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

1 Answer
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 = (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)#

enter image source here
Domain: #(-oo,-1),(-1,1),(1,oo)#
Range: #(-oo,oo)#