The document discusses first degree (linear) functions. It explains that most real-world mathematical functions can be composed of algebraic, trigonometric, or exponential-log formulas. Linear functions of the form f(x)=mx+b are especially important, where m is the slope and b is the y-intercept. The slope-intercept form allows expressions of the form Ax+By=C to be written as functions with y as the output. Examples are given of finding the slope and form of linear equations.
There are various reasons why we would want to find the extreme (maximum and minimum values) of a function. Fermat's Theorem tells us we can find local extreme points by looking at critical points. This process is known as the Closed Interval Method.
Basic Calculus 11 - Derivatives and Differentiation RulesJuan Miguel Palero
It is a powerpoint presentation that discusses about the lesson or topic of Derivatives and Differentiation Rules. It also encompasses some formulas, definitions and examples regarding the said topic.
There are various reasons why we would want to find the extreme (maximum and minimum values) of a function. Fermat's Theorem tells us we can find local extreme points by looking at critical points. This process is known as the Closed Interval Method.
Basic Calculus 11 - Derivatives and Differentiation RulesJuan Miguel Palero
It is a powerpoint presentation that discusses about the lesson or topic of Derivatives and Differentiation Rules. It also encompasses some formulas, definitions and examples regarding the said topic.
MENINGKATKAN HASIL BELAJAR SISWA MELALUI MODEL PEMBELAJARAN PICTURE AND PICTURE PADA MATA PELAJARAN IPA DENGAN TOPIK ORGAN PEREDARAN DARAH PADA MANUSIA KELAS V SEMESTER I SD NEGERI 17 KATOBU.
Apresentação sobre a importância da elaboração do trabalho escolar. Discute-se a temática acima citada com foco no uso efetivo das fontes de informação, enquanto mediadoras do saber, de modo a agregar valor, ampliando a qualidade da produção de conteúdos escolares. O público alvo desta apresentação são: alunos do ensino fundamental e médio.
Apresentação elaborada e apresentada por: Lucineia Silva, como contrapartida cultural ao projeto de atualização de uma biblioteca escolar comunitária de Contagem, patrocinado pelo FMIC 2016 de Contagem, MG.
DEFINICION DE LOS ANTINEOPLASICOS, CLASIFICACIÓN DE LOS ANTINEOPLÁSICOS
,FACTORES A TENER EN CUENTA A LA HORA DE ADMINISTRAR UN ANTINEOPLÁSICO,ALQUILANTES,ANTIMETABOLITOS, ANTIBIÓTICOS ANTITUMORALES, AGENTES HORMONALES
GraphRAG is All You need? LLM & Knowledge GraphGuy Korland
Guy Korland, CEO and Co-founder of FalkorDB, will review two articles on the integration of language models with knowledge graphs.
1. Unifying Large Language Models and Knowledge Graphs: A Roadmap.
https://arxiv.org/abs/2306.08302
2. Microsoft Research's GraphRAG paper and a review paper on various uses of knowledge graphs:
https://www.microsoft.com/en-us/research/blog/graphrag-unlocking-llm-discovery-on-narrative-private-data/
Securing your Kubernetes cluster_ a step-by-step guide to success !KatiaHIMEUR1
Today, after several years of existence, an extremely active community and an ultra-dynamic ecosystem, Kubernetes has established itself as the de facto standard in container orchestration. Thanks to a wide range of managed services, it has never been so easy to set up a ready-to-use Kubernetes cluster.
However, this ease of use means that the subject of security in Kubernetes is often left for later, or even neglected. This exposes companies to significant risks.
In this talk, I'll show you step-by-step how to secure your Kubernetes cluster for greater peace of mind and reliability.
Smart TV Buyer Insights Survey 2024 by 91mobiles.pdf91mobiles
91mobiles recently conducted a Smart TV Buyer Insights Survey in which we asked over 3,000 respondents about the TV they own, aspects they look at on a new TV, and their TV buying preferences.
