In 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. Show work

- anonymous

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- anonymous

Getting the table, one moment

- anonymous

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

- anonymous

how are you supposed to do this (just asking because you have to "show work")

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## More answers

- anonymous

I have no clue lol I typed everything the question asked and the table

- DecentNabeel

The general formula for quadratic eqn is y = ax^2 + bx + c
take any 3 points from the table
(0, 5.1)
(1, 3.03)
(5, 2.35)
plug these points into the above equation and you will get 3 different equations
use substitution or any method that you know to solve these 3 eqns with 3 unknowns

- anonymous

then use a calculator
http://www.wolframalpha.com/input/?i=quadratic+%280%2C5.1%29%2C%281%2C3.03%29%2C%282%2C1.72%29%2C%283%2C1.17%29%2C%284%2C1.38%29%2C%285%2C2.35%29%2C%286%2C4.08%29

- DecentNabeel

a point is written in the form (x,y) so plug 0 for x and 5.1 for y to get 1st eqn. follow same method to get other 2 eqns

- anonymous

@DecentNabeel i am not sure that would work, since it is not exact

- anonymous

im confused

- anonymous

So this part is correct?
The general formula for quadratic eqn is y = ax^2 + bx + c
take any 3 points from the table
(0, 5.1)
(1, 3.03)
(5, 2.35)
plug these points into the above equation and you will get 3 different equations
use substitution or any method that you know to solve these 3 eqns with 3 unknowns

- DecentNabeel

@satellite73 many way for solve i solve another but this is true

- anonymous

if it is exact you can do it using only three points
\[(0,5.1),(1,3.03),(2,2.35)\] or any other 3

- DecentNabeel

yes is correct

- anonymous

also you do not have to solve a system of three equations, only two

- anonymous

you want
\[ax^2+bx+c\] but sine \((0,5.1\) is on the graph, you have \(c=5.1\) right away, only need \(a\) and \(b\)

- anonymous

So lets write step one so i can start writing it so i can show our work lol

- anonymous

the part you wrote above is correct

- DecentNabeel

5.1 = a*0^2 + b*0 + c
3.03 = a*1^2 + b*1 + c
2.35 = a*5^2 + b*5 + c
solve of a, b and c

- anonymous

what @DecentNabeel said solve that system
but really it is not as hard as it looks since the first one tells you \(c=5.1\)

- anonymous

I dont know how to plug all that in

- anonymous

oops second one is
\[3.03=a+b=5.1\] so \[a+b=-2.07\]

- anonymous

do you see how @DecentNabeel got this one
\[5.1 = a*0^2 + b*0 + c\]

- anonymous

if the answer is "no" that is fine, just say so

- DecentNabeel

yes c=5.1 @satellite73

- anonymous

no i dont get it

- anonymous

ok do you get that you are trying to find \(a,b\) and \(c\) to make up
\[y=ax^2+bx+c\]?

- anonymous

yes

- anonymous

ok and from the table you see that \((0,5.1)\) is on the graph right?

- anonymous

correct

- anonymous

that means if \(x=0\) then \(y=5.1\) right ?

- anonymous

correct

- DecentNabeel

What @LynFran ????

- anonymous

@satellite73 are you back? The serve kicked me off

- anonymous

yea me too

- DecentNabeel

@EllenJaz17 what are you said @LynFran

- anonymous

they interrupted me learning lol

- anonymous

lol

- DecentNabeel

lol

- anonymous

ok so to resume our story, were were here:
we want \(a,b,c\) so make \(y=ax^2+bx+c\) and we know if \(x=0\) then \(y=5.1\)
plug them in and get
\[5.1=a\times 0^2+b\times 0+c\]

- anonymous

is that ok?

- anonymous

yes i get that one because its easy because x is 0 lol

- anonymous

right so that tells us right away that \(c=5.1\) yes?

- anonymous

correct

- anonymous

ok now \(a\) and \(b\) to go

- anonymous

@LynFran send him a message, im trying to get hw help, not get interrupted

- anonymous

ok im ready

- anonymous

the next point we have is \((1,3.03)\) right ?

- anonymous

yes

- anonymous

now don't forget, we already know \(c=5.1\) so we have
\[y=ax^2+bx+5.1\] put \(x=1\) and \(y=3.03\) to get
\[3.03=a\times 1^2+b\times 1+5.1\]

- anonymous

this also cleans up nicely, since \(1^2=1\) we have
\[3.03=a+b+5.1\]

- anonymous

good so far?

- anonymous

yes

- anonymous

now to make it neater, subtract \(5.1\) from both sides
that gives
\[-2.07=a+b\] or
\[a+b=-2.07\] and we put that aside for the moment

- anonymous

how we doing so far?

