Question

where is the vertical asymptote(s) for the simplified function $f(x) = \frac{x + 9}{x + 1}$? options: $x = -3, x = -1$; $x = -9, x = -3$; $x = -9$; $x = -1$

Explanation:

Step1: Recall Vertical Asymptote Rule

For a rational function \( f(x)=\frac{N(x)}{D(x)} \), vertical asymptotes occur where \( D(x) = 0 \) (and the numerator is not zero there, or after simplifying if there are common factors).

Step2: Analyze the Denominator

The function is \( f(x)=\frac{x + 9}{x + 1} \). The denominator is \( x + 1 \). Set \( x + 1=0 \), solving gives \( x=-1 \). The numerator \( x + 9 \) at \( x=-1 \) is \( -1 + 9 = 8
eq0 \), so no cancellation. Thus, the vertical asymptote is at \( x=-1 \).

Answer:

\( x = -1 \) (the last option: \( x = -1 \))