If `y=25-x^(2) and 1lexle5`, what is the smallest possible value of y?

Options

A. `0`
B. `1`
C. `5`
D. `10`

Answer

To find the smallest possible value of \( y \) given the equation \( y = 25 - x^2 \) and the constraint \( 1 \leq x \leq 5 \), we can follow these steps:

### Step 1: Understand the function
The function \( y = 25 - x^2 \) is a downward-opening parabola. As \( x \) increases, \( x^2 \) increases, which means \( y \) will decrease.

### Step 2: Identify the range of \( x \)
We are given that \( x \) can take values from 1 to 5, inclusive. Therefore, we will evaluate \( y \) at the endpoints of this interval.

### Step 3: Calculate \( y \) at the endpoints
1. **When \( x = 1 \)**:
\[ y = 25 - (1^2) = 25 - 1 = 24 \]

2. **When \( x = 5 \)**:
\[ y = 25 - (5^2) = 25 - 25 = 0 \]

### Step 4: Compare the values
Now we compare the values of \( y \) at the endpoints:
- At \( x = 1 \), \( y = 24 \)
- At \( x = 5 \), \( y = 0 \)

### Step 5: Determine the smallest value of \( y \)
The smallest value of \( y \) in the range \( 1 \leq x \leq 5 \) is:
\[ \text{Smallest value of } y = 0 \]

### Final Answer
The smallest possible value of \( y \) is \( \boxed{0} \).

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