Question

find the vertical asymptotes, if any, and the value of x corresponding to holes, if any, of the graph of the following rational function.

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

select the correct choice below and, if necessary, fill in the answer box to complete your choice. (type an integer or a fraction. use commas to separate answers as needed.)

a. vertical asymptote(s) at x = \square and hole(s) corresponding to x = \square
b. vertical asymptote(s) at x = \square. there are no holes.
c. there are no vertical asymptotes but there is (are) hole(s) corresponding to x = \square
d. there are no discontinuities.

Explanation:

Step1: Factor the numerator

The numerator \(x^2 - 25\) is a difference of squares, so it factors as \((x - 5)(x + 5)\). So the function becomes \(f(x)=\frac{(x - 5)(x + 5)}{x - 5}\).

Step2: Simplify the function (excluding \(x = 5\))

We can cancel out the common factor of \(x - 5\) from the numerator and the denominator, but we must note that \(x
eq5\) (since the original function is undefined at \(x = 5\)). So the simplified function is \(f(x)=x + 5\) for \(x
eq5\).

Step3: Analyze vertical asymptotes and holes

A vertical asymptote occurs where the denominator is zero and the numerator is not zero (after simplifying). A hole occurs where a factor cancels out from both the numerator and the denominator. Here, the factor \(x - 5\) cancels out, so there is a hole at \(x = 5\) (since that's where the original function is undefined but the simplified function is defined everywhere else except there). And since after canceling, the denominator has no remaining factors that make it zero (the simplified function is a linear function with no denominator issues), there are no vertical asymptotes.

Answer:

C. There are no vertical asymptotes but there is a hole corresponding to \(x = 5\)