- anonymous

good

- anonymous

|dw:1435898261868:dw|

- anonymous

ok now one more equation to go
you can pick any point you like but i will pick \((2,1.72)\)

- anonymous

Sounds good

- DecentNabeel

good @satellite73

- anonymous

again we have
\[y=ax^2+bx+5.1\] put \(x=2,y=1.72\) and get
\[1.72=a\times 2^2+b\times 2+5.1\]

- anonymous

ok with that?

- anonymous

yes

- anonymous

now we clean it up
actually you can
what do you get in terms of \(a,b\) and a number ?

- anonymous

\[-2.07^{2}+1.72+5.1\] ? Im sure that's wrong

- anonymous

quite wrong, it is much easier than that

- anonymous

haha oh jeez. I needed this class apparently very much lol and your help

- anonymous

lets go back to \[1.72=a\times 2^2+b\times 2+5.1\]

- anonymous

your equation will have an \(a\) and a \(b\) in it
all you have to do is compute \(2^2\)

- anonymous

So a=4?

- anonymous

no just \(a\times 4\) or \(4a\)

- anonymous

lets back up a second

- anonymous

remember that when we replace \(x\) by \(1\) and \(y\) by \(3.03\) we ended up with an equation with both an \(a\) and a \(b\) in it
it was \[a+b=-2.07\]

- anonymous

we are trying to accomplish the same thing here with \(x=2\) and \(y=1.72\)
first step is to write
\[1.72=a\times 2^2+b\times 2+5.1\]

- anonymous

so don't do too much work just see what the numbers give you
your equation should still have an \(a\) and a \(b\) in it

- anonymous

if you are still confused, say so and i will spell it out

- anonymous

Yeah im confused

- anonymous

thought so
i am just going to rewrite \[1.72=a\times 2^2+b\times 2+5.1\]

- anonymous

since \(2^2-4\) this is the same as
\[1.72=a\times 4+b\times 2+5.1\] clear?

- anonymous

yes

- anonymous

i meant "since \(2^2=4\)

- anonymous

ok now we just make it look better
\[1.72=4a+2b+5.1\] go that ?

- anonymous

yes

- anonymous

again we subtract \(5.1\) from both sides and arrive finally at
\[4a+2b=-3.38\]

- anonymous

ok

- anonymous

now lets get those two equations together
\[a+b=-2.07\\
4a+2b=-3.38\]

- anonymous

damn \[a+b=-2.07\\
4a+2b=-3.38\] thats better

- anonymous

we have to solve this "system of equations" to find \(a\) and \(b\)

- anonymous

sounds good

- anonymous

do you know how to solve this system?

- anonymous

no

- anonymous

i hate decimals, at this point i would cheat, but you can do whatever you like

- anonymous

do you have to show ALL your steps?

- anonymous

no it just says show work

- anonymous

well we have come this far, lets grind it out

- anonymous

probably the easiest thing to do is rewrite
\[a+b=-2.07\] as
\[b=-2.07-a\] and substitute that in the second equation get
\[4a+2(-2.07-a)=-3.38\]

- anonymous

ok im just writing as you go

- anonymous

multiply out, combine like terms, work all the magic, get
\[a=.33\] i will let you grind that out for yourself, but the answer is \(a=.33\)

- anonymous

so all thats left to find is b, than we do the equation, correct?

- anonymous

now we know that \(a=.33\) and
\[a+b=-2.07\] you have
\[.33+b=-2.07\] so \[b=-2.4\]

- anonymous

yeah that is it , find \(a\) and \(b\)
we get \(a=.33\) and \(b=-2.4\)

- anonymous

and we recall that \(c=5.1\)

- anonymous

dang i made a mistake somewheres \(b=-2.45\) actually but whatever

- anonymous

So .33^2-2.45+5.1

- anonymous

oh damn damn damn i made a stupid typo

- anonymous

everywhere you see \(.33\) replace it by \(.38\)

- anonymous

i told you i hate decimals
when you solve \[4a+2(-2.07-a)=-3.38\] you get \(a=.38\)

- anonymous

and since \(a=.38\) we get \(b=-2.45\)
final answer
\[y=.33x^2-2.45x+5.1\]

- anonymous

your math teacher must hate you to give you such a mess

- anonymous

at any rate, all the steps are there
you can copy them down and look at them at your leisure
took a while, hope you learned something

- anonymous

should it say y=.38 or y=.33 lol

- anonymous

Final Answer
\[\huge \color{red}{y=.38x^2-2.45x+5.1}\]

- anonymous

lol great thank you so much!

- anonymous

|dw:1435900099694:dw|

- anonymous

Do you wanna help me with another question that relates to this table lmao @satellite73

- anonymous

hell no
not without a shot of tequila

- anonymous

ok post i will take a look, maybe it will be easier

- anonymous

ooh to

- anonymous

THIS TABLE ok

- anonymous

|dw:1435900241301:dw|

- anonymous

There you go lmao

- anonymous

good, no lime or salt (for sissies)

- anonymous

bahaha

- anonymous

what is the question i am about to turn in to a pumpkin

- anonymous

@satellite73 sorry are you still there

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