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
Answers
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Answered by
GPT 3.5
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.
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.
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