To determine the vertical asymptotes of the function

$f(x) = \frac{x^2 + 7}{(x^2 - 1)(x^2 - 81)},$

we look for the values of $x$ that make the denominator equal to zero, as these are the points where the function may have vertical asymptotes (provided the numerator does not also become zero at those points).

Step 1: Analyze the denominator

The denominator is $(x^2 - 1)(x^2 - 81)$. Factorize each term: $x^2 - 1 = (x - 1)(x + 1),$ $x^2 - 81 = (x - 9)(x + 9).$ Thus, the denominator becomes: $(x - 1)(x + 1)(x - 9)(x + 9).$

Step 2: Identify the critical points

The critical points where the denominator equals zero are: $x = 1, \, x = -1, \, x = 9, \, x = -9.$

Step 3: Check the numerator

The numerator is $x^2 + 7$. This expression is always positive (as $x^2 \geq 0$ and $7 > 0$), so it does not become zero at the critical points of the denominator. Therefore, the function has vertical asymptotes at all these points.

Step 4: Final Answer

The function has four vertical asymptotes at: $x = -9, \, x = -1, \, x = 1, \, \text{and } x = 9.$

The correct choice is: D. The function has four vertical asymptotes. The asymptotes in order from leftmost to rightmost are: $x = -9, \, x = -1, \, x = 1, \, \text{and } x = 9.$

---

Would you like a detailed explanation of why vertical asymptotes occur or have any questions about the process? Here are 5 related questions to deepen your understanding:

1. What conditions must be satisfied for a function to have a vertical asymptote?
2. How do you determine whether a function has removable discontinuities instead of vertical asymptotes?
3. What happens to $f(x)$ as $x$ approaches a vertical asymptote?
4. How would the asymptotes change if the numerator also had a factor common with the denominator?
5. Can a vertical asymptote occur for a rational function if the degree of the numerator is greater than the denominator? Why or why not?

Tip: Always check both the numerator and denominator when determining vertical asymptotes to ensure no common factors simplify the expression.