DevOps and Testing slides at DASA ConnectKari Kakkonen
My and Rik Marselis slides at 30.5.2024 DASA Connect conference. We discuss about what is testing, then what is agile testing and finally what is Testing in DevOps. Finally we had lovely workshop with the participants trying to find out different ways to think about quality and testing in different parts of the DevOps infinity loop.
Unlocking Productivity: Leveraging the Potential of Copilot in Microsoft 365, a presentation by Christoforos Vlachos, Senior Solutions Manager – Modern Workplace, Uni Systems
GraphSummit Singapore | The Art of the Possible with Graph - Q2 2024Neo4j
Neha Bajwa, Vice President of Product Marketing, Neo4j
Join us as we explore breakthrough innovations enabled by interconnected data and AI. Discover firsthand how organizations use relationships in data to uncover contextual insights and solve our most pressing challenges – from optimizing supply chains, detecting fraud, and improving customer experiences to accelerating drug discoveries.
UiPath Test Automation using UiPath Test Suite series, part 5DianaGray10
Welcome to UiPath Test Automation using UiPath Test Suite series part 5. In this session, we will cover CI/CD with devops.
Topics covered:
CI/CD with in UiPath
End-to-end overview of CI/CD pipeline with Azure devops
Speaker:
Lyndsey Byblow, Test Suite Sales Engineer @ UiPath, Inc.
GridMate - End to end testing is a critical piece to ensure quality and avoid...ThomasParaiso2
End to end testing is a critical piece to ensure quality and avoid regressions. In this session, we share our journey building an E2E testing pipeline for GridMate components (LWC and Aura) using Cypress, JSForce, FakerJS…
UiPath Test Automation using UiPath Test Suite series, part 4DianaGray10
Welcome to UiPath Test Automation using UiPath Test Suite series part 4. In this session, we will cover Test Manager overview along with SAP heatmap.
The UiPath Test Manager overview with SAP heatmap webinar offers a concise yet comprehensive exploration of the role of a Test Manager within SAP environments, coupled with the utilization of heatmaps for effective testing strategies.
Participants will gain insights into the responsibilities, challenges, and best practices associated with test management in SAP projects. Additionally, the webinar delves into the significance of heatmaps as a visual aid for identifying testing priorities, areas of risk, and resource allocation within SAP landscapes. Through this session, attendees can expect to enhance their understanding of test management principles while learning practical approaches to optimize testing processes in SAP environments using heatmap visualization techniques
What will you get from this session?
1. Insights into SAP testing best practices
2. Heatmap utilization for testing
3. Optimization of testing processes
4. Demo
Topics covered:
Execution from the test manager
Orchestrator execution result
Defect reporting
SAP heatmap example with demo
Speaker:
Deepak Rai, Automation Practice Lead, Boundaryless Group and UiPath MVP
Generative AI Deep Dive: Advancing from Proof of Concept to ProductionAggregage
Join Maher Hanafi, VP of Engineering at Betterworks, in this new session where he'll share a practical framework to transform Gen AI prototypes into impactful products! He'll delve into the complexities of data collection and management, model selection and optimization, and ensuring security, scalability, and responsible use.
Alt. GDG Cloud Southlake #33: Boule & Rebala: Effective AppSec in SDLC using ...James Anderson
Effective Application Security in Software Delivery lifecycle using Deployment Firewall and DBOM
The modern software delivery process (or the CI/CD process) includes many tools, distributed teams, open-source code, and cloud platforms. Constant focus on speed to release software to market, along with the traditional slow and manual security checks has caused gaps in continuous security as an important piece in the software supply chain. Today organizations feel more susceptible to external and internal cyber threats due to the vast attack surface in their applications supply chain and the lack of end-to-end governance and risk management.
The software team must secure its software delivery process to avoid vulnerability and security breaches. This needs to be achieved with existing tool chains and without extensive rework of the delivery processes. This talk will present strategies and techniques for providing visibility into the true risk of the existing vulnerabilities, preventing the introduction of security issues in the software, resolving vulnerabilities in production environments quickly, and capturing the deployment bill of materials (DBOM).
