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]
