Find a polynomial f(x) of degree 2 where f(0)=8, f(1)=12, f(2)=18

Options

A. ` x^(2)+3x-8`
B. `x^(2)-3x+8`
C. `2x^(2)-x+3`
D. `x^(2)+3x+8`

Explanation

To find a polynomial \( f(x) \) of degree 2 that satisfies the conditions \( f(0) = 8 \), \( f(1) = 12 \), and \( f(2) = 18 \), we can follow these steps:

### Step 1: Assume the form of the polynomial
Let the polynomial be of the form:
\[ f(x) = ax^2 + bx + c \]

### Step 2: Use the first condition \( f(0) = 8 \)
Substituting \( x = 0 \) into the polynomial:
\[ f(0) = a(0)^2 + b(0) + c = c \]
Given that \( f(0) = 8 \), we have:
\[ c = 8 \]

### Step 3: Use the second condition \( f(1) = 12 \)
Substituting \( x = 1 \):
\[ f(1) = a(1)^2 + b(1) + c = a + b + c \]
Since \( c = 8 \), we can substitute this value:
\[ f(1) = a + b + 8 = 12 \]
This simplifies to:
\[ a + b = 12 - 8 = 4 \quad \text{(Equation 1)} \]

### Step 4: Use the third condition \( f(2) = 18 \)
Substituting \( x = 2 \):
\[ f(2) = a(2)^2 + b(2) + c = 4a + 2b + c \]
Again substituting \( c = 8 \):
\[ f(2) = 4a + 2b + 8 = 18 \]
This simplifies to:
\[ 4a + 2b = 18 - 8 = 10 \quad \text{(Equation 2)} \]

### Step 5: Solve the system of equations
We now have two equations:
1. \( a + b = 4 \)
2. \( 4a + 2b = 10 \)

From Equation 1, we can express \( b \) in terms of \( a \):
\[ b = 4 - a \]
Substituting this into Equation 2:
\[ 4a + 2(4 - a) = 10 \]
Expanding and simplifying:
\[ 4a + 8 - 2a = 10 \]
\[ 2a + 8 = 10 \]
\[ 2a = 10 - 8 = 2 \]
\[ a = 1 \]

### Step 6: Find \( b \)
Using the value of \( a \) in Equation 1:
\[ 1 + b = 4 \]
\[ b = 4 - 1 = 3 \]

### Step 7: Write the polynomial
Now we have:
- \( a = 1 \)
- \( b = 3 \)
- \( c = 8 \)

Thus, the polynomial is:
\[ f(x) = 1x^2 + 3x + 8 = x^2 + 3x + 8 \]

### Final Answer
The polynomial \( f(x) \) is:
\[ \boxed{x^2 + 3x + 8} \]