Fitting a Model to Data

1. SECTIONS 2-7 AND 2-8
FITTING A MODEL TO DATA

2. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k
b. y =
x
c. y = kx 2
k
d. y = 2
x

3. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k
b. y =
20 = 10k x
c. y = kx 2
k
d. y = 2
x

4. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k
b. y =
20 = 10k x
k=2
c. y = kx 2
k
d. y = 2
x

5. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k
b. y =
20 = 10k x
k=2
y = 2x
c. y = kx 2
k
d. y = 2
x

6. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k
b. y =
20 = 10k y = 2(30) x
k=2
y = 2x
c. y = kx 2
k
d. y = 2
x

7. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k
b. y =
20 = 10k y = 2(30) x
k=2 y = 60
y = 2x
c. y = kx 2
k
d. y = 2
x

8. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k
b. y =
20 = 10k y = 2(30) x
k
k=2 y = 60 20 =
10
y = 2x
c. y = kx 2
k
d. y = 2
x

9. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k
b. y =
20 = 10k y = 2(30) x
k
k=2 y = 60 20 =
10
y = 2x k = 200
c. y = kx 2
k
d. y = 2
x

10. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k 200
b. y = y=
20 = 10k y = 2(30) x x
k
k=2 y = 60 20 =
10
y = 2x k = 200
c. y = kx 2
k
d. y = 2
x

11. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k 200
b. y = y= 200
20 = 10k y = 2(30) x x y=
k 30
k=2 y = 60 20 =
10
y = 2x k = 200
c. y = kx 2
k
d. y = 2
x

12. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k 200
b. y = y= 200
20 = 10k y = 2(30) x x y=
k 30
k=2 y = 60 20 = 20
10 y=
y = 2x k = 200 3
c. y = kx 2
k
d. y = 2
x

13. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k 200
b. y = y= 200
20 = 10k y = 2(30) x x y=
k 30
k=2 y = 60 20 = 20
10 y=
y = 2x k = 200 3
c. y = kx 2
k
d. y = 2
20 = k(10) 2 x

14. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k 200
b. y = y= 200
20 = 10k y = 2(30) x x y=
k 30
k=2 y = 60 20 = 20
10 y=
y = 2x k = 200 3
c. y = kx 2
k
d. y = 2
20 = k(10) 2 x
k= 1
5

15. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k 200
b. y = y= 200
20 = 10k y = 2(30) x x y=
k 30
k=2 y = 60 20 = 20
10 y=
y = 2x k = 200 3
c. y = kx 2
k
d. y = 2
20 = k(10) 2 x
k= 1
5
y= x1
5
2

16. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k 200
b. y = y= 200
20 = 10k y = 2(30) x x y=
k 30
k=2 y = 60 20 = 20
10 y=
y = 2x k = 200 3
c. y = kx 2
k
d. y = 2
20 = k(10) 2
y = (30)
1 2 x
5
k= 1
5
y= x1
5
2

17. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k 200
b. y = y= 200
20 = 10k y = 2(30) x x y=
k 30
k=2 y = 60 20 = 20
10 y=
y = 2x k = 200 3
c. y = kx 2
k
d. y = 2
20 = k(10) 2
y = (30)
1 2 x
5
k= 1
5
y = 180
y= x1
5
2

18. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k 200
b. y = y= 200
20 = 10k y = 2(30) x x y=
k 30
k=2 y = 60 20 = 20
10 y=
y = 2x k = 200 3
c. y = kx 2
k
d. y = 2
20 = k(10) 2
y = 5 (30)
1 2
k x
20 = 2
k= 51
y = 180 10
y= x1
5
2

19. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k 200
b. y = y= 200
20 = 10k y = 2(30) x x y=
k 30
k=2 y = 60 20 = 20
10 y=
y = 2x k = 200 3
c. y = kx 2
k
d. y = 2
20 = k(10) 2
y = 5 (30)
1 2
k x
20 = 2
k= 51
y = 180 10
y= x k = 2000
1 2
5

20. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k 200
b. y = y= 200
20 = 10k y = 2(30) x x y=
k 30
k=2 y = 60 20 = 20
10 y=
y = 2x k = 200 3
c. y = kx 2
k 2000
d. y = 2 y = 2
20 = k(10) 2
y = 5 (30)
1 2
k x x
20 = 2
k= 51
y = 180 10
y= x k = 2000
1 2
5

21. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k 200
b. y = y= 200
20 = 10k y = 2(30) x x y=
k 30
k=2 y = 60 20 = 20
10 y=
y = 2x k = 200 3
c. y = kx 2
k 2000
d. y = 2 y = 2
20 = k(10) 2
y = 5 (30)
1 2
k x x
20 = 2 y = 2000
k= 51
y = 180 10 2
30
y= x k = 2000
1 2
5

