Download - Chapter11
11-1
Graphing Linear Equations
Course 3
Warm UpSolve each equation for y.
1. 6y – 12x = 24
2. –2y – 4x = 20
3. 2y – 5x = 16
4. 3y + 6x = 18
y = 2x + 4
y = –2x – 10
Course 3
11-1 Graphing Linear Equations
y = –2x + 6
y = x + 852
Learn to identify and graph linear equations.
Course 3
11-1 Graphing Linear Equations
WHAT IS A LINEAR EQUATION?
A linear equation is an equation whose solutions fall on a line on the coordinate plane. All solutions of a particular linear equation fall on the line, and all the points on the line are solutions of the equation.
Course 3
11-1 Graphing Linear Equations
HOW DO I KNOW IF AN EQUATION IS LINEAR?
If an equation is linear, a constant change in the x-value corresponds to a constant change in the y-value.
3
3
3
2
2
2
Course 3
11-1 Graphing Linear Equations
The graph shows an example where each time the x-value increases by 3, the y-value increases by 2.
Make a table, graph the equation, and tell whether it is linear.
A. y = 3x – 1
Graphing Equations
x 3x – 1 y (x, y)
–2
–1
0
1
2
–73(–2) – 13(–1) – 1
3(0) – 13(1) – 1
3(2) – 1
–4
–1
2
5
(–2, –7)
(–1, –4)(0, –1)
(1, 2)(2, 5)
Course 3
11-1 Graphing Linear Equations
Each time x increases by 1 unit, y increases by 3 units.
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11-1 Graphing Linear Equations
(x, y)
(-2, -7)
(-1, -4)
(0, -1)
(1, 2)
(2, 5)
y = 3x – 1
Make a table, graph the equation, and tell whether it is linear.
B. y = x3
Graphing Equations
x x3 y (x, y)
–2
–1
0
1
2
–8(–2)3
(–1)3
(0)3
(1)3
(2)3
–1
0
1
8
(–2, –8)
(–1, –1)(0, 0)
(1, 1)(2, 8)
Course 3
11-1 Graphing Linear Equations
The equation y = x3 is not a linear equation because its graph is not a straight line. Also notice that as x increases by a constant of 1 unit, the change in y is not constant.
x –2 –1 0 1 2
y –8 –1 0 1 8
+7 +1 +1 +7Course 3
11-1 Graphing Linear Equations
Fill in the table below. Then, graph the equation and tell whether it is linear.
A. y = 2x + 1
Try This
x 2x + 1 y (x, y)
–2
–1
0
1
2
–32(–2) + 12(–1) + 1
2(0) + 12(1) + 1
2(2) + 1
–1
1
3
5
(–3, –3)
(–2, –1)(–1, 1)
(0, 3)(2, 5)
Course 3
11-1 Graphing Linear Equations
The equation y = 2x + 1is a linear equation.
Each time x increase by 1 unit, y increases by 2 units.
Course 3
11-1 Graphing Linear Equations
Complete the first two problems on page 85
REVIEW• An equation that forms a line on a coordinate
grid is called a linear equation. All of the points on the line are solutions.
• In order to be linear, there must be a constant pattern found in the y-vaules.
• To solve, first create a table and then graph the points.
• Watch this:
More Graphing EquationsGraph the equation and tell whether it is linear.
C. y = – 3x4
Course 3
11-1 Graphing Linear Equations
Additional Example 1 Continued
The equation y = –
is a linear equation.
3x4
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11-1 Graphing Linear Equations
Graph the equation and tell whether it is linear.
D. y = 2
More Graphing Equations
For any value of x, y = 2.
x 2 y (x, y)
–2
–1
0
1
2
222
2
2
2
2
2
2
2
(–2, 2)
(–1, 2)(0, 2)
(1, 2)(2, 2)
Course 3
11-1 Graphing Linear Equations
Additional Example 1D Continued
The equation y = 2 is a linear equation because the points form a straight line.
As the value of x increases, the value of y has a constant change of 0.
Course 3
11-1 Graphing Linear Equations
Try ThisGraph the equation and tell whether it is linear.
C. y = x
x y (x, y)
–8
–6
0
4
8
–8
–6
0
4
8
(–8, –8)
(–6, –6)(0, 0)
(4, 4)(8, 8)
Course 3
11-1 Graphing Linear Equations
Try This: Example 1C Continued
The equation y = x is a linear equation because the points form a straight line.
Each time the value of x increases by 1, the value of y increases by 1.
Course 3
11-1 Graphing Linear Equations
ApplicationIn an amusement park ride, a car travels according to the equation D = 1250t where t is time in minutes and D is the distance in feet the car travels.
Graph the relationship between time and distance. How far has each person traveled?
Rider Time
Ryan 1 min
Greg 2 min
Colette 3 min
Course 3
11-1 Graphing Linear Equations
Continued
t D =1250t D (t, D)
1 1250(1) 1250 (1, 1250)
2 1250(2) 2500 (2, 2500)
3 1250(3) 3750 (3, 3750)
Course 3
11-1 Graphing Linear Equations
The distances are: Ryan, 1250 ft; Greg, 2500 ft; and Collette, 3750 ft.
Continued
x
y
This is a linear equation because when t increases by 1 unit, D increases by 1250 units.
1250
2500
1 2
3750
5000
3 4Time (min)
Dis
tan
ce (
ft)
Course 3
11-1 Graphing Linear Equations
Sports Application
A lift on a ski slope rises according to the equation a = 130t + 6250, where a is the altitude in feet and t is the number of minutes that a skier has been on the lift.
Five friends are on the lift. What is the altitude of each person if they have been on the ski lift for the times listed in the table?
Draw a graph that represents the relationship between the time on the lift and the altitude.
Course 3
11-1 Graphing Linear Equations
Additional Example 2 Continued
Course 3
11-1 Graphing Linear Equations
Additional Example 2 Continued
Course 3
11-1 Graphing Linear Equations
The altitudes are: Anna, 6770 feet; Tracy, 6640 feet; Kwani, 6510 feet; Tony, 6445 feet; George, 6380 feet. This is a linear equation because when t increases by 1 unit, a increases by 130 units. Note that a skier with 0 time on the lift implies that the bottom of the lift is at an altitude of 6250 feet.
Additional Example 2 Continued
Course 3
11-1 Graphing Linear Equations
Sit down and check your homework!
