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Sagot :
Certainly! Let's break down each step in detail.
### Part A: Setting Up the System of Equations
We are given three points that lie on the parabola: [tex]\((3,8)\)[/tex], [tex]\(\left(5, \frac{20}{3}\right)\)[/tex], and [tex]\((6,5)\)[/tex]. We want to find the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] of the parabola described by the equation [tex]\(y = ax^2 + bx + c\)[/tex].
Plugging each point into the parabola equation yields:
1. For the point [tex]\((3, 8)\)[/tex]:
[tex]\[ 8 = 3^2a + 3b + c = 9a + 3b + c \][/tex]
2. For the point [tex]\((5, \frac{20}{3})\)[/tex]:
[tex]\[ \frac{20}{3} = 5^2a + 5b + c = 25a + 5b + c \][/tex]
3. For the point [tex]\((6, 5)\)[/tex]:
[tex]\[ 5 = 6^2a + 6b + c = 36a + 6b + c \][/tex]
These equations create the following system of equations:
[tex]\[ \begin{array}{l} 8 = 9a + 3b + c \\ \frac{20}{3} = 25a + 5b + c \\ 5 = 36a + 6b + c \end{array} \][/tex]
### Part B: Solving Using Matrix Manipulation
We can represent this system of equations in matrix form [tex]\(AX = B\)[/tex] where:
[tex]\[ A = \begin{pmatrix} 9 & 3 & 1 \\ 25 & 5 & 1 \\ 36 & 6 & 1 \end{pmatrix}, \quad X = \begin{pmatrix} a \\ b \\ c \end{pmatrix}, \quad B = \begin{pmatrix} 8 \\ \frac{20}{3} \\ 5 \end{pmatrix} \][/tex]
To solve for [tex]\(X\)[/tex], we need to find [tex]\(A^{-1}\)[/tex] and compute [tex]\(X = A^{-1}B\)[/tex].
Let's find the inverse of [tex]\(A\)[/tex]:
[tex]\[ A^{-1} = \begin{pmatrix} 0.1667 & -0.5 & 0.3333 \\ -1.8333 & 4.5 & -2.6667 \\ 5 & -9 & 5 \end{pmatrix} \][/tex]
Now we multiply [tex]\(A^{-1}\)[/tex] by [tex]\(B\)[/tex] to find [tex]\(X\)[/tex]:
[tex]\[ X = A^{-1}B = \begin{pmatrix} 0.1667 & -0.5 & 0.3333 \\ -1.8333 & 4.5 & -2.6667 \\ 5 & -9 & 5 \end{pmatrix} \begin{pmatrix} 8 \\ \frac{20}{3} \\ 5 \end{pmatrix} = \begin{pmatrix} -0.3333 \\ 2.0000 \\ 5.0000 \end{pmatrix} \][/tex]
Hence, the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are:
[tex]\[ a = -0.3333, \quad b = 2.0000, \quad c = 5.0000 \][/tex]
### Final Equation of the Parabola
Using the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex], the equation of the parabola is:
[tex]\[ y = -0.3333x^2 + 2x + 5 \][/tex]
Thus, the final equation of the parabola passing through the points [tex]\((3,8)\)[/tex], [tex]\(\left(5, \frac{20}{3}\right)\)[/tex], and [tex]\((6,5)\)[/tex] is:
[tex]\[ y = -\frac{1}{3}x^2 + 2x + 5 \][/tex]
### Part A: Setting Up the System of Equations
We are given three points that lie on the parabola: [tex]\((3,8)\)[/tex], [tex]\(\left(5, \frac{20}{3}\right)\)[/tex], and [tex]\((6,5)\)[/tex]. We want to find the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] of the parabola described by the equation [tex]\(y = ax^2 + bx + c\)[/tex].
Plugging each point into the parabola equation yields:
1. For the point [tex]\((3, 8)\)[/tex]:
[tex]\[ 8 = 3^2a + 3b + c = 9a + 3b + c \][/tex]
2. For the point [tex]\((5, \frac{20}{3})\)[/tex]:
[tex]\[ \frac{20}{3} = 5^2a + 5b + c = 25a + 5b + c \][/tex]
3. For the point [tex]\((6, 5)\)[/tex]:
[tex]\[ 5 = 6^2a + 6b + c = 36a + 6b + c \][/tex]
These equations create the following system of equations:
[tex]\[ \begin{array}{l} 8 = 9a + 3b + c \\ \frac{20}{3} = 25a + 5b + c \\ 5 = 36a + 6b + c \end{array} \][/tex]
### Part B: Solving Using Matrix Manipulation
We can represent this system of equations in matrix form [tex]\(AX = B\)[/tex] where:
[tex]\[ A = \begin{pmatrix} 9 & 3 & 1 \\ 25 & 5 & 1 \\ 36 & 6 & 1 \end{pmatrix}, \quad X = \begin{pmatrix} a \\ b \\ c \end{pmatrix}, \quad B = \begin{pmatrix} 8 \\ \frac{20}{3} \\ 5 \end{pmatrix} \][/tex]
To solve for [tex]\(X\)[/tex], we need to find [tex]\(A^{-1}\)[/tex] and compute [tex]\(X = A^{-1}B\)[/tex].
Let's find the inverse of [tex]\(A\)[/tex]:
[tex]\[ A^{-1} = \begin{pmatrix} 0.1667 & -0.5 & 0.3333 \\ -1.8333 & 4.5 & -2.6667 \\ 5 & -9 & 5 \end{pmatrix} \][/tex]
Now we multiply [tex]\(A^{-1}\)[/tex] by [tex]\(B\)[/tex] to find [tex]\(X\)[/tex]:
[tex]\[ X = A^{-1}B = \begin{pmatrix} 0.1667 & -0.5 & 0.3333 \\ -1.8333 & 4.5 & -2.6667 \\ 5 & -9 & 5 \end{pmatrix} \begin{pmatrix} 8 \\ \frac{20}{3} \\ 5 \end{pmatrix} = \begin{pmatrix} -0.3333 \\ 2.0000 \\ 5.0000 \end{pmatrix} \][/tex]
Hence, the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are:
[tex]\[ a = -0.3333, \quad b = 2.0000, \quad c = 5.0000 \][/tex]
### Final Equation of the Parabola
Using the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex], the equation of the parabola is:
[tex]\[ y = -0.3333x^2 + 2x + 5 \][/tex]
Thus, the final equation of the parabola passing through the points [tex]\((3,8)\)[/tex], [tex]\(\left(5, \frac{20}{3}\right)\)[/tex], and [tex]\((6,5)\)[/tex] is:
[tex]\[ y = -\frac{1}{3}x^2 + 2x + 5 \][/tex]
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