1. ## Sketching Graphs

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??

2. Originally Posted by Jiyongie
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??
$\displaystyle x^2+4$ has no real roots, which means that the function is monotonous.

how did you get y=0 asymptote?

the function is discontinuous at x=2,-2.

3. If $\displaystyle y=f(x)$ is your curve then the vertical asymptotes are found by solving $\displaystyle \frac1y=0$, and the horizontal asymptotes by solving $\displaystyle \frac1x=0$.

In this case, $\displaystyle \frac1y=\frac{x^2-4}{-2x}=\frac{(x-2)(x+2)}{-2x}$, so vertical asymptotes are $\displaystyle x-2=0$ and $\displaystyle x+2=0$.

For the horizontal asymptote, write $\displaystyle y=\frac{-2x}{x^2-4}$ in the form $\displaystyle y=\frac{-2/x}{1-4/x^2}$ obtained by dividing numerator and denominator by $\displaystyle x^2$.

Now when you put $\displaystyle 1/x=0$ (strictly speaking, you let $\displaystyle 1/x$ tend to zero) you obtain $\displaystyle y=\frac{-2\times0}{1-4\times0^2}=0$, i.e. $\displaystyle y=0$ is the horizontal asymptote.

Another useful tip: try to find the sign of $\displaystyle f(x)$ -- it helps to sketch the curve when you know which parts lie above the $\displaystyle x$-axis and which parts below.

Since $\displaystyle f(x)=\frac{-2x}{(x-2)(x+2)}$, $\displaystyle f(x)$ changes sign at $\displaystyle x=0$, $\displaystyle x=2$ and at $\displaystyle x=-2$. You will discover (with the aid of a sign line for instance) that $\displaystyle f(x)$ is positive when $\displaystyle x<-2$ and when $\displaystyle 0<x<2$, and it is negative for $\displaystyle -2<x<0$ and for $\displaystyle x>2$.

Incidentally the curve passes through the origin and is symmetrical about it (since $\displaystyle f(-x)=-f(x)$ 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 $\displaystyle x<-2$: the curve is above the $\displaystyle x$-axis; as $\displaystyle x$ gets more negative the curve approaches the asymptote $\displaystyle y=0$ from above; as $\displaystyle x$ gets closer to $\displaystyle -2$ the curve approaches the vertical asymptote $\displaystyle x=-2$ and it must be going up to it.

2. In the region $\displaystyle -2<x<2$: going to the right, the curve lies below the $\displaystyle x$-axis, then passes through the origin, then lies above the x-axis; near the asymptote $\displaystyle x=-2$ the curve must be going down; near the asymptote $\displaystyle x=2$ the curve must go up.

3. In the region $\displaystyle x>2$: the curve lies below the $\displaystyle x$-axis and goes down to its asymptote $\displaystyle x=2$; as $\displaystyle x$ increases the curve approaches the asymptote $\displaystyle y=0$ from below.

The curve has no stationary points, as you found out. And although it may be monotonous the function is not monotonic.