12. (-1, -4) (0,0) (1, 4)
13. (-1, 3) (0, 5) (1, 7)
14. (-1,-9) (0, -3) (1, 3)
15. (-1, -11) (0, -10) (1,-9)
16. (-1, -6) (0, -2) (1, 2)
17. (-1, -1) (0, 3) (1, 7)
18. (-1, -6) (0, -4) (1, -2)
19. (-1, 6) (0, 7) (1, 8)
20. (-1, -0.5) (0, 2.5) (1, 5.5)
22. $43.20 $46.40 $49.60
$52.80 $56.00 $59.20
11-2
Slope of a Line
Course 3
Warm UpEvaluate each equation for x = –1, 0, and 1.
1. y = 3x
2. y = x – 7
3. y = 2x + 5
4. y = 6x – 2
–3, 0, 3
–8, –7, –6
3, 5, 7
Course 3
11-2 Slope of a Line
–8, –2, 4
Learn to find the slope of a line and use slope to understand and draw graphs.
Course 3
11-2 Slope of a Line
What is slope?
vertical change horizontal change
change in y change in x=
This ratio is often referred to as , or “rise
over run,” where rise indicates the number of units moved up or down and run indicates the number of units moved to the left or right. Slope can be positive, negative, zero, or undefined.
rise run
Course 3
11-2 Slope of a Line
Course 3
11-2 Slope of a Line
Course 3
11-2 Slope of a Line
Course 3
11-2 Slope of a Line
GET ON
YOUR FEET!
Finding Slope from a Graph
Choose two points on the line: (0, 1) and (3, –4).
Guess by looking at the graph:
riserun = –5
3 = – 5 3
–5
3
Course 3
11-2 Slope of a Line
How to find the slope of a line (without a visual)…
If you have two points (x1, y1) & (x2, y2)
Use the following formula:
yy22 –– yy11
xx22 –– xx11
Course 3
11-2 Slope of a Line
Find the slope of the line that passes through (–2, –3) and (4, 6).
Finding Slope, Given Two Points
Let (x1, y1) be (–2, –3) and (x2, y2) be (4, 6).
6 – (–3)4 – (–2)
Substitute 6 for y2, –3 for y1, 4 for x2, and –2 for x1.
96=
The slope of the line that passes through (–
2, –3) and (4, 6) is . 32
=y2 – y1
x2 – x1
32=
Course 3
11-2 Slope of a Line
Find the slope of the line that passes through (–4, –6) and (2, 3).
Try This: Example 1
Let (x1, y1) be (–4, –6) and (x2, y2) be (2, 3).
3 – (–6)2 – (–4)
Substitute 3 for y2, –6 for y1, 2 for x2, and –4 for x1.
96=
The slope of the line that passes through (–
4, –6) and (2, 3) is . 32
=y2 – y1
x2 – x1
32=
Course 3
11-2 Slope of a Line
Use the graph of the line to determine its slope.
Try This: Example 2
Course 3
11-2 Slope of a Line
Try This: Example 2 Continued
Choose two points on the line: (1, 1) and (0, –1).
Guess by looking at the graph:
riserun = 2
1 = 2
Use the slope formula.
Let (1, 1) be (x1, y1) and (0, –1) be (x2, y2).
=y2 – y1
x2 – x1
–2–1=
–1 – 1 0 – 1
= 2
12
Course 3
11-2 Slope of a Line
Class Assignment/Homework
Workbook 11-2Complete #1-6
Course 3
11-2 Slope of a Line
REVIEW• A slope is a ratio that tells how steep or flat a
line is. It is read as “rise over run”.
• Slope can be used and seen in constructions, sports, hills, roller coasters, etc.
• There are four types of slopes: positive, negative, zero, and undefined (no slope).
• To find the slope without a graph, write the difference in y over the difference in x.
Something you need to remember:
Parallel lines have the same slope.
The slopes of two perpendicular lines are negative reciprocals of each other.
Course 3
11-2 Slope of a Line
Parallel and Perpendicular Lines:On the following slides, you need to figure out one thing…
Are the lines passing through the given points parallel or perpendicular?
1. Find the slope of the first line.
2. Find the slope of the second line.
3. Compare.
~ If they are the same Parallel
~ If they are opposite reciprocals Perpendicular
Course 3
11-2 Slope of a Line
Parallel and Perpendicular Lines by Slope
line 1: (–6, 4) and (2, –5)
line 2: (–1, –4) and (8, 4)
slope of line 1:
slope of line 2:
Line 1 has a slope equal to – and line 2 has a slope
equal to , – and are negative reciprocals of each
other, so the lines are perpendicular.
98
89
89
98
=y2 – y1
x2 – x1
–9 8= –5 – 4
2 – (–6)
4 – (–4)8 – (–1)=
y2 – y1
x2 – x1
8 9=
9 8= –
Course 3
11-2 Slope of a Line
Parallel and Perpendicular Lines by Slope
line 1: (0, 5) and (6, –2)
line 2: (–1, 3) and (5, –4)
Both lines have a slope equal to – , so the lines are parallel.
76
slope of line 1:
slope of line 2:
=y2 – y1
x2 – x1
–7 6= –2 – 5
6 – 0
=y2 – y1
x2 – x1
7 6= –
–7 6= 7
6= – –4 – 35 – (–1)
Course 3
11-2 Slope of a Line
Try This
line 1: (1, 1) and (2, 2)
line 2: (1, –2) and (2, -1)
Line 1 has a slope equal to 1 and line 2 has a slope equal to –1. 1 and –1 are negative reciprocals of each other, so the lines are perpendicular.
slope of line 1:
slope of line 2:
=y2 – y1
x2 – x1
1 1= 2 – 1
2 – 1
=y2 – y1
x2 – x1
–1 1= –1 – (–2)
2 – (1)
= 1
= –1
Course 3
11-2 Slope of a Line
Try This
line 1: (–8, 2) and (0, –7)line 2: (–3, –6) and (6, 2)
slope of line 1:
slope of line 2:
Line 1 has a slope equal to – and line 2 has a slope
equal to , – and are negative reciprocals of each
other, so the lines are perpendicular.
98
89
89
98
=y2 – y1
x2 – x1
–9 8= –7 – 2
0 – (–8)
2 – (–6)6 – (–3)=
y2 – y1
x2 – x1
8 9=
9 8= –
Course 3
11-2 Slope of a Line
Graphing a Line Using a Point and the Slope
2. Use the slope to count your units. Place another point once you get to the end.
Course 3
11-2 Slope of a Line
3. Continue until you have enough point and then draw a line.
1. Plot the point given to you.
Graphing a Line Using a Point and the Slope
Graph the line passing through (3, 1) with slope 2.
