Answer to Question #150819 in Linear Algebra for shanto

Question

Determine the polynomial function whose graph passes through the points (2, 4), (3,6) and (5,10). Also sketch the graph of the polynomial function. (Using Cramer’s Method).

Answer 1

We have three points. So, we shall use second degree polynomial fitting.

$p(x)=ax^2+bx+c$

For $(2,4)$ ,

$4=a(2^2)+b(2)+c\\ 4=4a+2b+c......................(1)$

For $(3,6),$

$6=a(3^2)+b(3)+c\\ 6=9a+3b+c......................(2)$

For $(5,10),$

$10=a(5^2)+b(5)+c\\ 10=25a+5b+c...................(3)$

Bring the three equations together.

$4a+2b+c=4\\ 9a+3b+c=6\\ 25a+5b+c=10$

$\begin{pmatrix} 4 & 2 &1\\ 9 &3 &1\\ 25 &5 &1\\ \end{pmatrix}\begin{pmatrix}a\\b\\c\\\end{pmatrix}=\begin{pmatrix}4\\6\\10\\\end{pmatrix}$

Using Cramer's rule.

$\Delta=\begin{vmatrix} 4 & 2 &1\\ 9 &3 &1\\ 25 &5 &1\\ \end{vmatrix}=-6$

$\Delta_x=\begin{vmatrix} 4 & 2 &1\\ 6 &3 &1\\ 10 &5 &1\\ \end{vmatrix}=0\\ a=\frac{\Delta_x}{\Delta}=\frac{0}{-6}=0$

$\Delta_y=\begin{vmatrix} 4 & 4 &1\\ 9 &6 &1\\ 25 &10 &1\\ \end{vmatrix}=-12\\ b=\frac{\Delta_y}{\Delta}=\frac{-12}{-6}=2$

$\Delta_z=\begin{vmatrix} 4 & 2 &4\\ 9 &3 &6\\ 25 &5 &10\\ \end{vmatrix}=0\\ c=\frac{\Delta_z}{\Delta}=\frac{0}{-6}=0$

So,

$p(x)=0.x^2+2.x+0\\ p(x)=2x$

Here is the graph of $p(x)=2x$