To find the horizontal asymptote of the function [tex]\( f(x) = \frac{x^2 + 3x + 6}{x^2 + 1} \)[/tex], we need to compare the degrees of the polynomials in the numerator and the denominator.
1. Identify the degrees of the polynomials:
- The degree of the numerator [tex]\( x^2 + 3x + 6 \)[/tex] is 2.
- The degree of the denominator [tex]\( x^2 + 1 \)[/tex] is also 2.
2. Compare the degrees:
- Since the degrees of the numerator and the denominator are equal, the rule for finding the horizontal asymptote is to take the ratio of the leading coefficients of the highest degree terms.
3. Identify the leading coefficients:
- The leading coefficient of the numerator (the coefficient of [tex]\( x^2 \)[/tex]) is 1.
- The leading coefficient of the denominator (the coefficient of [tex]\( x^2 \)[/tex]) is also 1.
4. Calculate the horizontal asymptote:
- The horizontal asymptote is given by the ratio of these leading coefficients:
[tex]\[
y = \frac{1}{1} = 1
\][/tex]
Therefore, the horizontal asymptote of the function [tex]\( f(x) = \frac{x^2 + 3x + 6}{x^2 + 1} \)[/tex] is [tex]\( y = 1 \)[/tex].
The correct answer is:
[tex]\[ y = 1 \][/tex]
