Linear Algebra mixed with Calculus Let T: P^2-->R^3 be the linear transformation given by T(p(x)) = , B={1,x, x^2} the standard basis for P^2 and B' = {e1, e2, e3} be the standard basis for R^3. Use the inverse of the matrix [T]_B,B' to find a quadratic polynomial passing through (-3, 75) and (5, 99) and whose tangent line at x=2 has a slope of 13.

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Linear Algebra mixed with Calculus Let T: P^2-->R^3 be the linear transformation given by T(p(x)) = , B={1,x, x^2} the standard basis for P^2 and B' = {e1, e2, e3} be the standard basis for R^3. Use the inverse of the matrix [T]_B,B' to find a quadratic polynomial passing through (-3, 75) and (5, 99) and whose tangent line at x=2 has a slope of 13.

Linear Algebra
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|dw:1352007689346:dw|
ok, HUH???
uhh, wouldn't adding the three elementary matricies, and taking the rref, give you the inverse? I just don't know where to go from there

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ok, no idea what you're writing, but the numbers come out to be the same inverse that I got after doing the Gauss-Jordan Algorithm But a little English would be nice. If we don't interact, I have no clue what you're writing
Now second part !
I take that back, the numbers are a little different
I have just compute the inverse matrix with an individual method. Now we should follow the second part.
|dw:1352008344540:dw|
By adding these three columns: 1 0 0 0 1 0 0 0 1 to my original matrix and row reducing, I get that matrix as my inverse. I'm unfamiliar with your technique.
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|dw:1352008571359:dw|
MAHMIT!!!
|dw:1352008680377:dw|
While you have good intentions, just writing the answer with no explanation does me no good since I not only do not learn anything, but I do not know what you are doing. Could you please slow down to explain? Also, just writing out the solution is against the CoC.
|dw:1352008780107:dw|
Don't attend to inverse matrix. Just follow the second part. suppose I got it.
|dw:1352009050198:dw|
|dw:1352009161620:dw|
Is it clear?
I think Mahmit is being really painstaking in trying to explain this. Good work.
|dw:1352009295173:dw|
Very nice question. I was really enjoyed. And I hope you could understand it.
Erm, thanks.
I try to give you the same one.
|dw:1352009879051:dw|
|dw:1352010149750:dw|
I always thought how a software can find the polynomial with passing some points and some particular slope. So I figure it out. Thank you.
Umm, sure, although don't you think writing an actual program, the language (C++, Java, D, Ruby, Python, etc) doesn't matter, would be easier than doing this by hand every time?
I must find out some polynomial for a control system online. I have information with y, y' y'',... y(n). This problem help me a lot.
|dw:1352010808815:dw|
Indeed. very interesting.

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