Calc Question

[Jeenyuss]

FUCK I HAVE A CALC TEST TOMORROW THAT I HAVENT STUDIED FOR. FUCK.dsfsdf

[Fit of Rage]

Ugh, I'm in Civil Engineering 207 (Mechanics of Materials II) right now and I hate it. Get ready for some serious shit over the next couple years.

[Hero*]

Since this is calc help..i might as well ask.

Determine the equation of the polynomial function that matches the following data set.
x y
1 1
2 -3
3 5
4 37
5 105

This is an assignment and I have no idea where to begin..I tried finding the differences of Y but i never got a constant value..Any ideas?

[Troll King]

Do it again. Then a third time.

[Salubrious]

Since this is calc help..i might as well ask.

Determine the equation of the polynomial function that matches the following data set.
x y
1 1
2 -3
3 5
4 37
5 105

This is an assignment and I have no idea where to begin..I tried finding the differences of Y but i never got a constant value..Any ideas?

Graph the points and plot a best fit line using your graphing calculator?

I got y = 2x^3 - 6x^2 + 5

[Troll King]

Let me explain my last post a bit: When you took the derivatives, you didn't get a constant value, and if you take the second derivatives, you still won't get constants. You will however see a pattern. Take the third derivatives and you will finally get that constant.

[Reaver]

Let me explain my last post a bit: When you took the derivatives, you didn't get a constant value, and if you take the second derivatives, you still won't get constants. You will however see a pattern. Take the third derivatives and you will finally get that constant.

Are you talking to me or Hero? Cause if you're talking to me I don't follow. Are you saying it's wrong or offering another method?

[Troll King]

To Hero.

Salubrious' answer was correct, but I prefer doing it algebraically. I don't want to sound old, but in my day, the use of graphing calculators wasn't a regular part of the curriculum.

[Salubrious]

To Hero.

Salubrious' answer was correct, but I prefer doing it algebraically. I don't want to sound old, but in my day, the use of graphing calculators wasn't a regular part of the curriculum.

My guess is he would probably have to show work leading up to his solution anyways and can't just use his calculator.

[Edify]

To Hero.

Salubrious' answer was correct, but I prefer doing it algebraically. I don't want to sound old, but in my day, the use of graphing calculators wasn't a regular part of the curriculum.

We aren't allowed to use calculators at our calc tests. We have to derive everything and show it procedurally on paper. It's really oodles of fun. <_<

[Hero*]

yeah i cant use a graph for this..gotta be algebraically and I have no idea where to start. Like as you said TK, I tried doin that..but I never get a constant difference?

1
.. > 4
-3 .. >.. 4
.. > 8 .. > .. 20
5 .. > .. 24 .... > ..-32 ??????????????????
.. > 32 .. > .. 12
37 .. >.. 36
.. > 68
105

like i dont understand...i dont get a constant value.
ignore the ..'s

[Epinephrine]

To Hero.

Salubrious' answer was correct, but I prefer doing it algebraically. I don't want to sound old, but in my day, the use of graphing calculators wasn't a regular part of the curriculum.

You and your slide rules!! Boy am I glad I never have to do calculus ever again.

[Troll King]

yeah i cant use a graph for this..gotta be algebraically and I have no idea where to start. Like as you said TK, I tried doin that..but I never get a constant difference?

1
.. > 4
-3 .. >.. 4
.. > 8 .. > .. 20
5 .. > .. 24 .... > ..-32 ??????????????????
.. > 32 .. > .. -12
37 .. >.. 36
.. > 68
105

like i dont understand...i dont get a constant value.
ignore the ..'s

From x=1 to x=2, y goes from 1 to -3. How is that a change of positive 4?

[Hero*]

From x=1 to x=2, y goes from 1 to -3. How is that a change of positive 4?

well im a retard. wow i can't believe i made this mistake. whoops

i still have no idea what to do

[Fit of Rage]

It's like taking the derivative. Using TK's method, it shows that the difference after 3 differentiations is 12. So y''' = 12. Thus y'' = 12x + C, y' = 6x^2 + Cx + D, y = 2x^3 + (C/2)x^2 + Dx + E, where C, D, and E are constants. So you have the first term of the polynomial, 2x^3. I don't really know what to do beyond that though.