The given function is:

\[
f(x) = \sqrt{2x^2 + 5x + 2}
\]

Let’s break down a few key concepts about this function.

---

### 1. **Domain of the Function**
Since \( f(x) \) contains a square root, the expression inside the square root \( 2x^2 + 5x + 2 \) must be non-negative:

\[
2x^2 + 5x + 2 \geq 0
\]

To determine the domain, we solve the quadratic inequality.

1. **Find the roots:**  
   Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 5 \), and \( c = 2 \).  
   \[
   x = \frac{-5 \pm \sqrt{5^2 - 4(2)(2)}}{2(2)}
   = \frac{-5 \pm \sqrt{9}}{4}
   = \frac{-5 \pm 3}{4}
   \]
   So the roots are:
   \[
   x = -\frac{1}{2}, \quad x = -2
   \]

2. **Solve the inequality \( 2x^2 + 5x + 2 \geq 0 \):**  
   We need to determine the sign of the quadratic expression in the intervals formed by these roots:
   - \( (-\infty, -2) \)
   - \( (-2, -\frac{1}{2}) \)
   - \( (-\frac{1}{2}, \infty) \)

   For a quadratic of the form \( ax^2 + bx + c \), if \( a > 0 \), the expression is positive outside the roots. Therefore, the solution to the inequality is:

   \[
   (-\infty, -2] \cup \left[-\frac{1}{2}, \infty\right)
   \]

   **Domain:**  
   The domain of the function is:
   \[
   (-\infty, -2] \cup \left[-\frac{1}{2}, \infty\right)
   \]

---

### 2. **Range of the Function**
The function involves a square root, so the range will be non-negative, i.e., \( f(x) \geq 0 \). The minimum value of the square root occurs when the expression inside is minimized. This is tricky because it depends on \( 2x^2 + 5x + 2 \), which is non-negative across the domain.

Therefore, the range is:

\[
[0, \infty)
\]

---

### 3. **Derivative (Slope of the Function)**
To find the rate of change or slope of the function, we need the derivative using the **chain rule**.

\[
f'(x) = \frac{1}{2\sqrt{2x^2 + 5x + 2}} \cdot \frac{d}{dx}(2x^2 + 5x + 2)
\]

\[
f'(x) = \frac{1}{2\sqrt{2x^2 + 5x + 2}} \cdot (4x + 5)
\]

Thus, the derivative is:

\[
f'(x) = \frac{4x + 5}{2\sqrt{2x^2 + 5x + 2}}
\]

---

### 4. **Behavior of the Function (Increasing/Decreasing)**
- The function is increasing when \( f'(x) > 0 \).
- This happens when \( 4x + 5 > 0 \), i.e., \( x > -\frac{5}{4} \).
- The function is decreasing when \( x < -\frac{5}{4} \).

---

### 5. **Asymptotic Behavior**
There are no vertical asymptotes since the function is defined in terms of a square root polynomial. Also, as \( x \to \infty \) or \( x \to -\infty \), the function will grow without bound, meaning:

\[
\lim_{x \to \pm \infty} f(x) = \infty
\]

---

Would you like more details or have any questions?

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### Related Questions
1. How do you solve quadratic inequalities step by step?
2. What are the rules for applying the chain rule in differentiation?
3. How do we determine where a function is increasing or decreasing?
4. How do we find the range of complex functions involving square roots?
5. What is the meaning of asymptotic behavior for polynomial-root functions?

**Tip:** When solving quadratic inequalities, always test values within each interval to confirm the sign.

Explore the square root function f(x) = sqrt(2x^2 + 5x + 2). This detailed breakdown covers its domain, range, behavior using derivatives, and quadratic inequalities. Ideal for Grades 10-12 students studying advanced algebra and calculus.

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