Javon's Flower Growth – Linear Model (Weekly Growth)

Observed data - Weeks (x) and Flower Height (y): new observed points include (1, 2) and (2, 5). - These data show a constant increase in height as x increases by 1 each week, suggesting a linear relation. ### Determine the linear model - Goal: find a linear function in the form $y = mx + b$ that fits the new data. - Compute the slope using two points: - $m = \frac{y<em>2 - y</em>1}{x<em>2 - x</em>1} = \frac{5 - 2}{2 - 1} = \frac{3}{1} = 3.$ - Solve for the intercept using a point, e.g. (1,2): - $2 = 3 \cdot 1 + b \Rightarrow b = 2 - 3 = -1.$ - Therefore the new model is $y = 3x - 1.$ ### Verify the model against all data points - For (x, y) = (1, 2): $y = 3\cdot 1 - 1 = 2$

✓ - For (x, y) = (2, 5): $y = 3\cdot 2 - 1 = 5$

✓ ### Interpretation of the model - Slope (m) = 3 units per unit of x, representing the rate of change. - Intercept (b) = -1 units, representing the height when x = 0 (extrapolated; not observed in data). - Each additional unit of x increases y by 3 units. ### Answer choices cross-check - Based on the calculated model, $y = 3x - 1$ is the correct form. ### Final result - The equation that represents the relationship between x and y for these data points is $y = 3x - 1.$ ### Real-world considerations and notes - The data suggest linear growth over the observed points; this model may not hold for much larger x if growth changes rate. - Extrapolating to x = 0 yields y = -1, which may not be physically meaningful depending on the context, but is mathematically valid for the line through the observed points. - This is a practical example of using the slope-intercept form $y = mx + b$ to model a real-world process where the slope is the rate of change and the intercept is the initial condition (or extrapolated value).