To determine the value of [tex]\( b \)[/tex] in the standard quadratic equation [tex]\( f(x) = ax^2 + bx + c \)[/tex], we need to use the given points to set up a system of equations and solve for [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex].
Given the points [tex]\((x, f(x))\)[/tex]:
[tex]\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
-4 & 7 \\
\hline
-3 & 6 \\
\hline
-2 & 7 \\
\hline
-1 & 10 \\
\hline
0 & 15 \\
\hline
1 & 22 \\
\hline
2 & 31 \\
\hline
\end{array}
\][/tex]
We form the system of equations based on substituting each [tex]\( x \)[/tex] into the quadratic equation and setting it equal to [tex]\( f(x) \)[/tex].
1. For [tex]\( x = -4 \)[/tex]:
[tex]\[
16a - 4b + c = 7
\][/tex]
2. For [tex]\( x = -3 \)[/tex]:
[tex]\[
9a - 3b + c = 6
\][/tex]
3. For [tex]\( x = -2 \)[/tex]:
[tex]\[
4a - 2b + c = 7
\][/tex]
4. For [tex]\( x = -1 \)[/tex]:
[tex]\[
a - b + c = 10
\][/tex]
5. For [tex]\( x = 0 \)[/tex]:
[tex]\[
c = 15
\][/tex]
6. For [tex]\( x = 1 \)[/tex]:
[tex]\[
a + b + c = 22
\][/tex]
7. For [tex]\( x = 2 \)[/tex]:
[tex]\[
4a + 2b + c = 31
\][/tex]
Using the equation [tex]\( x = 0 \)[/tex]:
[tex]\[ c = 15 \][/tex]
We substitute [tex]\( c = 15 \)[/tex] into the remaining equations:
1. [tex]\( 16a - 4b + 15 = 7 \)[/tex]:
[tex]\[
16a - 4b = -8 \quad \Rightarrow \quad 4a - b = -2 \quad \text{(Equation 1)}
\][/tex]
2. [tex]\( 9a - 3b + 15 = 6 \)[/tex]:
[tex]\[
9a - 3b = -9 \quad \Rightarrow \quad 3a - b = -3 \quad \text{(Equation 2)}
\][/tex]
3. [tex]\( 4a - 2b + 15 = 7 \)[/tex]:
[tex]\[
4a - 2b = -8 \quad \Rightarrow \quad 2a - b = -4 \quad \text{(Equation 3)}
\][/tex]
4. [tex]\( a - b + 15 = 10 \)[/tex]:
[tex]\[
a - b = -5 \quad \text{(Equation 4)}
\][/tex]
5. [tex]\( a + b + 15 = 22 \)[/tex]:
[tex]\[
a + b = 7 \quad \text{(Equation 5)}
\][/tex]
6. [tex]\( 4a + 2b + 15 = 31 \)[/tex]:
[tex]\[
4a + 2b = 16 \quad \Rightarrow \quad 2a + b = 8 \quad \text{(Equation 6)}
\][/tex]
Now we solve these equations:
From Equations 4 and 5, we have:
[tex]\[ a - b = -5 \][/tex]
[tex]\[ a + b = 7 \][/tex]
We add these two equations:
[tex]\[ (a - b) + (a + b) = -5 + 7 \][/tex]
[tex]\[ 2a = 2 \][/tex]
[tex]\[ a = 1 \][/tex]
Next, we substitute [tex]\( a = 1 \)[/tex] into [tex]\( a + b = 7 \)[/tex]:
[tex]\[ 1 + b = 7 \][/tex]
[tex]\[ b = 6 \][/tex]
Thus, the value of [tex]\( b \)[/tex] is:
[tex]\[
\boxed{6}
\][/tex]
