For all the non-negative real x, what is the minimum and maximum non-negative value of y, where. `y= -x^2+8x+20`?

Options

A. 0,36
B. 18,36
C. 0,20
D. none of these

Answer

To find the minimum and maximum non-negative values of the function \( y = -x^2 + 8x + 20 \) for non-negative real values of \( x \), we can follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation is in the standard form \( y = ax^2 + bx + c \), where:
- \( a = -1 \)
- \( b = 8 \)
- \( c = 20 \)

### Step 2: Determine the vertex
Since \( a \) is negative, the parabola opens downwards, and the maximum value occurs at the vertex. The x-coordinate of the vertex can be found using the formula:
\[ x = -\frac{b}{2a} \]
Substituting the values of \( a \) and \( b \):
\[ x = -\frac{8}{2 \times -1} = \frac{8}{2} = 4 \]

### Step 3: Calculate the maximum value of \( y \)
Now, substitute \( x = 4 \) back into the equation to find the maximum value of \( y \):
\[ y = -4^2 + 8 \times 4 + 20 \]
Calculating this:
\[ y = -16 + 32 + 20 = 36 \]
Thus, the maximum non-negative value of \( y \) is \( 36 \).

### Step 4: Find the minimum non-negative value of \( y \)
To find the minimum non-negative value, we need to evaluate \( y \) at the endpoints of the domain of \( x \) (which is non-negative). We will check \( y \) at \( x = 0 \) and \( x = 4 \) (the vertex) and also consider \( x \) approaching infinity.

1. **At \( x = 0 \)**:
\[ y = -0^2 + 8 \times 0 + 20 = 20 \]

2. **At \( x = 4 \)** (already calculated):
\[ y = 36 \]

3. **As \( x \) approaches infinity**:
Since the parabola opens downwards, \( y \) will decrease indefinitely, tending towards negative infinity. However, since we are only interested in non-negative values, we can conclude that the minimum non-negative value occurs at the lowest point we calculated, which is \( 20 \).

### Conclusion
The maximum non-negative value of \( y \) is \( 36 \) and the minimum non-negative value of \( y \) is \( 20 \).