please help! I need to pass algebra to graduate!
widget wonders produces widgets. They have found that the cost, c(x), of making x widgets is a quadratic function in terms of x.
the company also discovered that it cost $16 to produce 4 widgets, ans $48 to produce 10 widgets.
find the total cost of producing 8 widgets.

- anonymous

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- schrodinger

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- Curry

So 16 = a(4)^2 - b4 -c
and
48 = a(10)^2 -b(10) - c

- Curry

Correction, those minus signs should be positive. so solve for those two equations, and the Cs cancel immediately. so your'e left with 16 =16a+4b. and 48 = 100a+10b.

- Curry

you can figure out a and b frmo there, and then you can find c once you have a and b. Lastly, when you have the three constants, (a,b,c), you can just plug in 8 for x into the quadratic eqauation. let me know if you need more help.

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## More answers

- anonymous

yeah Im so lost I don't even know what this means :(

- anonymous

@Curry

- anonymous

do you still need help?

- anonymous

yes!

- anonymous

ok so can you tell me what the general form of a quadratic function is?

- anonymous

if not thats ok, i need to see what you know and what you dont

- anonymous

I don't

- anonymous

ok, the general form of a quadratic function is this
ax^2 + bx + c
a = some number that is not 0
b = some number
c = some number
for example 2x^2 + 5x + 4

- anonymous

|dw:1433582094129:dw|

- anonymous

those two are equivalent, the one on the left is a way of writing it that we can easily type with a keyboard

- anonymous

so like
a= 1/2
b= -2
c= 18
this is what I got so far

- anonymous

ok so if
a = 1/2
b = -2
and c = 18
then what would the general form look like?

- anonymous

1/2x^2+(-2x)+18?

- anonymous

perfect

- anonymous

and simplying you get 1/2x^2 - 2x + 18

- anonymous

ok good i think you are ready to set up the problem

- anonymous

ok so now it says the cost, c(x), of making x widgets is quadratic, so lets write it out
c(x) = ax^2 + bx + c
x is the input of the function, c(x) is the output of the function

- anonymous

does this make sense?

- anonymous

so you will input the number of widgets, x, into the function and the output will be the cost, c(x)

- anonymous

does all of this make sense so far?

- anonymous

can you set it up for me this is a little confusing

- anonymous

you should set it up, so that when you are taking a test, you can do it yourself

- anonymous

cost = c(x) = output
number of widgets = x = input

- anonymous

cost is the same thing as c(x)
number of widgets is the same thing as x

- anonymous

ok so lets start with the first example

- anonymous

they said the cost was 16, so can you plug the 16 into this equation?
c(x) = ax^2 + bx + c

- anonymous

so c(x) = 16x^2 -2x +18

- anonymous

no remember cost is the same thing as c(x)

- anonymous

if cost is 16
and cost is c(x)
what is c(x)?

- anonymous

16

- anonymous

perfect, so c(x) = 16 lets plug it in

- anonymous

c(x) = ax^2 + bx + c
now im plugging in c(x) = 16
16 = ax^2 + bx + c

- anonymous

ok yes that makes sense

- anonymous

good

- anonymous

so now remember widgets is the same thing as x, and it says it costs 16 to produce 4 widgets

- anonymous

so we have
16 = ax^2 + bx + c
can you plug 4 in here in the correct place?

- anonymous

x is the number of widgets
the number of widgets is 4
so x must be?

- anonymous

and what about a b and c?

- anonymous

dont worry about those yet

- anonymous

step by step

- anonymous

so whats our equation look like now?

- anonymous

16=a(4)^2 -2(4) +18?

- anonymous

perfect

- anonymous

ok now set up another equatoin in the exact same way, with the other information
cost is 48 to produce 10 widgets

- anonymous

oh actually that last one should be 16 = a(4)^2 + b(4) + c sorry

- anonymous

yes I thought so too

- anonymous

can you set up the other equatoin now?

- anonymous

cost is 48 to produce 10 widgets, plug the given values into the equation
c(x) = ax^2 + bx + c

- anonymous

48= a(10^2) -b(10) +c

- anonymous

very close, except its +b(10) not -b(10)

- anonymous

sorry I got mixed up because b is -2

- anonymous

ok perfect so we have these two equations:
16 = a(4)^2 + b(4) + c
48 = a(10)^2 + b(10) + c

- anonymous

all good

- anonymous

can you simplify those two equatoins?

