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Presented by Mr. Widhalm

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Presentation on theme: "Presented by Mr. Widhalm"— Presentation transcript:

1 Presented by Mr. Widhalm
Sample project Presented by Mr. Widhalm

2 The Situation The Number 2 Pencil Company wants to know the mass of its pencils at various lengths. The company has hired me to develop a formula (a function) that will allow them to calculate the mass if the length of the pencil (only the painted portion) is known.

3 The data is in the following table.
Data Collection Starting with a whole, unsharpened pencil and I found its length ( cm) and mass ( g). (I used a ruler and a digital scale.) I then sharpened the pencil and measured both the mass and length of the painted portion. This process was repeated to get ten data sets. The data is in the following table.

4 Table 1 – Length of pencil versus mass of pencil
Length (cm) mass (g) 1.3 1.26 2.4 1.62 3.8 1.95 4.9 2.13 6.3 2.5 7.9 2.92 9.6 3.35 12.8 4.12 15 4.75 17.2 5.15

5 Figure 1-Graph of Pencil Data

6 Formula (Equation for the Function)
I noticed the data lined up really well. Visually, I could see a linear function (y = mx + b) would be best. I drew a line on the graph and extended it to meet the y-axis. (See next graph)

7 Figure 2- Graph of Pencil Data with Function
Slope = change in Y / change in X = (4 – 3)/(12 – 8) = ¼ = 0.25 (8, 3) (12, 4) y-intercept at y =1

8 Function (cont.) Using the slope (4) and y-intercept (1), I can write the formula (equation for the function) as: w(L) = 0.25L + 1 “b” is the minimum pencil mass (with no paint remaining). “m” is the rate at which the mass of the pencil decreases when the length decreases.

9 Summary of the Function
The company can use the following formula to find the mass, in grams, of the pencil given the length in centimeters: w(L) = 0.25L + 1 L – is the length w – is the mass L can be any value including and between 17.2 cm and 0 cm (domain) w can be any value including and between 5.3 g and 1 g (range)

10 Figure 3 - Checking Reasonableness of the Function
w(10) = 0.25*(10) + 1 = 3.5 = 3.5 (10, 3.5)


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