22. WARM-UP
The point with coordinates (10, 20) appears on the graph for
each variation function below. Determine the value of k in
each case, and then determine the value of y when x is 30.
a. y = kx k 200
b. y = y= 200
20 = 10k y = 2(30) x x y=
k 30
k=2 y = 60 20 = 20
10 y=
y = 2x k = 200 3
c. y = kx 2
k 2000
d. y = 2 y = 2
20 = k(10) 2
y = 5 (30)
1 2
k x x
20 = 2 y = 2000
k= 51
y = 180 10 2
30 20
y=
y= 5x1 2
k = 2000 9

23. MATHEMATICAL MODEL

24. MATHEMATICAL MODEL
A mathematical representation of a real-world situation

25. MATHEMATICAL MODEL
A mathematical representation of a real-world situation
A good model fits all possibilities as best it can; often is just
an approximation

26. EXAMPLE 1
Matt Mitarnowski and Fuzzy Jeff measured the intensity of
light at various distances from a lamp and got the data where
d = distance in meters and I = intensity in watts/m2.
d I
2 560
2.5 360
3 250
3.5 185
4 140

27. EXAMPLE 1
Matt Mitarnowski and Fuzzy Jeff measured the intensity of
light at various distances from a lamp and got the data where
d = distance in meters and I = intensity in watts/m2.
d I 600
2 560
450
2.5 360
3 250 300
3.5 185 150
4 140
0
0 1.25 2.50 3.75 5.00

28. EXAMPLE 1
This looks like either a hyperbola or an inverse-square curve.
Let’s test them out by checking some points in our equations.

29. EXAMPLE 1
This looks like either a hyperbola or an inverse-square curve.
Let’s test them out by checking some points in our equations.
k
I=
d

30. EXAMPLE 1
This looks like either a hyperbola or an inverse-square curve.
Let’s test them out by checking some points in our equations.
k k
I= 560 =
d 2

31. EXAMPLE 1
This looks like either a hyperbola or an inverse-square curve.
Let’s test them out by checking some points in our equations.
k k
I= 560 = k = 1120
d 2

32. EXAMPLE 1
This looks like either a hyperbola or an inverse-square curve.
Let’s test them out by checking some points in our equations.
k k 1120
I= 560 = k = 1120 I=
d 2 d

33. EXAMPLE 1
This looks like either a hyperbola or an inverse-square curve.
Let’s test them out by checking some points in our equations.
k k 1120 1120
I= 560 = k = 1120 I= I=
d 2 d 4

34. EXAMPLE 1
This looks like either a hyperbola or an inverse-square curve.
Let’s test them out by checking some points in our equations.
k k 1120 1120
I= 560 = k = 1120 I= I=
d 2 d 4
I = 280

35. EXAMPLE 1
This looks like either a hyperbola or an inverse-square curve.
Let’s test them out by checking some points in our equations.
k k 1120 1120
I= 560 = k = 1120 I= I=
d 2 d 4
I = 280 Not close

36. EXAMPLE 1
This looks like either a hyperbola or an inverse-square curve.
Let’s test them out by checking some points in our equations.
k k 1120 1120
I= 560 = k = 1120 I= I=
d 2 d 4
I = 280 Not close
k
I= 2
d

37. EXAMPLE 1
This looks like either a hyperbola or an inverse-square curve.
Let’s test them out by checking some points in our equations.
k k 1120 1120
I= 560 = k = 1120 I= I=
d 2 d 4
I = 280 Not close
k k
I= 2 560 = 2
d 2

38. EXAMPLE 1
This looks like either a hyperbola or an inverse-square curve.
Let’s test them out by checking some points in our equations.
k k 1120 1120
I= 560 = k = 1120 I= I=
d 2 d 4
I = 280 Not close
k k
I= 2 560 = 2 k = 2240
d 2

39. EXAMPLE 1
This looks like either a hyperbola or an inverse-square curve.
Let’s test them out by checking some points in our equations.
k k 1120 1120
I= 560 = k = 1120 I= I=
d 2 d 4
I = 280 Not close
k k 2240
I= 2 560 = 2 k = 2240 I= 2
d 2 d

40. EXAMPLE 1
This looks like either a hyperbola or an inverse-square curve.
Let’s test them out by checking some points in our equations.
k k 1120 1120
I= 560 = k = 1120 I= I=
d 2 d 4
I = 280 Not close
k k 2240 2240
I= 2 560 = 2 k = 2240 I= 2 I=
d 2 d 16

41. EXAMPLE 1
This looks like either a hyperbola or an inverse-square curve.
Let’s test them out by checking some points in our equations.
k k 1120 1120
I= 560 = k = 1120 I= I=
d 2 d 4
I = 280 Not close
k k 2240 2240
I= 2 560 = 2 k = 2240 I= 2 I=
d 2 d 16
I = 140

42. EXAMPLE 1
This looks like either a hyperbola or an inverse-square curve.
Let’s test them out by checking some points in our equations.
k k 1120 1120
I= 560 = k = 1120 I= I=
d 2 d 4
I = 280 Not close
k k 2240 2240
I= 2 560 = 2 k = 2240 I= 2 I=
d 2 d 16
I = 140 Right on!

