Question

12In an experiment, a petri dish with a colony of bacteria is exposed to cold temperatures and then warmed again.

Find a quadratic model for the data in the table. Type your answer below. Show your work.

Time (hours)
0
1
2
3
4
5
6
Population (1000s)
5.1
3.03
1.72
1.17
1.38
2.35
4.08

Answer 1

The required equation is
`P=0.38t^2-2.45t+5.1`.

Step-by-step explanation:

Consider the provided data.

We need to find a quadratic model.

Quadratic polynomial can be written as:

`at^(2)+bt+c=P`

Here, t represents time and P represents population.

Consider the given data,

At t = 0 the population P = 5.1.

Substitute t = 0 and P = 5.1 in above quadratic polynomial.

`a(0)^(2)+b(0)+c=5.1`

`c=5.1`

From the given data, at t = 1 the population P = 3.03.

Substitute t = 1, c = 5.1, and P = 3.03 in quadratic polynomial.

`a(1)^(2)+b(1)+5.1=3.03`

`a+b+5.1=3.03`

`a+b=-2.07`

`a=-2.07-b`

From the given data, at t = 2 the population P = 1.72.

Substitute t = 2, c = 5.1, and P = 1.72 in quadratic polynomial.

`a(2)^(2)+b(2)+5.1=1.72`

`4a+2b+5.1=1.72`

`4a+2b=-3.38`

Now, substitute the value of a in above equation.

`4(-2.07-b)+2b=-3.38`

`-8.28-4b+2b=-3.38`

`-2b=-3.38+8.28`

`-2b=4.9`

`b=-2.45`

Substitute
`b=-2.45` in
`a=-2.07-b`.

`a=-2.07-(-2.45)`

`a=-2.07+2.45`

`a=0.38`

Thus, the value of a = 0.38, b = -2.45, and c = 5.1.

Therefore, the required equation is
`P=0.38t^2-2.45t+5.1`.

Answer 2

Let the equation be : at^2 + bt + c = P,
where t = time (hrs),
P = population (1000's).
When t = 1, P = 3.03.
When t = 2, P = 1.72.
When t = 3, P = 1.17.
Substitute these into the equation to obtain these 3 simultaneous equations : a + b + c = 3.03
4a + 2b + c = 1.72
9a + 3b = c = 1.17

Solving gives :
a = 0.38,
b = -2.45,
c = 5.1.

The equation is therefore,
P = 0.38t^2 - 2.45t + 5.1
Testing with t = 0 to 6 gives the population values as provided, so it seems to be a valid model.
At t = 9 hrs,
P = 0.38*9^2 - 2.45*9 + 5.1
= 13.83.