Question

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

Answer 1

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 = (2(0))/((0)^2-1)=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=(2x)/(x^2-1)` means that the numerator of the fraction must = 0
   `0=2x`
   `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=(2x)/(x^2-1)`
   `y=(2x)/((x+1)(x-1))`
   Undefined when denominator = 0: `(x+1)(x-1)=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=(2x)/(x^2-1) -> y = "huge" / "HUGER" -> 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=(2(-2))/((-2)^2-1) = -4/(4-1) = -4/3` Point 1: `(-2, -4/3)`

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

• Pick a point to the right of the `x=1` asymptote, ie: `x=2`
  `y=(2(2))/((2)^2-1) = 4/(4-1) = 4/3` Point 3: `(2, -4/3)`

Domain: `(-oo,-1),(-1,1),(1,oo)`
Range: `(-oo,oo)`