The slope of a line
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Transcript The slope of a line
Level 0
Math 0
Chapter 1
Faculty of Engineering -
Basic Science Department-
Prof H N Agiza
The slope of a line
Faculty of Engineering -
Basic Science DeptN Agiza
Prof H
The Slope of a Line :
• We define run to be the distance we move to the right and rise to be the
corresponding distance that the line rises (or falls).
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Basic Science DeptN Agiza
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Finding the Slope of a Line Through Two Points
Example : Find the slope of the line that passes through the points P(2, 1)
and Q(8, 5).
Solution :
𝑦2 − 𝑦1 5 − 1 4 2
𝑚=
=
= =
𝑥2 − 𝑥1 8 − 2 6 3
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𝑥1 = 2
𝑥2 = 8
𝑦1 = 1
𝑦2 = 5
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Point-Slope Form of the Equation of a Line
• Finding the equation of the line that passes through a given point and
P(𝑥1 , 𝑦1 ) has slope m.
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Finding the Equation of a Line with Given Point
and Slope
Example : Find an equation of the line through (1, -3) with slope −
Solution :
1
𝑦+3=− 𝑥−1
2
2𝑦 + 6 = −𝑥 + 1
𝑥 + 2𝑦 + 5 = 0
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1
.
2
Slope-Intercept Form of the Equation of a Line
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Lines in Slope-Intercept Form
Example : Find the equation of the line with slope 3 and y-intercept -2.
Solution:
𝑦 = 3𝑥 − 2
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Vertical and Horizontal Lines
If a line is horizontal, its slope is 𝑚 = 0 , so its equation is 𝑦 = 𝑏 , where b is the y-intercept
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Vertical and Horizontal Lines
• An equation for the vertical line through (3, 5) is x = 3.
• An equation for the horizontal line through (8, 2) is y =2.
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Parallel and Perpendicular Lines
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Finding the Equation of a Line Parallel to a Given
Line
Example: Find an equation of the line through the point (5,2) that is parallel
to the line 4𝑥 + 6𝑦 + 5 = 0
Solution:
4𝑥 + 6𝑦 + 5 = 0
𝑦=
2
− 𝑥
3
−
6𝑦 = −4𝑥 − 5
5
6
𝑚 = −23
2
𝑦−2=− 𝑥−5
3
3𝑦 − 6 = −2𝑥 + 10
2𝑥 + 3𝑦 − 16 = 0
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PERPENDICULAR LINES
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• Example: Show that the points P(3,3), Q(8,17), and R(11,5)are the
vertices of a right triangle.
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Mathematical Notations
• ∈
• ∈
• ∀
Belongs to.
Does not belong to.
For all (Universal Quantifier).
• ∅
Empty set.
• P→Q
P implies Q.
• P↔𝑄
p if and only if Q.
• 𝑁
Natural Numbers {0,1,2,3,….}.
• 𝑍
• 𝑄
• R
The Integers {…,-3,-2,-1,0,1,2,3,}
The Rational Numbers.
The Real Numbers.
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𝐶
𝑎, 𝑏
The complex Numbers.
The open finite interval.
{X ∈ R : a < x < b}
[𝑎, 𝑏]
The closed interval.
{X ∈ R : a ≤ x ≤ b}
𝑎, 𝑏
The semi-closed interval.
{X ∈ R : a ≤ x < b}
𝑎, 𝑏
The semi-open interval.
{X ∈ R : a< x≤b}
𝑎, +∞
The infinite open interval.
{X ∈ R : x > a}
−∞, 𝑎
The infinite closed interval.
{X ∈ R : x < a}
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1.1 Sets and Notation
• Set : a collection of well defined members or elements.
• A subset: is a sub-collection of a set.
• Example :
• A = {x ∈ Z : 𝑥 2
1,0,1,2,3}
The sets
≤ 9} ,
B={x ∈ Z : 𝑥 ≤ 3} ,
C={-3,-2,-
Solution:
• The first set is the set of all integers whose square lies between
1 and 9 inclusive, which is precisely the second set, which again
is the third set.
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• The union of two sets A and B, is the set
𝐴 ∪ 𝐵 = {𝑥: 𝑥 ∈ 𝐴 𝑜𝑟 (𝑥 ∈ 𝐵)}
This is read “A union B.”
• The intersection of two sets A and B, is
𝐴 ∩ 𝐵 = {𝑥: 𝑥 ∈ 𝐴 𝑎𝑛𝑑 (𝑥 ∈ 𝐵)}
This is read “A intersection B.”
