Question

what is the horizontal asymptote of the function $f(x)=\frac{(x - 2)}{(x - 3)^2}$?
$\bigcirc$ $y = 0$
$\bigcirc$ $y = 1$
$\bigcirc$ $y = 2$
$\bigcirc$ $y = 3$

Explanation:

Step1: Recall horizontal asymptote rules

For a rational function \( f(x)=\frac{N(x)}{D(x)} \), where \( N(x) \) is the numerator and \( D(x) \) is the denominator:

• If the degree of \( N(x) \) (\( n \)) is less than the degree of \( D(x) \) (\( d \)), the horizontal asymptote is \( y = 0 \).
• If \( n = d \), the horizontal asymptote is \( y=\frac{\text{leading coefficient of } N(x)}{\text{leading coefficient of } D(x)} \).
• If \( n>d \), there is no horizontal asymptote (but there may be an oblique one).

Step2: Determine degrees of numerator and denominator

For \( f(x)=\frac{x - 2}{(x - 3)^2}=\frac{x - 2}{x^2-6x + 9} \):

• The numerator \( N(x)=x - 2 \) is a polynomial of degree \( n = 1 \) (the highest power of \( x \) is 1).
• The denominator \( D(x)=(x - 3)^2=x^2-6x + 9 \) is a polynomial of degree \( d = 2 \) (the highest power of \( x \) is 2).

Since \( n = 1

Answer:

\( y = 0 \) (corresponding to the option "y = 0")