Finite Differences Linear Modeling WorksheetAR.2C determine the function that models a given table of related values using finite differences Name: Date: a = slope • b = y-intercept • c = equation 1. Write an equation in \(y = mx + b\) form for the linear function described by the table. xy1-3.53-0.541.064.098.5 a. b. c. 2. Write an equation in \(y = mx + b\) form for the linear function described by the table. xy25.634.452.08-1.610-4.0 a. b. c. 3. Write an equation in \(y = mx + b\) form for the linear function described by the table. xy4-1.06-0.270.2101.4122.2 a. b. c. 4. Write an equation in \(y = mx + b\) form for the linear function described by the table. xy2-2.043.055.5813.01018.0 a. b. c.
Finite Differences Answer KeyTeacher Answer Key AR.2C • Finite Differences Linear Modeling Complete Solutions & Grading Guide Quick Scan: #1: \(y = 1.5x - 5\) • #2: \(y = -1.2x + 8\) • #3: \(y = 0.4x - 2.6\) • #4: \(y = 2.5x - 7\) a = slope \((m)\) • b = y-intercept \((b)\) • c = linear equation \((y = mx + b)\) 1. Linear Model Solution \(m > 0\) xy1-3.53-0.541.064.098.5 Finite Diffs: \(\frac{-0.5 - (-3.5)}{3 - 1} = \frac{3.0}{2} = 1.5\) \(\frac{1.0 - (-0.5)}{4 - 3} = \frac{1.5}{1} = 1.5\) \(\frac{4.0 - 1.0}{6 - 4} = \frac{3.0}{2} = 1.5\) a. \(1.5\) (or \(\frac{3}{2}\)) b. -5 \(1.0 = 1.5(4) + b \implies b = -5\) c. \(y = 1.5x - 5\) 2. Linear Model Solution \(m < 0\) xy25.634.452.08-1.610-4.0 Finite Diffs: \(\frac{4.4 - 5.6}{3 - 2} = \frac{-1.2}{1} = -1.2\) \(\frac{2.0 - 4.4}{5 - 3} = \frac{-2.4}{2} = -1.2\) \(\frac{-1.6 - 2.0}{8 - 5} = \frac{-3.6}{3} = -1.2\) a. \(-1.2\) (or \(-\frac{6}{5}\)) b. 8 \(2.0 = -1.2(5) + b \implies b = 8\) c. \(y = -1.2x + 8\) 3. Linear Model Solution Decimal Rate xy4-1.06-0.270.2101.4122.2 Finite Diffs: \(\frac{-0.2 - (-1.0)}{6 - 4} = \frac{0.8}{2} = 0.4\) \(\frac{0.2 - (-0.2)}{7 - 6} = \frac{0.4}{1} = 0.4\) \(\frac{1.4 - 0.2}{10 - 7} = \frac{1.2}{3} = 0.4\) a. \(0.4\) (or \(\frac{2}{5}\)) b. -2.6 \(0.2 = 0.4(7) + b \implies b = -2.6\) c. \(y = 0.4x - 2.6\) 4. Linear Model Solution Steeper Rate xy2-2.043.055.5813.01018.0 Finite Diffs: \(\frac{3.0 - (-2.0)}{4 - 2} = \frac{5.0}{2} = 2.5\) \(\frac{5.5 - 3.0}{5 - 4} = \frac{2.5}{1} = 2.5\) \(\frac{13.0 - 5.5}{8 - 5} = \frac{7.5}{3} = 2.5\) a. \(2.5\) (or \(\frac{5}{2}\)) b. -7 \(3.0 = 2.5(4) + b \implies b = -7\) c. \(y = 2.5x - 7\) Common Student Error: Forgetting to divide \(\Delta y\) by \(\Delta x\) when consecutive \(x\)-values skip intervals (e.g., from \(1\) to \(3\), \(\Delta x = 2\)). TEKS AR.2C
Finite Differences Questioning GuideTeacher Facilitation Guide • TEKS AR.2C DOK Levels 2 – 4 High-Level Mathematical Questions Use these targeted questions during whole-class discussion, small-group facilitation, or individual conferences. Category 1 Conceptual Understanding (Rate of Change) 1. "The \(\Delta y\) values in Problem 1 are \(3.0, 1.5, 3.0,\) and \(4.5\). Why is this table still linear even though the differences in \(y\) are not equal?" Target Reasoning: Linearity requires a constant ratio of \(\frac{\Delta y}{\Delta x}\), not just constant \(\Delta y\). Because \(\Delta x\) varies (\(2, 1, 2, 3\)), each ratio reduces to the same constant unit rate (\(1.5\)). 2. "What does the slope actually tell you about what happens between the rows of the table?" Target Reasoning: It represents the unit rate of change: for every \(1\) unit increase in \(x\), \(y\) changes by exactly \(m\) units, regardless of which two points are chosen. Category 2 Diagnostic & Error Analysis (DOK 3) 3. "A student uses row 1 and row 2 in Problem 2 and calculates the slope as \(m = \frac{3 - 2}{4.4 - 5.6} = -\frac{1}{1.2}\). What mistake was made, and what does this number actually represent?" Target Reasoning: The student inverted the formula (\(\frac{\Delta x}{\Delta y}\) instead of \(\frac{\Delta y}{\Delta x}\)). This gives the change in \(x\) per unit of \(y\) rather than slope. 4. "Does it matter which point from the table you substitute to solve for the \(y\)-intercept \(b\)? What would happen if two different points produced two different \(b\) values?" Target Reasoning: Any valid point must yield the identical \(b\). If different points yield different intercepts, either the calculation was flawed or the function is nonlinear. Category 3 Generalization & Extension (DOK 3 – 4) 5. "If we added a row where \(x = 0\) to any of these tables, how could we find its \(y\)-value using only finite differences without solving an algebraic equation?" Target Reasoning: Work backward from the closest known point by subtracting \(\Delta x \times m\) from the \(y\)-value (e.g., in Problem 1, from \(x=1, y=-3.5\), moving left \(1\) unit gives \(-3.5 - 1.5 = -5\)). 6. "How will our finite differences approach change when we encounter quadratic or exponential functions next?" Target Reasoning: For quadratic functions, first differences will vary linearly while second differences are constant. For exponential functions, consecutive \(y\)-values have a constant , not a constant difference.
