To find the vertical asymptotes of the function:

$f(x) = \frac{x^2 - 25}{(x - 5)(x^2 + 1)}$

Step 1: Identify Points that Make the Denominator Zero

Vertical asymptotes occur where the denominator is zero, provided the numerator does not also become zero at the same points (which would indicate a removable discontinuity).

The denominator is: $(x - 5)(x^2 + 1)$

Set each factor equal to zero:

1. $x - 5 = 0 \Rightarrow x = 5$
2. $x^2 + 1 = 0 \Rightarrow x^2 = -1$

The equation $x^2 = -1$ has no real solutions (since the square of a real number cannot be negative). Therefore, $x^2 + 1 = 0$ does not contribute any real vertical asymptotes.

Step 2: Check the Numerator for Removable Discontinuities

The numerator is: $x^2 - 25 = (x + 5)(x - 5)$

This shows that $x = 5$ is a root of both the numerator and the denominator. Therefore, $x = 5$ does not produce a vertical asymptote but rather a removable discontinuity.

Conclusion

Since neither $x = 5$ nor $x^2 + 1 = 0$ leads to a vertical asymptote in the real number system, there are no vertical asymptotes for this function.

Answer

The correct answer is: E. There is no vertical asymptote

Would you like a further breakdown of any part of this solution?

Here are some related questions to explore:

1. What are removable discontinuities, and how are they identified?
2. How do vertical asymptotes differ from horizontal asymptotes?
3. Can imaginary numbers ever contribute to vertical asymptotes?
4. How do you find horizontal asymptotes in rational functions?
5. What is the significance of holes in the graph of a function?

Tip: When factoring a rational function, always examine both the numerator and the denominator for shared factors, as they may indicate removable discontinuities rather than vertical asymptotes.