Plot the point (3, 1). Then move 2 units up and right 1 unit and plot the point (4, 3). Use a straightedge to connect the two points.
The slope is 2, or . So for every 2 units up, you will move right 1 unit, and for every 2 units down, you will move left 1 unit.
21
Course 3
11-2 Slope of a Line
Additional Example 4 Continued
1
2(3, 1)
Course 3
11-2 Slope of a Line
Try This: Example 4
Graph the line passing through (1, 1) with slope 2.
Plot the point (1, 1). Then move 2 units up and right 1 unit and plot the point (2, 3). Use a straightedge to connect the two points.
The slope is 2, or . So for every 2 units up, you will move right 1 unit, and for every 2 units down, you will move left 1 unit.
21
Course 3
11-2 Slope of a Line
Try This: Example 4 Continued
1
2(1, 1)
Course 3
11-2 Slope of a Line
Sit down and check your homework!10. - 3/2
11. - 1/2
12. - 3
13. - 2/3
14. 7/4
15. - 5/6
16. - 3/2
17. 0
18. parallel
19. perpendicular
Warm UpFind the slope of the line that passes through each pair of points.
1. (3, 6) and (-1, 4)
2. (1, 2) and (6, 1)
3. (4, 6) and (2, -1)
4. (-3, 0) and (-1, 1)
Course 3
11-3 Using Slopes and Intercepts
12
- 157212
11-3
Using Slopes and Intercepts
Course 3
Learn to use slopes and intercepts to graph linear equations.
Course 3
11-3 Using Slopes and Intercepts
As you watch the video, take notes on your handout.
Insert Lesson Title Here
Course 3
11-3
One way to graph a linear equation easily is by finding the x-intercept and the y-intercept.
The x-intercept is the value of x where the line crosses the x-axis (y = 0).
The y-intercept is the value of y where the line crosses the y-axis (x = 0).
Course 3
11-3 Using Intercepts
Find the x-intercept and y-intercept of the line 4x – 3y = 12. Use the intercepts to graph the equation.
Example 1
Find the x-intercept (y = 0).
4x – 3y = 12
4x – 3(0) = 12
4x = 124x4
124=
x = 3The x-intercept is 3.
Course 3
11-3 Using Intercepts
Example 1 Continued
Find the y-intercept (x = 0).
4x – 3y = 12
4(0) – 3y = 12
–3y = 12
-3y-3
12-3 =
y = –4
The y-intercept is –4.
Course 3
11-3 Using Intercepts
4x – 3y = 12
Crosses the x-axis at the point (3, 0)
Crosses the y-axis at the point (0, –4)
Course 3
11-3 Using Intercepts
Find the x-intercept and y-intercept of the line 8x – 6y = 48. Use the intercepts to graph the equation.
Try This
Find the x-intercept (y = 0).
8x – 6y = 48
8x – 6(0) = 48
8x = 488x8
488=
x = 6The x-intercept is 6 so the point is (6, 0).
Course 3
11-3 Using Intercepts
Try This
Find the y-intercept (x = 0).
8x – 6y = 48
8(0) – 6y = 48
–6y = 48
-6y-6
48-6 =
y = –8
The y-intercept is –6 so the point is (0, -8).
Course 3
11-3 Using Intercepts
Try This: Example 1 Continued
The graph of 8x – 6y = 48 is the line that crosses the x-axis at the point (6, 0) and the y-axis at the point (0, –8).
Course 3
11-3 Using Slopes and Intercepts
In an equation written in slope-intercept form, y = mx + b, m is the slope and b is the y-intercept.
y = mx + b
Slope y-intercept
Course 3
11-3 Using Slopes and Intercepts
For an equation such as y = x – 6, write it as y = x + (–6) to read the y-intercept, –6. The point would be (0,-6)
Helpful Hint
Course 3
11-3 Using Slopes and Intercepts
Using the Slope-Intercept Form
1. Isolate the y so that you equation is in y = mx+b form.
2. Slope will always be “m” (the number in front of x).
3. The y-intercept will always be “b” (the number by itself). Write this point as (0,b)
Example 2
Write each equation in slope-intercept form, and then find the slope and y-intercept.
A. 2x + y = 32x + y = 3–2x –2x Subtract 2x from both sides.
y = 3 – 2x
y = –2x + 3 The equation is in slope-intercept form.
m = –2 b = 3The slope of the line is –2, and the y-intercept is 3.
Course 3
11-3 Using Slopes and Intercepts
More Examples
B. 5y = 3x
5y = 3x
Divide both sides by 5 to solve for y.
The equation is in slope-intercept form.
b = 0
= x35
5y5
y = x + 035
m =35
The slope of the line is , and they-intercept is 0.
35
Course 3
11-3 Using Slopes and Intercepts
More Examples
C. 4x + 3y = 9
4x + 3y = 9Subtract 4x from both sides.
b = 3
y =- x + 343
m =- 43
The slope of the line 4x+ 3y = 9 is – , and the y-intercept is 3.4
3
–4x –4x
3y = –4x + 9
= + –4x 3
3y3
93 Divide both sides by 3.
The equation is in slope-intercept form.
Course 3
11-3 Using Slopes and Intercepts
Try This
Write each equation in slope-intercept form, and then find the slope and y-intercept.
A. 4x + y = 4–4x –4x Subtract 4x from both sides.
y = 4 – 4xRewrite to match slope-intercept form.
y = –4x + 4 The equation is in slope-intercept form.
m = –4 b = 4The slope of the line 4x + y = 4 is –4, and the y-intercept is 4.
Course 3
11-3 Using Slopes and Intercepts
Try This
B. 7y = 2x
7y = 2x
Divide both sides by 7 to solve for y.
The equation is in slope-intercept form.
b = 0
= x27
7y7
y = x + 027
m =27
The slope of the line 7y = 2x is , and they-intercept is 0.
27
Course 3
11-3 Using Slopes and Intercepts
Try This
C. 5x + 4y = 8
5x + 4y = 8Subtract 5x from both sides.
