Question

graph the function ( y=\frac{2x}{x^{2}-16} ) by identifying the domain and any symmetries, finding the derivatives ( y ) and ( y ), finding the critical points and identifying the functions behavior at each one, finding where the curve is increasing and where it is decreasing, finding the points of inflection, determining the concavity of the curve, identifying any asymptotes, and plotting any key points such as intercepts, critical points, and inflection points. then find coordinates of absolute extreme points, if any.
find any horizontal asymptotes. select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.
a. the function has one horizontal asymptote, (type an equation.)
b. the function has two horizontal asymptotes. the top asymptote is and the bottom asymptote is (type equations.)
c. the function has no horizontal asymptotes.

Explanation:

Step1: Recall the formula for horizontal asymptotes

For a rational function \(y = \frac{f(x)}{g(x)}\) where \(f(x)=2x\) (degree \(n = 1\)) and \(g(x)=x^{2}-16\) (degree \(m=2\)). The formula for horizontal asymptotes is based on the degrees of the numerator and denominator. If \(n

Step2: Apply the degree - comparison rule

Here, the degree of the numerator \(n = 1\) (since the highest power of \(x\) in \(2x\) is \(1\)) and the degree of the denominator \(m = 2\) (since the highest power of \(x\) in \(x^{2}-16\) is \(2\)). Since \(nightarrow\pm\infty}\frac{2x}{x^{2}-16}\). Divide numerator and denominator by \(x^{2}\): \(\lim_{x
ightarrow\pm\infty}\frac{\frac{2x}{x^{2}}}{\frac{x^{2}}{x^{2}}-\frac{16}{x^{2}}}=\lim_{x
ightarrow\pm\infty}\frac{\frac{2}{x}}{1 - \frac{16}{x^{2}}}\).
As \(x
ightarrow\pm\infty\), \(\lim_{x
ightarrow\pm\infty}\frac{2}{x}=0\) and \(\lim_{x
ightarrow\pm\infty}\frac{16}{x^{2}} = 0\). So, \(\lim_{x
ightarrow\pm\infty}\frac{2x}{x^{2}-16}=0\)

Answer:

A. The function has one horizontal asymptote, \(y = 0\)