If f(x) is a polynomial function of the second degree such that, `f(-3)=6,f(0)=6" and "f(2)=11`, then the graph of the function, f(x) cuts the ordinate x = 1 at the point

Options

A. (1, 8)
B. (1, 4)
C. (1, -2)
D. (1, 9)

Answer

To solve the problem step by step, we will follow the process of determining the coefficients of the polynomial function \( f(x) \) based on the given conditions, and then we will find the value of \( f(1) \).

### Step 1: Assume the form of the polynomial
Since \( f(x) \) is a polynomial function of the second degree, we can express it in the standard form:
\[ f(x) = ax^2 + bx + c \]

### Step 2: Use the first condition \( f(-3) = 6 \)
Substituting \( x = -3 \) into the polynomial:
\[ f(-3) = a(-3)^2 + b(-3) + c = 6 \]
This simplifies to:
\[ 9a - 3b + c = 6 \quad \text{(Equation 1)} \]

### Step 3: Use the second condition \( f(0) = 6 \)
Substituting \( x = 0 \):
\[ f(0) = c = 6 \quad \text{(Equation 2)} \]

### Step 4: Use the third condition \( f(2) = 11 \)
Substituting \( x = 2 \):
\[ f(2) = a(2)^2 + b(2) + c = 11 \]
This simplifies to:
\[ 4a + 2b + c = 11 \quad \text{(Equation 3)} \]

### Step 5: Substitute \( c = 6 \) into Equations 1 and 3
Substituting \( c = 6 \) into Equation 1:
\[ 9a - 3b + 6 = 6 \implies 9a - 3b = 0 \implies 3a = b \quad \text{(Equation 4)} \]

Substituting \( c = 6 \) into Equation 3:
\[ 4a + 2b + 6 = 11 \implies 4a + 2b = 5 \quad \text{(Equation 5)} \]

### Step 6: Substitute \( b = 3a \) from Equation 4 into Equation 5
Substituting \( b = 3a \) into Equation 5:
\[ 4a + 2(3a) = 5 \implies 4a + 6a = 5 \implies 10a = 5 \implies a = \frac{1}{2} \]

### Step 7: Find \( b \) using \( a \)
Using \( a = \frac{1}{2} \) in Equation 4:
\[ b = 3a = 3 \times \frac{1}{2} = \frac{3}{2} \]

### Step 8: Write the polynomial function
Now we have:
\[ a = \frac{1}{2}, \quad b = \frac{3}{2}, \quad c = 6 \]
Thus, the polynomial function is:
\[ f(x) = \frac{1}{2}x^2 + \frac{3}{2}x + 6 \]

### Step 9: Find \( f(1) \)
Now we need to find the value of \( f(1) \):
\[ f(1) = \frac{1}{2}(1)^2 + \frac{3}{2}(1) + 6 \]
Calculating this gives:
\[ f(1) = \frac{1}{2} + \frac{3}{2} + 6 = 2 + 6 = 8 \]

### Conclusion
The graph of the function \( f(x) \) cuts the ordinate at \( x = 1 \) at the point \( (1, 8) \).