# How do I graph the function: y=(2x)/(x^2-1)?

Feb 26, 2015

I like to identify the following things first, when asked to graph a rational function:
- y-intercept(s)
- x-intercept(s)
- vertical asymptote(s)
- horizontal asymptote(s)

1. To identify the y-intercept(s), ask yourself "what is the value of y when x=0"?
$y = \frac{2 \left(0\right)}{{\left(0\right)}^{2} - 1} = \frac{0}{-} 1 = 0$
y-intercept: (0,0)

2. To identify the x-intercept(s), ask yourself "what is the value of x when y=0"?
For this problem, since we've already identified that the graph goes through (0,0), we have both the x-int and y-int complete! But in case you didn't realize...
$0 = \frac{2 x}{{x}^{2} - 1}$ means that the numerator of the fraction must = 0
$0 = 2 x$
$0 = x$
x-intercept: (0,0)

3. To identify the vertical asymptotes, we first try and simplify the function as much as possible and then look at where it is undefined
$y = \frac{2 x}{{x}^{2} - 1}$
$y = \frac{2 x}{\left(x + 1\right) \left(x - 1\right)}$
Undefined when denominator = 0: $\left(x + 1\right) \left(x - 1\right) = 0$
Vertical asymptotes: $x = - 1 , x = 1$

4. To identify the horizontal asymptotes, we think of the limiting behavior (ie: what happens as x gets HUGE)
$y = \frac{2 x}{{x}^{2} - 1} \to y = \text{huge" / "HUGER} \to 0$
Horizontal asymptote: $y = 0$

Now you might pick a couple additional points to the left/right/between your horizontal asymptotes to get a sense of the graph shape.

• Pick a point to the left of the $x = - 1$ asymptote, ie: $x = - 2$
$y = \frac{2 \left(- 2\right)}{{\left(- 2\right)}^{2} - 1} = - \frac{4}{4 - 1} = - \frac{4}{3}$ Point 1: $\left(- 2 , - \frac{4}{3}\right)$

• Pick a point between the two asymptotes We already have the point (0,0) from above. Point 2: $\left(0 , 0\right)$

• Pick a point to the right of the $x = 1$ asymptote, ie: $x = 2$
$y = \frac{2 \left(2\right)}{{\left(2\right)}^{2} - 1} = \frac{4}{4 - 1} = \frac{4}{3}$ Point 3: $\left(2 , - \frac{4}{3}\right)$

Domain: $\left(- \infty , - 1\right) , \left(- 1 , 1\right) , \left(1 , \infty\right)$
Range: $\left(- \infty , \infty\right)$