which graph represents the function $f(x)=\frac{2x}{x^{2}-1}$?

Explanation:

Step1: Find vertical asymptotes

Set the denominator equal to zero: $x^{2}-1 = 0$. Factoring gives $(x + 1)(x - 1)=0$. So $x=-1$ and $x = 1$ are vertical asymptotes.

Step2: Find horizontal asymptote

Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is $y = 0$.

Step3: Analyze the sign of the function

Let $f(x)=\frac{2x}{x^{2}-1}$. We can use test - points in the intervals $(-\infty,-1)$, $(-1,1)$ and $(1,\infty)$. For $x=-2$, $f(-2)=\frac{2\times(-2)}{(-2)^{2}-1}=\frac{-4}{3}<0$. For $x = 0$, $f(0)=\frac{2\times0}{0^{2}-1}=0$. For $x = 2$, $f(2)=\frac{2\times2}{2^{2}-1}=\frac{4}{3}>0$.

Answer:

The graph with vertical asymptotes at $x=-1$ and $x = 1$ and horizontal asymptote at $y = 0$, negative on $(-\infty,-1)$, zero at $x = 0$ and positive on $(1,\infty)$ (the one shown in the image if it matches these characteristics).