Answer to Question #150955 in Algebra for Noman

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