Graph it!

We follow the procedure listed out in Graphing Rational Equations

Step 0) There are no holes.

Step 1) The y-intercept is $y = \frac{ 1}{ 0-1 } + \frac{ 1}{ 0+ 1 } = 0$ .

Step 2) The x-intercept occurs when the numerator is 0. $\frac{ 1}{ x-1} + \frac{ 1}{x+1} = \frac{ (x+1) + (x-1) } { (x-1)(x+1) } = \frac{ 2x} { (x-1)(x+1) }$ . Hence, the x-intercept occurs when $2x = 0$ , or $x = 0$ .

Step 3) $\frac{2x}{ (x-1)(x+1) }$ has a horizontal asymptote of $y = 0$ .

Step 4) Vertical asymptotes at $x = 1, x = -1$ .
As $x \rightarrow 1 ^ +$ , $y = \frac{ 2x } { (x-1)( x+1) } \sim \frac{ + } { (+) ( +) }$ thus $y \rightarrow + \infty$ .
As $x \rightarrow 1 ^ -$ , $y = \frac{ 2x } { (x-1)( x+1) } \sim \frac{ + } { (-) ( +) }$ thus $y \rightarrow - \infty$ .

As $x \rightarrow -1 ^ +$ , $y = \frac{ 2x } { (x-1)( x+1) } \sim \frac{ - } { (- ) ( +) }$ thus $y \rightarrow + \infty$ .
As $x \rightarrow -1 ^ -$ , $y = \frac{ 2x } { (x-1)( x+1) } \sim \frac{ - } { (-) ( -) }$ thus $y \rightarrow - \infty$ .

Step 5) Putting it all together, we see that the answer is C.

the Graph consists of 2 graphs added together: 1/x-1 and 1/x+1 I drew both graphs which are the graphs of 1/x but the first is translated 1 to the right and the second is translated to the left and observed what would happen if both where added. The outer most curves (right and left) would remain unchanged since they are added to very small values of the other graph therfore they retain their shape so this eliminates (A) and (D), Next, adding the area between -1 and 1 gave me the shape in C therfore the answer is C