# Homework Help: Linear Algebra - Find a Polynomial

1. Jan 18, 2013

### dkotschessaa

1. The problem statement, all variables and given/known data

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

2. Relevant 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

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

2. Jan 18, 2013

### CompuChip

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

3. Jan 18, 2013

### 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...

4. Jan 18, 2013

### CompuChip

Is $1 \cdot a + 1 \cdot b + 4 \cdot c = -1$ the same as $4c = -1$?
Or is $0 \cdot a + 0 \cdot b + 4 \cdot c = -1$ the same as $4c = -1$?

5. Jan 18, 2013

### 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

6. Jan 18, 2013

### 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

7. Jan 18, 2013

### Ray Vickson

Your expression for f '(2) is incorrect.

Last edited: Jan 18, 2013
8. Jan 21, 2013

### CompuChip

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

9. Jan 22, 2013

### 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. :)