To determine which equation best models the distance of the rock over time, we need to analyze the data from the table and identify the quadratic nature of the relationship, as the distance decreases after reaching a peak.
Let's examine the provided options:
- y = −1.5x² + 4.6x + 14.7
- y = −0.4x + 5.3
- y = −2.4x² + 5.6x + 14.9
- y = −1.5x² + 4.6x + 14.8
Given that the rock was thrown from a height of 15 feet, the starting point (at x = 0 seconds) should correspond to 15 feet, which matches the y-intercept in the quadratic equations.
To find the correct model, we should look for the quadratic equation that fits the points in the table.
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Initial Distance (t = 0): All quadratic equations have similar intercepts (15 feet).
- Verify how well these equations track the data at points t = 1, t = 2, t = 4, and t = 5.
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After substituting x values from the table into the equations provided, we can check if the calculated distance matches or closely aligns with the known distances.
For example:
- For the first equation, substituting x = 1 yields: \( y = -1.5(1)^2 + 4.6(1) + 14.7 = 17.8 \) (does not match)
- Substituting x = 2 gives: \( y = -1.5(2)^2 + 4.6(2) + 14.7 = 19.6 \) (does not match)
After testing each equation with the values in the table: It seems that the most plausible quadratic model should decrease in a similar fashion to the points provided up until the rock hits the water (0 feet), which is represented well by the first or fourth functions.
After checking closely and rounding properly, option: y = -1.5x² + 4.6x + 14.8 matches adequately through calculation and trend.
Thus, the answer is:
y = −1.5x² + 4.6x + 14.8