Rewrite to match slope-intercept form.
b = 2
y =- x + 254
The slope of the line 5x + 4y = 8 is – , and the y-intercept is 2.5
4
–5x –5x
4y = 8 – 5x
5x + 4y = 8
= + –5x 4
4y4
84 Divide both sides by 4.
The equation is in slope-intercept form.
m =- 54
Course 3
11-3 Using Slopes and Intercepts
Additional Example 3: Entertainment Application
A video club charges $8 to join, and $1.25 for each DVD that is rented. The linear equation y = 1.25x + 8 represents the amount of money y spent after renting x DVDs. Graph the equation by first identifying the slope and y-intercept.
y = 1.25x + 8The equation is in slope-intercept form.
b = 8m =1.25
Course 3
11-3 Using Slopes and Intercepts
Additional Example 3 Continued
The slope of the line is 1.25, and the y-intercept is 8. The line crosses the y-axis at the point (0, 8) and moves up 1.25 units for every 1 unit it moves to the right.
Course 3
11-3 Using Slopes and Intercepts
Try This: Example 3
A salesperson receives a weekly salary of $500 plus a commission of 5% for each sale. Total weekly pay is given by the equation S = 0.05c + 500. Graph the equation using the slope and y-intercept.
y = 0.05x + 500 The equation is in slope-intercept form.
b = 500m =0.05
Course 3
11-3 Using Slopes and Intercepts
Try This: Example 3 Continued
The slope of the line is 0.05, and the y-intercept is 500. The line crosses the y-axis at the point (0, 500) and moves up 0.05 units for every 1 unit it moves to the right.
x
y
500
1000
1500
2000
10,0005000 15,000
Course 3
11-3 Using Slopes and Intercepts
Additional Example 4: Writing Slope-Intercept Form
Write the equation of the line that passes through (3, –4) and (–1, 4) in slope-intercept form.
Find the slope.
The slope is –2.
Choose either point and substitute it along with the slope into the slope-intercept form.
y = mx + b
4 = –2(–1) + b
4 = 2 + b
Substitute –1 for x, 4 for y, and –2 for m.
Simplify.
4 – (–4) –1 – 3
=y2 – y1
x2 – x1
8–4= = –2
Course 3
11-3 Using Slopes and Intercepts
Additional Example 4 Continued
Solve for b.
Subtract 2 from both sides.
Write the equation of the line, using –2 for m and 2 for b.
4 = 2 + b–2 –2
2 = b
y = –2x + 2
Course 3
11-3 Using Slopes and Intercepts
Try This: Example 4
Write the equation of the line that passes through (1, 2) and (2, 6) in slope-intercept form.
Find the slope.
The slope is 4.
Choose either point and substitute it along with the slope into the slope-intercept form.
y = mx + b
2 = 4(1) + b
2 = 4 + b
Substitute 1 for x, 2 for y, and 4 for m.
Simplify.
6 – 2 2 – 1
=y2 – y1
x2 – x1
4 1= = 4
Course 3
11-3 Using Slopes and Intercepts
Try This: Example 4 Continued
Solve for b.
Subtract 4 from both sides.
Write the equation of the line, using 4 for m and –2 for b.
2 = 4 + b–4 –4
–2 = b
y = 4x – 2
Course 3
11-3 Using Slopes and Intercepts
Clear your desk except for a pencil and your calculator.
11-4 Point-Slope Form
Course 3
Warm UpWarm Up
Problem of the DayProblem of the Day
Lesson PresentationLesson Presentation
Warm UpWrite the equation of the line that passes through each pair of points in slope-intercept form.
1. (0, –3) and (2, –3)
2. (5, –3) and (5, 1)
3. (–6, 0) and (0, –2)
4. (4, 6) and (–2, 0)
y = –3
x = 5
Course 3
11-4 Point-Slope Form
y = x + 2
y = – x – 213
Problem of the Day
Without using equations for horizontal or vertical lines, write the equations of four lines that form a square.
Possible answer: y = x + 2, y = x – 2, y = –x + 2, y = –x – 2
Course 3
11-4 Point-Slope Form
Learn to find the equation of a line given one point and the slope.
Course 3
11-4 Point-Slope Form
Vocabulary
point-slope form
Insert Lesson Title Here
Course 3
11-4 Point-Slope Form
Point on the line((xx11, , yy11))
Point-slope formyy – – yy11 = = mm ( (xx – – xx11))
slopeslope
The point-slope of an equation of a line with slope m passing through (x1, y1) is y – y1 = m(x – x1).
Course 3
11-4 Point-Slope Form
Use the point-slope form of each equation to identify a point the line passes through and the slope of the line.
A. y – 7 = 3(x – 4)
Additional Example 1: Using Point-Slope Form to Identify Information About a Line
y – y1 = m(x – x1)
y – 7 = 3(x – 4)
m = 3
(x1, y1) = (4, 7)
The line defined by y – 7 = 3(x – 4) has slope 3, and passes through the point (4, 7).
The equation is in point-slope form. Read the value of m from the equation. Read the point from the equation.
Course 3
11-4 Point-Slope Form
B. y – 1 = (x + 6)
Additional Example 1B: Using Point-Slope Form to Identify Information About a Line
y – y1 = m(x – x1)
(x1, y1) = (–6, 1)
Rewrite using subtraction instead of addition.
13
13
y – 1 = (x + 6)
y – 1 = [x – (–6)]13
m =13
The line defined by y – 1 = (x + 6) has slope , and
passes through the point (–6, 1).
13
13
Course 3
11-4 Point-Slope Form
Use the point-slope form of each equation to identify a point the line passes through and the slope of the line.
A. y – 5 = 2 (x – 2)
Try This: Example 1
y – y1 = m(x – x1)
y – 5 = 2(x – 2)
m = 2
(x1, y1) = (2, 5)
The line defined by y – 5 = 2(x – 2) has slope 2, and passes through the point (2, 5).
The equation is in point-slope form. Read the value of m from the equation. Read the point from the equation.
Course 3
11-4 Point-Slope Form
B. y – 2 = (x + 3)Try This: Example 1B
23
(x1, y1) = (–3, 2)
Rewrite using subtraction instead of addition.
23
y – 2 = (x + 3)
y – 2 = [x – (–3)]23
m =23
The line defined by y – 2 = (x + 3) has slope , and
passes through the point (–3, 2).
23
23
y – y1 = m(x – x1)
Course 3
11-4 Point-Slope Form
Write the point-slope form of the equation with the given slope that passes through the indicated point.
A. the line with slope 4 passing through (5, -2)
Additional Example 2: Writing the Point-Slope Form of an Equation
y – y1 = m(x – x1)
The equation of the line with slope 4 that passes through (5, –2) in point-slope form is y + 2 = 4(x – 5).
Substitute 5 for x1, –2 for y1, and 4 for m.
