Question

1. what is the vertical asymptote of $y = \frac{3}{x + 5}$?
2. what is the horizontal asymptote of $y = \frac{3}{x + 5}$?

Explanation:

Sub - question 1:

Step 1: Recall the rule for vertical asymptotes of rational functions

For a rational function \(y=\frac{f(x)}{g(x)}\), the vertical asymptotes occur where \(g(x) = 0\) (provided that \(f(x)\) and \(g(x)\) have no common factors). In the function \(y=\frac{3}{x + 5}\), the denominator is \(g(x)=x + 5\).

Step 2: Solve for \(x\) when the denominator is zero

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

Step 1: Recall the rule for horizontal asymptotes of rational functions

For a rational function \(y=\frac{f(x)}{g(x)}\), if the degree of the numerator \(f(x)\) is less than the degree of the denominator \(g(x)\), the horizontal asymptote is \(y = 0\). In the function \(y=\frac{3}{x + 5}\), the degree of the numerator (which is a constant, so degree 0) is less than the degree of the denominator (degree 1).

Step 2: Determine the horizontal asymptote

Based on the rule, the horizontal asymptote is \(y = 0\).

Answer:

The vertical asymptote is \(x = - 5\)

Sub - question 2: