So I'm supposed to sketch f(x)= (-2x)/(x^2-4)
I've found that domain is all real numbers except +/-2
Intercepts are (0,0)
Asymptotes - y=0 is the line of the horizontal one
I'm not sure how to get the vertical asymptote???
Then I found the derivative of f(x), which is f'(x)= 2(x^2+4) / (x^2-4)^2
Now I need to find the critical points.. but I'm really confused at this part because x^2 can't equal -4 right? But critical points are also points that don't exist.. so do I use +/-2 as critical points??
has no real roots, which means that the function is monotonous.
Originally Posted by Jiyongie
how did you get y=0 asymptote?
the function is discontinuous at x=2,-2.
If is your curve then the vertical asymptotes are found by solving , and the horizontal asymptotes by solving .
In this case, , so vertical asymptotes are and .
For the horizontal asymptote, write in the form obtained by dividing numerator and denominator by .
Now when you put (strictly speaking, you let tend to zero) you obtain , i.e. is the horizontal asymptote.
Another useful tip: try to find the sign of -- it helps to sketch the curve when you know which parts lie above the -axis and which parts below.
Since , changes sign at , and at . You will discover (with the aid of a sign line for instance) that is positive when and when , and it is negative for and for .
Incidentally the curve passes through the origin and is symmetrical about it (since we are dealing with the graph of an odd function which always has this property.)
So what does the graph look like?
1. In the region : the curve is above the -axis; as gets more negative the curve approaches the asymptote from above; as gets closer to the curve approaches the vertical asymptote and it must be going up to it.
2. In the region : going to the right, the curve lies below the -axis, then passes through the origin, then lies above the x-axis; near the asymptote the curve must be going down; near the asymptote the curve must go up.
3. In the region : the curve lies below the -axis and goes down to its asymptote ; as increases the curve approaches the asymptote from below.
The curve has no stationary points, as you found out. And although it may be monotonous the function is not monotonic.