# [SOLVED] Polynomial Curve Fitting

• Sep 5th 2009, 04:16 PM
Alterah
[SOLVED] Polynomial Curve Fitting
Hello, the problem I need to do is an even problem, therefore the answer isn't in the back, meaning I won't at least be able to know I did the work right. Anyhow, the problem is:

The graph of a parabola passes through the points (0,1) and (0.5,0.5) and has a horizontal tangent at (0.5,0.5). Fin an equation for the parabola and sketch its graph.
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I need to use matrices to come up with a function that fits the points. So, the general equation for a line is:

$\displaystyle p(x) = a_0 + a_1x + a_2x^2$

Using the points I can come up with three equations and then use matrices and elementary row operations to find values for $\displaystyle a_0, a_1, and a_2$. The problem being, I only have two point. But, because I know the function is a parabola, and there is a horizontal tangent at (0.5,0.5). I can safely assume that (0.5,0.5). I can then use another point that should also be on the polynomial (1,1). I just want to make sure my line of thought here is correct before I put the following matrix in reduced row-echelon form:

$\displaystyle \left(\begin{array}{cccc}a_0&0&0&1\\a_0&0.5a_1&.25 a_2&0.5\\a_0&a_1&a_2&1\end{array}\right)$

Thanks for the input. Basically I am wondering if my initial line of thought and setup of the matrix is correct for the problem. I'd hate to do the problem and get it wrong and not really find out why.
• Sep 5th 2009, 11:23 PM
CaptainBlack
Quote:

Originally Posted by Alterah
Hello, the problem I need to do is an even problem, therefore the answer isn't in the back, meaning I won't at least be able to know I did the work right. Anyhow, the problem is:

The graph of a parabola passes through the points (0,1) and (0.5,0.5) and has a horizontal tangent at (0.5,0.5). Fin an equation for the parabola and sketch its graph.
---------------------------

I need to use matrices to come up with a function that fits the points. So, the general equation for a line is:

$\displaystyle p(x) = a_0 + a_1x + a_2x^2$

Using the points I can come up with three equations and then use matrices and elementary row operations to find values for $\displaystyle a_0, a_1, and a_2$. The problem being, I only have two point. But, because I know the function is a parabola, and there is a horizontal tangent at (0.5,0.5). I can safely assume that (0.5,0.5). I can then use another point that should also be on the polynomial (1,1). I just want to make sure my line of thought here is correct before I put the following matrix in reduced row-echelon form:

$\displaystyle \left(\begin{array}{cccc}a_0&0&0&1\\a_0&0.5a_1&.25 a_2&0.5\\a_0&a_1&a_2&1\end{array}\right)$

Thanks for the input. Basically I am wondering if my initial line of thought and setup of the matrix is correct for the problem. I'd hate to do the problem and get it wrong and not really find out why.

That the tangent is horizontal at $\displaystyle (0.5,0.5)$ tells you that:

$\displaystyle 0=a_1+a_2$

CB
• Sep 6th 2009, 08:26 AM
Alterah
Quote:

Originally Posted by CaptainBlack
That the tangent is horizontal at $\displaystyle (0.5,0.5)$ tells you that:

$\displaystyle 0=a_1+a_2$

CB

That's the case because if you take the derivative of:
$\displaystyle p(x) = a_0 + a_1x + a_2x^2$
You get:
$\displaystyle p'(x) = a_1x + 2a_2$

You then plug in 0.5 for x and set equation equal to zero. If that's not how you got the expression, let me know, because it's the only thing that makes sense...Was my line of thought not valid? Or simply just setting me up for more work? Thanks!
• Sep 6th 2009, 10:02 AM
CaptainBlack
Quote:

Originally Posted by Alterah
That's the case because if you take the derivative of:
$\displaystyle p(x) = a_0 + a_1x + a_2x^2$
You get:
$\displaystyle p'(x) = a_1x + 2a_2$

You then plug in 0.5 for x and set equation equal to zero. If that's not how you got the expression, let me know, because it's the only thing that makes sense...Was my line of thought not valid? Or simply just setting me up for more work? Thanks!

That is the correct argument. Your previous argument was wrong because you did not know that (1,1) was on the solution parabola.

CB
• Sep 6th 2009, 10:43 AM
Alterah
Quote:

Originally Posted by CaptainBlack
That is the correct argument. Your previous argument was wrong because you did not know that (1,1) was on the solution parabola.

CB

True...I simply guessed, though it was an educated guess. Thanks!