Answer:
The required equation is
.
Explanation:
Consider the provided data.
We need to find a quadratic model.
Quadratic polynomial can be written as:
![at^{2}+bt+c=P](https://tex.z-dn.net/?f=at%5E%7B2%7D%2Bbt%2Bc%3DP)
Here, <em>t</em> represents time and <em>P</em> represents population.
Consider the given data,
At <em>t</em> = 0 the population <em>P</em> = 5.1.
Substitute <em>t</em> = 0 and <em>P</em> = 5.1 in above quadratic polynomial.
![a(0)^{2}+b(0)+c=5.1](https://tex.z-dn.net/?f=a%280%29%5E%7B2%7D%2Bb%280%29%2Bc%3D5.1)
![c=5.1](https://tex.z-dn.net/?f=c%3D5.1)
From the given data, at <em>t</em> = 1 the population <em>P</em> = 3.03.
Substitute <em>t</em> = 1, <em>c </em>= 5.1, and <em>P</em> = 3.03 in quadratic polynomial.
![a(1)^{2}+b(1)+5.1=3.03](https://tex.z-dn.net/?f=a%281%29%5E%7B2%7D%2Bb%281%29%2B5.1%3D3.03)
![a+b+5.1=3.03](https://tex.z-dn.net/?f=a%2Bb%2B5.1%3D3.03)
![a+b=-2.07](https://tex.z-dn.net/?f=a%2Bb%3D-2.07)
![a=-2.07-b](https://tex.z-dn.net/?f=a%3D-2.07-b)
From the given data, at <em>t</em> = 2 the population <em>P</em> = 1.72.
Substitute <em>t</em> = 2, <em>c </em>= 5.1, and <em>P</em> = 1.72 in quadratic polynomial.
![a(2)^{2}+b(2)+5.1=1.72](https://tex.z-dn.net/?f=a%282%29%5E%7B2%7D%2Bb%282%29%2B5.1%3D1.72)
![4a+2b+5.1=1.72](https://tex.z-dn.net/?f=4a%2B2b%2B5.1%3D1.72)
![4a+2b=-3.38](https://tex.z-dn.net/?f=4a%2B2b%3D-3.38)
Now, substitute the value of <em>a</em> in above equation.
![4(-2.07-b)+2b=-3.38](https://tex.z-dn.net/?f=4%28-2.07-b%29%2B2b%3D-3.38)
![-8.28-4b+2b=-3.38](https://tex.z-dn.net/?f=-8.28-4b%2B2b%3D-3.38)
![-2b=-3.38+8.28](https://tex.z-dn.net/?f=-2b%3D-3.38%2B8.28)
![-2b=4.9](https://tex.z-dn.net/?f=-2b%3D4.9)
![b=-2.45](https://tex.z-dn.net/?f=b%3D-2.45)
Substitute
in
.
![a=-2.07-(-2.45)](https://tex.z-dn.net/?f=a%3D-2.07-%28-2.45%29)
![a=-2.07+2.45](https://tex.z-dn.net/?f=a%3D-2.07%2B2.45)
![a=0.38](https://tex.z-dn.net/?f=a%3D0.38)
Thus, the value of <em>a</em> = 0.38, b = -2.45, and c = 5.1.
Therefore, the required equation is
.