Question

given the function below:
$f(x) = \frac{x - 1}{x + 3}$
which of the following is true?
\bigcirc the horizontal asymptote is $y=-1$ and the vertical asymptote is $x=3$.
\bigcirc the horizontal asymptote is $y=1$ and the vertical asymptote is $x=-3$.
\bigcirc the horizontal asymptote is $y=3$ and the vertical asymptote is $x=-1$.
\bigcirc the horizontal asymptote is $y=-3$ and the vertical asymptote is $x=1$.

Explanation:

Step1: Find Vertical Asymptote

For a rational function \( f(x)=\frac{N(x)}{D(x)} \), vertical asymptotes occur where \( D(x) = 0 \) (and \( N(x)
eq0 \) at those points). Here, \( D(x)=x + 3 \), so set \( x+3=0 \), which gives \( x=-3 \).

Step2: Find Horizontal Asymptote

For a rational function where the degree of \( N(x) \) (degree of numerator) is equal to the degree of \( D(x) \) (degree of denominator), the horizontal asymptote is \( y=\frac{\text{leading coefficient of }N(x)}{\text{leading coefficient of }D(x)} \). The numerator \( x - 1 \) has leading coefficient 1, and the denominator \( x + 3 \) has leading coefficient 1. So horizontal asymptote is \( y=\frac{1}{1}=1 \).

Answer:

The horizontal asymptote is \( y = 1 \) and the vertical asymptote is \( x=-3 \) (the second option: The horizontal asymptote is \( y = 1 \) and the vertical asymptote is \( x=-3 \))