SECTIONS 2-7 AND 2-8

FITTING A MODEL TO DATA

WARM-UPThe 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 b. y =

kx

d. y =

kx2 c. y = kx2

WARM-UPThe 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 b. y =

kx

d. y =

kx2

20 = 10k

c. y = kx2

WARM-UPThe 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 b. y =

kx

d. y =

kx2

20 = 10kk = 2

c. y = kx2

WARM-UPThe 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 b. y =

kx

d. y =

kx2

20 = 10kk = 2

y = 2x

c. y = kx2

WARM-UPThe 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 b. y =

kx

d. y =

kx2

20 = 10kk = 2

y = 2x

y = 2(30)

c. y = kx2

WARM-UPThe 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 b. y =

kx

d. y =

kx2

20 = 10kk = 2

y = 2x

y = 2(30)y = 60

c. y = kx2

WARM-UPThe 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 b. y =

kx

d. y =

kx2

20 = 10kk = 2

y = 2x

y = 2(30)y = 60

20 =

k10

c. y = kx2

WARM-UPThe 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 b. y =

kx

d. y =

kx2

20 = 10kk = 2

y = 2x

y = 2(30)y = 60

20 =

k10

k = 200

c. y = kx2

WARM-UPThe 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 b. y =

kx

d. y =

kx2

20 = 10kk = 2

y = 2x

y = 2(30)y = 60

20 =

k10

k = 200

y =

200x

c. y = kx2

WARM-UPThe 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 b. y =

kx

d. y =

kx2

20 = 10kk = 2

y = 2x

y = 2(30)y = 60

20 =

k10

k = 200

y =

200x

y =

20030

c. y = kx2

WARM-UPThe 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 b. y =

kx

d. y =

kx2

20 = 10kk = 2

y = 2x

y = 2(30)y = 60

20 =

k10

k = 200

y =

200x

y =

20030

y =

203

c. y = kx2

WARM-UPThe 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 b. y =

kx

d. y =

kx2

20 = 10kk = 2

y = 2x

y = 2(30)y = 60

20 =

k10

k = 200

y =

200x

y =

20030

y =

203

c. y = kx2

20 = k(10)2

WARM-UPThe 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 b. y =

kx

d. y =

kx2

20 = 10kk = 2

y = 2x

y = 2(30)y = 60

20 =

k10

k = 200

y =

200x

y =

20030

y =

203

c. y = kx2

20 = k(10)2

k = 15

WARM-UPThe 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 b. y =

kx

d. y =

kx2

20 = 10kk = 2

y = 2x

y = 2(30)y = 60

20 =

k10

k = 200

y =

200x

y =

20030

y =

203

c. y = kx2

20 = k(10)2

k = 15

y = 15 x2

WARM-UPThe 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 b. y =

kx

d. y =

kx2

20 = 10kk = 2

y = 2x

y = 2(30)y = 60

20 =

k10

k = 200

y =

200x

y =

20030

y =

203

c. y = kx2

20 = k(10)2

k = 15

y = 15 x2

y = 15 (30)2

WARM-UPThe 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 b. y =

kx

d. y =

kx2

20 = 10kk = 2

y = 2x

y = 2(30)y = 60

20 =

k10

k = 200

y =

200x

y =

20030

y =

203

c. y = kx2

20 = k(10)2

k = 15

y = 15 x2

y = 15 (30)2

y = 180

WARM-UPThe 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 b. y =

kx

d. y =

kx2

20 = 10kk = 2

y = 2x

y = 2(30)y = 60

20 =

k10

k = 200

y =

200x

y =

20030

y =

203

c. y = kx2

20 = k(10)2

k = 15

y = 15 x2

y = 15 (30)2

y = 180 20 =

k102

WARM-UPThe 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 b. y =

kx

d. y =

kx2

20 = 10kk = 2

y = 2x

y = 2(30)y = 60

20 =

k10

k = 200

y =

200x

y =

20030

y =

203

c. y = kx2

20 = k(10)2

k = 15

y = 15 x2

y = 15 (30)2

y = 180 20 =

k102

k = 2000

WARM-UPThe 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 b. y =

kx

d. y =

kx2

20 = 10kk = 2

y = 2x

y = 2(30)y = 60

20 =

k10

k = 200

y =

200x

y =

20030

y =

203

c. y = kx2

20 = k(10)2

k = 15

y = 15 x2

y = 15 (30)2

y = 180 20 =

k102

k = 2000

y =

2000x2

WARM-UPThe 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 b. y =

kx

d. y =

kx2

20 = 10kk = 2

y = 2x

y = 2(30)y = 60

20 =

k10

k = 200

y =

200x

y =

20030

y =

203

c. y = kx2

20 = k(10)2

k = 15

y = 15 x2

y = 15 (30)2

y = 180 20 =

k102

k = 2000

y =

2000x2

y =

2000302

WARM-UPThe 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 b. y =

kx

d. y =

kx2

20 = 10kk = 2

y = 2x

y = 2(30)y = 60

20 =

k10

k = 200

y =

200x

y =

20030

y =

203

c. y = kx2

20 = k(10)2

k = 15

y = 15 x2

y = 15 (30)2

y = 180 20 =

k102

k = 2000

y =

2000x2

y =

2000302

y =

209

MATHEMATICAL MODEL

MATHEMATICAL MODEL