43. EXAMPLE 1
This looks like either a hyperbola or an inverse-square curve.
Let’s test them out by checking some points in our equations.
k k 1120 1120
I= 560 = k = 1120 I= I=
d 2 d 4
I = 280 Not close
k k 2240 2240
I= 2 560 = 2 k = 2240 I= 2 I=
d 2 d 16
I = 140 Right on!
2240
I= 2
d

44. EXAMPLE 1
This looks like either a hyperbola or an inverse-square curve.
Let’s test them out by checking some points in our equations.
k k 1120 1120
I= 560 = k = 1120 I= I=
d 2 d 4
I = 280 Not close
k k 2240 2240
I= 2 560 = 2 k = 2240 I= 2 I=
d 2 d 16
I = 140 Right on!
2240
I= 2 Make sure to check that all points are close
d

45. STEPS TO FINDING A
MODEL FROM DATA

46. STEPS TO FINDING A
MODEL FROM DATA
1. Graph the data

47. STEPS TO FINDING A
MODEL FROM DATA
1. Graph the data
2. Describe the graph

48. STEPS TO FINDING A
MODEL FROM DATA
1. Graph the data
2. Describe the graph
3. State the equation

49. STEPS TO FINDING A
MODEL FROM DATA
1. Graph the data
2. Describe the graph
3. State the equation
4. Find k, plug it in

50. STEPS TO FINDING A
MODEL FROM DATA
1. Graph the data
2. Describe the graph
3. State the equation
4. Find k, plug it in
5. Check other points

51. STEPS TO FINDING A
MODEL FROM DATA
1. Graph the data
2. Describe the graph
3. State the equation
4. Find k, plug it in
5. Check other points
6. State the model

52. EXAMPLE 2
Matt Mitarnowski measured how far a marble rolled down a
ramp over a period of time. He obtained the following data
where t is time in seconds and D is distance in inches.
t D
1 0.6
2 2.4
3 5.2
4 9.8
5 15.2

53. EXAMPLE 2
Matt Mitarnowski measured how far a marble rolled down a
ramp over a period of time. He obtained the following data
where t is time in seconds and D is distance in inches.
20
t D
15
1 0.6
2 2.4 10
3 5.2
5
4 9.8
5 15.2 0
0 1.25 2.50 3.75 5.00

54. EXAMPLE 2
Matt Mitarnowski measured how far a marble rolled down a
ramp over a period of time. He obtained the following data
where t is time in seconds and D is distance in inches.
20
t D
15
1 0.6
2 2.4 10
3 5.2
5
4 9.8
5 15.2 0
0 1.25 2.50 3.75 5.00
Should be a parabola

56. D = kt 2

57. D = kt 2
.6 = k(1) 2

58. D = kt 2
.6 = k(1) 2
k = .6

59. D = kt 2
.6 = k(1) 2
k = .6
D = .6t 2

60. D = kt 2
.6 = k(1) 2
k = .6
D = .6t 2
D = .6(2) 2

61. D = kt 2
.6 = k(1) 2
k = .6
D = .6t 2
D = .6(2) 2
D = 2.4

62. D = kt 2
.6 = k(1) 2
k = .6
D = .6t 2
D = .6(2) 2
D = 2.4
D = .6t 2

63. D = kt 2
.6 = k(1) 2
k = .6
D = .6t 2
D = .6(2) 2
D = 2.4
D = .6t 2
Check the rest of the points to make sure satisfy the model

64. CONVERSE OF THE
FUNDAMENTAL THEOREM OF
VARIATION

65. CONVERSE OF THE
FUNDAMENTAL THEOREM OF
VARIATION
a. If multiplying every x-value of a function by c results in
multiplying the corresponding y-value by cn, then y varies
directly as the nth power of x, or y = kxn.

66. CONVERSE OF THE
FUNDAMENTAL THEOREM OF
VARIATION
a. If multiplying every x-value of a function by c results in
multiplying the corresponding y-value by cn, then y varies
directly as the nth power of x, or y = kxn.
b. If multiplying every x-value of a function by c results in
dividing the corresponding y-value by cn, then y varies
inversely as the n th power of x, or k .
y= n
x

67. HOMEWORK

68. HOMEWORK
Make sure to read Sections 2-7 and 2-8
p. 113 #1-11, p. 119 #1-11