https://file.aiquickdraw.com/mathbot/file/1730370787754jozq14kd.jpg

The problem in the image remains the same:

\[
f(x) = \frac{1}{2x^2 - x - 6}
\]

The task is to determine the **domain** and **range** of this function. Let's summarize and verify the steps for each.

---

### Domain of \( f(x) \)

To find the domain, we need to identify values of \( x \) that make the denominator zero, as these are the points where \( f(x) \) is undefined. We start by setting the denominator equal to zero and solving for \( x \):

\[
2x^2 - x - 6 = 0
\]

#### Solving the Quadratic Equation

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = -1 \), and \( c = -6 \):

1. Substitute the values:
   \[
   x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-6)}}{2 \cdot 2}
   \]

2. Simplify under the square root:
   \[
   x = \frac{1 \pm \sqrt{1 + 48}}{4} = \frac{1 \pm \sqrt{49}}{4}
   \]

3. Since \( \sqrt{49} = 7 \), we get:
   \[
   x = \frac{1 + 7}{4} = 2 \quad \text{and} \quad x = \frac{1 - 7}{4} = -\frac{3}{2}
   \]

Therefore, the function is undefined at \( x = 2 \) and \( x = -\frac{3}{2} \).

#### Domain Conclusion

The domain of \( f(x) \) is all real numbers except \( x = 2 \) and \( x = -\frac{3}{2} \). In interval notation:

\[
\text{Domain of } f(x) = (-\infty, -\frac{3}{2}) \cup \left(-\frac{3}{2}, 2\right) \cup (2, \infty)
\]

---

### Range of \( f(x) \)

To find the range, we examine the output values of \( f(x) \), particularly by looking at its behavior at infinity and near the points where it is undefined.

#### Behavior as \( x \to \pm \infty \)

- As \( x \) becomes very large in either the positive or negative direction, the \( 2x^2 \) term in the denominator dominates, making \( f(x) \approx \frac{1}{2x^2} \), which approaches 0.
- Thus, \( f(x) \) gets closer to zero but never actually reaches it. Therefore, \( y = 0 \) is a **horizontal asymptote**.

#### Behavior Near the Vertical Asymptotes

- **At \( x = 2 \)** and **\( x = -\frac{3}{2} \)**: The function goes to \( \pm \infty \) near these values due to division by zero, indicating vertical asymptotes at these points.
- This behavior allows \( f(x) \) to take on both very large positive and negative values, suggesting that the function can reach any value except zero.

#### Range Conclusion

Since \( f(x) \) approaches all real values except zero, the range of \( f(x) \) is:

\[
\text{Range of } f(x) = (-\infty, 0) \cup (0, \infty)
\]

---

### Final Answer

- **Domain**: \( (-\infty, -\frac{3}{2}) \cup \left(-\frac{3}{2}, 2\right) \cup (2, \infty) \)
- **Range**: \( (-\infty, 0) \cup (0, \infty) \)

Would you like additional clarification on any part, or help with similar problems?

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Here are five questions that further explore related concepts:

1. How would the domain change if the function were \( f(x) = \frac{1}{x^2 - x - 6} \)?
2. What is the significance of horizontal and vertical asymptotes in determining the range?
3. Can a rational function have a range that includes zero?
4. How does factoring help in determining the points where a rational function is undefined?
5. How do transformations affect the asymptotes of rational functions?

**Tip:** When dealing with rational functions, always identify where the denominator is zero to determine the domain, then examine limits to understand the range.

Determine the domain and range of the function f(x) = 1 / (2x^2 - x - 6).

Explore how to determine the domain and range of the rational function f(x) = 1 / (2x^2 - x - 6). This solution covers essential algebraic concepts such as solving quadratic equations, identifying asymptotes, and understanding the behavior of rational functions. Ideal for high school students in Grades 10-12.

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