For the non negative real x, such that `xge-11` what is the minimium and maximum value of y, where `y=x^2+12x+27`?

Options

A. `-9,16`
B. `9,27`
C. 0,27
D. none of these

Answer

To solve the problem, we need to find the minimum and maximum values of the function \( y = x^2 + 12x + 27 \) for non-negative real values of \( x \) such that \( x \geq -11 \).

### Step-by-Step Solution:

1. **Identify the function**:
We have the quadratic function \( y = x^2 + 12x + 27 \).

2. **Determine the vertex of the parabola**:
The vertex of a quadratic function \( ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \).
Here, \( a = 1 \) and \( b = 12 \).
\[ x = -\frac{12}{2 \cdot 1} = -6 \]

3. **Evaluate the function at the vertex**:
Now, we substitute \( x = -6 \) into the function to find the corresponding value of \( y \):
\[ y = (-6)^2 + 12(-6) + 27 \]
\[ y = 36 - 72 + 27 = -9 \]

4. **Check the constraints**:
Since we are only considering non-negative values of \( x \) (i.e., \( x \geq 0 \)), we need to evaluate the function at the endpoints of our constraint. The minimum value of \( x \) is \( 0 \).

5. **Evaluate the function at \( x = 0 \)**:
\[ y = 0^2 + 12(0) + 27 = 27 \]

6. **Determine the maximum value**:
Since the parabola opens upwards (as the coefficient of \( x^2 \) is positive), the minimum value occurs at \( x = 0 \) and the maximum value occurs at the vertex \( x = -6 \) which is outside our range. Thus, the maximum value for \( x \geq 0 \) is \( y = 27 \).

7. **Conclusion**:
The minimum value of \( y \) is \( -9 \) (at \( x = -6 \), but not in the range) and the maximum value of \( y \) is \( 27 \) (at \( x = 0 \)).

### Final Answer:
- Minimum Value: \( -9 \) (not valid in the range)
- Maximum Value: \( 27 \)