Question

find the horizontal asymptote of $f(x) = \frac{2x^2 + 3x + 6}{x^2 + 1}$.
options:
$y = -2$
$y = -1$
$y = 1$
$y = 2$

Explanation:

Step1: Recall Horizontal Asymptote Rule

For a rational function \( f(x)=\frac{a_nx^n + \dots + a_0}{b_mx^m + \dots + b_0} \), if \( n = m \), the horizontal asymptote is \( y=\frac{a_n}{b_m} \).

Step2: Identify Degrees and Leading Coefficients

In \( f(x)=\frac{2x^2 + 3x + 6}{x^2 + 1} \), the degree of numerator (\( n \)) and denominator (\( m \)) is 2 (so \( n = m \)). The leading coefficient of numerator \( a_n = 2 \), leading coefficient of denominator \( b_m = 1 \).

Step3: Calculate Horizontal Asymptote

Using the rule, \( y=\frac{2}{1}=2 \).

Answer:

\( y = 2 \) (corresponding to the option with \( y = 2 \))