5.1 equation of lines using slope-intercept · 5.1 equation of lines using slope-intercept mr....

5.1 Equation of Lines Using Slope-Intercept

Mr. Noyes, Akimel A-al Middle School 1 Heath Algebra 1 - An Integrated Approach

Objectives:

• Learn how to use slope intercept form to write an equation of a line

• Learn how to model a real-life situation with a linear equation

Slope Intercept Form: y = mx + b Where m = slope b = y-intercept Write an equation when given the slope and y-intercept:

m = –2 m = 5

b = 4 b = 12−

y = –2x + 4 y = 5x – 12



Write an equation from the graph

y = mx + b

y = 32 x +(–3)

y = mx + b

y = –x –1

y = mx + b

y = 3 Real-life: the Phone Company charges a flat fee of $0.75 for the first minute of long

distance plus $0.10 per minute after that. Write an equation to use to figure out each call.

1. Verbal Model:

Total cost = first minute cost + rateminute · number of minutes

2. Labels:

y = total cost

c = cost for first minute ($0.75)

x = # of additional minutes

r = rate per additional minute ($0.10)



3. Algebraic Model:

Fill in what you know and write in slope-intercept form (y = mx + b)

y = .75 + .10x 4. Solve/Give Answer:

Find values for extra minutes: 0, 5, 10 and graph

x y

0 .75

5 1.25

10 1.75

1 2 3 4 5 6 7 8 9 10 11 x

1

2

3

y

minutes

Cost ($)

5.2 Equations of Lines Given the Slope and a Point


Objectives:

• Learn how to use slope and any point to write an equation of the line

• Learn how to model a real-life situation with a linear equation

Slope intercept form: y = mx + b Need to know m and b Given the slope, m = –2 and the point (6, –3), find equation of the line:

y = mx + b

–3 = –2(6) + b substitute what you know into the slope-intercept form of the equation, and then solve for b

–3 = –12 + b +12 +12 9 = b

y = –2x + 9 graph check:

1 2 3 4 5 6 7 8 910–1–2–3–4–5–6–7–8–9–10 x

123456789

10

–1–2–3–4–5–6–7–8–9

–10

y



Write the equation for a line that passes through point (6, 7) and has a slope 23

y = mx + b graph:

7 = 6(32 ) + b

7 = 4 + b –4 –4 3 = b

y = 23 x + 3

Real-life: Find an equation for vacation trips y (in millions) in terms of the

year, t. Let t = 0 correspond to 1980. From 1980 – 1990, vacation trips increased by about 15 million per year. In 1985, Americans went on 340 million vacation trips.

Because change is constant, you can model this as a linear equation y = mt + b Constant is 15 trips per year, so you know slope. In 1985, the year would be t = 0 + 5 and the number of vacation trips (y) would be 340 million.

y = 15t + b

340 = 15(5) + b

340 = 75 + b –75 –75 265 = b

1 2 3 4 5 6 7 8 9–1–2–3–4–5–6–7–8–9 x

123456789

–1–2–3–4–5–6–7–8–9

y



y = 15t + 265

In 1998, the value of t = 1998 - 1980 = 18

So the number of vacations taken in 1998 would be:

y = 15(18) + 265

y = 270 + 265

y = 535

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 t

100

200

300

400

500

600

y

Year ( 0 ↔ 1980)

5.3 Equations of Lines Given 2 Points


Objectives:

• Learn how to write an equation of a line given 2 points on the line

• Learn how to model a real-life problem with a linear equation

Equation of a line: y = mx + b Need slope and y-intercept Given 2 points, can you find the slope and y-intercept?

Points (3, 5) and (–7, 2)

1. Find the slope: m = 2 1

2 1

riserun

y yx x−

=−

m = 2 57 3−

− − = 3

10

y = mx + b Substitute the value of m back into the slope-intercept form of the linear equation

y = 310 x + b

2. Substitute the x and y values from one point to find the y-intercept (the value of b):

y = mx + b (3, 5)

5 = 310 (3) + b

5 = 910 + b

– 910 – 9

10



4 110 = b

3. Substitute both m and b back into the slope-intercept form of the linear equation and solve:

y = 3

10 x + 4 110 This is the equation of the line passing through points

(3, 5) and (–7, 2)

(3, 5)

(–7, 2)

1 2 3 4 5 6 7 8 910–1–2–3–4–5–6–7–8–9–10 x

123456789

10

–1–2–3–4–5–6–7–8–9

–10

y

Example: Find linear equation with points (–3, 2) and (5,–2), and then graph

Let’s say (x1, y1) = (–3, 2) and (x2, y2) = (5,–2) (Though it really doesn’t matter)