• The difference of two sets A and B, is
𝐴 /𝐵 = {𝑥: 𝑥 ∈ 𝐴 𝑎𝑛𝑑 (𝑥 ∈ 𝐵)}
This is read “A set minus B.”
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Basic Science DeptN Agiza
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Example :
If S ={1, 2, 3, 4, 5} , T={4, 5, 6, 7}, and V = {6, 7, 8}, find the sets S ∪ T, S ∩
T, and S ∩ V.
S ∪ T = {1, 2, 3, 4, 5,6,7}
S ∩ T = {4 , 5}
S∩v= Φ
Example :
Let A={1,2,3,4,5} , B={1,3,5,7,9} .Find the sets A ∪ B
A ∪ B = {1,2,3,4,5,6,7,9}
A ∩ B = {1, 3, 5}
A / B = {2, 4, 6}
B / A = {7, 9}
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1.2 Intervals
An interval is a subset of the real numbers.
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Example :
Graphing Intervals
Express each interval in terms of inequalities, and then graph the
interval .
(a) −1, 2 = 𝑥 − 1 ≤ 𝑥 ≤ 2 }
(b)[1.5, 4]= 𝑥 1.5 ≤ 𝑥 ≤ 4 }
(C) −3 , ∞ = 𝑥 − 3 < 𝑥}
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• Example :
Find the Intersection of this Interval, Graph each set.
𝟏, 𝟑 ∩ [𝟐, 𝟕]
• The intersection of two intervals consists of the numbers that are in both intervals. Therefore
𝟏, 𝟑 ∩ 𝟐, 𝟕 = 𝑥 1 < 𝑥 < 3 𝑎𝑛𝑑 2 𝑥 ≤ 7}
= 𝑥 2 ≤ 𝑥 < 3 } = 2,3
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Absolute Value
• The absolute value of a number a, denoted by 𝑎 , is the
distance from a to 0 on the real number line.
• Distance is always positive or zero, so we have 𝑎 ≥ 0
• Definition of Absolute value
If a is a real number , then the absolute value of a is
𝑎 𝑖𝑓 𝑎 ≥ 0
𝑎 =
−𝑎 𝑖𝑓 𝑎 < 0
Example: Evaluating Absolute Values of Numbers
3 =3
−3 = − −3
=3
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Basic Science DeptN Agiza
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Distance between points
• If a and b are real numbers, then the distance between the points a
and b on the real line is
𝑑 𝑎, 𝑏 = 𝑏 − 𝑎
𝑏−𝑎 = 𝑎−𝑏
• Note that
• Example :
Distance Between Points
The distance between the numbers 8 and 2 is
𝑑 𝑎, 𝑏 = −8 − 2 = −10 = 10
Faculty of Engineering -
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Integer Exponents
• A product of identical numbers is usually written in exponential notation.
• Exponential notation
If a is any real number and n is a positive integer , then the 𝑛𝐭𝐡 power of a is
𝒂𝒏 = 𝒂. 𝒂. … . 𝒂
The number 𝒂 is called the base , and 𝒏 is called the exponent
Example:
𝟏
𝟏
𝟏
𝟏
𝟏
𝟏
𝟏
I. (𝟐)𝟓 = 𝟐 𝟐 𝟐 𝟐 𝟐 = 𝟑𝟐
II. (−𝟑)𝟒 = −𝟑 . −𝟑 . −𝟑 . −𝟑 = 𝟖𝟏
𝟒 𝟎
III. (𝟕) = 𝟏
𝟏
𝟏
𝟏
IV. (−𝟐)−𝟑 = (−𝟐)−𝟑 = −𝟖 = −𝟖
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Laws of Exponents
• 𝑎𝑚 𝑎𝑛 = 𝑎𝑚+𝑛
•
•
32 35 = 38
𝑎𝑚
𝑚−𝑛
=
𝑎
𝑎𝑛
(𝑎𝑚 )𝑛 = 𝑎𝑚𝑛
35
32
(32 )5 = 310
𝑏
𝑎
4
3
• (𝑎𝑏)−𝑛 = ( )𝑛
(34)−2 = ( )2
• Example :
I. 𝑥 4 𝑥 7 = 𝑥 11
II. 𝑦 4 𝑦 −7 = 𝑦13
5
III. ( ) =
𝑥
2
𝑥5
25
=
= 35−2 = 33
𝑥2
25
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Basic Science DeptN Agiza
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Adding and Subtracting Polynomials
• We add and subtract polynomials using the properties of real
numbers .
