What is the maximum value of the function
`y=(2x^(2)+3x+4)/(x^(2)+x+3)`

Answer

To find the maximum value of the function

\[ y = \frac{2x^2 + 3x + 4}{x^2 + x + 3} \]

we will follow these steps:

### Step 1: Identify the Domain
First, we need to ensure that the denominator is not equal to zero.

The denominator is \(x^2 + x + 3\). To check if it can be zero, we calculate the discriminant:

\[ D = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 3 = 1 - 12 = -11 \]

Since the discriminant is negative, \(x^2 + x + 3\) is always positive for all real values of \(x\). Therefore, the function is defined for all real numbers.

**Hint:** Always check the domain of the function to ensure it is defined.

### Step 2: Set Up the Equation
We can rewrite the function as:

\[ y = \frac{2x^2 + 3x + 4}{x^2 + x + 3} \]

To find the maximum value, we will manipulate this equation.

### Step 3: Cross-Multiply
We can cross-multiply to eliminate the fraction:

\[ y(x^2 + x + 3) = 2x^2 + 3x + 4 \]

This leads to:

\[ yx^2 + yx + 3y = 2x^2 + 3x + 4 \]

### Step 4: Rearrange the Equation
Rearranging gives us:

\[ (y - 2)x^2 + (y - 3)x + (3y - 4) = 0 \]

This is a quadratic equation in \(x\). For \(x\) to have real solutions, the discriminant must be non-negative:

\[ D = (y - 3)^2 - 4(y - 2)(3y - 4) \geq 0 \]

### Step 5: Expand the Discriminant
Expanding the discriminant:

\[ D = (y - 3)^2 - 4[(y - 2)(3y - 4)] \]

Calculating the second term:

\[ = (y - 3)^2 - 4(3y^2 - 4y - 6y + 8) \]
\[ = (y - 3)^2 - 4(3y^2 - 10y + 8) \]
\[ = (y - 3)^2 - 12y^2 + 40y - 32 \]

### Step 6: Combine Like Terms
Combining the terms gives:

\[ D = y^2 - 6y + 9 - 12y^2 + 40y - 32 \geq 0 \]
\[ = -11y^2 + 34y - 23 \geq 0 \]

### Step 7: Solve the Quadratic Inequality
To find the roots of the quadratic equation:

\[ -11y^2 + 34y - 23 = 0 \]

Using the quadratic formula:

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-34 \pm \sqrt{34^2 - 4 \cdot (-11) \cdot (-23)}}{2 \cdot (-11)} \]

Calculating the discriminant:

\[ = \frac{34 \pm \sqrt{1156 - 1012}}{-22} = \frac{34 \pm \sqrt{144}}{-22} \]
\[ = \frac{34 \pm 12}{-22} \]

Calculating the two roots:

1. \(y = \frac{46}{-22} = -\frac{23}{11}\)
2. \(y = \frac{22}{-22} = -1\)

### Step 8: Determine the Maximum Value
The quadratic opens downwards (since the coefficient of \(y^2\) is negative), so the maximum value occurs at the vertex. The vertex can be found using:

\[ y = -\frac{b}{2a} = -\frac{34}{2 \cdot -11} = \frac{34}{22} = \frac{17}{11} \]

### Conclusion
Thus, the maximum value of the function \(y\) is:

\[ \boxed{\frac{23}{11}} \]