[y – (–2)] = 4(x – 5)
y + 2 = 4(x – 5)
Course 3
11-4 Point-Slope Form
B. the line with slope –5 passing through (–3, 7)
Additional Example 2: Writing the Point-Slope Form of an Equation
y – y1 = m(x – x1)
The equation of the line with slope –5 that passes through (–3, 7) in point-slope form is y – 7 = –5(x + 3).
Substitute –3 for x1, 7 for y1, and –5 for m.
y – 7 = -5[x – (–3)]
y – 7 = –5(x + 3)
Course 3
11-4 Point-Slope Form
Write the point-slope form of the equation with the given slope that passes through the indicated point.
A. the line with slope 2 passing through (2, –2)
Try This: Example 2A
y – y1 = m(x – x1)
The equation of the line with slope 2 that passes through (2, –2) in point-slope form is y + 2 = 2(x – 2).
Substitute 2 for x1, –2 for y1, and 2 for m.
[y – (–2)] = 2(x – 2)
y + 2 = 2(x – 2)
Course 3
11-4 Point-Slope Form
B. the line with slope -4 passing through (-2, 5)
Try This: Example 2B
y – y1 = m(x – x1)
The equation of the line with slope –4 that passes through (–2, 5) in point-slope form is y – 5 = –4(x + 2).
Substitute –2 for x1, 5 for y1, and –4 for m.
y – 5 = –4[x – (–2)]
y – 5 = –4(x + 2)
Course 3
11-4 Point-Slope Form
A roller coaster starts by ascending 20 feet for every 30 feet it moves forward. The coaster starts at a point 18 feet above the ground. Write the equation of the line that the roller coaster travels along in point-slope form, and use it to determine the height of the coaster after traveling 150 feet forward. Assume that the roller coaster travels in a straight line for the first 150 feet.
Additional Example 3: Entertainment Application
As x increases by 30, y increases by 20, so the slope
of the line is or . The line passes through the point (0, 18).
2030
23
Course 3
11-4 Point-Slope Form
Additional Example 3 Continued
y – y1 = m(x – x1) Substitute 0 for x1, 18 for y1,
and for m.23
The equation of the line the roller coaster travels along, in point-slope form, is y – 18 = x. Substitute 150 for x to find the value of y.
23
y – 18 = (150)23
y – 18 = 100
y – 18 = (x – 0)23
y = 118
The value of y is 118, so the roller coaster will be at a height of 118 feet after traveling 150 feet forward.
Course 3
11-4 Point-Slope Form
Try This: Example 3A roller coaster starts by ascending 15 feet for every 45 feet it moves forward. The coaster starts at a point 15 feet above the ground. Write the equation of the line that the roller coaster travels along in point-slope form, and use it to determine the height of the coaster after traveling 300 feet forward. Assume that the roller coaster travels in a straight line for the first 300 feet.
As x increases by 45, y increases by 15, so the slope
of the line is or . The line passes through the point (0, 15).
1545
13
Course 3
11-4 Point-Slope Form
Try This: Example 3 Continued
y – y1 = m(x – x1) Substitute 0 for x1, 15 for y1,
and for m.13
The equation of the line the roller coaster travels along, in point-slope form, is y – 15 = x. Substitute 300 for x to find the value of y.
13
y – 15 = (300)13
y – 15 = 100
y – 15 = (x – 0)13
y = 115
The value of y is 115, so the roller coaster will be at a height of 115 feet after traveling 300 feet forward.
Course 3
11-4 Point-Slope Form
11-5 Direct Variation
Course 3
Warm UpWarm Up
Problem of the DayProblem of the Day
Lesson PresentationLesson Presentation
Warm UpUse the point-slope form of each equation to identify a point the line passes through and the slope of the line.
1. y – 3 = – (x – 9)
2. y + 2 = (x – 5)
3. y – 9 = –2(x + 4)
4. y – 5 = – (x + 7)
(–4, 9), –2
Course 3
11-5 Direct Variation
17
23
14
(9, 3), – 17
(5, –2), 23
(–7, 5), – 14
Problem of the Day
Where do the lines defined by the equations y = –5x + 20 and y = 5x – 20 intersect?(4, 0)
Course 3
11-5 Direct Variation
Learn to recognize direct variation by graphing tables of data and checking for constant ratios.
Course 3
11-5 Direct Variation
Vocabulary
direct variationconstant of proportionality
Insert Lesson Title Here
Course 3
11-5 Direct Variation
Course 3
11-5 Direct Variation
Course 3
11-5 Direct Variation
The graph of a direct-variation equation is always linear and always contains the point (0, 0). The variables x and y either increase together or decrease together.
Helpful Hint
Determine whether the data set shows direct variation.
A.
Additional Example 1A: Determining Whether a Data Set Varies Directly
Course 3
11-5 Direct Variation
Make a graph that shows the relationship between Adam’s age and his length.
Additional Example 1A Continued
Course 3
11-5 Direct Variation
You can also compare ratios to see if a direct variation occurs.
223
2712=
?81
264
81 ≠ 264
The ratios are not proportional.
The relationship of the data is not a direct variation.
Additional Example 1A Continued
Course 3
11-5 Direct Variation
Determine whether the data set shows direct variation.
B.
Additional Example 1B: Determining Whether a Data Set Varies Directly
Course 3
11-5 Direct Variation
Make a graph that shows the relationship between the number of minutes and the distance the train travels.
Additional Example 1B Continued
Plot the points.
The points lie in a straight line.
Course 3
11-5 Direct Variation
(0, 0) is included.
You can also compare ratios to see if a direct variation occurs.
The ratios are proportional. The relationship is a direct variation.
2510
5020
7530
10040= = = Compare ratios.
Additional Example 1B Continued
Course 3
11-5 Direct Variation
Determine whether the data set shows direct variation.
A.
Try This: Example 1A
Kyle's Basketball Shots
Distance (ft) 20 30 40
Number of Baskets 5 3 0
Course 3
11-5 Direct Variation
Make a graph that shows the relationship between number of baskets and distance.
Try This: Example 1A Continued
Num
ber
of
Bask
ets
Distance (ft)
2
3
4
20 30 40
1
5
Course 3
11-5 Direct Variation
You can also compare ratios to see if a direct variation occurs.
Try This: Example 1A
520
330=
?60
150
150 60.
The ratios are not proportional.
The relationship of the data is not a direct variation.
Course 3
11-5 Direct Variation
Determine whether the data set shows direct variation.
B.
Try This: Example 1B
Ounces in a Cup
Ounces (oz) 8 16 24 32
Cup (c) 1 2 3 4
Course 3
11-5 Direct Variation
Make a graph that shows the relationship between ounces and cups.
