The minimum value of `(x^(2)+2x+4)/(x+2)`, is

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To find the minimum value of the function \( y = \frac{x^2 + 2x + 4}{x + 2} \), we can follow these steps:

### Step 1: Set up the equation
We start with the function:
\[ y = \frac{x^2 + 2x + 4}{x + 2} \]

### Step 2: Cross multiply
To eliminate the fraction, we can cross multiply:
\[ y(x + 2) = x^2 + 2x + 4 \]
This simplifies to:
\[ xy + 2y = x^2 + 2x + 4 \]

### Step 3: Rearrange the equation
Rearranging the equation gives us:
\[ x^2 + 2x - xy + 4 - 2y = 0 \]
This can be rewritten as:
\[ x^2 + (2 - y)x + (4 - 2y) = 0 \]

### Step 4: Identify coefficients
In this quadratic equation in terms of \( x \):
- Coefficient of \( x^2 \) (a) = 1
- Coefficient of \( x \) (b) = \( 2 - y \)
- Constant term (c) = \( 4 - 2y \)

### Step 5: Use the discriminant condition
For \( x \) to have real solutions, the discriminant must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ (2 - y)^2 - 4(1)(4 - 2y) \geq 0 \]

### Step 6: Simplify the discriminant
Expanding the discriminant:
\[ (2 - y)^2 - 16 + 8y \geq 0 \]
This simplifies to:
\[ 4 - 4y + y^2 - 16 + 8y \geq 0 \]
Combining like terms gives:
\[ y^2 + 4y - 12 \geq 0 \]

### Step 7: Factor the quadratic
Factoring the quadratic:
\[ (y + 6)(y - 2) \geq 0 \]

### Step 8: Solve the inequality
To find the intervals where this inequality holds, we find the roots:
- \( y + 6 = 0 \) gives \( y = -6 \)
- \( y - 2 = 0 \) gives \( y = 2 \)

### Step 9: Test intervals
We test the intervals determined by the roots:
1. \( y < -6 \): both factors negative, product positive.
2. \( -6 < y < 2 \): one factor negative, one positive, product negative.
3. \( y > 2 \): both factors positive, product positive.

Thus, the solution to the inequality is:
\[ y \leq -6 \quad \text{or} \quad y \geq 2 \]

### Step 10: Determine the minimum value
Since we are looking for the minimum value, we find:
\[ \text{Minimum value of } y = 2 \]

### Final Answer
The minimum value of \( \frac{x^2 + 2x + 4}{x + 2} \) is \( \boxed{2} \).