Speakers:
Bob Boule
Robert Boule is a technology enthusiast with PASSION for technology and making things work along with a knack for helping others understand how things work. He comes with around 20 years of solution engineering experience in application security, software continuous delivery, and SaaS platforms. He is known for his dynamic presentations in CI/CD and application security integrated in software delivery lifecycle.
Gopinath Rebala
Gopinath Rebala is the CTO of OpsMx, where he has overall responsibility for the machine learning and data processing architectures for Secure Software Delivery. Gopi also has a strong connection with our customers, leading design and architecture for strategic implementations. Gopi is a frequent speaker and well-known leader in continuous delivery and integrating security into software delivery.
In his public lecture, Christian Timmerer provides insights into the fascinating history of video streaming, starting from its humble beginnings before YouTube to the groundbreaking technologies that now dominate platforms like Netflix and ORF ON. Timmerer also presents provocative contributions of his own that have significantly influenced the industry. He concludes by looking at future challenges and invites the audience to join in a discussion.
2. First Degree Functions
Most mathematical functions used in the real world are
“composed” with members from the following three
groups of formulas.
3. Most mathematical functions used in the real world are
“composed” with members from the following three
groups of formulas.
* The algebraic family – these are polynomials,
rational expressions and roots, etc..
First Degree Functions
4. Most mathematical functions used in the real world are
“composed” with members from the following three
groups of formulas.
* The algebraic family – these are polynomials,
rational expressions and roots, etc..
* The trigonometric family – these are sin(x),
cos(x), .. etc that come from line measurements.
First Degree Functions
5. Most mathematical functions used in the real world are
“composed” with members from the following three
groups of formulas.
* The algebraic family – these are polynomials,
rational expressions and roots, etc..
* The trigonometric family – these are sin(x),
cos(x), .. etc that come from line measurements.
* The exponential–log family – these are ex and ln(x)
that come from exponential contexts.
First Degree Functions
6. Most mathematical functions used in the real world are
“composed” with members from the following three
groups of formulas.
* The algebraic family – these are polynomials,
rational expressions and roots, etc..
* The trigonometric family – these are sin(x),
cos(x), .. etc that come from line measurements.
* The exponential–log family – these are ex and ln(x)
that come from exponential contexts.
Degree 1 or linear functions: f(x) = mx + b and
degree 2 or quadratic functions: f(x) = ax2 + bx + c
are especially important.
First Degree Functions
7. Most mathematical functions used in the real world are
“composed” with members from the following three
groups of formulas.
* The algebraic family – these are polynomials,
rational expressions and roots, etc..
* The trigonometric family – these are sin(x),
cos(x), .. etc that come from line measurements.
* The exponential–log family – these are ex and ln(x)
that come from exponential contexts.
Degree 1 or linear functions: f(x) = mx + b and
degree 2 or quadratic functions: f(x) = ax2 + bx + c
are especially important.
First Degree Functions
We review below the basics of linear equations and
linear functions.
8. First Degree Functions
The algebraic family
The exponential–log familyThe trigonometric family
Below is the MS Win-10 desktop scientific calculator,
a typical scientific calculator input panel:
9. The graphs of the equations Ax + By = C are straight
lines.
First Degree Functions
10. The graphs of the equations Ax + By = C are straight
lines. It's easy to graph lines by graphing the x and y
intercepts,
First Degree Functions
11. a.2x – 3y = 12
The graphs of the equations Ax + By = C are straight
lines. It's easy to graph lines by graphing the x and y
intercepts,
First Degree Functions
12. a.2x – 3y = 12
The graphs of the equations Ax + By = C are straight
lines. It's easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get the y–intercept,
(0,–4)
First Degree Functions
13. a.2x – 3y = 12
The graphs of the equations Ax + By = C are straight
lines. It's easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get the y–intercept, and set
y = 0 for the x–intercept.
(6,0)
(0,–4)
First Degree Functions
14. a.2x – 3y = 12
The graphs of the equations Ax + By = C are straight
lines. It's easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get the y–intercept, and set
y = 0 for the x–intercept.
