Answer :
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]
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]