1. Find the slope, m:

m = 2 25 ( 3)− −− −

= 48− = 1

2−

y = mx + b Substitute m into the slope-intercept form of the linear equation

y = 12− x + b



2. Substitute the x and y values from one of the two points given in the question to find the y-intercept (the value of b):

y = mx + b (–3, 2)

2 = 12− (–3) + b

2 = 32 + b

– 32 – 3

2

12 = b

3. Substitute both m and b back into the slope-intercept form of the linear equation and solve:

y = 1

2− x + 12

(–3, 2)

(5, –2)

1 2 3 4 5 6 7 8 910–1–2–3–4–5–6–7–8–9–10 x

123456789

10

–1–2–3–4–5–6–7–8–9

–10

y



Can we find the equation of a line if we know both the x, and y-intercepts?

Write an equation that has points whose y-intercept is –4 and x-intercept is –6

What are your 2 points? (0, –4) (–6, 0)

m = 2 1

2 1

y yx x−−

= 0 ( 4)6 0− −− −

= 46−

= – 23

y = mx + b 0 = – 2

3 (–6) + b

0 = +4 + b –4 –4

–4 = b

y = mx + b

y = – 23 x – 4

(0, –4)

(–6, 0)

1 2 3 4 5 6 7 8–1–2–3–4–5–6–7–8 x

12345678

–1–2–3–4–5–6–7–8

y



Real–Life: (see textbook, page 253, Question #29) The Tunnel (aka, “The Chunnel”) from Calais, France to Dover, England Write a linear equation of the line formed from A to B

Point A: (0, 60) Point B: (15, – 70)

m = 2 1

2 1

y yx x−−

= 70 6015,000 0− −

−

= 130 1315000 1500− = −

y = mx + b

60 = 131500− (0) + b

60 = b

y = mx + b

y = – 131500 x + 60

5.4 Exploring Data: Fitting a Line to Data


Objectives:

• Learn how to find a linear equation that approximates a set of data points

• Learn how to use scatter plots to determine positive, negative or no correlation

The winning Olympic times for the 100-meter run from 1928 to 1988 are plotted in the graph below. Approximate the best-fitting line for these times. Let y represent the winning time and x the year (x = 0, corresponding to 1928).

20 40 60 x

10

11

12

13

y

Looking at the graph, do you see the trend in the info given?

Can we write a linear equation to match?

Often in life the data collected (no matter how carefully done) to analyze whether a relationship exists between two variables will not appear as a nice and neat straight line. However, while all the data may not fall on one line, they may still exhibit a trend that can best be described as linear. When this happens we draw a single line that best represents (or approximates) the set of data points. This line is called the line of best fit.



Draw a line of best fit and pick any 2 points along the line (they don’t have to be actual points plotted on the scatter plot).

Let’s say points (0, 12) and (50, 11) Use the method described in Chapter 5.3

m= 2 1

2 1

riserun

y yx x−

=−

=11 1250 0−−

m = 150−

y = mx + b

11 = 150− (50) + b

11 = –1 + b +1 +1

12 = b

y = 150− x + 12

Correlation: A quantitative assessment of whether a relationship exists between

two variables. Describes the slope of the line of best fit.

x

y

x

y

x

y

Positive correlation Negative correlation No correlation When a line cannot be drawn to represent the set of data points, we say there is no correlation.



Real-life: Make a graph, find the equation of the line of best fit (if possible), and make a prediction.

Find approximate wing area of a 400 g bird

Bird Weight (g) Wing Area (cm2)

Sparrow 25 87 Martin 47 186 Blackbird 78 245 Starling 93 190 Dove 143 357 Crow 607 1344 Gull 840 2006 Blue Heron 2090 4436 Let x represent weight and y wing area

200 400 600 800 1000 1200 1400 1600 1800 2000 2200 x

400800

1200160020002400280032003600400044004800

y

Pick any two points: Let’s say (x1, y1) is (140, 360) and (x2, y2) is (1400, 3200)



y = mx + b

m= 2 1

2 1

y yx x−−

= 3200 360 28401400 140 1260

− =−

= 14263 ≈ 2.25

y = mx + b

360 = 2.25 (140) + b

360 = 315 + b

45 = b y ≈ 2.25x + 45

5.5 Standard Form of a Linear Equation


Objectives:

• Learn how to transform a linear equation into standard form

• Learn how to model a real-life situation using the standard form of a linear equation

So far, we have focused on the Slope-Intercept Form of the Linear Equation, though

the Standard Form of the Linear Equation is also commonly used.

