The function given in the image is:

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

Domain of $f(x)$

To find the domain, we need to identify the values of $x$ for which the function is undefined. The function becomes undefined when the denominator is zero. So, we set:

$2x^2 - x - 6 = 0$

Solving this quadratic equation will give the values of $x$ that make the function undefined.

Solving for $x$:

1. Use the quadratic formula: $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-6)}}{2 \cdot 2}$ Simplifying this: $x = \frac{1 \pm \sqrt{1 + 48}}{4}$ $x = \frac{1 \pm \sqrt{49}}{4}$ $x = \frac{1 \pm 7}{4}$

2. This gives: $x = \frac{8}{4} = 2 \quad \text{and} \quad x = \frac{-6}{4} = -\frac{3}{2}$

Thus, the function is undefined at $x = 2$ and $x = -\frac{3}{2}$. Therefore, the domain of $f(x)$ is all real numbers except $x = 2$ and $x = -\frac{3}{2}$.

Range of $f(x)$

The range requires analyzing the behavior of the function as $x$ approaches certain values.

1. As $x \to \pm \infty$, $f(x)$ approaches 0 because the degree of the polynomial in the denominator is higher than that in the numerator.
2. Near the values $x = 2$ and $x = -\frac{3}{2}$, $f(x)$ will tend toward $\pm \infty$, indicating vertical asymptotes.

Thus, the range of $f(x)$ is all real values $y$, $y \neq 0$.

Would you like a more detailed breakdown or have any questions?

Here are five related questions for further exploration:

1. How would you find vertical and horizontal asymptotes for more complex functions?
2. Can a rational function have holes in its graph, and how would you identify them?
3. What impact does the degree of the numerator and denominator have on the function's behavior?
4. How would you determine the end behavior of polynomial functions?
5. How do you apply limits to find the asymptotic behavior of rational functions?

Tip: When analyzing rational functions, always start by factoring the numerator and denominator if possible; it can simplify finding zeros and undefined points.