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20. What is the horizontal asymptote for [tex]y=\frac{10x^2}{5x^2+1}[/tex]?

A. [tex]y=10[/tex]

B. [tex]y=0[/tex]

C. No horizontal asymptotes

D. [tex]y=2[/tex]


Sagot :

To determine the horizontal asymptote for the rational function [tex]\( y = \frac{10x^2}{5x^2 + 1} \)[/tex]:

1. Identify the degrees of the numerator and the denominator:
- The degree of the numerator ([tex]\(10x^2\)[/tex]) is 2.
- The degree of the denominator ([tex]\(5x^2 + 1\)[/tex]) is also 2.

2. Rule for horizontal asymptotes of rational functions:
- If the degrees of the numerator and the denominator are the same, the horizontal asymptote is found by taking the ratio of the leading coefficients.

3. Determine the leading coefficients:
- The leading coefficient of the numerator is 10 (from [tex]\(10x^2\)[/tex]).
- The leading coefficient of the denominator is 5 (from [tex]\(5x^2\)[/tex]).

4. Calculate the horizontal asymptote:
- The horizontal asymptote is given by the ratio of the leading coefficients:
[tex]\[ y = \frac{\text{leading coefficient of the numerator}}{\text{leading coefficient of the denominator}} = \frac{10}{5} = 2 \][/tex]

Therefore, the horizontal asymptote for the function [tex]\( y = \frac{10x^2}{5x^2 + 1} \)[/tex] is [tex]\( y = 2 \)[/tex].

So the correct answer is [tex]\( y = 2 \)[/tex].