Linear Algebra - Find a Polynomial

[dkotschessaa]

Homework Statement

Find all polynomials of the form a + bx + cx^2 that:

Goes through the points (1,1) and (3,3)

and such that f'(2) = 1

Homework Equations

a + bx + cx^2
f'(x) = x+2cx
f'(2) = 2 + 4c

polynomial through (1,1) = a + b1 + c1 = 1
polynomial through (3,3) = a + b3+ c3^2 = 3

The Attempt at a Solution

I have the general idea that this should result in a series of equations that I need to do gauss Jordan on. Similar problems like this resulted in 3 similar equations and were quite simple.

My problem here is that, since I have taken the derivative of f(x) I have lost my constant a. So I'm not sure what my matrix should look like. I've tried:

1 1 1 1
1 3 9 3

But for the one with the derivative, my f"(2) isn't in the same form - it's 2+4c = 1

so I"m not sure whether to use

1 2 4 1
0 2 4 1

or something more general. If I use something more general though like:

a 2 4 1

I can't get a pivot in my first column...

'elp!

-Dave K

 

[CompuChip]

You can rewrite 2+4c = 1 to 4c = -1.
That's of the form ... a + ... b + ... c = ...

 

[dkotschessaa]

That helps, but I'm still not sure whether my a's and b's for that equation should be 0s or 1's. I can try both, I suppose, but clearly my understanding of the problem is muddled or I wouldn't have to guess...

 

[CompuChip]

Is [itex]1 \cdot a + 1 \cdot b + 4 \cdot c = -1[/itex] the same as [itex]4c = -1[/itex]?
Or is [itex]0 \cdot a + 0 \cdot b + 4 \cdot c = -1[/itex] the same as [itex]4c = -1[/itex]?

 

[dkotschessaa]

Thanks for that clarification. I'm still getting some weird stuff (fractions I shouldn't be getting) but there must be another problem, since I'm pretty sure I'm not supposed to get those and that I should wind up with a free variable instead.

I'll check back later if I'm still working on it. Thanks again!
-Dave K

 

[lurflurf]

1 1 1 1
1 3 9 3
0 2 4 1
do you know why?

Since x is such a polynomial it is easier to let
f(x)=a + bx + cx^2
g(x)=f(x)-x
f(1)=1
f(3)=3
f'(2)=1
imply
g(1)=0
g(3)=0
g'(2)=0
the matrix is
1 1 1 0
1 3 9 0
0 2 4 0

 

[Ray Vickson]

Homework Statement

Find all polynomials of the form a + bx + cx^2 that:

Goes through the points (1,1) and (3,3)

and such that f'(2) = 1

Homework Equations

a + bx + cx^2
f'(x) = x+2cx
f'(2) = 2 + 4c

polynomial through (1,1) = a + b1 + c1 = 1
polynomial through (3,3) = a + b3+ c3^2 = 3

The Attempt at a Solution

I have the general idea that this should result in a series of equations that I need to do gauss Jordan on. Similar problems like this resulted in 3 similar equations and were quite simple.

My problem here is that, since I have taken the derivative of f(x) I have lost my constant a. So I'm not sure what my matrix should look like. I've tried:

1 1 1 1
1 3 9 3

But for the one with the derivative, my f"(2) isn't in the same form - it's 2+4c = 1

so I"m not sure whether to use

1 2 4 1
0 2 4 1

or something more general. If I use something more general though like:

a 2 4 1

I can't get a pivot in my first column...

'elp!

-Dave K

Your expression for f '(2) is incorrect.

 

[CompuChip]

Good catch Ray! It took three people to see that :)

 

[dkotschessaa]

Yes, indeed. That was a problem too. But your question "Is 1⋅a+1⋅b+4⋅c=−1 the same as 4c=−1?" was also instructive.

Thanks to the whole team. :)