To determine the equation of the parabola y = ax² + bx + c that the ball's path follows through the given points (3, 8), (5, 2), and (6, 5), we need to set up a system of three equations with three unknowns (a, b, and c).
Let's plug the coordinates of these points into the general equation of a parabola:
1. For the point (3, 8):
[tex]\[ y = ax^2 + bx + c \][/tex]
Substitute [tex]\( x = 3 \)[/tex] and [tex]\( y = 8 \)[/tex]:
[tex]\[ 8 = a(3)^2 + b(3) + c \][/tex]
[tex]\[ 8 = 9a + 3b + c \][/tex]
2. For the point (5, 2):
[tex]\[ y = ax^2 + bx + c \][/tex]
Substitute [tex]\( x = 5 \)[/tex] and [tex]\( y = 2 \)[/tex]:
[tex]\[ 2 = a(5)^2 + b(5) + c \][/tex]
[tex]\[ 2 = 25a + 5b + c \][/tex]
3. For the point (6, 5):
[tex]\[ y = ax^2 + bx + c \][/tex]
Substitute [tex]\( x = 6 \)[/tex] and [tex]\( y = 5 \)[/tex]:
[tex]\[ 5 = a(6)^2 + b(6) + c \][/tex]
[tex]\[ 5 = 36a + 6b + c \][/tex]
Now, we have a system of three equations:
1. [tex]\( 9a + 3b + c = 8 \)[/tex]
2. [tex]\( 25a + 5b + c = 2 \)[/tex]
3. [tex]\( 36a + 6b + c = 5 \)[/tex]
This system can be expressed in matrix form [tex]\( A \cdot X = B \)[/tex], where [tex]\( A \)[/tex] is the coefficient matrix, [tex]\( X \)[/tex] is the vector of unknowns, and [tex]\( B \)[/tex] is the vector of constants:
[tex]\[ A \begin{pmatrix} 9 & 3 & 1 \\ 25 & 5 & 1 \\ 36 & 6 & 1 \end{pmatrix} \][/tex]
[tex]\[ X = \begin{pmatrix} a \\ b \\ c \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} 8 \\ 2 \\ 5 \end{pmatrix} \][/tex]
So the final system of equations is represented in matrix form as:
[tex]\[
\begin{pmatrix}
9 & 3 & 1 \\
25 & 5 & 1 \\
36 & 6 & 1
\end{pmatrix}
\begin{pmatrix}
a \\
b \\
c
\end{pmatrix}
=
\begin{pmatrix}
8 \\
2 \\
5
\end{pmatrix}
\][/tex]
