If `f(x)` is a quadratic function such that `f(0)=3,f(1)=6 and f(2)=11`, then: `f(x)=`

Options

A. `2x^3+x+3`
B. `x^3+x+3`
C. `x^3+2x+3`
D. none of these.

Answer

To find the quadratic function \( f(x) \) given the values \( f(0) = 3 \), \( f(1) = 6 \), and \( f(2) = 11 \), we can follow these steps:

### Step 1: Assume the form of the quadratic function
We know that a quadratic function can be expressed in the form:
\[ f(x) = ax^2 + bx + c \]
where \( a \), \( b \), and \( c \) are constants.

### Step 2: Use the given values to create equations
1. From \( f(0) = 3 \):
\[ f(0) = a(0)^2 + b(0) + c = c = 3 \]
Thus, we have:
\[ c = 3 \]

2. From \( f(1) = 6 \):
\[ f(1) = a(1)^2 + b(1) + c = a + b + c = 6 \]
Substituting \( c = 3 \):
\[ a + b + 3 = 6 \implies a + b = 3 \quad \text{(Equation 1)} \]

3. From \( f(2) = 11 \):
\[ f(2) = a(2)^2 + b(2) + c = 4a + 2b + c = 11 \]
Substituting \( c = 3 \):
\[ 4a + 2b + 3 = 11 \implies 4a + 2b = 8 \implies 2a + b = 4 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have two equations:
1. \( a + b = 3 \) (Equation 1)
2. \( 2a + b = 4 \) (Equation 2)

We can solve these equations simultaneously.

Subtract Equation 1 from Equation 2:
\[ (2a + b) - (a + b) = 4 - 3 \]
This simplifies to:
\[ 2a + b - a - b = 1 \implies a = 1 \]

### Step 4: Substitute back to find \( b \)
Now that we have \( a = 1 \), we can substitute it back into Equation 1:
\[ 1 + b = 3 \implies b = 3 - 1 = 2 \]

### Step 5: Write the final quadratic function
Now we have all the coefficients:
- \( a = 1 \)
- \( b = 2 \)
- \( c = 3 \)

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

### Final Answer:
\[ f(x) = x^2 + 2x + 3 \]

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