Minimum value of `y = x ^(2) -3x+5, x in [-4, 1]` is:

Options

A. 3
B. `11/4`
C. `0`
D. 9

Answer

To find the minimum value of the function \( y = x^2 - 3x + 5 \) for \( x \) in the interval \([-4, 1]\), we can follow these steps:

### Step 1: Rewrite the quadratic function in vertex form
The given quadratic function can be rewritten in vertex form by completing the square.

\[ y = x^2 - 3x + 5 \]

To complete the square, we take the coefficient of \( x \), which is \(-3\), halve it to get \(-\frac{3}{2}\), and then square it to get \(\left(-\frac{3}{2}\right)^2 = \frac{9}{4}\).

Now we can rewrite the function:

\[ y = \left(x^2 - 3x + \frac{9}{4}\right) + 5 - \frac{9}{4} \]

This simplifies to:

\[ y = \left(x - \frac{3}{2}\right)^2 + \frac{11}{4} \]

### Step 2: Identify the vertex
The vertex of the parabola represented by the function \( y = \left(x - \frac{3}{2}\right)^2 + \frac{11}{4} \) occurs at \( x = \frac{3}{2} \). The minimum value of \( y \) occurs at this vertex, which is \( y = \frac{11}{4} \).

### Step 3: Check the interval
However, we need to find the minimum value of \( y \) within the interval \([-4, 1]\). Since \( \frac{3}{2} \) (or \( 1.5 \)) is not in the interval \([-4, 1]\), we need to evaluate the function at the endpoints of the interval.

### Step 4: Evaluate at the endpoints
1. Evaluate at \( x = -4 \):
\[ y(-4) = (-4)^2 - 3(-4) + 5 = 16 + 12 + 5 = 33 \]

2. Evaluate at \( x = 1 \):
\[ y(1) = (1)^2 - 3(1) + 5 = 1 - 3 + 5 = 3 \]

### Step 5: Compare the values
Now we compare the values obtained from the evaluations:
- \( y(-4) = 33 \)
- \( y(1) = 3 \)

The minimum value of \( y \) in the interval \([-4, 1]\) is:

\[ \text{Minimum value} = 3 \quad \text{(at } x = 1\text{)} \]

### Final Answer
Thus, the minimum value of \( y = x^2 - 3x + 5 \) for \( x \) in the interval \([-4, 1]\) is \( 3 \).

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