Question

find the horizontal asymptote, if any, of the graph of the rational function
f(x)=\frac{16 x}{7 x^{2}+1}
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the horizontal asymptote is
(type an equation.)
b. there is no horizontal asymptote.

Explanation:

Step1: Determine the degrees of numerator and denominator

For the function \( f(x)=\frac{16x}{7x^{2}+1} \), the degree of the numerator \( n = 1 \) (since the highest - power of \( x \) in the numerator \(16x\) is \(x^1\)) and the degree of the denominator \( m=2 \) (since the highest - power of \( x \) in the denominator \(7x^{2}+1\) is \(x^2\)).

Step2: Apply the horizontal asymptote rule

The rule for horizontal asymptotes of a rational function \(y = \frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots + b_0}\) is:

• If \(n\lt m\), then \(y = 0\) is the horizontal asymptote.
• If \(n=m\), then \(y=\frac{a_n}{b_m}\) is the horizontal asymptote.
• If \(n\gt m\), then there is no horizontal asymptote.

Since \(n = 1\) and \(m = 2\) (\(n\lt m\)), we consider the limit as \(x
ightarrow\pm\infty\).

We can also use the limit formula \(\lim_{x
ightarrow\pm\infty}\frac{16x}{7x^{2}+1}\). Divide both the numerator and denominator by \(x^{2}\):

$$
LATEXBLOCK0
$$

As \(x
ightarrow\pm\infty\), \(\lim_{x
ightarrow\pm\infty}\frac{16}{x}=0\) and \(\lim_{x
ightarrow\pm\infty}\frac{1}{x^{2}} = 0\). So \(\lim_{x
ightarrow\pm\infty}\frac{\frac{16}{x}}{7+\frac{1}{x^{2}}}=0\)

Answer:

A. The horizontal asymptote is \(y = 0\)