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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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 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
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
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.
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
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
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
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
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
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
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
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
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
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
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
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
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
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!
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
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
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 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
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
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
D = kt 2
D = kt 2
.6 = k(1) 2
D = kt 2
.6 = k(1) 2
k = .6
D = kt 2
.6 = k(1) 2
k = .6
D = .6t 2
D = kt 2
.6 = k(1) 2
k = .6
D = .6t 2
D = .6(2) 2
D = kt 2
.6 = k(1) 2
k = .6
D = .6t 2
D = .6(2) 2
D = 2.4
D = kt 2
.6 = k(1) 2
k = .6
D = .6t 2
D = .6(2) 2
D = 2.4
D = .6t 2
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
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 n th power of x, or k .
y= n
x
HOMEWORK
HOMEWORK
Make sure to read Sections 2-7 and 2-8
p. 113 #1-11, p. 119 #1-11
Be the first to comment