(6,0)
(0,–4)
First Degree Functions
15. a.2x – 3y = 12
The graphs of the equations Ax + By = C are straight
lines. It's easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get the y–intercept, and set
y = 0 for the x–intercept. If there is only one variable
in the equation, we get a vertical or a horizontal line.
(6,0)
(0,–4)
First Degree Functions
16. a.2x – 3y = 12 b. –3y = 12
The graphs of the equations Ax + By = C are straight
lines. It's easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get the y–intercept, and set
y = 0 for the x–intercept. If there is only one variable
in the equation, we get a vertical or a horizontal line.
(6,0)
(0,–4)
First Degree Functions
17. a.2x – 3y = 12 b. –3y = 12
The graphs of the equations Ax + By = C are straight
lines. It's easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get the y–intercept, and set
y = 0 for the x–intercept. If there is only one variable
in the equation, we get a vertical or a horizontal line.
(6,0)
(0,–4) y = –4
First Degree Functions
18. a.2x – 3y = 12 b. –3y = 12 c. 2x = 12
The graphs of the equations Ax + By = C are straight
lines. It's easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get the y–intercept, and set
y = 0 for the x–intercept. If there is only one variable
in the equation, we get a vertical or a horizontal line.
(6,0)
(0,–4) y = –4
First Degree Functions
19. a.2x – 3y = 12 b. –3y = 12 c. 2x = 12
The graphs of the equations Ax + By = C are straight
lines. It's easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get the y–intercept, and set
y = 0 for the x–intercept. If there is only one variable
in the equation, we get a vertical or a horizontal line.
(6,0)
(0,–4) y = –4
x = 6
First Degree Functions
20. a.2x – 3y = 12 b. –3y = 12 c. 2x = 12
If both x and y are
present, we get a
tilted line.
The graphs of the equations Ax + By = C are straight
lines. It's easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get the y–intercept, and set
y = 0 for the x–intercept. If there is only one variable
in the equation, we get a vertical or a horizontal line.
(6,0)
(0,–4) y = –4
x = 6
First Degree Functions
21. a.2x – 3y = 12 b. –3y = 12 c. 2x = 12
If both x and y are
present, we get a
tilted line.
If the equation is
y = c, we get a
horizontal line.
The graphs of the equations Ax + By = C are straight
lines. It's easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get the y–intercept, and set
y = 0 for the x–intercept. If there is only one variable
in the equation, we get a vertical or a horizontal line.
(6,0)
(0,–4) y = –4
x = 6
First Degree Functions
22. a.2x – 3y = 12 b. –3y = 12 c. 2x = 12
If both x and y are
present, we get a
tilted line.
If the equation is
y = c, we get a
horizontal line.
The graphs of the equations Ax + By = C are straight
lines. It's easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get the y–intercept, and set
y = 0 for the x–intercept. If there is only one variable
in the equation, we get a vertical or a horizontal line.
(6,0)
(0,–4) y = –4
x = 6
If the equation is
x = c, we get a
vertical line.
First Degree Functions
24. First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
25. First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
m is called the slope and b is the y intercept,
26. First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
m is called the slope and b is the y intercept, and the
form is called the slope–intercept form.
27. First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
m is called the slope and b is the y intercept, and the
form is called the slope–intercept form.
From the examples above,
a. 2x – 3y = 12
28. First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
m is called the slope and b is the y intercept, and the
form is called the slope–intercept form.
2
3
2
3
From the examples above,
a. 2x – 3y = 12 y = x – 4, so the slope =
29. First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
m is called the slope and b is the y intercept, and the
form is called the slope–intercept form.
2
3
2
3
From the examples above,
a. 2x – 3y = 12 y = x – 4, so the slope =
b. –3y = 12
30. First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
m is called the slope and b is the y intercept, and the
form is called the slope–intercept form.
2
3
2
3
From the examples above,
a. 2x – 3y = 12 y = x – 4, so the slope =
b. –3y = 12 y = 0x – 4, so the slope = 0
31. First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
m is called the slope and b is the y intercept, and the
form is called the slope–intercept form.
