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So 16 = a(4)^2 - b4 -c and 48 = a(10)^2 -b(10) - c
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.
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.
yeah Im so lost I don't even know what this means :(
do you still need help?
ok so can you tell me what the general form of a quadratic function is?
if not thats ok, i need to see what you know and what you dont
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
those two are equivalent, the one on the left is a way of writing it that we can easily type with a keyboard
so like a= 1/2 b= -2 c= 18 this is what I got so far
ok so if a = 1/2 b = -2 and c = 18 then what would the general form look like?
and simplying you get 1/2x^2 - 2x + 18
ok good i think you are ready to set up the problem
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
does this make sense?
so you will input the number of widgets, x, into the function and the output will be the cost, c(x)
does all of this make sense so far?
can you set it up for me this is a little confusing
you should set it up, so that when you are taking a test, you can do it yourself
cost = c(x) = output number of widgets = x = input
cost is the same thing as c(x) number of widgets is the same thing as x
ok so lets start with the first example
they said the cost was 16, so can you plug the 16 into this equation? c(x) = ax^2 + bx + c
so c(x) = 16x^2 -2x +18
no remember cost is the same thing as c(x)
if cost is 16 and cost is c(x) what is c(x)?
perfect, so c(x) = 16 lets plug it in
c(x) = ax^2 + bx + c now im plugging in c(x) = 16 16 = ax^2 + bx + c
ok yes that makes sense
so now remember widgets is the same thing as x, and it says it costs 16 to produce 4 widgets
so we have 16 = ax^2 + bx + c can you plug 4 in here in the correct place?
x is the number of widgets the number of widgets is 4 so x must be?
and what about a b and c?
dont worry about those yet
step by step
so whats our equation look like now?
16=a(4)^2 -2(4) +18?
ok now set up another equatoin in the exact same way, with the other information cost is 48 to produce 10 widgets
oh actually that last one should be 16 = a(4)^2 + b(4) + c sorry
yes I thought so too
can you set up the other equatoin now?
cost is 48 to produce 10 widgets, plug the given values into the equation c(x) = ax^2 + bx + c
48= a(10^2) -b(10) +c
very close, except its +b(10) not -b(10)
sorry I got mixed up because b is -2
ok perfect so we have these two equations: 16 = a(4)^2 + b(4) + c 48 = a(10)^2 + b(10) + c
can you simplify those two equatoins?
I don't know how
ok i will do one you do the other
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
alright that was too easy all you did was square the 4
ok so we have these two equations: 16a + 4b + c = 16 100a + 10b + c = 48
well actually, hold on.
48= a(10^2) +b10 +c 48= a100 +b10 +c
so now we have the two equations that @Curry posted in the first post
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?
wait, are we missing the initial condition value?
that's only 2 equations. we have three unknowns.
hey @billj5, i calculated it ignoring the c too. And i got 36.2 as i posted earlier.
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
ok are you still here sarah?
brb i need to grab something to eat
let me know if you are still here sarah and we can finish the problem
ok great, are you familiar with solving systems of equations?
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?
so we have these two equations: 16a + 4b = 16 100a + 10b = 48 so can you solve that system for a and b?
@Curry you brought up a good point, sorry, I think we can assume c = 0
let me know the values you get for a and b
ok so what do I solve for a?
you solve the system of those 2 equations for a and b
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.
yes, i agree @Curry
sarah please let me know what you get for the values of a and b
a=-b/4+1 b=-4a+4 now the seconf one
yes what you wrote is correct, however you need to find the number value for both a and b
are you familiar with solving systems of equations by substitution or elimination or both?
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
well i think @Curry posted the answer above
i didnt check if it was correct though
can you check and tell me because this test is gonna end and I need to pass. then we can continue learning
what did he say was the answer?
it was wrong... I failed.. and its 3 am im tired im just gonna cut and go to sleep ;c
problem with rounding probably...
unless they assumed a value for c or something.
@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
@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
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.
@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
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.
@math&ing001, no you are giving wrong information, but thank you
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 !
@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