Question

A person dives into a pool from its edge to swim to the other side. The table shows the depth in feet of the person from the surface of the water after x seconds. The data can be modeled by a quadratic function.



Which function best models the data?
heres the table
time,x(seconds) depth of person frrom surface of water,d(x) (feet)
1 -1.85
4 -7.28
6 -8.3
8.5 -6.65
10 -4.1
11.5 -0.38

answer choices
d(x)=0.05x^2-0.74x
d(x) =0.05x^2+0.74x+9.17
d(x) =0.26x^2-4.11x
d(x)=0.26x^2-3.11x+1

Answer 1

To find the quadratic function that best models the data, we substitute the given data points into each of the answer choices and choose the one that gives the closest values:

For d(x) = 0.05x^2 - 0.74x:
When x = 1, d(1) = 0.05(1)^2 - 0.74(1) = 0.05 - 0.74 = -0.69
When x = 4, d(4) = 0.05(4)^2 - 0.74(4) = 0.05(16) - 0.74(4) = 0.8 - 2.96 = -2.16
When x = 6, d(6) = 0.05(6)^2 - 0.74(6) = 0.05(36) - 0.74(6) = 1.8 - 4.44 = -2.64
When x = 8.5, d(8.5) = 0.05(8.5)^2 - 0.74(8.5) = 0.05(72.25) - 0.74(8.5) = 3.6125 - 6.29 = -2.6775
When x = 10, d(10) = 0.05(10)^2 - 0.74(10) = 0.05(100) - 0.74(10) = 5 - 7.4 = -2.4
When x = 11.5, d(11.5) = 0.05(11.5)^2 - 0.74(11.5) = 0.05(132.25) - 0.74(11.5) = 6.625 - 8.51 = -1.885

For d(x)= 0.05x^2 + 0.74x + 9.17:
When x = 1, d(1) = 0.05(1)^2 + 0.74(1) + 9.17 = 0.05 + 0.74 + 9.17 = 9.96
When x = 4, d(4) = 0.05(4)^2 + 0.74(4) + 9.17 = 0.05(16) + 0.74(4) + 9.17 = 0.8 + 2.96 + 9.17 = 12.93
When x = 6, d(6) = 0.05(6)^2 + 0.74(6) + 9.17 = 0.05(36) + 0.74(6) + 9.17 = 1.8 + 4.44 + 9.17 = 15.41
When x = 8.5, d(8.5) = 0.05(8.5)^2 + 0.74(8.5) + 9.17 = 0.05(72.25) + 0.74(8.5) + 9.17 = 3.6125 + 6.29 + 9.17 = 18.0725
When x = 10, d(10) = 0.05(10)^2 + 0.74(10) + 9.17 = 0.05(100) + 0.74(10) + 9.17 = 5 + 7.4 + 9.17 = 21.57
When x = 11.5, d(11.5) = 0.05(11.5)^2 + 0.74(11.5) + 9.17 = 0.05(132.25) + 0.74(11.5) + 9.17 = 6.625 + 8.51 + 9.17 = 24.305

For d(x)=0.26x^2-4.11x:
When x = 1, d(1) = 0.26(1)^2 - 4.11(1) = 0.26 - 4.11 = -3.85
When x = 4, d(4) = 0.26(4)^2 - 4.11(4) = 0.26(16) - 4.11(4) = 4.16 - 16.44 = -12.28
When x = 6, d(6) = 0.26(6)^2 - 4.11(6) = 0.26(36) - 4.11(6) = 9.36 - 24.66 = -15.3
When x = 8.5, d(8.5) = 0.26(8.5)^2 - 4.11(8.5) = 0.26(72.25) - 4.11(8.5) = 18.77 - 34.935 = -16.165
When x = 10, d(10) = 0.26(10)^2 - 4.11(10) = 0.26(100) - 4.11(10) = 26 - 41.1 = -15.1
When x = 11.5, d(11.5) = 0.26(11.5)^2 - 4.11(11.5) = 0.26(132.25) - 4.11(11.5) = 34.495 - 47.265 = -12.77

For d(x)=0.26x^2-3.11x+1:
When x = 1, d(1) = 0.26(1)^2 - 3.11(1) + 1 = 0.26 - 3.11 + 1 = -1.85
When x = 4, d(4) = 0.26(4)^2 - 3.11(4) + 1 = 0.26(16) - 3.11(4) + 1 = 4.16 - 12.44 + 1 = -7.28
When x = 6, d(6) = 0.26(6)^2 - 3.11(6) + 1 = 0.26(36) - 3.11(6) + 1 = 9.36 - 18.66 + 1 = -8.3
When x = 8.5, d(8.5) = 0.26(8.5)^2 - 3.11(8.5) + 1 = 0.26(72.25) - 3.11(8.5) + 1 = 18.77 - 26.335 + 1 = -6.565
When x = 10, d(10) = 0.26(10)^2 - 3.11(10) + 1 = 0.26(100) - 3.11(10) + 1 = 26 - 31.1 + 1 = -4.1
When x = 11.5, d(11.5) = 0.26(11.5)^2 - 3.11(11.5) + 1 = 0.26(132.25) - 3.11(11.5) + 1 = 34.495 - 35.865 + 1 = -0.37

Based on the results, the function d(x)=0.26x^2-3.11x+1 best models the data as it provides values closest to the given data points.