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.