2
3
2
3
From the examples above,
a. 2x – 3y = 12 y = x – 4, so the slope =
b. –3y = 12 y = 0x – 4, so the slope = 0
c. 2x = 12,
32. First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
m is called the slope and b is the y intercept, and the
form is called the slope–intercept form.
2
3
2
3
From the examples above,
a. 2x – 3y = 12 y = x – 4, so the slope =
b. –3y = 12 y = 0x – 4, so the slope = 0
c. 2x = 12, the slope is undefined since we can't
solve for y.
33. First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
m is called the slope and b is the y intercept, and the
form is called the slope–intercept form.
The slope m is also the ratio of the change in the
output vs. the change in the input.
2
3
2
3
From the examples above,
a. 2x – 3y = 12 y = x – 4, so the slope =
b. –3y = 12 y = 0x – 4, so the slope = 0
c. 2x = 12, the slope is undefined since we can't
solve for y.
34. First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
m is called the slope and b is the y intercept, and the
form is called the slope–intercept form.
The slope m is also the ratio of the change in the
output vs. the change in the input.
If two points on the line are given,
the slope is defined via the following formula.
2
3
2
3
From the examples above,
a. 2x – 3y = 12 y = x – 4, so the slope =
b. –3y = 12 y = 0x – 4, so the slope = 0
c. 2x = 12, the slope is undefined since we can't
solve for y.
35. Slope Formula: Let (x1, y1) and (x2, y2) be two points
on a line,
First Degree Functions
(x1, y1)
(x2, y2)
36. Slope Formula: Let (x1, y1) and (x2, y2) be two points
on a line, then the slope
Δy
Δxm =
First Degree Functions
(x1, y1)
(x2, y2)
37. Slope Formula: Let (x1, y1) and (x2, y2) be two points
on a line, then the slope
Δy
Δxm =
First Degree Functions
(The Greek letter Δ means "the difference".)
(x1, y1)
(x2, y2)
38. Slope Formula: Let (x1, y1) and (x2, y2) be two points
on a line, then the slope
Δy
Δx
y2 – y1
x2 – x1
m = =
First Degree Functions
(The Greek letter Δ means "the difference".)
(x1, y1)
(x2, y2)
39. Slope Formula: Let (x1, y1) and (x2, y2) be two points
on a line, then the slope
Δy
Δx
y2 – y1
x2 – x1
m =
rise
run
=
=
First Degree Functions
(The Greek letter Δ means "the difference".)
(x1, y1)
(x2, y2)
40. Slope Formula: Let (x1, y1) and (x2, y2) be two points
on a line, then the slope
Δy
Δx
y2 – y1
x2 – x1
m =
rise
run
=
= (x1, y1)
(x2, y2)
First Degree Functions
(The Greek letter Δ means "the difference".)
41. Slope Formula: Let (x1, y1) and (x2, y2) be two points
on a line, then the slope
Δy
Δx
y2 – y1
x2 – x1
m =
rise
run
=
= (x1, y1)
(x2, y2)
Δy=y2–y1=rise
First Degree Functions
(The Greek letter Δ means "the difference".)
42. Slope Formula: Let (x1, y1) and (x2, y2) be two points
on a line, then the slope
Δy
Δx
y2 – y1
x2 – x1
m =
rise
run
=
= (x1, y1)
(x2, y2)
Δy=y2–y1=rise
Δx=x2–x1=run
First Degree Functions
(The Greek letter Δ means "the difference".)
43. Slope Formula: Let (x1, y1) and (x2, y2) be two points
on a line, then the slope
Δy
Δx
y2 – y1
x2 – x1
m =
rise
run
=
= (x1, y1)
(x2, y2)
Δy=y2–y1=rise
Δx=x2–x1=run
The Point Slope Formula: Let y = f(x) be a first
degree equation with slope m, and (x1, y1) is a point
on the line, then y = f(x) = m(x – x1) + y1
First Degree Functions
(The Greek letter Δ means "the difference".)
