To determine the horizontal asymptote of the rational function [tex]\( g(x) = \frac{10x^2}{2x^2 + 1} \)[/tex], we should analyze the behavior of [tex]\( g(x) \)[/tex] as [tex]\( x \)[/tex] approaches infinity, both positively and negatively.
A rational function [tex]\( \frac{P(x)}{Q(x)} \)[/tex] will have different types of horizontal asymptotes depending on the degrees of the polynomial in the numerator (top, [tex]\( P(x) \)[/tex]) and the polynomial in the denominator (bottom, [tex]\( Q(x) \)[/tex]):
1. If the degree of [tex]\( P(x) \)[/tex] is less than the degree of [tex]\( Q(x) \)[/tex], the horizontal asymptote is [tex]\( y = 0 \)[/tex].
2. If the degree of [tex]\( P(x) \)[/tex] is equal to the degree of [tex]\( Q(x) \)[/tex], the horizontal asymptote is found by dividing the leading coefficients of [tex]\( P(x) \)[/tex] and [tex]\( Q(x) \)[/tex].
3. If the degree of [tex]\( P(x) \)[/tex] is greater than the degree of [tex]\( Q(x) \)[/tex], there is no horizontal asymptote (though there may be an oblique asymptote).
In the given function [tex]\( g(x) = \frac{10x^2}{2x^2 + 1} \)[/tex]:
- The degree of the numerator [tex]\( 10x^2 \)[/tex] is 2.
- The degree of the denominator [tex]\( 2x^2 + 1 \)[/tex] is also 2.
Since the degrees of the numerator and the denominator are equal, we need to find the horizontal asymptote by dividing the leading coefficients of the numerator and denominator. The leading coefficient of the numerator is 10, and the leading coefficient of the denominator is 2.
Therefore, the horizontal asymptote is given by:
[tex]\[ y = \frac{10}{2} = 5 \][/tex]
So, the horizontal asymptote of the function [tex]\( g(x) = \frac{10x^2}{2x^2 + 1} \)[/tex] is [tex]\( y = 5 \)[/tex].
The correct answer is:
[tex]\[ \boxed{y = 5} \][/tex]
This corresponds to option B.
