Find Polynomial of Degree <2 to Satisfy (1, p), (2, q), (3, r)

[VinnyCee]

Find a polynomial of degree < 2 [a polynomial of the form $f\left(t\right)\,=\,a\,+\,bt\,+\,ct^2$] whose graph goes through the points (1, p), (2, q), (3, r), where p, q, r are arbitrary constants. Does such a polynomial exist for all values of p, q, r?


Here is what I have so far:

(1, p) ==> $p\,=\,a\,+\,b\,+\,c$
(2, q) ==> $q\,=\,a\,+\,2b\,+\,4c$
(3, r) ==> $r\,=\,a\,+\,3b\,+\,9c$

$$\begin{displaymath} \left[ \begin{array}{ccc|c} 1 & 1 & 1 & p \\ 1 & 2 & 4 & q \\ 1 & 3 & 9 & r \end{array} \right] \end{displaymath}$$

$$R_2\,=\,R_2\,-\,R_1$$
$$R_3\,=\,R_3\,-\,R_1$$

$$\begin{displaymath} \left[ \begin{array}{ccc|c} 1 & 1 & 1 & p \\ 0 & 1 & 3 & q - p \\ 0 & 2 & 8 & r - p \end{array} \right] \end{displaymath}$$

$$R_3\,=\,R_3\,-\,2R_2$$

$$\begin{displaymath} \left[ \begin{array}{ccc|c} 1 & 1 & 1 & p \\ 0 & 1 & 3 & q - p \\ 0 & 0 & 2 & r - q \end{array} \right] \end{displaymath}$$

$$R_3\,=\frac{1}{2}\,R_3$$

$$\begin{displaymath} \left[ \begin{array}{ccc|c} 1 & 1 & 1 & p \\ 0 & 1 & 3 & q - p \\ 0 & 0 & 1 & \frac{1}{2}\left(r - q\right) \end{array} \right] \end{displaymath}$$

$$R_2\,=\,R_2\,-\,3R_3$$

$$\begin{displaymath} \left[ \begin{array}{ccc|c} 1 & 1 & 1 & p \\ 0 & 1 & 0 & \frac{5}{2}\,q\,-\,p\,-\,\frac{3}{2}\,r \\ 0 & 0 & 1 & \frac{1}{2}\left(r - q\right) \end{array} \right] \end{displaymath}$$

$$R_1\,=\,R_1\,-\,R_2\,-\,R_3$$

$$\begin{displaymath} \left[ \begin{array}{ccc|c} 1 & 0 & 0 & 2p\,-\,2q\,+\,r \\ 0 & 1 & 0 & \frac{1}{2}\left(5q\,-\,2p\,-\,3r\right) \\ 0 & 0 & 1 & \frac{1}{2}\left(r - q\right) \end{array} \right] \end{displaymath}$$

$$a\,=\,2p\,-\,2q\,+\,r$$
$$b\,=\,\frac{1}{2}\,\left(5q\,-\,2p\,-\,3r\right)$$
$$c\,=\,\frac{1}{2}\,\left(r\,-\,q\right)$$


$$f\left(t\right)\,=\,a\,+\,bt\,+\,ct^2$$
$$f\left(t\right)\,=\,\left(2p\,-\,2q\,+\,r\right)\,+\,\left[\frac{1}{2}\,\left(5q\,-\,2p\,-\,3r\right)\right]\,t\,+\,\left[\frac{1}{2}\,\left(r\,-\,q\right)\right]\,t^2$$
$$f\left(t\right)\,=\frac{1}{2}\,r\,t^2\,-\,\frac{1}{2}\,q\,t^2\,+\,\frac{5}{2}\,q\,t\,-\,\frac{3}{2}\,r\,t\,-\,p\,t\,+\,2\,p\,+\,r\,-\,2\,q$$

But when I try a set of arbitrary numbers [(1, 8), (2, 12), (3, 3)], I get an equation that does not match the points:

$$f\left(t\right)\,=\,-\frac{9}{2}\,t^2\,+\,\frac{35}{2}\,t\,-\,5$$

$$f\left(1\right)\,=\,-\frac{9}{2}\,(1)^2\,+\,\frac{35}{2}\,(1)\,-\,5\,=\,8$$

$$f\left(2\right)\,=\,-\frac{9}{2}\,(2)^2\,+\,\frac{35}{2}\,(2)\,-\,5\,=\,12$$

$$f\left(3\right)\,=\,-\frac{9}{2}\,(3)^2\,+\,\frac{35}{2}\,(3)\,-\,5\,=\,7$$

However, $f\left(3\right)$ should be equal to r, which is 3, not 7.

Please help:)

 

[HallsofIvy]

Find a polynomial of degree < 2 [a polynomial of the form ] whose graph goes through the points (1, p), (2, q), (3, r), where p, q, r are arbitrary constants. Does such a polynomial exist for all values of p, q, r?

Of degree < 2? Then it is linear, of the form y= mx+ b and you can only determine only the two numberss m and b. In general 2 equations will do that. You can choose m and b so that 2 of the requirements (1, p), (2, q), (3, r) are satisfied but not all 3.

 

[radou]

For (p, q, r) = (0, 0, 0) we have (a, b, c) = (0, 0, 0), and therefore a zero polynomial, which has a degree undefined or $$-\infty$$, so it's less than 2.

 

[HallsofIvy]

The title says "Find a >2 degree polynomial" and the post itself says
"Find a polynomial of degree < 2".

Is it any wonder I'm confused!