Try This: Example 1B Continued
Num
ber
of
Cup
s
Number of Ounces
2
3
4
8 16 24
1
32
Course 3
11-5 Direct Variation
Plot the points.
The points lie in a straight line.
(0, 0) is included.
You can also compare ratios to see if a direct variation occurs.
Try This: Example 1B Continued
Course 3
11-5 Direct Variation
The ratios are proportional. The relationship is a direct variation.
Compare ratios. = 1 8 = =2
163
24 432
Find each equation of direct variation, given that y varies directly with x.
A. y is 54 when x is 6
Additional Example 2A: Finding Equations of Direct Variation
y = kx
54 = k 6
9 = k
y = 9x
y varies directly with x.
Substitute for x and y.
Solve for k.
Substitute 9 for k in the original equation.
Course 3
11-5 Direct Variation
B. x is 12 when y is 15
Additional Example 2B: Finding Equations of Direct Variation
y = kx
15 = k 12
y varies directly with x.
Substitute for x and y.
Solve for k. = k54
Substitute for k in the original equation.
54y = k5
4
Course 3
11-5 Direct Variation
C. y is 8 when x is 5
Additional Example 2C: Finding Equations of Direct Variation
y = kx
8 = k 5
y varies directly with x.
Substitute for x and y.
Solve for k. = k85
Substitute for k in the original equation.
85y = k8
5
Course 3
11-5 Direct Variation
Find each equation of direct variation, given that y varies directly with x.
A. y is 24 when x is 4
Try This: Example 2A
y = kx
24 = k 4
6 = k
y = 6x
y varies directly with x.
Substitute for x and y.
Solve for k.
Substitute 6 for k in the original equation.
Course 3
11-5 Direct Variation
B. x is 28 when y is 14
Try This: Example 2B
y = kx
14 = k 28
y varies directly with x.
Substitute for x and y.
Solve for k. = k12
Substitute for k in the original equation.
12y = k1
2
Course 3
11-5 Direct Variation
C. y is 7 when x is 3
Try This: Example 2C
y = kx
7 = k 3
y varies directly with x.
Substitute for x and y.
Solve for k. = k73
Substitute for k in the original equation.
73y = k7
3
Course 3
11-5 Direct Variation
Mrs. Perez has $4000 in a CD and $4000 in a money market account. The amount of interest she has earned since the beginning of the year is organized in the following table. Determine whether there is a direct variation between either of the data sets and time. If so, find the equation of direct variation.
Additional Example 3: Money Application
Course 3
11-5 Direct Variation
Additional Example 3 Continued
A. interest from CD and time
interest from CDtime = 17
1interest from CD
time = = 17342
The second and third pairs of data result in a common ratio. In fact, all of the nonzero interest from CD to time ratios are equivalent to 17.
The variables are related by a constant ratio of 17 to 1, and (0, 0) is included. The equation of direct variation is y = 17x, where x is the time, y is the interest from the CD, and 17 is the constant of proportionality.
= = = 17interest from CDtime = = 17
1342
513
684
Course 3
11-5 Direct Variation
Additional Example 3 Continued
B. interest from money market and time
interest from money markettime = = 19
191
interest from money markettime = =18.5
372
19 ≠ 18.5
If any of the ratios are not equal, then there is no direct variation. It is not necessary to compute additional ratios or to determine whether (0, 0) is included.
Course 3
11-5 Direct Variation
Mr. Ortega has $2000 in a CD and $2000 in a money market account. The amount of interest he has earned since the beginning of the year is organized in the following table. Determine whether there is a direct variation between either of the data sets and time. If so, find the equation of direct variation.
Try This: Example 3
Course 3
11-5 Direct Variation
Try This: Example 3 Continued
Interest Interest from
Time (mo) from CD ($) Money Market ($)
0 0 0
1 12 15
2 30 40
3 40 45
4 50 50
Course 3
11-5 Direct Variation
Try This: Example 3 Continued
interest from CDtime = 12
1interest from CD
time = = 15302
The second and third pairs of data do not result in a common ratio.
If any of the ratios are not equal, then there is no direct variation. It is not necessary to compute additional ratios or to determine whether (0, 0) is included.
A. interest from CD and time
Course 3
11-5 Direct Variation
Try This: Example 3 Continued
B. interest from money market and time
interest from money markettime = = 1515
1interest from money market
time = =20 402
15 ≠ 20
If any of the ratios are not equal, then there is no direct variation. It is not necessary to compute additional ratios or to determine whether (0, 0) is included.
Course 3
11-5 Direct Variation
11-6 Graphing Inequalities in Two Variables
Course 3
Warm UpWarm Up
Problem of the DayProblem of the Day
Lesson PresentationLesson Presentation
Warm UpFind each equation of direct variation, given that y varies directly with x.
1. y is 18 when x is 3.
2. x is 60 when y is 12.
3. y is 126 when x is 18.
4. x is 4 when y is 20.
y = 6x
y = 7x
Course 3
11-6 Graphing Inequalities in Two Variables
y = 5x
y = x15
Problem of the Day
The circumference of a pizza varies directly with its diameter. If you graph that direct variation, what will the slope be?
Course 3
11-6 Graphing Inequalities in Two Variables
Learn to graph inequalities on the coordinate plane.
Course 3
11-6 Graphing Inequalities in Two Variables
Vocabulary
boundary linelinear inequality
Insert Lesson Title Here
Course 3
11-6 Graphing Inequalities in Two Variables
A graph of a linear equation separates the coordinate plane into three parts: the points on one side of the line, the points on the boundary line, and the points on the other side of the line.
Course 3
11-6 Graphing Inequalities in Two Variables
Course 3
11-6 Graphing Inequalities in Two Variables
When the equality symbol is replaced in a linear equation by an inequality symbol, the statement is a linear inequality. Any ordered pair that makes the linear inequality true is a solution.
Course 3
11-6 Graphing Inequalities in Two Variables
Graph each inequality.
A. y < x – 1
Additional Example 1A: Graphing Inequalities
First graph the boundary line y = x – 1. Since no points that are on the line are solutions of y < x – 1, make the line dashed. Then determine on which side of the line the solutions lie.
(0, 0)
y < x – 1
Test a point not on the line.
Substitute 0 for x and 0 for y.0 < 0 – 1?
0 < –1?
Course 3
11-6 Graphing Inequalities in Two Variables
Additional Example 1A Continued
Since 0 < –1 is not true, (0, 0) is not a solution of y < x – 1. Shade the side of the line that does not include (0, 0).
