We are asked to find the horizontal asymptote of the function $f(x) = \frac{1-6x^2-x}{2x^2-x+2}$.

1. Problem Description

We are asked to find the horizontal asymptote of the function $f(x) = \frac{1-6x^2-x}{2x^2-x+2}$.

2. Solution Steps

To find the horizontal asymptote of a rational function, we need to examine the limit of the function as $x$ approaches infinity. We can do this by dividing both the numerator and denominator by the highest power of $x$ that appears in the function, which in this case is $x^2$.

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

Now, we take the limit as $x$ approaches infinity:

$$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{\frac{1}{x^2} - 6 - \frac{1}{x}}{2 - \frac{1}{x} + \frac{2}{x^2}}$$

As $x$ approaches infinity, $\frac{1}{x}$ and $\frac{1}{x^2}$ approach

0. Therefore,

$$\lim_{x \to \infty} f(x) = \frac{0 - 6 - 0}{2 - 0 + 0} = \frac{-6}{2} = -3$$

Thus, the horizontal asymptote is $y = -3$.

3. Final Answer

The horizontal asymptote is $y = -3$.