Answered

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Find the horizontal asymptote, if any, of the graph of the rational function:

[tex]\[ h(x) = \frac{4x^3}{2x^2 + 1} \][/tex]

A. [tex]\( y = \frac{1}{2} \)[/tex]
B. [tex]\( y = 0 \)[/tex]
C. [tex]\( y = 2 \)[/tex]
D. No horizontal asymptote

Sagot :

GinnyAnsweridn
To determine the horizontal asymptote of the given rational function [tex]\( h(x) = \frac{4x^3}{2x^2 + 1} \)[/tex], we need to compare the degrees of the polynomials in the numerator and the denominator.

The general approach is as follows:

1. Identify the degrees of the numerator and the denominator:
- The degree of the numerator [tex]\( P(x) = 4x^3 \)[/tex] is 3.
- The degree of the denominator [tex]\( Q(x) = 2x^2 + 1 \)[/tex] is 2.

2. Compare the degrees of the numerator and the denominator:
- If the degree of the numerator is greater than the degree of the denominator, the function has no horizontal asymptote.
- If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is [tex]\( y = 0 \)[/tex].
- If the degrees of the numerator and the denominator are equal, the horizontal asymptote is [tex]\( y = \)[/tex] the ratio of the leading coefficients.

In this case:
- The degree of the numerator (3) is greater than the degree of the denominator (2).

Therefore, the function [tex]\( h(x) = \frac{4x^3}{2x^2 + 1} \)[/tex] has no horizontal asymptote.

Thus, the correct answer is:
D. no horizontal asymptote
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