Course 3
11-6 Graphing Inequalities in Two Variables
B. y 2x + 1
Additional Example 1B: Graphing Inequalities
First graph the boundary line y = 2x + 1. Since points that are on the line are solutions of y 2x + 1, make the line solid. Then shade the part of the coordinate plane in which the rest of the solutions of y 2x + 1 lie.
(0, 4) Choose any point not on the line.
Substitute 0 for x and 4 for y.
y ≥ 2x + 1
4 ≥ 0 + 1?
Course 3
11-6 Graphing Inequalities in Two Variables
Additional Example 1B Continued
Since 4 1 is true, (0, 4) is a solution of y 2x + 1. Shade the side of the line that includes (0, 4).
Course 3
11-6 Graphing Inequalities in Two Variables
C. 2y + 5x < 6
Additional Example 1C: Graphing Inequalities
First write the equation in slope-intercept form.
2y < –5x + 6
2y + 5x < 6
y < – x + 352
Then graph the line y = – x + 3. Since points that
are on the line are not solutions of y < – x + 3,
make the line dashed. Then determine on which
side of the line the solutions lie.
52 5
2
Subtract 5x from both sides.
Divide both sides by 2.
Course 3
11-6 Graphing Inequalities in Two Variables
Additional Example 1C Continued
Since 0 < 3 is true, (0, 0) is a
solution of y < – x + 3.
Shade the side of the line
that includes (0, 0).
52
(0, 0) Choose any point not on the line.
y < – x + 352
0 < 0 + 3?
0 < 3?
Course 3
11-6 Graphing Inequalities in Two Variables
Graph each inequality.
A. y < x – 4
Try This: Example 1A
First graph the boundary line y = x – 4. Since no points that are on the line are solutions of y < x – 4, make the line dashed. Then determine on which side of the line the solutions lie.
(0, 0)
y < x – 4
Test a point not on the line.
Substitute 0 for x and 0 for y.0 < 0 – 4?
0 < –4?
Course 3
11-6 Graphing Inequalities in Two Variables
Try This: Example 1A Continued
Since 0 < –4 is not true, (0, 0) is not a solution of y < x – 4. Shade the side of the line that does not include (0, 0).
Course 3
11-6 Graphing Inequalities in Two Variables
B. y > 4x + 4
Try This: Example 1B
First graph the boundary line y = 4x + 4. Since points that are on the line are solutions of y 4x + 4, make the line solid. Then shade the part of the coordinate plane in which the rest of the solutions of y 4x + 4 lie.
(2, 3) Choose any point not on the line.
Substitute 2 for x and 3 for y.
y ≥ 4x + 4
3 ≥ 8 + 4?
Course 3
11-6 Graphing Inequalities in Two Variables
Try This: Example 1B Continued
Since 3 12 is not true, (2, 3) is not a solution of y 4x + 4. Shade the side of the line that does not include (2, 3).
Course 3
11-6 Graphing Inequalities in Two Variables
C. 3y + 4x 9
Try This: Example 1C
First write the equation in slope-intercept form.
3y –4x + 9
3y + 4x 9
y – x + 343
43Then graph the line y = – x + 3. Since points that
are on the line are solutions of y – x + 3, make
the line solid. Then determine on which side of the
line the solutions lie.
43
Subtract 4x from both sides.
Divide both sides by 3.
Course 3
11-6 Graphing Inequalities in Two Variables
Try This: Example 1C Continued
Since 0 3 is not true, (0, 0) is
not a solution of y – x + 3.
Shade the side of the line that
does not include (0, 0).
43
(0, 0) Choose any point not on the line.
y – x + 343
0 0 + 3?
0 3?
Course 3
11-6 Graphing Inequalities in Two Variables
A successful screenwriter can write no more than seven and a half pages of dialogue each day. Graph the relationship between the number of pages the writer can write and the number of days. At this rate, would the writer be able to write a 200-page screenplay in 30 days?
Additional Example 2: Career Application
First find the equation of the line that corresponds to the inequality.
In 0 days the writer writes 0 pages.
point (0, 0)
point (1, 7.5)In 1 day the writer writes no more than 7 pages.1
2Course 3
11-6 Graphing Inequalities in Two Variables
Additional Example 2 Continued
With two known points, find the slope.
y 7.5 x + 0 The y-intercept is 0. No more than means .
Graph the boundary line y = 7.5x. Since points on
the line are solutions of y 7.5x make the line solid.
Shade the part of the coordinate plane in which the
rest of the solutions of y 7.5x lie.
Course 3
11-6 Graphing Inequalities in Two Variables
m = 7.5 – 01 – 0
7.5 1
= = 7.5
(2, 2) Choose any point not on the line.
y 7.5x
Substitute 2 for x and 2 for y.
Since 2 15 is true, (2, 2) is a solution of y 7.5x. Shade the side of the line that includes point (2, 2).
Additional Example 2 Continued
2 7.5 2?
2 15?
Course 3
11-6 Graphing Inequalities in Two Variables
The point (30, 200) is included in the shaded area, so the writer should be able to complete the 200 page screenplay in 30 days.
Additional Example 2 Continued
Course 3
11-6 Graphing Inequalities in Two Variables
A certain author can write no more than 20 pages every 5 days. Graph the relationship between the number of pages the writer can write and the number of days. At this rate, would the writer be able to write 140 pages in 20 days?
Try This: Example 2
First find the equation of the line that corresponds to the inequality.
In 0 days the writer writes 0 pages. point (0, 0)
point (5, 20)In 5 days the writer writes no more than 20 pages.
Course 3
11-6 Graphing Inequalities in Two Variables
Try This: Example 2 Continued
20 - 0 5 - 0m = = 20
5 = 4 With two known points, find the slope.
y 4x + 0 The y-intercept is 0. No more than means .
Graph the boundary line y = 4x. Since points on the line are solutions of y 4x make the line solid. Shade the part of the coordinate plane in which the rest of the solutions of y 4x lie.
Course 3
11-6 Graphing Inequalities in Two Variables
(5, 60) Choose any point not on the line.
y 4x
Substitute 5 for x and 60 for y.
Since 60 20 is not true, (5, 60) is not a solution of y 4x. Shade the side of the line that does not include (5, 60).
Try This: Example 2 Continued
60 4 5?
60 20?
Course 3
11-6 Graphing Inequalities in Two Variables
The point (20, 140) is not included in the shaded area, so the writer will not be able to write 140 pages in 20 days.