A mathematical representation of a real-world situation

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

EXAMPLE 1Matt 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 I2 560

2.5 3603 250

3.5 1854 140

EXAMPLE 1Matt 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 I2 560

2.5 3603 250

3.5 1854 140

0

150

300

450

600

0 1.25 2.50 3.75 5.00

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

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

I = k

d

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

I = k

d 560 =

k2

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

I = k

d 560 =

k2 k = 1120

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

I = k

d 560 =

k2 k = 1120

I = 1120

d

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

I = k

d 560 =

k2 k = 1120

I = 1120

d I = 1120

4

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

I = k

d 560 =

k2 k = 1120

I = 1120

d I = 1120

4

I = 280

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

I = k

d 560 =

k2 k = 1120

I = 1120

d I = 1120

4

I = 280 Not close

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

I = k

d 560 =

k2 k = 1120

I = 1120

d I = 1120

4

I = 280 Not close

I = k

d2

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

I = k

d 560 =

k2 k = 1120

I = 1120

d I = 1120

4

I = 280 Not close

I = k

d2 560 =

k22

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

I = k

d 560 =

k2 k = 1120

I = 1120

d I = 1120

4

I = 280 Not close

I = k

d2 560 =

k22 k = 2240

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

I = k

d 560 =

k2 k = 1120

I = 1120

d I = 1120

4

I = 280 Not close

I = k

d2 560 =

k22 k = 2240

I = 2240

d2

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

I = k

d 560 =

k2 k = 1120

I = 1120

d I = 1120

4

I = 280 Not close

I = k

d2 560 =

k22 k = 2240

I = 2240

d2 I = 2240

16

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

I = k

d 560 =

k2 k = 1120

I = 1120

d I = 1120

4

I = 280 Not close

I = k

d2 560 =

k22 k = 2240

I = 2240

d2 I = 2240

16

I = 140

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

I = k

d 560 =

k2 k = 1120

I = 1120

d I = 1120

4

I = 280 Not close

I = k

d2 560 =

k22 k = 2240

I = 2240

d2 I = 2240

16

I = 140 Right on!

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

I = k

d 560 =

k2 k = 1120

I = 1120

d I = 1120

4

I = 280 Not close

I = k

d2 560 =

k22 k = 2240

I = 2240

d2 I = 2240

16

I = 140 Right on!

I = 2240

d2

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

I = k

d 560 =

k2 k = 1120

I = 1120

d I = 1120

4

I = 280 Not close

I = k

d2 560 =

k22 k = 2240

I = 2240

d2 I = 2240

16

I = 140 Right on!

I = 2240

d2 Make sure to check that all points are close

STEPS TO FINDING A MODEL FROM DATA

STEPS TO FINDING A MODEL FROM DATA

1. Graph the data

STEPS TO FINDING A MODEL FROM DATA

1. Graph the data

2. Describe the graph

STEPS TO FINDING A MODEL FROM DATA

1. Graph the data

2. Describe the graph

3. State the equation

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

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

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

EXAMPLE 2Matt 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 D1 0.62 2.43 5.24 9.85 15.2

EXAMPLE 2Matt 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 D1 0.62 2.43 5.24 9.85 15.2 0

5

10

15

20

0 1.25 2.50 3.75 5.00

EXAMPLE 2Matt 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 D1 0.62 2.43 5.24 9.85 15.2 0

5

10

15

20

0 1.25 2.50 3.75 5.00

Should be a parabola

D = kt2

D = kt2

.6 = k(1)2

D = kt2

.6 = k(1)2

k = .6

D = kt2

.6 = k(1)2

k = .6

D = .6t2

D = kt2

.6 = k(1)2

k = .6

D = .6t2

D = .6(2)2

D = kt2

.6 = k(1)2

k = .6

D = .6t2

D = .6(2)2

D = 2.4

D = kt2

.6 = k(1)2

k = .6

D = .6t2

D = .6(2)2

D = 2.4

D = .6t2

D = kt2

.6 = k(1)2

k = .6

D = .6t2

D = .6(2)2

D = 2.4

D = .6t2

Check the rest of the points to make sure satisfy the model

CONVERSE OF THE FUNDAMENTAL THEOREM OF

VARIATION

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.

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 nth power of x, or . y =

kxn

HOMEWORK

HOMEWORK

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