We need to find the quadratic equation coefficients a, b, c, where:

F(t)=at²+bt+c.

QUADRATIC REGRESSION
**t** |
**F(t)** |
**t²** |
**t³** |
**t⁴** |
**tF(t)** |
**t²F(t)** |

0.75 |
0.8 |
0.5625 |
0.421875 |
0.31640625 |
0.6 |
0.45 |

2 |
1.3 |
4 |
8 |
16 |
2.6 |
5.2 |

2.5 |
1.2 |
6.25 |
15.625 |
39.0625 |
3 |
7.5 |

4 |
1.6 |
16 |
64 |
256 |
6.4 |
25.6 |

6 |
1.7 |
36 |
216 |
1296 |
10.2 |
61.2 |

8 |
1.8 |
64 |
512 |
4096 |
14.4 |
115.2 |

8.5 |
1.7 |
72.25 |
614.125 |
5220.0625 |
14.45 |
122.825 |

31.75 |
10.1 |
199.0625 |
1430.171875 |
10923.44140625 |
51.65 |
337.975 |

System of equations is:

(1) 10923.44140625a+1430.171875b+199.0625c=337.975, where

a∑t⁴+b∑t³+c∑t²=∑(t²F(t));

(2) 1430.171875a+199.0625b+31.75c=51.65, where

a∑t³+b∑t²+c∑t=∑(tF(t));

(3) 199.0625a+31.75b+7c=10.1, where

a∑t²+b∑t+nc=∑F(t), and n=7, the data size.

The arithmetic is very tedious to solve for a, b, c, but the technique is to eliminate c between (1) and (2) to create two equations (4) and (5) containing a and b only. From these a and b are found by eliminating one or other and then using that value to calculate the remaining value. Equation (3) can then be used to calculate c. This done we get:

F(t)=-0.0243t²+0.3391t+0.5968.

F(9.25)=1.6516 approx.