Question

1. what is the vertical asymptote of \\(\frac{3x + 1}{x + 5}\\)?

1. what is the horizontal asymptote of \\(\frac{3x + 1}{x + 5}\\)?

Explanation:

Question 3

Step1: Recall vertical asymptote rule

For a rational function \( \frac{f(x)}{g(x)} \), vertical asymptotes occur where \( g(x) = 0 \) (and \( f(x)
eq 0 \) at those points, which is true here as \( 3(-5)+1=-14
eq0 \)).

Step2: Solve denominator for zero

Set \( x + 5 = 0 \). Solving gives \( x=-5 \).

Step1: Recall horizontal asymptote rule

For a rational function \( \frac{ax^n + \dots}{bx^m + \dots} \), if \( n = m \), the horizontal asymptote is \( y=\frac{a}{b} \). Here, \( n = m = 1 \), \( a = 3 \), \( b = 1 \).

Step2: Apply the rule

So horizontal asymptote is \( y=\frac{3}{1}=3 \).

Answer:

\( x = -5 \)

Question 4