Math 103 Section 1.2: Linear Equations and ... - Bruce E. Shapiro

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Application: Linear Interpolation two points on the line: (q, p) = (8, 2.10), (20, 3.30). Price 5.00 4.50 4.00 3.50 3.00 2.50 2.00 1.50 1.00 .50 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 Ounces Slope: We know that the price of an 8-ounce cup is $2.10 and the price of a 20-ounce cup is $3.30. So the change in q is $3.30−$2.10 = $1.20 and the change in p is 20oz.−8oz. = 12oz.. So the slope is: m = 3.30−2.10 20−8 = 1.20 12 = .10 Units for the slope m: dollars per ounce 18
Application: Linear Interpolation Slope is $.10 per ounce. How much does the price increase if the size of the cup increases by 4 ounces? Point-slope form: p − 2.10 = .10(q − 8) Slope-intercept form: p = .10q + 1.30. y-intercept is $1.30. Interpretation: The price of a cup of coffee is $1.30 plus ten cents an ounce. 19
Page 1 and 2: Math 103 Section 1.2: Linear Equati
Page 3 and 4: x = y x 2 − y = 3 3x − 4y = 6 y
Page 5 and 6: The shape of the graph of a linear
Page 7 and 8: x− and y− intercepts of an equa
Page 9 and 10: Intercepts for the Price-demand equ
Page 11 and 12: Slope of a line: The slope or steep
Page 13 and 14: Special forms for linear equations:
Page 15 and 16: Special forms for linear equations:
Page 17 and 18: Application: Depreciation. Linear D
Page 19: Application: Linear Interpolation T

Math 103 Section 1.2: Linear Equations and ... - Bruce E. Shapiro

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