Let `f (x) =x ^(2) + bx + c AA in R, (b,c, in R) ` attains its least value at `x =-1` and the graph of `f (x)` cuts y-axis at `y =2.`
The value of `f (-2) + f(0) + f(1)=`

Options

A. 3
B. 5
C. 7
D. 9

Answer

To solve the problem, we need to find the values of \( f(-2) \), \( f(0) \), and \( f(1) \) for the quadratic function \( f(x) = x^2 + bx + c \), given that it attains its least value at \( x = -1 \) and cuts the y-axis at \( y = 2 \).

### Step-by-step Solution:

1. **Identify the y-intercept**:
Since the graph of \( f(x) \) cuts the y-axis at \( y = 2 \), we have:
\[ f(0) = c = 2 \]

2. **Find the vertex**:
The vertex of the quadratic function \( f(x) = x^2 + bx + c \) occurs at \( x = -\frac{b}{2} \). Given that the least value occurs at \( x = -1 \), we set:
\[ -\frac{b}{2} = -1 \]
Solving for \( b \):
\[ b = 2 \]

3. **Write the quadratic function**:
Now that we have \( b \) and \( c \), we can write the function:
\[ f(x) = x^2 + 2x + 2 \]

4. **Calculate \( f(-2) \)**:
Substitute \( x = -2 \) into the function:
\[ f(-2) = (-2)^2 + 2(-2) + 2 = 4 - 4 + 2 = 2 \]

5. **Calculate \( f(0) \)**:
We already found that \( f(0) = c = 2 \).

6. **Calculate \( f(1) \)**:
Substitute \( x = 1 \) into the function:
\[ f(1) = (1)^2 + 2(1) + 2 = 1 + 2 + 2 = 5 \]

7. **Sum the values**:
Now, we add the values we found:
\[ f(-2) + f(0) + f(1) = 2 + 2 + 5 = 9 \]

### Final Answer:
\[ f(-2) + f(0) + f(1) = 9 \]