Please display the steps of how you solved a and b

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

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

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

o

b. Use the model to estimate the population of bacteria at 9 hours. Type your answer below.

Question

Please display the steps of how you solved a and b

In an experiment, a petri dish with a colony of bacteria is exposed to cold temperatures and then warmed again.

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


a.	Find a quadratic model for the data in the table. Type your answer below.

o	

b. Use the model to estimate the population of bacteria at 9 hours. Type your answer below.

whpalmer4 · Answer

General form for a quadratic is $$y = ax^2 +bx + c$$although in this case we might do $$p = at^2+bt+c$$
We need to find values for $a,b,c$ to fit our data table.  We'll plug in select values from the table to create a system of equations we can solve to find those values.

The most obvious value to plug in is $t=0$:

$$5.1 = a(0)^2+b(0)+c$$That gives us a value for $c$.

Now plug in $t = 1$ and our value for $c$:
$$3.03 = a(1)^2+b(1) + c$$
That gives us an equation in terms of $a$ and $b$ (remember that we've replaced $c$ with the numeric value we found in the first step)

Repeat with another point in the table to get a second equation in terms of $a$ and $b$.  Solve that system of two equations for the values of $a$ and $b$, write out your completed equation, and use it to predict the population at $t = 9$.

anonymous · Answer

I don't know what's going on for a but 


for B do I basically just plug in 9 as x or t to get the estimated population?

whpalmer4 · Answer

Let's work through it, then.

at t = 0, the population is 5.1, so
$$5.1 = at^2+bt+c$$$$5.1=a(0)^2+b(0)+c$$$$5.1=0+0+c$$$$5.1=c$$Agreed?

anonymous · Answer

Agreed

whpalmer4 · Answer

That makes our new, improved formula
$$p(t) = at^2+bt+5.1$$Let's plug in another value, how about t=1, p(1) = 3.03:
$$3.03 = a(1)^2+b(1)+5.1$$$$3.03= a+b+5.1$$
Set that equation aside for the moment, but we'll use it again shortly.

Plug in yet another value, how about t = 2, p(2) = 1.72
$$1.72 = a(2)^2+b(2) + 5.1$$$$1.72=4a+2b+5.1$$

Now we have a system of two equations in two unknowns to solve:

$$3.03 = a + b + 5.1$$$$1.72=4a+2b+5.1$$I would suggest multiplying the first equation by -2, and adding them together.  Then you'll have an equation in only one variable, which is easily solved.

whpalmer4 · Answer

Do that and report what you get...

anonymous · Answer

>.< I don't know what exactly I'm multiplying... I mean you said to multiply the first equation by -2 but I don't see where that... Am I multiplaying like this? 3.03(-2) = a(-2) + b(-2) + 5.1(-2)

Because I really don't see any other way to do it...

whpalmer4 · Answer

Yes, that's exactly what I meant.   We're solving by elimination.  You could also solve the first equation for $a$, and then substitute what you got for $a$ in the other equation, if that's more familiar (seems like more work to me, but I'm not the one doing it!)

If you went ahead as I suggested, you would get:
$$-6.06 -2a-2b-10.2$$$$1.72=4a+2b+5.1$$Adding them together
$$1.72-6.06 = 4a-2a+2b-2b+5.1-10.2$$$$1.72-6.06=2a+5.1-10.2$$and I know you can solve that.  Take the value you get for $a$ and plug it back into either of the equations (I would suggest $3.03 = a + b + 5.1$ is the easiest to use) to find the value of $b$.

Write out $p(t) = at^2+bt+c$ with your values of $a,b,c$ in place of the letters and you're done with the first part.

For the second part, you just evaluate $p(9) = a(9)^2+b(9) +c$ with your values of $a,b,c$.