44. Slope Formula: Let (x1, y1) and (x2, y2) be two points
on a line, then the slope
Δy
Δx
y2 – y1
x2 – x1
m =
rise
run
=
= (x1, y1)
(x2, y2)
Δy=y2–y1=rise
Δx=x2–x1=run
The Point Slope Formula: Let y = f(x) be a first
degree equation with slope m, and (x1, y1) is a point
on the line, then y = f(x) = m(x – x1) + y1
First degree functions are also called linear functions
because their graphs are straight lines.
First Degree Functions
(The Greek letter Δ means "the difference".)
45. Slope Formula: Let (x1, y1) and (x2, y2) be two points
on a line, then the slope
Δy
Δx
y2 – y1
x2 – x1
m =
rise
run
=
= (x1, y1)
(x2, y2)
Δy=y2–y1=rise
Δx=x2–x1=run
The Point Slope Formula: Let y = f(x) be a first
degree equation with slope m, and (x1, y1) is a point
on the line, then y = f(x) = m(x – x1) + y1
First degree functions are also called linear functions
because their graphs are straight lines. We will use
linear functions to approximate other functions just like
we use line segments to approximate a curve.
First Degree Functions
(The Greek letter Δ means "the difference".)
46. Linear Equations and Lines
Example A. A river floods regularly, and on a rock by
the river there is a mark indicating the highest point
the water level ever recorded.
At 12 pm July 11, the water level is 28 inches below
this mark. At 8 am July 12 the water is 18 inches
below this mark.
47. Linear Equations and Lines
Example A. A river floods regularly, and on a rock by
the river there is a mark indicating the highest point
the water level ever recorded.
At 12 pm July 11, the water level is 28 inches below
this mark. At 8 am July 12 the water is 18 inches
below this mark. Let x = time,
y = distance between the water level and the mark.
Find the linear function y = f(x) = mx + b
of the distance y in terms of time x.
48. Linear Equations and Lines
Example A. A river floods regularly, and on a rock by
the river there is a mark indicating the highest point
the water level ever recorded.
At 12 pm July 11, the water level is 28 inches below
this mark. At 8 am July 12 the water is 18 inches
below this mark. Let x = time,
y = distance between the water level and the mark.
Find the linear function y = f(x) = mx + b
of the distance y in terms of time x.
Since how the time was measured is not specified,
we may select the stating time 0 to be time of the
first observation.
49. Linear Equations and Lines
Example A. A river floods regularly, and on a rock by
the river there is a mark indicating the highest point
the water level ever recorded.
At 12 pm July 11, the water level is 28 inches below
this mark. At 8 am July 12 the water is 18 inches
below this mark. Let x = time,
y = distance between the water level and the mark.
Find the linear function y = f(x) = mx + b
of the distance y in terms of time x.
Since how the time was measured is not specified,
we may select the stating time 0 to be time of the
first observation.
By setting x = 0 (hr) at 12 pm July 11,
50. Equations of Lines
In particular, we are given that at x = 0 →y = 28,
and at x = 20 → y = 18
51. Equations of Lines
In particular, we are given that at x = 0 →y = 28,
and at x = 20 → y = 18 and that we want the
equation y = m(x – x1) + y1 of the line that contains
the points (0, 28) and (20, 18).
52. Δy
Δx
28 – 18
0 – 20
Equations of Lines
In particular, we are given that at x = 0 →y = 28,
and at x = 20 → y = 18 and that we want the
equation y = m(x – x1) + y1 of the line that contains
the points (0, 28) and (20, 18).
=The slope m = = –1/2
53. Using the point (0, 28) and the point–slope formula,
y = – ½ (x – 0) + 28 or that
Δy
Δx
28 – 18
0 – 20
Equations of Lines
In particular, we are given that at x = 0 →y = 28,
and at x = 20 → y = 18 and that we want the
equation y = m(x – x1) + y1 of the line that contains
the points (0, 28) and (20, 18).
=The slope m = = –1/2
– xy = + 28
2
54. Using the point (0, 28) and the point–slope formula,
y = – ½ (x – 0) + 28 or
Δy
Δx
28 – 18
0 – 20
Equations of Lines
In particular, we are given that at x = 0 →y = 28,
and at x = 20 → y = 18 and that we want the
equation y = m(x – x1) + y1 of the line that contains
the points (0, 28) and (20, 18).