• The idea is to combine like terms ,using the Distributive Property.
• For instance,
5𝑥 7 + 3𝑥 7 = 5 + 3 𝑥 7 = 8𝑥 7
• Example:
Adding and Subtracting Polynomials
a)Find the sum (𝒙𝟑 − 𝟔𝒙𝟐 + 𝟐𝒙 + 𝟒)+(𝒙𝟑 + 𝟓𝒙𝟐 − 𝟕𝒙)
= 𝒙𝟑 + 𝒙𝟑 + −6𝒙𝟐 + 5𝒙𝟐 + 𝟐𝒙 − 𝟕𝒙 + 4
= 2𝒙𝟑 − 𝒙𝟐 − 5𝒙 + 4
b) Find the difference (𝒙𝟑 − 𝟔𝒙𝟐 + 𝟐𝒙 + 𝟒) −(𝒙𝟑 + 𝟓𝒙𝟐 − 𝟕𝒙)
= 𝒙𝟑 − 𝟔𝒙𝟐 + 𝟐𝒙 + 𝟒 − 𝒙𝟑 − 𝟓𝒙𝟐 + 𝟕𝒙
= 𝒙𝟑 − 𝒙𝟑 − 𝟔𝒙𝟐 − 𝟓𝒙𝟐 + 𝟐𝒙 + 𝟕𝒙 + 4
𝟐 + 9𝒙 + 4
=
−11𝒙
Faculty of Engineering - Basic Science DeptProf H
N Agiza
Multiplying Polynomials
• To find the product of polynomials or other algebraic expressions, we
need to use the Distributive Property repeatedly.
𝑎 + 𝑏 𝑐 + 𝑑 = 𝑎𝑐 + 𝑎𝑑 + 𝑏𝑐 + 𝑏𝑑
Example:
Find the product
(𝟐𝒙𝟐 + 𝟑)(𝒙𝟐 − 𝟓𝒙 + 𝟒)
𝟐𝒙𝟐 + 𝟑 𝒙𝟐 − 𝟓𝒙 + 𝟒 = 2𝑥 𝒙𝟐 − 5𝑥 + 4 + 3 𝒙𝟐 − 5𝑥 + 4
= 2𝑥. 𝒙𝟐 − 2𝑥. 5𝑥 + 2𝑥. 4 + 3. 𝒙𝟐 − 3.5𝑥 + 4
= 2𝑥 3 − 10𝒙𝟐 + 8𝑥 + 3𝑥 2 − 15𝑥 + 12
= 2𝑥 3 − 7𝒙𝟐 − 7𝑥 + 12
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Special Product Formulas
•
•
•
•
•
𝐴 + 𝐵 𝐴 − 𝐵 = 𝐴2 − 𝐵2
𝐴 + 𝐵 2 = 𝐴2 + 2𝐴𝐵 + 𝐵2
𝐴 − 𝐵 2 = 𝐴2 − 2𝐴𝐵 + 𝐵2
𝐴 + 𝐵 3 = 𝐴3 + 3𝐴2 𝐵 + 3𝐴𝐵2 + 𝐵3
𝐴 − 𝐵 3 = 𝐴3 − 3𝐴2 𝐵 + 3𝐴𝐵2 − 𝐵3
Example :
Using the Special Product Formulas
2
3
2
2
2
2
3
3
2
(3𝑥 + 𝑦 ) = (3𝑥 ) +2(3𝑥 ) 𝑦 + (𝑦 )
4
2
3
6
= 9𝑥 + 6𝑥 𝑦 + 𝑦
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Factorization
• We use the Distributive Property to expand algebraic
expressions. We sometimes need to reverse this process by
factoring an expression as a product of simpler ones.
• We say that 𝑥 − 2 and 𝑥 + 2 are factors of 𝑥 2 − 4
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Factoring Trinomials
• To factor a trinomial of the form 𝑥 2 + 𝑏𝑥 + 𝑐 , we note that
𝑥 + 𝑟 𝑥 + 𝑠 = 𝑥 2 + 𝑟 + 𝑠 𝑥 + 𝑟𝑠
• so we need to choose numbers r and s so that r + s=b and rs = c.
Example :
Factor
6𝑥 2 + 7𝑥 − 5
6𝑥 2 + 7𝑥 − 5 = (3𝑥 + 5)(2𝑥 − 1)
Factor
𝑥 2 − 2𝑥 − 3
𝑥 2 − 2𝑥 − 3Faculty
= (𝑥
− 3)(𝑥 + 1)
of Engineering - Basic Science DeptN Agiza
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