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Sagot :
To solve this problem, we need to determine the constants [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] such that the quadratic function [tex]\(y = ax^2 + bx + c\)[/tex] passes through the points [tex]\((-1, 6)\)[/tex], [tex]\((1, 0)\)[/tex], and [tex]\((2, 3)\)[/tex].
First, let's set up the system of equations by substituting each point into the quadratic equation.
1. For the point [tex]\((-1, 6)\)[/tex]:
[tex]\[ 6 = a(-1)^2 + b(-1) + c \implies 6 = a - b + c \][/tex]
2. For the point [tex]\((1, 0)\)[/tex]:
[tex]\[ 0 = a(1)^2 + b(1) + c \implies 0 = a + b + c \][/tex]
3. For the point [tex]\((2, 3)\)[/tex]:
[tex]\[ 3 = a(2)^2 + b(2) + c \implies 3 = 4a + 2b + c \][/tex]
We now have the following system of linear equations:
[tex]\[ \begin{cases} 6 = a - b + c \\ 0 = a + b + c \\ 3 = 4a + 2b + c \end{cases} \][/tex]
Let's solve this system step-by-step.
From the second equation:
[tex]\[ 0 = a + b + c \implies c = -a - b \][/tex]
We can substitute [tex]\(c\)[/tex] in the other two equations.
Substituting [tex]\(c = -a - b\)[/tex] into the first equation:
[tex]\[ 6 = a - b + (-a - b) \implies 6 = -2b \implies b = -3 \][/tex]
Now that we have [tex]\(b = -3\)[/tex], we substitute [tex]\(b\)[/tex] into [tex]\(c = -a - b\)[/tex]:
[tex]\[ c = -a - (-3) \implies c = -a + 3 \][/tex]
Substituting [tex]\(b = -3\)[/tex] and [tex]\(c = -a + 3\)[/tex] into the third equation:
[tex]\[ 3 = 4a + 2(-3) + (-a + 3) \implies 3 = 4a - 6 - a + 3 \implies 3 = 3a - 3 \implies 6 = 3a \implies a = 2 \][/tex]
Now we have values [tex]\(a = 2\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = -a + 3\)[/tex]:
[tex]\[ c = -2 + 3 \implies c = 1 \][/tex]
Thus, the values for the constants [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are:
[tex]\[ a = 2, b = -3, c = 1 \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{B} \][/tex]
First, let's set up the system of equations by substituting each point into the quadratic equation.
1. For the point [tex]\((-1, 6)\)[/tex]:
[tex]\[ 6 = a(-1)^2 + b(-1) + c \implies 6 = a - b + c \][/tex]
2. For the point [tex]\((1, 0)\)[/tex]:
[tex]\[ 0 = a(1)^2 + b(1) + c \implies 0 = a + b + c \][/tex]
3. For the point [tex]\((2, 3)\)[/tex]:
[tex]\[ 3 = a(2)^2 + b(2) + c \implies 3 = 4a + 2b + c \][/tex]
We now have the following system of linear equations:
[tex]\[ \begin{cases} 6 = a - b + c \\ 0 = a + b + c \\ 3 = 4a + 2b + c \end{cases} \][/tex]
Let's solve this system step-by-step.
From the second equation:
[tex]\[ 0 = a + b + c \implies c = -a - b \][/tex]
We can substitute [tex]\(c\)[/tex] in the other two equations.
Substituting [tex]\(c = -a - b\)[/tex] into the first equation:
[tex]\[ 6 = a - b + (-a - b) \implies 6 = -2b \implies b = -3 \][/tex]
Now that we have [tex]\(b = -3\)[/tex], we substitute [tex]\(b\)[/tex] into [tex]\(c = -a - b\)[/tex]:
[tex]\[ c = -a - (-3) \implies c = -a + 3 \][/tex]
Substituting [tex]\(b = -3\)[/tex] and [tex]\(c = -a + 3\)[/tex] into the third equation:
[tex]\[ 3 = 4a + 2(-3) + (-a + 3) \implies 3 = 4a - 6 - a + 3 \implies 3 = 3a - 3 \implies 6 = 3a \implies a = 2 \][/tex]
Now we have values [tex]\(a = 2\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = -a + 3\)[/tex]:
[tex]\[ c = -2 + 3 \implies c = 1 \][/tex]
Thus, the values for the constants [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are:
[tex]\[ a = 2, b = -3, c = 1 \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{B} \][/tex]
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