- anonymous

I don't know how

- anonymous

ok i will do one you do the other

- anonymous

16 = a(4)^2 + b(4) + c
simplifies to this:
16 = a(16) + b(4) + c
written in a standard way looks like this:
16 = 16a + 4b + c

- anonymous

alright that was too easy all you did was square the 4

- anonymous

exactly

- anonymous

ok so we have these two equations:
16a + 4b + c = 16
100a + 10b + c = 48

- Curry

well actually, hold on.

- anonymous

48= a(10^2) +b10 +c
48= a100 +b10 +c

- anonymous

perfect :)

- anonymous

so now we have the two equations that @Curry posted in the first post

- anonymous

so now we have to find the values of a, b, and c that make those two equations true, do you know how to do that?

- anonymous

nope

- Curry

wait, are we missing the initial condition value?

- Curry

that's only 2 equations. we have three unknowns.

- Curry

hey @billj5, i calculated it ignoring the c too. And i got 36.2 as i posted earlier.

- Curry

But strictly speaking, we can't jsut assume that c is 0. if you did, then yes, you can just ignore c. and the answer would be 36.2

- anonymous

ok are you still here sarah?

- anonymous

brb i need to grab something to eat

- anonymous

let me know if you are still here sarah and we can finish the problem

- anonymous

yes

- anonymous

ok great, are you familiar with solving systems of equations?

- anonymous

yes!

- anonymous

since this is a cost function and they didnt give us any information about initial cost before you take into account the number of widgets, lets assume that c = 0 ok?

- anonymous

so we have these two equations:
16a + 4b = 16
100a + 10b = 48
so can you solve that system for a and b?

- anonymous

@Curry you brought up a good point, sorry, I think we can assume c = 0

- anonymous

let me know the values you get for a and b

- anonymous

ok so what do I solve for a?

- anonymous

you solve the system of those 2 equations for
a
and
b

- Curry

ye, that's what i did too. Assumed c = 0. This is a bad problem, considering they didn't give us an initial condition. Hence, we're basically forced to omit C. which defeats half the purpose of a quadratic equation.

- anonymous

yes, i agree @Curry

- anonymous

sarah please let me know what you get for the values of
a
and
b

- anonymous

a=-b/4+1
b=-4a+4
now the seconf one

- anonymous

yes what you wrote is correct, however you need to find the number value for both
a and b

- anonymous

are you familiar with solving systems of equations by substitution or elimination or both?

- anonymous

hey im running out of time can you tell me the answer and then we can continue working on it so I can learn? and no im not familiar with those

- anonymous

well i think @Curry posted the answer above

- anonymous

i didnt check if it was correct though

- anonymous

can you check and tell me because this test is gonna end and I need to pass. then we can continue learning

- anonymous

what did he say was the answer?

- anonymous

it was wrong... I failed.. and its 3 am im tired im just gonna cut and go to sleep ;c

- anonymous

fill in

- Curry

problem with rounding probably...

- Curry

unless they assumed a value for c or something.

- math&ing001

@Curry @billj5 Here the c is the cost we're looking for.
A quadratic equation looks like this ax^2+bx+c=0 hence c=ax^2+bx
Since c is a function of x, we can write it like this c(x)=ax^2+bx.
The rest is pretty much what you said:
16a + 4b = 16
100a + 10b = 48
When solved you get a=2/15 and b=54/15
Calculating the cost for 8 widgets:
c(8) = 2/15 * 8^2 + 54/15 * 8 = $37.33

- anonymous

@math&ing001, b is 52/15 and c is not the cost we are looking for, c(x) is the cost we are looking for, c is a constant in the quadratic function which is not what we are looking for, also, if you did what you said it would be -c=ax^2+bx not c=ax^2+bx

- math&ing001

Here the c is not a constant and is a function of x hence the notation c(x). You're right b=52/15 which makes it $36.27. As for the minus sign, because it's a variable, writing c or -c is absolutely the same.

- anonymous

@math&ing001, no you are confusing c in the general form of a quadratic which is a constant term and c(x) which is the function itself, and c is not the same as -c

- math&ing001

I'm just trying to teach you something new, but if you're not open to learning then there's not much I can do about it.

- anonymous

@math&ing001, no you are giving wrong information, but thank you

- math&ing001

Just because you don't understand it, doesn't make it wrong. Anyways, I'm not going to keep this conversation going, it's not taking us anywhere. Have a good day !

- anonymous

@math&ing001, please have any of the mods come in here and verify what you are saying.. i promise that you are wrong, and im very open to learning by the way

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