Question

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

Explanation:

Step1: Recall Horizontal Asymptote Rule

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

• If \( \text{deg}(N(x)) < \text{deg}(D(x)) \), horizontal asymptote is \( y = 0 \).
• If \( \text{deg}(N(x))=\text{deg}(D(x)) \), horizontal asymptote is \( y=\frac{\text{leading coefficient of } N(x)}{\text{leading coefficient of } D(x)} \).
• If \( \text{deg}(N(x)) > \text{deg}(D(x)) \), no horizontal asymptote (oblique or none).

Step2: Determine Degrees of Numerator and Denominator

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

• Numerator \( N(x)=x - 2 \), degree \( \text{deg}(N(x)) = 1 \).
• Denominator \( D(x)=(x - 3)^3=x^3-9x^2 + 27x-27 \), degree \( \text{deg}(D(x)) = 3 \).

Since \( 1<3 \) (i.e., \( \text{deg}(N(x))<\text{deg}(D(x)) \)), by the rule, horizontal asymptote is \( y = 0 \).

Answer:

\( y = 0 \)