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 $p(x)=2x$