Cubic help

- anonymous

Cubic help

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- anonymous

Graph doesn't look very cubic to me though which makes me thing I did it wrong.

- anonymous

what's the exact question?

- anonymous

as in what is the exact wording of the question you're trying to answer.
positive rate of change → function is increasing
negative rate of change → function is decreasing

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## More answers

- mathmate

Is it possible to post an image of the original question?
That's the fastest way to get help!

- anonymous

You're giving a lot of information, but there's no context

- anonymous

So I can do cubic regression or do l*w*h where (12-x) represents the sides and (x) represents the height of the boxes... whichever one you guys are able to help me with.

- anonymous

Kinda wanted to know how to do it by hand, but maybe that isn't good to know as a pre-algebra student.

- mathmate

Hints:
"where (12-x) represents the sides and (x) represents the height of the box)"
means:
So if the height (x)=1, then the sides are (12-1)=11, or 1*11*11, V=121
If height is 2, sides are 12-2=10, volume = 2*10*10=200
in general, if the height equals x, the volume is x(12-x)(12-x), which is evidently a cubic.
Do you agree with the above calculations?

- anonymous

Yeah that makes sense.

- mathmate

Yes, a cubic is a polynomial, which is any expression in the form of "the sum of positive integer powers of a variable, each multiplied by a constant".

- mathmate

I do not know how you got
y=-12x^2 + 49.6x + 67.2
but the volume itself is already a cubic (third degree polynomial), and will not need to have a cubic regression, because it is exactly a cubic polynomial.

- mathmate

But your table does not correspond to the given constraints of x=height, (12-x)=sides, so it cannot be used to "generate" your regression cubic.

- mathmate

But your table does not correspond to the given constraints of x=height, (12-x)=sides, so it cannot be used to "generate" your regression cubic that represent the results of the problem.

- anonymous

But if it's wrong then I will just go with the way you showed me.

- mathmate

In your table, I do not understand what you meant by "size of corner cut ". I use x=height, as per instructions.

- anonymous

size of corner cut meants (remember I used grid paper to make cubes) I had to cut off one square on each corner. Let me draw it for you okay. I'm really sorry for confusing you. I did a very bad job explaining everything.

- anonymous

Give me a few minutes. please.

- mathmate

So that means the height.
|dw:1440381788262:dw|

- anonymous

right but in the first drawing it's 10 by 10

- mathmate

So you need to make a table:
|dw:1440381900195:dw|

- mathmate

Can you please look at my table and see if you agree with it?

- anonymous

for that table you draw for
3 would be 12-3 = 9 3*9*9 or 243

- mathmate

exactly!

- anonymous

Yeah that makes sense to me.

- anonymous

So was my equation wrong then? I can't get an equation from your table you drew right?

- mathmate

So use this new table and try to fit a cubic and see what you get.
"Our" volume is x(x-2)^2=x^3-24x^2+144x.
Your "regression" cubic should give exactly the same expression.

- mathmate

Remember to fit a cubic you need at least 4 points, and preferably more if you're doing regression.

- anonymous

wait what do I put in for x?

- anonymous

Are you there @mathmate

- mathmate

Sorry, I was afk.
a bunch of numbers, like 1,2,3,4,5,6, or 2,4,6,8,10, ...
If everything is calculated correctly, they should all give the same cubic polynomial.

- mathmate

The numbers don't have to be evenly spaced like my examples, they could be
x=1,2,5,10,21 if you want.

- mathmate

Oops, cannot be greater than 12! So forget the 21!

- anonymous

I graphed this on my paper I did my y axis by 10's and my x axis 12345 but skipped every two lines

- anonymous

not sure if that's relevant

- mathmate

Trick to graphing by hand is to get the smallest and largest y to start.
So start with the table.
|dw:1440383506458:dw|
so make a y-scale high enough for 256, and x from 1 to 10.

- anonymous

okay.

- anonymous

okay thanks for your help. Really appreciate you sitting here and going over this with me :)))

- mathmate

No problem! :)

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