=The slope m = = –1/2
– xy = + 28
2
The linear equation that we found is also called the
trend line.
55. Using the point (0, 28) and the point–slope formula,
y = – ½ (x – 0) + 28 or
Δy
Δx
28 – 18
0 – 20
Equations of Lines
In particular, we are given that at x = 0 →y = 28,
and at x = 20 → y = 18 and that we want the
equation y = m(x – x1) + y1 of the line that contains
the points (0, 28) and (20, 18).
=The slope m = = –1/2
– xy = + 28
2
The linear equation that we found is also called the
trend line. So if at 4 pm July 12, i.e. when x = 28,
we measured that y = 12”
56. Using the point (0, 28) and the point–slope formula,
y = – ½ (x – 0) + 28 or
Δy
Δx
28 – 18
0 – 20
Equations of Lines
In particular, we are given that at x = 0 →y = 28,
and at x = 20 → y = 18 and that we want the
equation y = m(x – x1) + y1 of the line that contains
the points (0, 28) and (20, 18).
=The slope m = = –1/2
– xy = + 28
2
The linear equation that we found is also called the
trend line. So if at 4 pm July 12, i.e. when x = 28,
we measured that y = 12” but based on the formula
prediction that y should be – 28/2 + 28 = 14”,
57. Using the point (0, 28) and the point–slope formula,
y = – ½ (x – 0) + 28 or
Δy
Δx
28 – 18
0 – 20
Equations of Lines
In particular, we are given that at x = 0 →y = 28,
and at x = 20 → y = 18 and that we want the
equation y = m(x – x1) + y1 of the line that contains
the points (0, 28) and (20, 18).
=The slope m = = –1/2
– xy = + 28
2
The linear equation that we found is also called the
trend line. So if at 4 pm July 12, i.e. when x = 28,
we measured that y = 12” but based on the formula
prediction that y should be – 28/2 + 28 = 14”, we may
conclude that the flood is intensifying.
59. More Facts on Slopes:
• Parallel lines have the same slope.
First Degree Functions
60. More Facts on Slopes:
• Parallel lines have the same slope.
• Slopes of perpendicular lines are the negative
reciprocal of each other.
First Degree Functions
61. More Facts on Slopes:
• Parallel lines have the same slope.
• Slopes of perpendicular lines are the negative
reciprocal of each other.
Example B. Find the equation of the line L that
passes through (2, –4) and is perpendicular to
4x – 3y = 5.
First Degree Functions
62. More Facts on Slopes:
• Parallel lines have the same slope.
• Slopes of perpendicular lines are the negative
reciprocal of each other.
Example B. Find the equation of the line L that
passes through (2, –4) and is perpendicular to
4x – 3y = 5.
Solve for y to find the slope of 4x – 3y = 5
First Degree Functions
63. More Facts on Slopes:
• Parallel lines have the same slope.
• Slopes of perpendicular lines are the negative
reciprocal of each other.
Example B. Find the equation of the line L that
passes through (2, –4) and is perpendicular to
4x – 3y = 5.
Solve for y to find the slope of 4x – 3y = 5
4x – 5 = 3y
First Degree Functions
64. More Facts on Slopes:
• Parallel lines have the same slope.
• Slopes of perpendicular lines are the negative
reciprocal of each other.
Example B. Find the equation of the line L that
passes through (2, –4) and is perpendicular to
4x – 3y = 5.
Solve for y to find the slope of 4x – 3y = 5
4x – 5 = 3y
4x/3 – 5/3 = y
First Degree Functions
65. More Facts on Slopes:
• Parallel lines have the same slope.
• Slopes of perpendicular lines are the negative
reciprocal of each other.
Example B. Find the equation of the line L that
passes through (2, –4) and is perpendicular to
4x – 3y = 5.
Solve for y to find the slope of 4x – 3y = 5
4x – 5 = 3y
4x/3 – 5/3 = y
Hence the slope of 4x – 2y = 5 is 4/3.
