The function in the image is $y = \frac{2x^2 + 1}{x(x - 2)}$.

Here are the steps to solve each part:

a. Horizontal Asymptote(s)

To find the horizontal asymptote(s), we analyze the behavior of the function as $x \to \infty$ and $x \to -\infty$.

1. Degree of the numerator: The numerator $2x^2 + 1$ has degree 2 (the highest power of $x$ is $x^2$).
2. Degree of the denominator: The denominator $x(x - 2) = x^2 - 2x$ also has degree 2 (the highest power of $x$ is $x^2$).

When the degrees of the numerator and denominator are equal, the horizontal asymptote is given by the ratio of the leading coefficients of the highest degree terms.

The leading coefficient of the numerator is 2, and the leading coefficient of the denominator is 1. Therefore, the horizontal asymptote is:

$y = \frac{2}{1} = 2$

b. Right-hand side and left-hand side limits as $x \to 2$

We now examine the limits of the function as $x$ approaches 2 from both sides:

• Right-hand limit: As $x \to 2^+$, the denominator $x(x - 2)$ approaches 0 from the positive side because $x - 2$ is positive but small.
• Left-hand limit: As $x \to 2^-$, the denominator approaches 0 from the negative side because $x - 2$ is negative but small.

In both cases, since the numerator remains finite, the limits approach infinity but with opposite signs:

$\lim_{x \to 2^+} y = +\infty \quad \text{and} \quad \lim_{x \to 2^-} y = -\infty$

So there is a vertical asymptote at $x = 2$.

Would you like more details on any part of this?

Here are 5 related questions you might want to consider:

1. What happens at other critical points like $x = 0$?
2. How would you find vertical asymptotes for other rational functions?
3. Can you explain why there are no horizontal asymptotes at other values of $x$?
4. How do the right and left-hand limits differ for a discontinuity?
5. What is the significance of the leading term when finding horizontal asymptotes?

Tip: When analyzing rational functions, always compare the degrees of the numerator and denominator first to find horizontal asymptotes.