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

we have three points so we use second polynomial fitting like that

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

the points are (2, 4); (3, 6) and (5, 10).

$y_1 =$ $ax_1^2 + bx_1+c$

$y_2 = ax_2^2 + bx_2 + c$

$y_3 = ax_3^2 + bx_3 + c$

The task is to find a, b and c. Start by substituting each of the points into the equation, we have

4 = a(2)² + b(2) + c;

6 = a(3)² + b(3) + c;

10 = a(5)² + b(5) + c;

We can write this more compactly as a matrix equation

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

Or we use Cramer's rule for:

4a + 2b + c = 4

9a + 3b + c = 6

25a + 5b + c = 10

D = $\begin{bmatrix} 4&2&1\\ 9&3&1\\ 25&5&1 \end{bmatrix}$= (4*3*1 + 2*1*25 + 1*9*5) - (25*3*1 + 5*1*4 + 1*9*2) = -6;

D_x = $\begin{bmatrix} 4&2&1\\ 6&3&1\\ 10&5&1 \end{bmatrix}$ = (4*3*1 + 2*1*10 + 1*6*5) - (10 * 3 *1 + 5*1*4 + 1*6*2) = 0;

x = $\frac{D_x}{D} = \frac{0}{-6} = 0$

D_y = $\begin{bmatrix} 4&4&1\\ 9&6&1\\ 25&10&1 \end{bmatrix}$ = (4*6*1 + 4*1*25 + 1*9*10) - (25*6 + 10*1*4+1*9*4) = -12

$b = \frac{D_y}{D} = \frac{-12}{--6} = 2$

D$_z = \begin{bmatrix} 4&2&4\\ 9&3&6\\ 25&5&10\end{bmatrix}$ = 0

$c = \frac{D_z}{D} = \frac{0}{-12} = 0$

4 + c = 4 -> c = 0;

y = 2x