Slope-Intercept Form Standard Form y = mx + b Ax + By = C With the Standard Form of the Linear Equation, notice that the two variables are shown

on the left side of the equation, while a constant is found on the right side. One

advantage of the Standard Form is that it can be used for any type of line, including

vertical lines.

Changing/Transforming Slope-Intercept Form to Standard Form:

y = 34 x + 2 1) Multiply by 4 to clear the fraction

4y = 3x + 8 –3x –3x 2) Subtract 3x from both sides

–1 · (–3x + 4y) = 8 · (–1) 3) Standard Form

3x – 4y = –8 4) Multiply by –1 to make x positive

Same Standard Form -- acceptable in either form

5.5 Standard Form of a Linear Equation


Example:

y = –.23x + 5.2 1) multiply by 100 to clear the decimal

100y = –23x + 520 2) move the term containing x to the left side +23x +23x

23x + 100y = 520 Standard Form - it’s now easy to find both the x- and y-intercepts by setting each to zero.

Real-life: You have $6 to buy bananas and apples. Bananas cost $0.49 per pound,

and apples cost $0.34 per pound. Write a linear equation that represents

the different amounts of fruit you can buy.

Let x represent bananas and y represent apples

.49x + .34y = 6

2 4 6 8 10 12 14 16 x

2468

10121416182022

y

5.6 Point-Slope Form of the Equation of a Line


Objectives:

• Learn how to use the point-slope form to write an equation of a line

• Learn how to model a real-life situation using point-slope form

Consider a line containing the point (x1, y1) = (2, 5) with a slope of 23 . Let (x, y) be

any other point on the line.

Because you know (2, 5) and (x, y) are 2 points on the line, you can find the slope of

the line m= 5riserun 2

yx−=−

and since we know the slope was given as 23 , we can

write:

m= 52

yx−−

= 23 by cross-multiplying and rewriting we can say

y – 5 = 23 (x – 2) This is called point-slope form

Point-Slope Form: y – y1 = m(x – x1)

When using point-slope form, be sure you see the difference between

(x1, y1) and (x, y).

(x1, y1) = the given point

(x, y) = any point on the line

5.7 Problem Solving Using Linear Models


Objectives:

• Learn how to create and use Linear Models to solve problems

• Learn how to make linear models that are accurate but simple to use

Linear Models are used to Solve 2 Basic Types of Real-Life Problems: 1. Constant rate of change

2. Two variables such that the sum Ax + By = C is a constant

Rate of Change Example: The cost of renting a lawn mower is $10 for the first hour and $5 for each additional

hour. Write a linear model that gives the cost of renting a lawn mower in terms of the

number of hours rented, then determine how much it would cost to rent a lawn mower

for 3 hours

1. Verbal Model:

Total cost = # of additional hours · (additional hourly cost) + 1st hour cost

2. Label: identify what you know and assign variables for what you don’t know

y = total cost

1st hour cost = $10

Additional hourly cost = $5/hour

x = the total number of hours needed (including the 1st hour), so . . .

x – 1 = number of additional hours needed

3. Algebraic Model:

y = 5(x – 1) + 10



4. Solve:

y = 5(x – 1) + 10 when x = 3 hours

y = 5(3 – 1) + 10

y = 5(2) + 10

y = 20 It would cost $20 to rent a lawn mower for 3 hours

Constant Addition Example: The total income from the sale of raffle tickets was $202.50. If student tickets sold for

$0.75 and adult tickets sold for $1.25, write a linear equation that describes the

relationship between the number of student and adult tickets and construct a graph to

provide a visual model

Let: s = the number of student tickets sold

a = the number of adult tickets sold

Linear Equation: .75s + 1.25a = 202.50

Graph:

If only adult tickets were sold, s = 0

.75s + 1.25a = 202.50

.75(0) + 1.25a = 202.50

1.25a = 202.50 1.25 1.25

a = 162

If only student tickets were sold, a = 0

.75s + 1.25a = 202.50

.75s + 1.25(0) = 202.50

.75s = 202.50 .75 .75

s = 270



(270, 0)

(0, 162)

25 50 75 100 125 150 175 200 225 250 275 s

25

50

75

100

125

150

175 a

Student Tickets

Adult Tickets

5.1 equation of lines using slope-intercept · 5.1 equation of lines using slope-intercept mr....

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