Answer to Question #151001 in Linear Algebra for Shahrin

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 Cramers Method).

Answer 1

These tree points lie on the straight line $y=2x$. The power of polynomial is 1. If we will find polynomial with power 2 as $y=ax^2+bx+c$ we will

receive $a=0, b=2, c=0$, because parabola can have only two roots,

and $y=2x$:

$\begin{cases} 4a+2b+c=4\\ 9a+3b+c=6\\ 25a+5b+c=10 \end{cases}$

$a=det\begin{Vmatrix} 4 & 2 & 1 \\ 6 & 3 & 1\\ 10&5 &1 \end{Vmatrix} / det\begin{Vmatrix} 4 & 2 & 1\\ 9 & 3 & 1\\ 25 & 5 & 1 \end{Vmatrix} =$ $=(4(3-5)-2(6-10)+1(6*5-3*10))/$

$(4(3-5)--2(9-25)+1(9*5-25*3))=$

$=(-8+8+0)/(-8+32-30)=0/(-6)=0$

$b=det\begin{Vmatrix} 4 & 4 & 1 \\ 9 & 6 & 1\\ 25&10 &1 \end{Vmatrix} / det\begin{Vmatrix} 4 & 2 & 1\\ 9 & 3 & 1\\ 25 & 5 & 1 \end{Vmatrix} =$

$=(4(6-10)-4(9-25)+(9*10-25*6))/(-6)=$

$=(-16+64-60)/(-6)=(-12)/(-6)=2$

$c=det\begin{Vmatrix} 4 & 2 & 4 \\ 9 & 3 & 6\\ 25&5 &10 \end{Vmatrix} / det\begin{Vmatrix} 4 & 2 & 1\\ 9 & 3 & 1\\ 25 & 5 & 1 \end{Vmatrix} =$

$=(4(3*10-5*6)-2(9*10-6*25)+$

$+4(9*5-3*25))/(-6)=$

$=(4*0+2*60-4*30)/(-6)=0$

If we want to find polynomial with power 3, we need solve system:

$\begin{cases} 8a+4b+2c+d=4\\ 27a+9b+3c+d=6\\ 125a+25b+5c+d=10 \end{cases}$

and this system will have infinitely many solutions.