Investigate1. Draw a dot on a piece of

paper2. Draw a second dot3. Connect the dots with a

straight line4. Draw a third dot – draw as

many straight lines that go through the dots as you can

1. You should have 3 lines

5. How many lines can you draw using 100

points?

3.5 Quadratic Models Using Factored Form

Try to find a relationship between the number of points and number of lines to predict how many lines we can draw with 100 points.

Investigate

Number of Points, x 0 1 2 3 4 5 6

Maximum Number of Lines, y 0 0

Number of Points, x 0 1 2 3 4 5 6

Maximum Number of Lines, y 0 0 1 3 6 10 15

Is this what you got?

Let’s graph it:Investigate cont’d

Number of Points, x 0 1 2 3 4 5 6

Maximum Number of Lines, y 0 0 1 3 6 10 15

0 1 2 3 4 5 6 70

2

4

6

8

10

12

14

16

Predicting Max Number of Lines Using 100 Points

Number of Points

Maxim

um

Num

ber

of

Lin

es,

y

Is the curve approximately linear, quadratic or something else?◦ Quadratic, since 2nd differences are the same

Investigate cont’d

0 1 2 3 4 5 6 70

2

4

6

8

10

12

14

16

Predicting Max Number of Lines Using 100 Points

Number of Points

Maxim

um

Num

ber

of

Lin

es,

y

What are the zeros of the curve?◦ Recall, zeros are the x-values when y = 0:

◦ Thus, zeros occur at (0,0) and (1,0)◦ y = a(x-r)(x-s) becomes y = a(x-0)(x-1) = ax(x-1)◦ We can use the point (2,1) to find ‘a’:

1 = a(2)(2-1) 1 = a(2)(1) 1 = 2a a = 0.5, so y = 0.5x(x-1)

Investigate cont’d

Number of Points, x 0 1 2 3 4 5 6

Maximum Number of Lines, y 0 0 1 3 6 10 15

Using the equation, y = 0.5x(x-1), how many lines can you draw with 100 points? Translate to mathematical terms: what is

y when x = 100?y = 0.5(100)(100-1)y = (50)(99)y = 4950Therefore, you can draw a maximum of 4950 lines with 100 dots.

Investigate cont’d

Data from the flight of a golf ball are shown. If the max height of the ball is 30.0m, determine an equation for a curve of good fit.

Example #1

Horizontal Distance (m) 0 30 60 80 90

Height (m) 0.0 22.0 30.0 27.0 22.5

• We can draw the curve using the points

Shape of the curve seems to fit a parabolaThe vertex is the maximum height: 30.0mWe can see one zero at (0,0)Where is the other zero?

Example #1 cont’d

0 10 20 30 40 50 60 70 80 90 1000

5

10

15

20

25

30

35

Flight of a Golf Ball

Horizontal Distance (m)

Heig

ht

(m)

Since the parabola is a symmetric

shape, the other zero must be 60 units to the right of the vertex, at x = 120

Start with general equation:y = a(x-r)(x-s)Plug in zeros: y = a(x-0)(x-120)Plug in a point (60, 30): 30 = a(60-0)(60-120)30 = a(60)(-60)30 = a(-3600)

a = Therefore the equation describing the golf ball’s flight is y =

Example #1 cont’d

A competitive diver does a handstand dive from a 10m platform. The table of values below shows the time in seconds and the height of the diver, relative to the surface of the water, in meters.

Example #2

Time (s) 0 0.3 0.6 0.9 1.2 1.5

Height (m) 10.00 9.56 8.24 6.03 2.94 -1.03

Determine an equation that models the height of the diver above the surface of the water

during the dive.

We can assume that the maximum height (diving board height) is the vertex (10.00m).We can also estimate the value of the zeros.We see one zero must occur between 1.2 and 1.5 seconds, so estimate 1.4 seconds.Since a parabola is symmetric and vertex is at x=0, the other zero must be at -1.4.

Example #2 cont’dTime (s) 0 0.3 0.6 0.9 1.2 1.5

Height (m) 10.00 9.56 8.24 6.03 2.94 -1.03

Equation of the form: y = a(x-s)(x-r)y = a(x-1.4)(x-(-1.4)) = a(x-1.4)(x+1.4)Plug in a known point (x,y) = (0, 10) to find ‘a’10=a(0-1.4)(0+1.4)10=a(-1.4)(1.4)10=-1.96a

a -5.1 therefore, y =-5.1(x-1.4)(x+1.4)

Example #2 cont’dTime (s) 0 0.3 0.6 0.9 1.2 1.5

Height (m) 10.00 9.56 8.24 6.03 2.94 -1.03

Calculate or in some cases estimate the x-intercepts, zeros for a curve in the form y=a(x-r)(x-s)

The value of ‘a’ can be determined algebraically by substituting coordinates of a point (other than 0) that lies on or close to the line

In Summary…