Question

the function $f(x) = \frac{x - 10}{x^2 - 100}$ is a rational function.
a. find all vertical asymptotes.
b. create a rough sketch of the graph near each asymptote.
a. select the correct choice and, if necessary, fill in the answer box(es) to complete your choice.
\\(\bigcirc\\) a. the function has three vertical asymptotes. the leftmost asymptote is \\(\square\\), the middle asympt (type an equations. use integers or fractions for any numbers in the equations.)
\\(\bigcirc\\) b. the function has one vertical asymptote, \\(\square\\).
(type an equation. use integers or fractions for any numbers in the equation.)
\\(\bigcirc\\) c. the function has two vertical asymptotes. the leftmost asymptote is \\(\square\\) and the rightmost as (type equations. use integers or fractions for any numbers in the equations.)
\\(\bigcirc\\) d. the function has no vertical asymptotes.
b. which graph best represents the behavior of $f(x)$ near the vertical asymptote(s)? choose the correc \\(\bigcirc\\) a. \\(\bigcirc\\) b. \\(\bigcirc\\) c.

Explanation:

Step1: Factor the denominator

The denominator of the function \( f(x)=\frac{x - 10}{x^{2}-100} \) can be factored using the difference of squares formula \( a^{2}-b^{2}=(a + b)(a - b) \). Here, \( a=x \) and \( b = 10 \), so \( x^{2}-100=(x + 10)(x - 10) \). So the function becomes \( f(x)=\frac{x - 10}{(x + 10)(x - 10)} \).

Step2: Simplify the function (remove common factors)

We can cancel out the common factor of \( (x - 10) \) from the numerator and the denominator, but we have to note that \( x
eq10 \) (since the original function is undefined at \( x = 10 \)). So the simplified function is \( f(x)=\frac{1}{x + 10} \), with a hole at \( x = 10 \).

Step3: Find vertical asymptotes

Vertical asymptotes occur where the denominator of the simplified function is zero (and the numerator is not zero there). For the simplified function \( f(x)=\frac{1}{x + 10} \), the denominator is zero when \( x+10 = 0 \), which gives \( x=-10 \). Also, we check the point \( x = 10 \): at \( x = 10 \), the original function has a hole (a removable discontinuity), not a vertical asymptote. So there is one vertical asymptote at \( x=-10 \).

Answer:

a. B. The function has one vertical asymptote, \( \boldsymbol{x=-10} \).

(For part b, since the graphs are not fully visible in a detailed way to describe the behavior, but based on the function \( f(x)=\frac{1}{x + 10} \) (with a hole at \( x = 10 \)), the graph near \( x=-10 \) should have the behavior of \( y=\frac{1}{x} \) shifted left by 10 units. So if we assume the options are typical for \( y=\frac{1}{x + 10} \) near \( x=-10 \), but since the graphs are not fully shown, we focus on part a as per the main calculation here.)