To solve for the coefficient [tex]\( b \)[/tex] in the standard form of the quadratic function [tex]\( f(x) = ax^2 + bx + c \)[/tex], we start by creating a system of equations using the given points on the function. We will use three points to create our equations.
The points given are:
- (-4, 7)
- (-3, 6)
- (-2, 7)
Using these points, we can write out the following system of equations:
1. For [tex]\( x = -4 \)[/tex] and [tex]\( f(x) = 7 \)[/tex]:
[tex]\[
a(-4)^2 + b(-4) + c = 7
\][/tex]
This simplifies to:
[tex]\[
16a - 4b + c = 7
\][/tex]
2. For [tex]\( x = -3 \)[/tex] and [tex]\( f(x) = 6 \)[/tex]:
[tex]\[
a(-3)^2 + b(-3) + c = 6
\][/tex]
This simplifies to:
[tex]\[
9a - 3b + c = 6
\][/tex]
3. For [tex]\( x = -2 \)[/tex] and [tex]\( f(x) = 7 \)[/tex]:
[tex]\[
a(-2)^2 + b(-2) + c = 7
\][/tex]
This simplifies to:
[tex]\[
4a - 2b + c = 7
\][/tex]
Now we have the following system of linear equations:
[tex]\[
\begin{cases}
16a - 4b + c = 7 \\
9a - 3b + c = 6 \\
4a - 2b + c = 7
\end{cases}
\][/tex]
To solve for [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex], we can first eliminate [tex]\( c \)[/tex]. Subtract the second equation from the first equation:
[tex]\[
(16a - 4b + c) - (9a - 3b + c) = 7 - 6
\][/tex]
Simplifies to:
[tex]\[
7a - b = 1 \quad \text{(Equation 4)}
\][/tex]
Next, subtract the third equation from the second equation:
[tex]\[
(9a - 3b + c) - (4a - 2b + c) = 6 - 7
\][/tex]
Simplifies to:
[tex]\[
5a - b = -1 \quad \text{(Equation 5)}
\][/tex]
Now, we have two simpler equations to solve:
[tex]\[
\begin{cases}
7a - b = 1 \\
5a - b = -1
\end{cases}
\][/tex]
Subtract the fifth equation from the fourth equation to eliminate [tex]\( b \)[/tex]:
[tex]\[
(7a - b) - (5a - b) = 1 - (-1)
\][/tex]
This simplifies to:
[tex]\[
2a = 2
\][/tex]
Solving for [tex]\( a \)[/tex], we get:
[tex]\[
a = 1
\][/tex]
Substitute [tex]\( a = 1 \)[/tex] back into Equation 4:
[tex]\[
7(1) - b = 1
\][/tex]
Simplifies to:
[tex]\[
7 - b = 1
\][/tex]
Solving for [tex]\( b \)[/tex], we get:
[tex]\[
b = 6
\][/tex]
Thus, the value of [tex]\( b \)[/tex] is:
[tex]\[
\boxed{6}
\][/tex]