 

[Mute]

$$\begin{displaymath} \left[ \begin{array}{ccc|c} 1 & 1 & 1 & p \\ 0 & 1 & 3 & q - p \\ 0 & 2 & 8 & r - p \end{array} \right] \end{displaymath}$$

$$R_3\,=\,R_3\,-\,2R_2$$

$$\begin{displaymath} \left[ \begin{array}{ccc|c} 1 & 1 & 1 & p \\ 0 & 1 & 3 & q - p \\ 0 & 0 & 2 & r - q \end{array} \right] \end{displaymath}$$

Confusion about <2 vs >2 aside, your derivation has an error in the quoted section. You substract $2 R_2$ from $R_3$, but in the augmented part of the matrix you've only subtracted $1 R_2$, resulting in r - q instead of r + p - 2q. I didn't check to see if everything else was correct, but this is certainly a place to start with.

 

[VinnyCee]

Sorry for the confusion!

The polynomial needs to have a degree of GREATER THAN OR EQUAL TO 2.

So it needs to have at least one squared term.

Here is the corrected derivation with help from Mute (Thanks!):

(1, p) ==> $p\,=\,a\,+\,b\,+\,c$
(2, q) ==> $q\,=\,a\,+\,2b\,+\,4c$
(3, r) ==> $r\,=\,a\,+\,3b\,+\,9c$

$$\begin{displaymath} \left[ \begin{array}{ccc|c} 1 & 1 & 1 & p \\ 1 & 2 & 4 & q \\ 1 & 3 & 9 & r \end{array} \right] \end{displaymath}$$

$$R_2\,=\,R_2\,-\,R_1$$
$$R_3\,=\,R_3\,-\,R_1$$

$$\begin{displaymath} \left[ \begin{array}{ccc|c} 1 & 1 & 1 & p \\ 0 & 1 & 3 & q - p \\ 0 & 2 & 8 & r - p \end{array} \right] \end{displaymath}$$

$$R_3\,=\,R_3\,-\,2R_2$$

$$\begin{displaymath} \left[ \begin{array}{ccc|c} 1 & 1 & 1 & p \\ 0 & 1 & 3 & q - p \\ 0 & 0 & 2 & r - 2q + p \end{array} \right] \end{displaymath}$$

$$R_3\,=\frac{1}{2}\,R_3$$

$$\begin{displaymath} \left[ \begin{array}{ccc|c} 1 & 1 & 1 & p \\ 0 & 1 & 3 & q - p \\ 0 & 0 & 1 & \frac{1}{2} r - q + \frac{1}{2} p \end{array} \right] \end{displaymath}$$

$$R_2\,=\,R_2\,-\,3R_3$$

$$\begin{displaymath} \left[ \begin{array}{ccc|c} 1 & 1 & 1 & p \\ 0 & 1 & 0 & -\frac{5}{2}\,p\,+\,4\,q\,-\,\frac{3}{2}\,r \\ 0 & 0 & 1 & \frac{1}{2} r - q + \frac{1}{2} p \end{array} \right] \end{displaymath}$$

$$R_1\,=\,R_1\,-\,R_2\,-\,R_3$$

$$\begin{displaymath} \left[ \begin{array}{ccc|c} 1 & 0 & 0 & 3\,p\,-\,3\,q\,+\,r \\ 0 & 1 & 0 & -\frac{5}{2}\,p\,+\,4\,q\,-\,\frac{3}{2}\,r \\ 0 & 0 & 1 & \frac{1}{2} r - q + \frac{1}{2} p \end{array} \right] \end{displaymath}$$

$$a\,=\,3\,p\,-\,3\,q\,+\,r$$
$$b\,=\,-\frac{5}{2}\,p\,+\,4\,q\,-\,\frac{3}{2}\,r$$
$$c\,=\,\frac{1}{2} r - q + \frac{1}{2} p$$


$$f\left(t\right)\,=\,a\,+\,bt\,+\,ct^2$$
$$f\left(t\right)\,=\,\left(3\,p\,-\,3\,q\,+\,r\right)\,+\,\left[-\frac{5}{2}\,p\,+\,4\,q\,-\,\frac{3}{2}\,r\right]\,t\,+\,\left[\frac{1}{2} r - q + \frac{1}{2} p\right]\,t^2$$
$$f\left(t\right)\,=\frac{1}{2}\,r\,t^2\,+\,\frac{1}{2}\,p\,t^2\,-\,q\,t^2\,+\,4\,q\,t\,-\,\frac{5}{2}\,p\,t\,-\,\frac{3}{2}\,r\,t\,+\,3\,p\,+\,r\,-\,3\,q$$

$$p\,=\,8$$
$$q\,=\,12$$
$$r\,=\,3$$

So now it works for these numbers!

$$f\,(\,t\,)\,=\,-\frac{13}{2}\,t^2\,+\,\frac{47}{2}\,t\,-\,9$$

$$f\,(\,1\,)\,=\,-\frac{13}{2}\,(\,1\,)^2\,+\,\frac{47}{2}\,(\,1\,)\,-\,9\,=\,8$$
$$f\,(\,2\,)\,=\,-\frac{13}{2}\,(\,2\,)^2\,+\,\frac{47}{2}\,(\,2\,)\,-\,9\,=\,12$$
$$f\,(\,3\,)\,=\,-\frac{13}{2}\,(\,3\,)^2\,+\,\frac{47}{2}\,(\,3\,)\,-\,9\,=\,3$$

But how can I figure (or prove) whether or not a polynomial exists for all values of p, q, and r? There does exist such a polynomial for all p, q, and r right?