Certainly! Let's find the horizontal asymptote for the function [tex]\( f(x) = \frac{x^2 + 3x + 6}{x^2 + 1} \)[/tex].
To determine the horizontal asymptote for a rational function, we need to compare the degrees of the polynomial in the numerator and the polynomial in the denominator.
1. Identify the degrees of the polynomials:
- The degree of the numerator [tex]\( x^2 + 3x + 6 \)[/tex] is [tex]\( 2 \)[/tex] (since the highest power of [tex]\( x \)[/tex] is [tex]\( x^2 \)[/tex]).
- The degree of the denominator [tex]\( x^2 + 1 \)[/tex] is [tex]\( 2 \)[/tex] (since the highest power of [tex]\( x \)[/tex] is [tex]\( x^2 \)[/tex]).
2. Compare the degrees:
- When the degrees of the numerator and the denominator are the same, the horizontal asymptote is given by the ratio of the leading coefficients.
3. Determine the leading coefficients:
- The leading coefficient of the numerator [tex]\( x^2 + 3x + 6 \)[/tex] is [tex]\( 1 \)[/tex].
- The leading coefficient of the denominator [tex]\( x^2 + 1 \)[/tex] is [tex]\( 1 \)[/tex].
4. Calculate the horizontal asymptote:
- The horizontal asymptote is [tex]\( \frac{1}{1} = 1 \)[/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]
