Question

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

Answer 1

Getting the table, one moment

Answer 2

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 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

im confused

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

yes is correct

Answer 14

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

Answer 15

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 \(b\)

Answer 16

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

Answer 17

the part you wrote above is correct

Answer 18

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

Answer 19

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

Answer 20

I dont know how to plug all that in

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

yes c=5.1 @satellite73

Answer 25

no i dont get it

Answer 26

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

Answer 27

yes

Answer 28

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

Answer 29

correct

Answer 30

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

Answer 31

correct

Answer 32

What @LynFran ????

Answer 33

@satellite73 are you back? The serve kicked me off

Answer 34

yea me too

Answer 35

@EllenJaz17 what are you said @LynFran

Answer 36

they interrupted me learning lol

Answer 37

lol

Answer 38

lol

Answer 39

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
\[5.1=a\times 0^2+b\times 0+c\]

Answer 40

is that ok?

Answer 41

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

Answer 42

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

Answer 43

correct

Answer 44

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

Answer 45

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

Answer 46

ok im ready

Answer 47

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

Answer 48

yes

Answer 49

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
\[3.03=a\times 1^2+b\times 1+5.1\]

Answer 50

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

Answer 51

good so far?

Answer 52

yes

Answer 53

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

Answer 54

how we doing so far?

Answer 55

good

Answer 56

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

Answer 57

Sounds good

Answer 58

good @satellite73

Answer 59

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

Answer 60

ok with that?

Answer 61

yes

Answer 62

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

Answer 63

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

Answer 64

quite wrong, it is much easier than that

Answer 65

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

Answer 66

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

Answer 67

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

Answer 68

So a=4?

Answer 69

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

Answer 70

lets back up a second

Answer 71

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 \[a+b=-2.07\]

Answer 72

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

Answer 73

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

Answer 74

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

Answer 75

Yeah im confused

Answer 76

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

Answer 77

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

Answer 78

yes