Question

Determine whether a quadratic model exists for each set of values. If so, write the model.
f(0)=5, f(2)=3, f(-1)=0

Answer 1

why not?

Answer 2

you want an answer or a method?

Answer 3

i just want to know how to get the answer. I was in class and i understood but its just the way this question is set up.

Answer 4

ok we can go through it carefully if you like
start with
\[f(x)=ax^2+bx+c\] and use the info you got to find \(a,b, c\)

Answer 5

So far I have f(0)=a(0)^2+b(0)^2+c=5

Answer 6

is that how i start?

Answer 7

ok so now you know \(c=5\)

Answer 8

yes

Answer 9

this is where i stopped because i dont know where to go

Answer 10

repeat for the other two, but you will have to use two equations and two unknowns \(a\) and \(b\)

Answer 11

\[f(x)=ax^2+bx+5\] since we got rid of the \(c\) on the first step
\[f(2)=a\times 2^2+b\times 2+5=3\\
4a+2b+5=3\\
4a+2b=-2\] is one equation

Answer 12

repeat for \(x=-1\) and get another equation in \(a\) and \(b\)
clear?

Answer 13

So..
a(-1)^2+b(-1)^2+5=0?

Answer 14

right

Answer 15

oh not right !

Answer 16

\[a(-1)^2+b(-1)+5=0\] is more like it
no square on the second term

Answer 17

?

Answer 18

you wrote \[a(-1)^2+b(-1)^2+5=0\] but that is not right, because you had a square on the second term

Answer 19

it is
\[f(x)=ax^2+bx+5\]

Answer 20

no square on the second term

Answer 21

Only on the first part because the standard form is ax^2+bx+c?

Answer 22

right

Answer 23

so the second equation is
\[a-b=-5\]

Answer 24

and the system is
\[4a+2b=-2\\
a-b=-5\] which is not too hard to solve, either by elimination or substitution

Answer 25

btw i hope it is clear that
\[a(-1)^2+b(-1)+5=0\]is the same as
\[a-b=-5\]

Answer 26

O

Answer 27

is this the third part or the second?

Answer 28

nevermind

Answer 29

so what happens to the 3 and 0?

Answer 30

what do you mean?
oh i see hold on

Answer 31

we got this \[a(-1)^2+b(-1)+5=0\] right?

Answer 32

the f(2)=3 , f(-1)=0 .... what happens to the 3 and 0?

Answer 33

the zero is in \[a(-1)^2+b(-1)+5=0\] on the right
then compute and get
\[a-b+5=0\]

Answer 34

subtracting \(5\) gives
\[a-b=-5\]

Answer 35

\[f(x)=ax^2+bx+5\] \[f(2)=a\times 2^2+b\times 2+5=3\] i.e.
\[4a+2b+5=3\] the 3 is on the right
subtract 5 get
\[4a+2b=-2\]

Answer 36

im going to look at it for a minute..

Answer 37

take your time
maybe i skipped on step

Answer 38

\[f(2)=a\times 2^2+b\times 2+5=3\]
\[a\times 4+b\times 2+5=3\\
4a+2b+5=3\\
4a+2b=-2\]

Answer 39

OK that makes better sense

Answer 40

lol it is the old commutative law that confuses every time!

Answer 41

so what are the equations supposed to look like?

Answer 42

it was so long ago i forget
we just got
\[4a+2b=-2\] as one of them right ?

Answer 43

yes

Answer 44

aah the other is
\[a(-1)^2+b(-1)+5=0\] and i am going to let you figure out why that is the same as
\[a-b=-5\]

Answer 45

system to solve is
\[4a+2b=-2\\

a-b=-5\]

Answer 46

subtract 5 on both sides so it becomes a - and -1 x -1 =1 2 which is just 'a' and do the same for b then im left with a-b=-5... right?

Answer 47

yes

Answer 48

can i use substitution?

Answer 49

can i use substitution?

Answer 50

wait hold up am i solving for a and b?

Answer 51

im stuck

Answer 52

you can use substitution or elimination, your choice

Answer 53

its 10 at night and im only on the 1st question out of 32....

Answer 54

they get quicker

Answer 55

\[4a+2b=-2\\
a-b=-5\] the second equation is the same as \(a=b-5\) so you can substitute in the first one and solve
\[4(b-5)+2b=-5\]

Answer 56

doing it right now ;)

Answer 57

b=3

Answer 58

damn i made a typo

\[4(b-5)+2b=-2\] is more like it
yes \(b=3\)

Answer 59

and so \(a=-2\) and your answer is
\[f(x)=-2x^2+3x+5\] want to check it?

Answer 60

just wait

Answer 61

where am i plugging the 3 at

Answer 62

either of these \[4a+2b=-2\\
a-b=-5\]

Answer 63

nevermind i just got confused cause i got to0 much on my paper

Answer 64

\[a-3=-5\iff a=-2\]

Answer 65

so the question was asking me to determine whether a quadratic model exists, how do i know if it does?

Answer 66

that is a dumb question
unless it was a line or not a function
you have three equations and three unknowns
you can almost always solve that

Answer 67

i don't mean you asked a dumb question, i mean the question that you were asked was dumb

Answer 68

if i said \[f(0)-5, f(1)=4,f(3)=6\] i could make a quadratic, out of it
the coefficients might be annoying fractions, not like the simple answer we got, but i could do it

Answer 69

for the next problem its f(-1)=-4, f(1)=-2, f(2)=-1 and the answer says that there is no model.

Answer 70

hmm let me see if i can figure out why

Answer 71

ok

Answer 72

first lets cheat and see that our last answer was correct

Answer 73

http://www.wolframalpha.com/input/?i=quadratic+%280%2C5%29%2C%282%2C3%29%2C%28-1%2C0%29

Answer 74

nice!

Answer 75

now that you know how to cheat, lets do it for the one it says can't be done

Answer 76

k

Answer 77

oooh doe it is a line!!

Answer 78

yeah

Answer 79

so in order to know would i just graph it?

Answer 80

i guess
i wasn't thinking
i suppose you could try to make it
\[f(x)=ax^2+bx+c\] as before and discover that \(a=0\)

Answer 81

now if you really have 30 of these to do (that seems like an awful lot) i would cheat my brains out

Answer 82

when i graph and it turns out as a parabola then a model exists?

Answer 83

lol

Answer 84

hate being an honors student

Answer 85

yes

Answer 86

think of all the honors you'll get
not to mention the fact that you avoid sitting in a class full of "i don't get it"s

Answer 87

yeah lol well just one last question how would i graph the one we spent a long time working out? i know its supposed to become a parabola but when i graph it it becomes funky looking.

Answer 88

there is a picture in the link i sent
use that

Answer 89

O well thank you! Cant believe you were willing to sit here for an hour+ showing me to work a problem. Again appreciate the help now im going to get these other 31 problems done lol

Answer 90

@satellite73