First Degree Functions
66. More Facts on Slopes:
• Parallel lines have the same slope.
• Slopes of perpendicular lines are the negative
reciprocal of each other.
Example B. Find the equation of the line L that
passes through (2, –4) and is perpendicular to
4x – 3y = 5.
Solve for y to find the slope of 4x – 3y = 5
4x – 5 = 3y
4x/3 – 5/3 = y
Hence the slope of 4x – 2y = 5 is 4/3.
Therefore L has slope –3/4.
First Degree Functions
67. More Facts on Slopes:
• Parallel lines have the same slope.
• Slopes of perpendicular lines are the negative
reciprocal of each other.
Example B. Find the equation of the line L that
passes through (2, –4) and is perpendicular to
4x – 3y = 5.
Solve for y to find the slope of 4x – 3y = 5
4x – 5 = 3y
4x/3 – 5/3 = y
Hence the slope of 4x – 2y = 5 is 4/3.
Therefore L has slope –3/4. So the equation of L is
First Degree Functions
y = (–3/4)(x – 2) + (–4) or y = –3x/4 – 5/2.
68. Linear Equations and Lines
Exercise A. Estimate the slope by eyeballing two points then
find an equation of each line below.
1. 2. 3. 4.
5. 6. 7. 8.
69. Linear Equations and Lines
B. Draw each line that passes through the given two points.
Find the slope and an equation of the line.
Again Identify the vertical lines and the horizontal lines by
inspection and solve for them first.
(Fraction Review: slide 99 of 1.2)
1. (0, –1), (–2, 1) 2. (3, –1), (3, 1)
4. (1, –2), (–2, 3)
3. (2, –1), (3, –1)
6. (4, –2), (4, 0)5. (7, –2), (–2, –6)
7. (3/2, –1), (3/2, 1) 8. (3/4, –1/3), (1/3, 3/2)
9. (–1/4, –3/2), (2/3, –3/2) 10. (–1/3, –1/6), (–3/4, 1/2)
11. (2/5, –3/10), (–1/2, –3/5) 12. (–3/4, 5/6), (–3/4, –4/3)
70. Linear Equations and Lines
2. It’s perpendicular to 2x – 4y = 1 and passes through (–2, 1)
6. It’s perpendicular to 3y = x with x–intercept at x = –3.
12. It has y–intercept at y = 3 and is parallel to 3y + 4x = 1.
8. It’s perpendicular to the y–axis with y–intercept at 4.
9. It has y–intercept at y = 3 and is parallel to the x axis.
10. It’s perpendicular to the x– axis containing the point (4, –3).
11. It is parallel to the y axis has x–intercept at x = –7.
5. It is parallel to the x axis has y–intercept at y = 7.
C. Find the equations of the following lines.
1. The line that passes through (0, 1) and has slope 3.
7. The line that passes through (–2 ,1) and has slope –1/2.
3. The line that passes through (5, 2) and is parallel to y = x.
4. The line that passes through (–3, 2) and is perpendicular
to –x = 2y.
71. Linear Equations and Lines
The cost y of renting a tour boat consists of a base–cost plus
the number of tourists x. With 4 tourists the total cost is $65,
with 11 tourists the total is $86.
1. What is the base cost and what is the charge per tourist?
2. Find the equation of y in terms of x.
3. What is the total cost if there are 28 tourists?
The temperature y of water in a glass is rising slowly.
After 4 min. the temperature is 30 Co, and after 11 min. the
temperature is up to 65 Co. Answer 42–44 assuming the
temperature is rising linearly.
4. What is the temperature at time 0 and what is the rate of
the temperature rise?
5. Find the equation of y in terms of time.
6. How long will it take to bring the water to a boil at 100 Co?
D. Find the equations of the following lines.
72. Linear Equations and Lines
The cost of gas y on May 3 is $3.58 and on May 9 is $4.00.
Answer 45–47 assuming the price is rising linearly.
7. Let x be the date in May, what is the rate of increase in
price in terms of x?
8. Find the equation of the price in term of the date x in May.
9. What is the projected price on May 20?