.6𝑥 + 7.52
So, price-demand equation is: p = -0.6x+7.52
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Example (solution)
3. Find the equilibrium point algebraically.
0.3𝑥 − 1 = −0.6𝑥 + 7.52
0.9𝑥 = 8.52
𝑥 ≈ 9.47
𝑦 = 0.3𝑥 − 1
𝑦 = 0.3(9.47) − 1
𝑦 = 1.84
So the equilibrium point is (9.47, 1.84). This means the
equilibrium price is $1.84 and the equilibrium quantity is
9.47 billion bushels.
When the supply and demand are both 9.47 billion bushels,
then the price per bushel is $1.84.
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Example (solution)
4. Graph the equations on your graphing calculator and
verify your answer in step 3.
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Business – Markup Policy
Urban Outfitters sells a shirt costing $20 for $33 and a
jacket costing $60 for $93.
A) If the markup policy of the store is assumed to be linear,
write an equation that expresses retail price R in terms
of wholesale cost C.
B) What does the store pay for an outfit that retails for
$240?
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Business (solution)
Urban Outfitters sells a shirt costing $20 for $33 and a
jacket costing $60 for $93.
Ordered pairs (wholesale, retail): (20, 33), (60, 93)
93 − 33
𝑚=
= 1.5
60 − 20
𝑦 − 33 = 1.5(𝑥 − 20)
𝑦 = 1.5𝑥 + 3
𝑅 = 1.5𝐶 + 3
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Business (solution)
What does the store pay for an outfit that retails
for $240?
𝑅 = 1.5𝐶 + 3
𝑅 = $240
$240 = 1.5𝐶 + 3
C = $158
The store pays $158 for an outfit that retails for $240.
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Homework
#1-2B: Pg 24
(61-79 odd)
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