Answer:
The correct answer is B
Step-by-step explanation:
Step 1
The first step in identifying the graph of the function is to determine where the vertical asymptotes occur. The vertical asymptotes occurs where the expression in the numerator is zero,
[tex]x^2-1 =0\\(x+1)(x-1)=0\\\implies x=-1, x=1[/tex] .
The next step is to calculate the [tex]x[/tex] intercept. The [tex]x[/tex] intercept occurs where [tex]y=0[/tex]. We determine intercept as shown below,
[tex]f(x)=\frac{2x}{x^2-1}=0\\ \implies 2x=0\\\\\implies x=0[/tex]
Step 2
The next step is to find the [tex]y[/tex] intercept. The [tex]y[/tex] intercept occurs when [tex]x=0[/tex]. We determine the [tex]y[/tex] intercept as shown below,
[tex]\frac{2(0)}{x^2+1} =0.[/tex]
The [tex]y[/tex] intercept occurs at [tex]y=0.[/tex]
Step 3
We now investigate the behavior of the function for different values of [tex]x[/tex]. We can tell that
,[tex]x>1,f(x)>0\\0<x<2,f(x)<0\\-1<x<0,f(x)>0\\x<-2,f(x)<0.[/tex]
Step 4
The only graph that has vertical asymptotes at [tex]x=1,x=-1[/tex] and that crosses goes through [tex](0,0)[/tex] and meets all the conditions in Step 3 is the second graph B.
