Let's solve the problem shown in the image step by step.

Problem:

You are asked to simplify the function: $f(x) = \frac{x^2 - 25}{x + 5}$ and find any vertical asymptotes.

Step 1: Factor the numerator

The numerator $x^2 - 25$ is a difference of squares, so we can factor it as: $x^2 - 25 = (x - 5)(x + 5)$ Thus, the function becomes: $f(x) = \frac{(x - 5)(x + 5)}{x + 5}$

Step 2: Simplify the function

As long as $x + 5 \neq 0$ (i.e., $x \neq -5$), we can cancel out the $x + 5$ terms: $f(x) = x - 5$ This is the simplified form of the function, but we must note the domain restriction that $x \neq -5$, since the original function had a denominator of $x + 5$, which is undefined at $x = -5$.

Step 3: Identify the vertical asymptote

Vertical asymptotes occur where the function is undefined, which happens when the denominator is zero. In this case, the denominator $x + 5 = 0$ when $x = -5$. Therefore, there is a vertical asymptote at: $x = -5$

Conclusion:

The simplified function is $f(x) = x - 5$ with a vertical asymptote at $x = -5$.

Looking at the answer choices:

• Option d: $f(x) = x + 5$, $x \neq -5$, vertical asymptote: $x = -5$.

Thus, option d is the correct answer.

Would you like further clarification on any part of this solution?

Here are 5 related questions you can explore:

1. How do you find vertical asymptotes for rational functions in general?
2. What is the horizontal asymptote of a rational function, and does this problem have one?
3. Can you explain how the difference of squares works for factoring?
4. Why does cancelling terms change the domain of the function?
5. What happens if both the numerator and denominator approach zero at the same point?

Tip: When simplifying rational functions, always keep an eye on domain restrictions caused by the original denominator!