The function given is:

$y = \frac{x^2 + 4}{x^2 - 1}$

Vertical Asymptotes:

Vertical asymptotes occur where the denominator is zero (i.e., where the function is undefined), provided the numerator is not zero at the same points.

The denominator is $x^2 - 1$, which can be factored as: $x^2 - 1 = (x - 1)(x + 1)$ Setting the denominator equal to zero gives the vertical asymptotes: $x = 1 \quad \text{and} \quad x = -1$

Horizontal Asymptotes:

To find the horizontal asymptote, we analyze the degree of the numerator and denominator. Both the numerator and denominator are degree 2 polynomials.

For rational functions where the degree of the numerator and denominator are equal, the horizontal asymptote is found by taking the ratio of the leading coefficients. The leading coefficient of $x^2$ in both the numerator and denominator is 1, so the horizontal asymptote is: $y = \frac{1}{1} = 1$

Conclusion:

• Vertical asymptotes: $x = -1$ and $x = 1$
• Horizontal asymptote: $y = 1$

Would you like more details or have any questions?

Here are 5 related questions for practice:

1. How would the asymptotes change if the numerator were $x^3 + 4$?
2. What happens to the graph near a vertical asymptote?
3. Can a rational function have no vertical asymptotes? If so, when?
4. How can you determine a slant asymptote for a rational function?
5. How do holes in the graph of a rational function relate to vertical asymptotes?

Tip: To identify vertical asymptotes, always check where the denominator equals zero, as these points indicate where the function is undefined.