The table represents a quadratic function [tex]\( f(x) \)[/tex].

[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $f(x)$ \\
\hline
-4 & 7 \\
\hline
-3 & 6 \\
\hline
-2 & 7 \\
\hline
-1 & 10 \\
\hline
0 & 15 \\
\hline
1 & 22 \\
\hline
2 & 31 \\
\hline
\end{tabular}
\][/tex]

If the equation of the function [tex]\( f(x) \)[/tex] is written in standard form [tex]\( f(x) = ax^2 + bx + c \)[/tex], what is the value of [tex]\( b \)[/tex]?

A. 3
B. 6
C. 16
D. 22



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]