If `f(x) = ax^(2)+bx+c, f(-1)=-18`, and `f(1)=10`, what is the value of b ?

Options

A. `-12`
B. `-4`
C. 14
D. 21

Explanation

To solve for the value of \( b \) in the function \( f(x) = ax^2 + bx + c \) given the conditions \( f(-1) = -18 \) and \( f(1) = 10 \), we can follow these steps:

### Step 1: Set up the equations from the given conditions
We know:
1. \( f(-1) = -18 \)
2. \( f(1) = 10 \)

Using the function \( f(x) = ax^2 + bx + c \):

For \( f(-1) \):
\[ f(-1) = a(-1)^2 + b(-1) + c = a - b + c = -18 \]
This gives us our first equation:
\[ a - b + c = -18 \quad \text{(1)} \]

For \( f(1) \):
\[ f(1) = a(1)^2 + b(1) + c = a + b + c = 10 \]
This gives us our second equation:
\[ a + b + c = 10 \quad \text{(2)} \]

### Step 2: Subtract the two equations
Now, we will subtract equation (1) from equation (2) to eliminate \( a \) and \( c \):
\[ (a + b + c) - (a - b + c) = 10 - (-18) \]
This simplifies to:
\[ a + b + c - a + b - c = 10 + 18 \]
\[ 2b = 28 \]

### Step 3: Solve for \( b \)
Now, divide both sides by 2:
\[ b = \frac{28}{2} = 14 \]

### Final Answer
Thus, the value of \( b \) is:
\[ \boxed{14} \]