Upload
lawrence-hunter
View
215
Download
1
Embed Size (px)
Citation preview
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
Lesson Quiz: Part 1
Find each equation of direct variation, given that y varies directly with x.
1. y is 78 when x is 3.
2. x is 45 when y is 5.
3. y is 6 when x is 5.
y = 26x
Insert Lesson Title Here
y = x19
y = x65
Course 3
11-5 Direct Variation
Lesson Quiz: Part 2
4. The table shows the amount of money Bob
makes for different amounts of time he works.
Determine whether there is a direct variation
between the two sets of data. If so, find the
equation of direct variation.
Insert Lesson Title Here
direct variation; y = 12x
Course 3
11-5 Direct Variation