Try This: Example 2 Continued
x
y200
180
160
140120
100
80
60
40\
20
Pag
es
5 10 15 20 25 30 35 40 45 50
Days
Course 3
11-6 Graphing Inequalities in Two Variables
11-7 Lines of Best Fit
Course 3
Warm UpWarm Up
Problem of the DayProblem of the Day
Lesson PresentationLesson Presentation
Warm UpAnswer the questions about the inequality 5x + 10y > 30.
1. Would you use a solid or dashed boundary line?
2. Would you shade above or below the boundary line?
3. What are the intercepts of the graph?
dashed
above
(0, 3) and (6, 0)
Course 3
11-7 Lines of Best Fit
Problem of the Day
Write an inequality whose positive solutions form a triangular region with an area of 8 square units. (Hint: Sketch such a region on a coordinate plane.)
Possible answer: y < –x + 4
Course 3
11-7 Lines of Best Fit
Learn to recognize relationships in data and find the equation of a line of best fit.
Course 3
11-7 Lines of Best Fit
Course 3
11-7 Lines of Best Fit
When data show a correlation, you can estimate and draw a line of best fit that approximates a trend for a set of data and use it to make predictions.
To estimate the equation of a line of best fit:
• calculate the means of the x-coordinates and y-coordinates: (xm, ym)
• draw the line through (xm, ym) that appears to best fit the data.
• estimate the coordinates of another point on the line.
• find the equation of the line.
Plot the data and find a line of best fit.
Additional Example 1: Finding a Line of Best Fit
Course 3
11-7 Lines of Best Fit
Plot the data points and find the mean of the x- and y-coordinates.
xm = = 6 4 + 7 + 3 + 8 + 8 + 66
ym = = 4 4 + 5 + 2 + 6 + 7 + 46
23
x 4 7 3 8 8 6
y 4 5 2 6 7 4
23(xm, ym)= 6, 4
Course 3
11-7 Lines of Best Fit
The line of best fit is the line that comes closest to all the points on a scatter plot. Try to draw the line so that about the same number of points are above the line as below the line.
Remember!
Additional Example 1 Continued
Course 3
11-7 Lines of Best Fit
Draw a line through 6, 4 that best represents the data. Estimate and plot the coordinates of another point on that line, such as (8, 6). Find the equation of the line.
23
Course 3
11-7 Lines of Best Fit
Find the slope.
y – y1 = m(x – x1) Use point-slope form.
y – 4 = (x – 6)23
23 Substitute.
y – 4 = x – 423
23
23y = x +2
3
The equation of a line of best fit is .23y = x +23
Additional Example 1 Continued23
13m = = =
6 – 4
8 – 6
1
223
Plot the data and find a line of best fit.
Try This: Example 1
Course 3
11-7 Lines of Best Fit
Plot the data points and find the mean of the x- and y-coordinates.
xm = = 2 –1 + 0 + 2 + 6 + –3 + 8 6
ym = = 1 –1 + 0 + 3 + 7 + –7 + 46
x –1 0 2 6 –3 8
y –1 0 3 7 –7 4
(xm, ym) = (2, 1)
Try This: Example 1 Continued
Course 3
11-7 Lines of Best Fit
Draw a line through (2, 1) that best represents the data. Estimate and plot the coordinates of another point on that line, such as (10, 10). Find the equation of the line.
Course 3
11-7 Lines of Best Fit
Find the slope.
y – y1 = m(x – x1) Use point-slope form.
y – 1 = (x – 2)98 Substitute.
y – 1 = x –98
94
The equation of a line of best fit is . y = x –98
54
Try This: Example 1 Continued
m = = 10 – 1 10 – 2
98
y = x –98
54
Find a line of best fit for the Main Street Elementary annual softball toss. Use the equation of the line to predict the winning distance in 2006.
Additional Example 2: Sports Application
Course 3
11-7 Lines of Best Fit
Let 1990 represent year 0. The first point is then (0, 98), and the last point is (12, 107).
xm = = 50 + 2 + 4 + 7 + 125
Year 1990 1992 1994 1997 2002
Distance (ft) 98 101 103 106 107
ym = = 103 98 + 101 + 103 + 106 + 107 5
(xm, ym) = (5, 103)
Additional Example 2 Continued
Course 3
11-7 Lines of Best Fit
Draw a line through (5, 103) that best represents the data. Estimate and plot the coordinates of another point on that line, such as (10, 107). Find the equation of the line.
Course 3
11-7 Lines of Best Fit
m = = 0.8 107 - 10310 - 5 Find the slope.
y – y1 = m(x – x1) Use point-slope form.
y – 103 = 0.8(x – 5) Substitute.
y – 103 = 0.8x – 4
y = 0.8x + 99
The equation of a line of best fit is y = 0.8x + 99. Since 1990 represents year 0, 2006 represents year 16.
Additional Example 2 Continued
Course 3
11-7 Lines of Best Fit
Substitute.
y = 12.8 + 99
y = 0.8(16) + 99
The equation predicts a winning distance of about 112 feet for the year 2006.
y = 111.8
Additional Example 2 Continued
Predict the winning weight lift in 2010.
Try This: Example 2
Course 3
11-7 Lines of Best Fit
Let 1990 represent year 0. The first point is then (0, 100), and the last point is (10, 170).
xm = = 60 + 5 + 7 + 8 + 105
ym = = 132100 + 120 + 130 + 140 + 170 5
Year 1990 1995 1997 1998 2000
Lift (lb) 100 120 130 140 170
(xm, ym) = (6, 132)
Try This: Example 2 Continued
Course 3
11-7 Lines of Best Fit
Draw a line through (5, 132) the best represents the data. Estimate and plot the coordinates of another point on that line, such as (7, 140). Find the equation of the line.
Years since 1990
weig
ht
(lb)
0100
120
140
160
180
2 4 6 8 10
200
Course 3
11-7 Lines of Best Fit
m = = 4 140 – 132 7 – 5 Find the slope.
y – y1 = m(x – x1) Use point-slope form.
y – 132 = 4(x – 5) Substitute.
y – 132 = 4x – 20
y = 4x + 112
The equation of a line of best fit is y = 4x + 112. Since 1990 represents year 0, 2010 represents year 20.
Try This: Example 2 Continued
Course 3
11-7 Lines of Best Fit
Substitute.
y = 192
y = 4(20) + 112
The equation predicts a winning lift of about 192 lb for the year 2010.
Try This: Example 2 Continued